NCERT Solutions · Class 9 Mathematics Measuring Space: Perimeter and Area

Build a second, upside-down copy of the same trapezium, and glue it to the first one — together they form a parallelogram whose area you already know how to find.Draw the trapezium \(\displaystyle ABCD\) with \(\displaystyle AB \parallel DC\). Say the two parallel sides are \[AB = a, \qquad DC = b, \] and let \(\displaystyle h\) be the height of the trapezium — the perpendicular (straight up-and-down) distance between the two parallel sides \(\displaystyle AB\) and \(\displaystyle DC\). \(\displaystyle h\) is not the length of the slanted side \(\displaystyle AD\) or \(\displaystyle BC\) — that slanted length is always longer than the actual height, so using it here is the mistake to watch for.Step $\displaystyle 1$ — Make a copy and turn it upside down. Trace another trapezium exactly congruent to \(\displaystyle ABCD\) — call it \(\displaystyle A'B'C'D'\), with \(\displaystyle A'B' = a\) and \(\displaystyle D'C' = b\), matching \(\displaystyle ABCD\) side for side. Now rotate this copy by \(\displaystyle 180^\circ\) (spin it half a turn, so it is upside down compared to the original).Step $\displaystyle 2$ — Glue the copy onto the original along a slanted side. Attach the rotated copy to the original trapezium along side \(\displaystyle BC\), so the two shapes sit next to each other, one right-side up and one upside down, sharing that slanted edge.Step $\displaystyle 3$ — Check that the combined shape is a parallelogram. Look at the long bottom edge of the combined figure. On one side of the join you have \(\displaystyle DC = b\) (from the original), and right next to it, in a straight line, you now have the copy's side of length \(\displaystyle a\) (because the copy is upside down, its short/long sides have swapped which end they're on). So the whole bottom edge has length \[a + b. \] The same thing happens on the top edge: it is now the copy's side \(\displaystyle b\) sitting next to the original's side \(\displaystyle a\), again giving a straight edge of length \(\displaystyle a+b\).So the combined figure has: Both pairs of opposite sides are equal and parallel — this is exactly the definition of a parallelogram. Its base is \(\displaystyle a+b\).Step $\displaystyle 4$ — Check the height didn't change. A \(\displaystyle 180^\circ\) rotation only spins the shape; it doesn't stretch, shrink, or tilt it. So the perpendicular distance between the new top edge and new bottom edge is still \(\displaystyle h\), the same height as the original trapezium.Step $\displaystyle 5$ — Use the parallelogram area formula. Formula: Area of a parallelogram \(\displaystyle = \) base \(\displaystyle \times\) height, where the base is one side and the height is the perpendicular distance to the opposite side.Here the base is \(\displaystyle a+b\) and the height is \(\displaystyle h\), so \[\text{Area of the parallelogram} = (a+b) \times h. \]Step $\displaystyle 6$ — Split that area back into the two trapeziums. The parallelogram is made of exactly two non-overlapping pieces: the original trapezium \(\displaystyle ABCD\) and the rotated copy \(\displaystyle A'B'C'D'\), and the copy is congruent to (has the exact same area as) the original. So \[\text{Area}(ABCD) + \text{Area}(A'B'C'D') = (a+b)h \] \[2 \times \text{Area}(ABCD) = (a+b)h. \]Dividing both sides by \(\displaystyle 2\): \[\text{Area}(ABCD) = \frac{1}{2}(a+b)h. \]This is exactly the formula we set out to prove, and it holds for any trapezium — a is one parallel side, b is the other parallel side, and h is the perpendicular height between them.Answer: Area of a trapezium \(\displaystyle = \dfrac{1}{2}(a+b)h\), where \(\displaystyle a\) and \(\displaystyle b\) are the lengths of the two parallel sides and \(\displaystyle h\) is the perpendicular height between them — shown by joining a \(\displaystyle 180^\circ\)-rotated copy of the trapezium to itself to form a parallelogram of base \(\displaystyle (a+b)\) and height \(\displaystyle h\), whose area \(\displaystyle (a+b)h\) is exactly twice the trapezium's area.

Split the trapezium into two triangles using one diagonal — then show both triangles secretly share the same height.Let the trapezium be \(\displaystyle ABCD\), with \(\displaystyle AB \parallel DC\). Call the two parallel sides \(\displaystyle AB = a\) and \(\displaystyle DC = b\), and let \(\displaystyle h\) be the height of the trapezium — the perpendicular (straight up-and-down) distance between the two parallel sides \(\displaystyle AB\) and \(\displaystyle DC\).Step $\displaystyle 1$: Draw one diagonal.Draw the diagonal \(\displaystyle AC\). This single line cuts the trapezium into exactly two triangles: Together, these two triangles fill up the whole trapezium exactly once, with no overlap. So:\[\text{Area}(ABCD) = \text{Area}(ABC) + \text{Area}(ACD) \]Step $\displaystyle 2$: Find the height of triangle \(\displaystyle ABC\).Triangle \(\displaystyle ABC\) has base \(\displaystyle AB\). Its height is the perpendicular distance from the opposite vertex \(\displaystyle C\) down to the line \(\displaystyle AB\).But \(\displaystyle C\) lies on the line \(\displaystyle DC\) — and \(\displaystyle DC\) is parallel to \(\displaystyle AB\), at a constant distance \(\displaystyle h\) from it (that is exactly what \(\displaystyle h\), the height of the trapezium, means). So the perpendicular distance from \(\displaystyle C\) to line \(\displaystyle AB\) is just \(\displaystyle h\).This is the step people miss: it looks like triangle \(\displaystyle ABC\) should have its "own" height, different from the trapezium's height. It doesn't — because \(\displaystyle C\) sits on the other parallel side, its distance from \(\displaystyle AB\) IS the trapezium's height \(\displaystyle h\).Using the triangle area formula, Area of a triangle \(\displaystyle = \dfrac{1}{2} \times \text{base} \times \text{height}\):\[\text{Area}(ABC) = \frac{1}{2} \times a \times h \]Step $\displaystyle 3$: Find the height of triangle \(\displaystyle ACD\).Triangle \(\displaystyle ACD\) has base \(\displaystyle DC = b\). Its height is the perpendicular distance from the opposite vertex \(\displaystyle A\) down to the line \(\displaystyle DC\).By the exact same reasoning: \(\displaystyle A\) lies on line \(\displaystyle AB\), which is parallel to \(\displaystyle DC\) at distance \(\displaystyle h\). So the perpendicular distance from \(\displaystyle A\) to line \(\displaystyle DC\) is also \(\displaystyle h\) — the same \(\displaystyle h\) as before, not a new number.\[\text{Area}(ACD) = \frac{1}{2} \times b \times h \]Step $\displaystyle 4$: Add the two areas.\[\text{Area}(ABCD) = \frac{1}{2}ah + \frac{1}{2}bh \]Now pull out the common factor \(\displaystyle \dfrac{1}{2}h\) from both terms:\[\text{Area}(ABCD) = \frac{1}{2}h(a+b) = \frac{1}{2}(a+b) \times h \]Since \(\displaystyle a\) and \(\displaystyle b\) are exactly the two parallel sides of the trapezium, \(\displaystyle (a+b)\) is their sum, and \(\displaystyle \dfrac{1}{2}(a+b)\) is half that sum. So this says:\[\text{Area}(ABCD) = \left(\text{half the sum of the parallel sides}\right) \times \left(\text{height}\right) \]which is exactly the formula we were asked to show.Common trap to watch for: \(\displaystyle h\) here must be the perpendicular height between the parallel sides — not the length of a slanting side \(\displaystyle AD\) or \(\displaystyle BC\). If the trapezium is slanted, the slant side is always longer than \(\displaystyle h\), and using it in place of \(\displaystyle h\) gives a wrong, inflated area.Answer: Splitting trapezium \(\displaystyle ABCD\) (with parallel sides \(\displaystyle a\) and \(\displaystyle b\), height \(\displaystyle h\)) along diagonal \(\displaystyle AC\) gives two triangles of areas \(\displaystyle \frac{1}{2}ah\) and \(\displaystyle \frac{1}{2}bh\); adding them gives \(\displaystyle \text{Area} = \frac{1}{2}(a+b)h\) — half the sum of the parallel sides, multiplied by the height.

Turn one copy of the trapezium upside down and slide it against the other along a slanted side — the two copies snap together into a parallelogram whose area you already know how to find.Step $\displaystyle 1$ — Label the trapezium. Draw trapezium ABCD with AB \(\displaystyle \parallel\) DC. Step $\displaystyle 2$ — Make an identical copy and rotate it \(\displaystyle 180^\circ\). Trace the exact same trapezium onto tracing paper and cut it out — call it A'B'C'D'. It is congruent to ABCD (same shape, same size, same angles). Now rotate this copy by \(\displaystyle 180^\circ\) — a half turn in the flat plane, the way you'd rotate a photo \(\displaystyle 180^\circ\) in an image editor, not the way you'd flip it over like a pancake (that would be a mirror image, which is not what we want).Step $\displaystyle 3$ — Attach the rotated copy along side BC. Slide the rotated copy so the side that used to be BC lands exactly on the original's side BC. Since the copy is identical, this side matches perfectly, end to end, because both pieces have that same side length.Step $\displaystyle 4$ — Check the join is a straight line, not a bent corner. Look at vertex B. Since \(\displaystyle AB \parallel DC\) and BC crosses both of them, \(\displaystyle \angle ABC\) and \(\displaystyle \angle BCD\) are co-interior angles (they sit on the same side of the transversal BC, between the two parallel lines), so \[\angle ABC + \angle BCD = 180^\circ . \] When the rotated copy is attached along BC, the copy's angle at that point equals \(\displaystyle \angle BCD\) (rotation doesn't change angle sizes). That angle sits right next to \(\displaystyle \angle ABC\), and together they make exactly \(\displaystyle 180^\circ\) — a straight line, with no kink. The same thing happens at vertex C. So side AB (length \(\displaystyle a\)) and the copy's corresponding side (length \(\displaystyle b\)) fuse into one single straight edge of length \(\displaystyle a+b\). The same fusing happens on the other long side too.Step $\displaystyle 5$ — Name the new shape. The combined figure now has: A four-sided shape with both pairs of opposite sides equal and parallel is a parallelogram. Its height — the gap between its two long parallel sides — is still \(\displaystyle h\), because rotating and sliding the copy along BC never stretched or squashed anything; the perpendicular distance between the parallel sides is untouched.Step $\displaystyle 6$ — Use the parallelogram's area formula, then divide by 2. Area of a parallelogram = base \(\displaystyle \times\) height, so \[\text{Area of the parallelogram} = (a+b)\times h . \] This parallelogram is made of exactly two identical copies of the trapezium sitting side by side, so one trapezium is exactly half of it: \[\text{Area of one trapezium} = \frac{1}{2}\times(a+b)\times h , \] where \(\displaystyle a\) and \(\displaystyle b\) are the lengths of the trapezium's two parallel sides and \(\displaystyle h\) is the height (perpendicular distance) between them — never the length of a slanted side.Answer: Two identical copies of a trapezium, one rotated \(\displaystyle 180^\circ\) and joined to the other along a slanted side, fit together exactly into a parallelogram of base \(\displaystyle (a+b)\) and height \(\displaystyle h\) (the trapezium's own height). Since the parallelogram's area is \(\displaystyle (a+b)\times h\) and it consists of $\displaystyle 2$ equal trapeziums, the area of one trapezium is \(\displaystyle \dfrac{1}{2}(a+b)h\), i.e. half the sum of the parallel sides times the height.

Split the kite along one diagonal into two triangles that share that diagonal as a base — the pieces of the OTHER diagonal turn out to be exactly the two heights you need.Setting upA kite is a quadrilateral with two pairs of equal adjacent sides. Call the kite \(\displaystyle ABCD\), with \[AB = AD \quad \text{and} \quad CB = CD. \] So \(\displaystyle A\) and \(\displaystyle C\) are the two "special" corners where a pair of equal sides meets. Let the diagonals \(\displaystyle AC\) and \(\displaystyle BD\) cross at a point \(\displaystyle O\), and call their lengths \[AC = d_1, \qquad BD = d_2. \]A fact about kite diagonals we'll need firstCompare triangle \(\displaystyle ABC\) and triangle \(\displaystyle ADC\): So triangle \(\displaystyle ABC \cong\) triangle \(\displaystyle ADC\) (SSS). Matching angles give \(\displaystyle \angle BAC = \angle DAC\) — line \(\displaystyle AC\) cuts angle \(\displaystyle A\) exactly in half.Now compare triangle \(\displaystyle ABO\) and triangle \(\displaystyle ADO\): So triangle \(\displaystyle ABO \cong\) triangle \(\displaystyle ADO\) (SAS). This gives two things: Watch the trap here: only \(\displaystyle BD\) gets bisected. There's no reason \(\displaystyle AO\) should equal \(\displaystyle OC\) — that only happens for special kites like a rhombus. Keep \(\displaystyle AO\) and \(\displaystyle OC\) as two different lengths throughout.(i) Using algebraCut the kite along diagonal \(\displaystyle BD\) into two triangles, \(\displaystyle ABD\) and \(\displaystyle CBD\). Both share the same base, \(\displaystyle BD = d_2\).Formula: Area of a triangle \(\displaystyle = \dfrac{1}{2} \times \text{base} \times \text{height}\), where the height is the perpendicular distance from the opposite vertex down to the base line.For triangle \(\displaystyle ABD\): the base is line \(\displaystyle BD\). Since \(\displaystyle AC \perp BD\), the perpendicular distance from \(\displaystyle A\) to line \(\displaystyle BD\) is just the length \(\displaystyle AO\) (segment \(\displaystyle AO\) lies along \(\displaystyle AC\), which already meets \(\displaystyle BD\) at a right angle). So \[\text{Area}(ABD) = \frac{1}{2} \times BD \times AO = \frac{1}{2} d_2 \cdot AO. \]For triangle \(\displaystyle CBD\): by the same reasoning, the perpendicular distance from \(\displaystyle C\) to line \(\displaystyle BD\) is \(\displaystyle CO\). So \[\text{Area}(CBD) = \frac{1}{2} \times BD \times CO = \frac{1}{2} d_2 \cdot CO. \]Add the two triangles to get the whole kite: \[\text{Area}(ABCD) = \frac{1}{2} d_2 \cdot AO + \frac{1}{2} d_2 \cdot CO = \frac{1}{2} d_2 \left( AO + CO \right). \]Now \(\displaystyle AO + OC = AC = d_1\) (the two pieces of diagonal \(\displaystyle AC\) add back up to the whole diagonal), so \[\text{Area}(ABCD) = \frac{1}{2} d_2 \cdot d_1 = \frac{1}{2} d_1 d_2. \]That's the whole algebra proof: two triangle-area formulas, added, then one substitution.(ii) Using geometryThis time we don't compute anything — we build a rectangle around the kite and match up congruent triangles.Construction. Through \(\displaystyle A\) and \(\displaystyle C\), draw two lines parallel to \(\displaystyle BD\) (these become the left and right sides of a rectangle). Through \(\displaystyle B\) and \(\displaystyle D\), draw two lines parallel to \(\displaystyle AC\) (top and bottom sides). Because \(\displaystyle AC \perp BD\), all four lines meet at right angles, so they close up into a rectangle. Call it \(\displaystyle PQRS\) — \(\displaystyle P\) top-left, \(\displaystyle Q\) top-right, \(\displaystyle R\) bottom-right, \(\displaystyle S\) bottom-left — with \(\displaystyle A\) on side \(\displaystyle PS\), \(\displaystyle C\) on side \(\displaystyle QR\), \(\displaystyle B\) on side \(\displaystyle PQ\), \(\displaystyle D\) on side \(\displaystyle SR\). So rectangle \(\displaystyle PQRS\) has length \(\displaystyle d_1\) and breadth \(\displaystyle d_2\). Using Area of a rectangle \(\displaystyle = \text{length} \times \text{breadth}\): \[\text{Area}(PQRS) = d_1 \times d_2. \]Look inside this rectangle now. Because \(\displaystyle BO = OD\) (shown above), line \(\displaystyle AC\) sits exactly halfway between the top and bottom sides — it splits \(\displaystyle PQRS\) into an upper strip and a lower strip, each \(\displaystyle d_1\) wide and \(\displaystyle \frac{d_2}{2}\) tall.Also, since \(\displaystyle OB \perp AC\) and \(\displaystyle OB\) runs in the same direction as the rectangle's left/right sides, \(\displaystyle B\) sits directly above \(\displaystyle O\). So the vertical line through \(\displaystyle B\) and \(\displaystyle O\) cuts the upper strip into two smaller rectangles: a left one \(\displaystyle AO\) wide (between corner \(\displaystyle P\) and point \(\displaystyle A\)) and a right one \(\displaystyle OC\) wide (between point \(\displaystyle C\) and corner \(\displaystyle Q\)). The same thing happens in the lower strip using \(\displaystyle D\), giving two more small rectangles of the same widths \(\displaystyle AO\) and \(\displaystyle OC\). That's $\displaystyle 4$ small rectangles in total, each of height \(\displaystyle \frac{d_2}{2}\).Here's the key step: in each small rectangle, one side of the kite is exactly its diagonal. A diagonal always cuts a rectangle into two congruent right triangles — they share the diagonal, and the legs of one match the legs of the other because opposite sides of a rectangle are equal. So in every one of the $\displaystyle 4$ small rectangles: \[\text{(triangle inside the kite)} \;\cong\; \text{(triangle in the leftover corner, outside the kite)}. \]Congruent triangles have equal area, and this holds in all $\displaystyle 4$ small rectangles at once, so: \[\text{Area(kite)} = \text{Area(the four corner triangles left over outside the kite)}. \]But the kite together with those four corner triangles fills the whole rectangle \(\displaystyle PQRS\) exactly. So: \[\text{Area(kite)} + \text{Area(kite)} = \text{Area}(PQRS) = d_1 d_2 \] \[2 \times \text{Area(kite)} = d_1 d_2 \] \[\text{Area(kite)} = \frac{1}{2} d_1 d_2. \]Same result — this time without adding a single number, just by matching congruent triangles.Why both proofs matter: the algebra proof is what you actually use when you're handed real diagonal lengths — plug in \(\displaystyle d_1\) and \(\displaystyle d_2\), get a number. The geometry proof is the one that shows why the formula has to hold for every kite shape, not just the one you happened to draw.Answer: For any kite with diagonals of length \(\displaystyle d_1\) and \(\displaystyle d_2\), Area \(\displaystyle = \dfrac{1}{2} d_1 d_2\) — proved by splitting the kite into two triangles on the shared base \(\displaystyle BD\) (algebra), and by comparing it, corner triangle by corner triangle, to the rectangle built around it (geometry).

When every side of a shape is scaled by a factor \(\displaystyle k\), the area scales by \(\displaystyle k^2\) — but whether the same number of small copies can be physically packed into the big shape depends on how the shape's sides subdivide it. Check each part by actually cutting the big shape up.(i) Rectangles ABCD (sides \(\displaystyle a,b\)) and PQRS (sides \(\displaystyle 2a,2b\))The formula for the area of a rectangle is \[\text{Area} = \text{length} \times \text{breadth}. \] So \[\text{Area(ABCD)} = a \times b = ab, \] \[\text{Area(PQRS)} = 2a \times 2b = 4ab = 4 \times \text{Area(ABCD)}. \] That proves PQRS has $\displaystyle 4$ times the area of ABCD.Does that mean $\displaystyle 4$ copies of ABCD fit inside PQRS? Yes — check by cutting. Since \(\displaystyle 2a = a+a\) and \(\displaystyle 2b = b+b\), make one cut through the middle of PQRS parallel to its breadth (splitting the \(\displaystyle 2a\) side into two pieces of length \(\displaystyle a\)) and one cut parallel to its length (splitting the \(\displaystyle 2b\) side into two pieces of length \(\displaystyle b\)). This chops PQRS into a \(\displaystyle 2\times2\) grid of four rectangles, each \(\displaystyle a \times b\) — that is, four exact copies of ABCD, with nothing left over.Watch out: doubling the sides does not double the area — a very common slip is to think "sides doubled, so area doubled." The area formula multiplies two lengths together, so doubling both dimensions multiplies the area by \(\displaystyle 2 \times 2 = 4\), not by \(\displaystyle 2\).(ii) Triangle ABC (sides \(\displaystyle a,b,c\)) and triangle PQR (sides \(\displaystyle 2a,2b,2c\))Use Heron's formula: if a triangle has sides \(\displaystyle a, b, c\) and semi-perimeter \[s = \frac{a+b+c}{2}, \] then \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}. \]For triangle PQR the sides are \(\displaystyle 2a, 2b, 2c\), so its semi-perimeter is \[s' = \frac{2a+2b+2c}{2} = 2\left(\frac{a+b+c}{2}\right) = 2s. \] Then \(\displaystyle s' - 2a = 2s - 2a = 2(s-a)\), and similarly \(\displaystyle s'-2b = 2(s-b)\), \(\displaystyle s'-2c = 2(s-c)\). So \[\text{Area(PQR)} = \sqrt{(2s)\big(2(s-a)\big)\big(2(s-b)\big)\big(2(s-c)\big)} = \sqrt{16\,s(s-a)(s-b)(s-c)} = 4\sqrt{s(s-a)(s-b)(s-c)}. \] That last square root is exactly Area(ABC), so \[\text{Area(PQR)} = 4 \times \text{Area(ABC)}. \]Does that mean $\displaystyle 4$ copies of ABC fit inside PQR? Yes — and here's why, not just "it works out." Mark the midpoint of each side of PQR and join each midpoint to the other two. By the midpoint theorem, the segment joining the midpoints of two sides of a triangle is parallel to the third side and exactly half its length. Since PQR's sides are \(\displaystyle 2a, 2b, 2c\), half of each is \(\displaystyle a, b, c\) — precisely ABC's side lengths. This construction splits PQR into $\displaystyle 4$ smaller triangles: three sitting in the corners the same way up as ABC, and one in the middle sitting upside-down relative to them. All four have side lengths \(\displaystyle a, b, c\), so all four are congruent to ABC — the "upside-down" one is just ABC rotated \(\displaystyle 180^\circ\), which is still the same triangle, not a different or bigger one.Watch out: "congruent" allows flipping and rotating. Don't assume the middle, upside-down triangle must be a different shape just because it looks turned around — matching side lengths is what makes two triangles congruent, not which way they're pointing.(iii) Triangle ABC (sides \(\displaystyle a,b,c\)) and triangle PQR (sides \(\displaystyle 3a,3b,3c\))Same method, new scale factor. Semi-perimeter of PQR: \[s' = \frac{3a+3b+3c}{2} = 3\left(\frac{a+b+c}{2}\right) = 3s. \] So \(\displaystyle s' - 3a = 3(s-a)\), and likewise for the other two terms. By Heron's formula, \[\text{Area(PQR)} = \sqrt{(3s)\big(3(s-a)\big)\big(3(s-b)\big)\big(3(s-c)\big)} = \sqrt{81\,s(s-a)(s-b)(s-c)} = 9\sqrt{s(s-a)(s-b)(s-c)} = 9 \times \text{Area(ABC)}. \]Does that mean $\displaystyle 9$ copies of ABC fit inside PQR? Yes — extend the same cutting idea. Divide each side of PQR into $\displaystyle 3$ equal parts (each part has length \(\displaystyle a\), \(\displaystyle b\), or \(\displaystyle c\), matching ABC's sides), and through these division points draw lines parallel to each of PQR's three sides. This lays a triangular grid over PQR made of $\displaystyle 9$ small triangles — some pointing the same way as ABC, some flipped upside-down — but every single one has side lengths \(\displaystyle a, b, c\), so every one is congruent to ABC. The $\displaystyle 9$ copies tile PQR exactly, with no gaps and no overlaps.This is the general pattern behind all three parts: scaling a shape's sides by a whole number \(\displaystyle n\) multiplies its area by \(\displaystyle n^2\), and for rectangles and triangles you can always physically slice the bigger shape into exactly \(\displaystyle n^2\) copies of the smaller one — a \(\displaystyle n \times n\) grid of cuts for a rectangle, or the parallel-line grid for a triangle.Answer: (i) Area(PQRS) \(\displaystyle = 4 \times\) Area(ABCD); yes, $\displaystyle 4$ copies of ABCD fit into PQRS (cut it into a \(\displaystyle 2\times2\) grid). (ii) Area(PQR) \(\displaystyle = 4 \times\) Area(ABC); yes, $\displaystyle 4$ copies of ABC fit into PQR (join the midpoints of PQR's sides). (iii) Area(PQR) \(\displaystyle = 9 \times\) Area(ABC); yes, $\displaystyle 9$ copies of ABC fit into PQR (divide each side of PQR into $\displaystyle 3$ equal parts and draw lines parallel to the sides).

Everything you need is in the tick marks — count them before you calculate anything. Equal marks mean equal lengths, and that is the only "measurement" either picture gives you. Nothing here has a length in centimetres, so both answers come out as pure fractions with no units.Fig. $\displaystyle 6.43$ — the triangleRead the marks: the left side is cut in half, but the long right side is cut into three equal pieces. Then get the shaded part by taking the two white corners away from the whole triangle.Name the corners: \(\displaystyle A\) at the bottom-left, \(\displaystyle B\) at the bottom-right, \(\displaystyle C\) at the top.Now count what is drawn on each slanted side. \[AD = DC = \tfrac{1}{2}\,AC \] \[CE = EF = FB = \tfrac{1}{3}\,CB \]Watch out — this is the step people get wrong here. It is tempting to glance at the picture, see a dot on each slanted side, and write "midpoint, midpoint". Count again: the left side has one dot, the right side has two. Two dots make three pieces, not two, so there is no midpoint on \(\displaystyle CB\) at all. And the different stroke counts (three on the left, two on the right) are not saying "the left pieces equal the right pieces" — a tick mark only says the pieces wearing that same mark are equal to each other.The shaded piece is the quadrilateral \(\displaystyle ADEF\): from \(\displaystyle A\) up to \(\displaystyle D\), across to \(\displaystyle E\), down to \(\displaystyle F\), then straight back to \(\displaystyle A\). Look again at the picture and notice two things: the base \(\displaystyle AB\) is not an edge of the shaded piece, and the corner \(\displaystyle B\) is not inside it. The bottom-right of the picture is one big white triangle.So the whole triangle is made of exactly three pieces: \[\triangle ABC \;=\; \underbrace{ADEF}_{\text{shaded}} \;+\; \underbrace{\triangle CDE}_{\text{white, top}} \;+\; \underbrace{\triangle AFB}_{\text{white, bottom-right}} \]I will work out the two white pieces and subtract them.The one tool I need — the "same height" rule.Formula used — Area of a triangle: \(\displaystyle \text{Area} = \tfrac{1}{2} \times b \times h\), where \(\displaystyle b\) is any one side you choose as the base, and \(\displaystyle h\) is the perpendicular (straight-down) distance from the opposite corner to that base.What follows from it: if two triangles have the same height, their areas are in the same ratio as their bases. The \(\displaystyle \tfrac{1}{2}\) is the same in both and the \(\displaystyle h\) is the same in both, so only \(\displaystyle b\) can change the answer.Step $\displaystyle 1$ — the white triangle \(\displaystyle CDE\) at the top.Get there in two hops.Hop 1. Compare \(\displaystyle \triangle CDE\) with \(\displaystyle \triangle CAE\). Their bases \(\displaystyle CD\) and \(\displaystyle CA\) both lie along the same straight line \(\displaystyle AC\), and both triangles have their third corner at the same point \(\displaystyle E\). Same corner and same line means the perpendicular distance from \(\displaystyle E\) to line \(\displaystyle AC\) is one single height serving both. So \[\frac{\text{area}(CDE)}{\text{area}(CAE)}=\frac{CD}{CA}=\frac{1}{2} \]Hop 2. Compare \(\displaystyle \triangle CAE\) with the whole \(\displaystyle \triangle CAB\). Their bases \(\displaystyle CE\) and \(\displaystyle CB\) both lie along the same straight line \(\displaystyle CB\), and both have their third corner at the same point \(\displaystyle A\). Same height again. So \[\frac{\text{area}(CAE)}{\text{area}(CAB)}=\frac{CE}{CB}=\frac{1}{3} \]Multiply the two hops: \[\text{area}(CDE)=\frac{1}{2}\times\frac{1}{3}\times\text{area}(ABC)=\frac{1}{6}\,\text{area}(ABC) \]Step $\displaystyle 2$ — the white triangle \(\displaystyle AFB\) at the bottom right.Compare \(\displaystyle \triangle AFB\) with the whole \(\displaystyle \triangle ACB\). Their bases \(\displaystyle FB\) and \(\displaystyle CB\) lie along the same straight line \(\displaystyle CB\), and both have their third corner at the same point \(\displaystyle A\). Same height, so \[\frac{\text{area}(AFB)}{\text{area}(ACB)}=\frac{FB}{CB}=\frac{1}{3} \qquad\Longrightarrow\qquad \text{area}(AFB)=\frac{1}{3}\,\text{area}(ABC) \]Step $\displaystyle 3$ — subtract the two white pieces.\[\text{area}(ADEF)=\text{area}(ABC)-\text{area}(CDE)-\text{area}(AFB) \] \[=\text{area}(ABC)\left(1-\frac{1}{6}-\frac{1}{3}\right) \]Put the fractions over a common denominator of \(\displaystyle 6\), using \(\displaystyle 1=\dfrac{6}{6}\) and \(\displaystyle \dfrac{1}{3}=\dfrac{2}{6}\): \[1-\frac{1}{6}-\frac{1}{3}=\frac{6}{6}-\frac{1}{6}-\frac{2}{6}=\frac{3}{6}=\frac{1}{2} \]So the shaded quadrilateral is \(\displaystyle \dfrac{1}{2}\) of the triangle.Step $\displaystyle 4$ — check it a completely different way.A fraction of a shape does not change if you stretch or shrink the whole picture, so I am allowed to pick convenient corners and just count. Take \[A(0,0),\qquad B(6,0),\qquad C(0,6) \] Then \(\displaystyle D\), the midpoint of \(\displaystyle AC\), is \(\displaystyle (0,3)\). Going from \(\displaystyle C(0,6)\) towards \(\displaystyle B(6,0)\), each third step adds \(\displaystyle 2\) to \(\displaystyle x\) and takes \(\displaystyle 2\) off \(\displaystyle y\), so \(\displaystyle E=(2,4)\) and \(\displaystyle F=(4,2)\).Formula used — Area of a right-angled triangle: \(\displaystyle \tfrac{1}{2}\times\text{base}\times\text{height}\). Here the base \(\displaystyle AB=6\) and the height \(\displaystyle AC=6\), so \[\text{area}(ABC)=\tfrac{1}{2}\times 6\times 6=18 \text{ square units} \] Cut the shaded quadrilateral \(\displaystyle A(0,0),D(0,3),E(2,4),F(4,2)\) into two triangles along the line \(\displaystyle AE\): \[\triangle ADE:\ \text{base } AD = 3 \text{ (along the } y\text{-axis)},\ \text{height} = 2 \text{ (the } x\text{-distance of } E) \] \[\text{area}(ADE)=\tfrac{1}{2}\times 3\times 2 = 3 \text{ square units} \] \[\triangle AEF:\ \text{area}=\tfrac{1}{2}\bigl|x_E\,y_F-x_F\,y_E\bigr|=\tfrac{1}{2}\bigl|2\times 2-4\times 4\bigr|=\tfrac{1}{2}\times 12=6 \text{ square units} \] \[\text{area}(ADEF)=3+6=9 \text{ square units} \] \[\frac{9}{18}=\frac{1}{2}\quad\checkmark \]Sense-check on the picture: the shaded quadrilateral looks like about half, and the two white corners together look like the other half. If you had wrongly treated \(\displaystyle CB\) as merely bisected, you would get \(\displaystyle \tfrac{3}{4}\) — and one look at how much white there is shows three-quarters is far too big.Fig. $\displaystyle 6.44$ — the squarePut the square on a grid, write down the equation of each of the four slanted lines, solve them in pairs to find the four corners of the tilted shaded square, then compare its area with the big square's.First read the marks. Every side has one dot with equal ticks (two strokes) on both sides of it, so each dot is the midpoint of its side.Next read where the lines go. Each line starts at a corner and ends at the midpoint of a side that corner does not touch, and all four go round the same way.Since the question asks only for a fraction, the real size does not matter — enlarging the picture multiplies every area by the same number and leaves the ratio alone. I will pick a side of \(\displaystyle 10\) units to keep the numbers whole.\[A(0,0)\ \text{bottom-left},\quad B(10,0)\ \text{bottom-right},\quad C(10,10)\ \text{top-right},\quad D(0,10)\ \text{top-left} \] Midpoints: \[P(5,0)\ \text{on } AB,\quad Q(10,5)\ \text{on } BC,\quad R(5,10)\ \text{on } CD,\quad S(0,5)\ \text{on } DA \] The four drawn lines are \(\displaystyle A\to Q\), \(\displaystyle B\to R\), \(\displaystyle C\to S\) and \(\displaystyle D\to P\).Step $\displaystyle 1$ — write each line as \(\displaystyle y=mx+c\).Formula used — Gradient (steepness): \(\displaystyle m=\dfrac{y_2-y_1}{x_2-x_1}\), the "rise" divided by the "run" between two points on the line. Then \(\displaystyle c\) is the \(\displaystyle y\)-value where the line crosses the \(\displaystyle y\)-axis; find it by putting one known point into \(\displaystyle y=mx+c\).Line \(\displaystyle AQ\), from \(\displaystyle A(0,0)\) to \(\displaystyle Q(10,5)\): \[m=\frac{5-0}{10-0}=\frac{1}{2},\qquad \text{it passes through } (0,0) \text{ so } c=0 \qquad\Longrightarrow\qquad y=\tfrac{1}{2}x \] Line \(\displaystyle BR\), from \(\displaystyle B(10,0)\) to \(\displaystyle R(5,10)\): \[m=\frac{10-0}{5-10}=\frac{10}{-5}=-2,\qquad 0=-2(10)+c \Rightarrow c=20 \qquad\Longrightarrow\qquad y=-2x+20 \] Line \(\displaystyle CS\), from \(\displaystyle C(10,10)\) to \(\displaystyle S(0,5)\): \[m=\frac{5-10}{0-10}=\frac{-5}{-10}=\frac{1}{2},\qquad \text{it passes through } (0,5) \text{ so } c=5 \qquad\Longrightarrow\qquad y=\tfrac{1}{2}x+5 \] Line \(\displaystyle DP\), from \(\displaystyle D(0,10)\) to \(\displaystyle P(5,0)\): \[m=\frac{0-10}{5-0}=-2,\qquad \text{it passes through } (0,10) \text{ so } c=10 \qquad\Longrightarrow\qquad y=-2x+10 \]Step $\displaystyle 2$ — find the four corners of the shaded shape.Two lines cross where their \(\displaystyle y\)-values are equal, so set the right-hand sides equal and solve for \(\displaystyle x\).Corner \(\displaystyle W\), where \(\displaystyle AQ\) meets \(\displaystyle DP\): \[\tfrac{1}{2}x=-2x+10 \] Multiply every term by \(\displaystyle 2\) to clear the fraction: \[x=-4x+20 \;\Rightarrow\; 5x=20 \;\Rightarrow\; x=4,\qquad y=\tfrac{1}{2}(4)=2 \qquad\Longrightarrow\qquad W(4,2) \] Corner \(\displaystyle X\), where \(\displaystyle AQ\) meets \(\displaystyle BR\): \[\tfrac{1}{2}x=-2x+20 \;\Rightarrow\; x=-4x+40 \;\Rightarrow\; 5x=40 \;\Rightarrow\; x=8,\qquad y=\tfrac{1}{2}(8)=4 \qquad\Longrightarrow\qquad X(8,4) \] Corner \(\displaystyle Y\), where \(\displaystyle BR\) meets \(\displaystyle CS\): \[-2x+20=\tfrac{1}{2}x+5 \;\Rightarrow\; -4x+40=x+10 \;\Rightarrow\; 30=5x \;\Rightarrow\; x=6,\qquad y=-2(6)+20=8 \qquad\Longrightarrow\qquad Y(6,8) \] Corner \(\displaystyle Z\), where \(\displaystyle CS\) meets \(\displaystyle DP\): \[\tfrac{1}{2}x+5=-2x+10 \;\Rightarrow\; x+10=-4x+20 \;\Rightarrow\; 5x=10 \;\Rightarrow\; x=2,\qquad y=\tfrac{1}{2}(2)+5=6 \qquad\Longrightarrow\qquad Z(2,6) \]Watch out: never read a crossing point off the drawing with a ruler and your eye. Solve the two equations together. The drawing is a guide; the algebra is the proof.Step $\displaystyle 3$ — check the shape really is a square, and find its side.Formula used — Distance between two points: \(\displaystyle \text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). This is just Pythagoras' theorem, with the horizontal gap and the vertical gap as the two short sides of a right-angled triangle.\[WX:\ \Delta x=8-4=4,\ \Delta y=4-2=2 \;\Rightarrow\; \sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20} \] \[XY:\ \Delta x=6-8=-2,\ \Delta y=8-4=4 \;\Rightarrow\; \sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20} \] \[YZ:\ \Delta x=2-6=-4,\ \Delta y=6-8=-2 \;\Rightarrow\; \sqrt{16+4}=\sqrt{20} \] \[ZW:\ \Delta x=4-2=2,\ \Delta y=2-6=-4 \;\Rightarrow\; \sqrt{4+16}=\sqrt{20} \] All four sides are the same length.Are the corners right angles? Two lines are perpendicular when their gradients multiply to \(\displaystyle -1\). Side \(\displaystyle WX\) lies along line \(\displaystyle AQ\), gradient \(\displaystyle \tfrac{1}{2}\); side \(\displaystyle XY\) lies along line \(\displaystyle BR\), gradient \(\displaystyle -2\). \[\tfrac{1}{2}\times(-2)=-1 \] So they meet at a right angle, and the same pair of gradients meets at every corner. Four equal sides and four right angles: it is a square.Step $\displaystyle 4$ — the two areas.Formula used — Area of a square: \(\displaystyle \text{Area}=(\text{side})^2\). \[\text{area(shaded square)}=\left(\sqrt{20}\right)^2=20 \text{ square units} \] (The square root and the squaring undo each other, so no decimals are needed anywhere.) \[\text{area(big square)}=10\times 10=100 \text{ square units} \]Watch out: \(\displaystyle \sqrt{20}\) is the side, not the area. And do not measure from \(\displaystyle W(4,2)\) straight across to \(\displaystyle Y(6,8)\) — that is a diagonal, not a side. Squaring the diagonal would give \(\displaystyle 40\), which is double the true area.Step $\displaystyle 5$ — the fraction. \[\frac{\text{area(shaded)}}{\text{area(big square)}}=\frac{20}{100}=\frac{1}{5} \] because dividing the top and the bottom by \(\displaystyle 20\) gives \(\displaystyle \dfrac{1}{5}\).Sense-check: \(\displaystyle \tfrac{1}{5}=0.2\), so the tilted square should look like about a fifth of the big one — and it does.Both answers are fractions of an area, so they have no units.Answer: In Fig. $\displaystyle 6.43$, \(\displaystyle \dfrac{1}{2}\) of the triangle is shaded; in Fig. $\displaystyle 6.44$, \(\displaystyle \dfrac{1}{5}\) of the square is shaded.

Cut each rectangle into squares — one square around each circle. Every circle is drawn inside its own square and touches all four of its sides, and a circle drawn that way always covers \(\displaystyle \dfrac{\pi}{4} \) of its square. So the whole picture is covered in the same proportion, however many squares there happen to be.First, read each picture on its own. They are two separate drawings of two separate rectangles, and the question never says the two rectangles are the same size — in fact the Fig. $\displaystyle 6.46$ rectangle is drawn longer than the Fig. $\displaystyle 6.45$ one. So work each figure out by itself, and only compare the two answers at the end.In Fig. $\displaystyle 6.45$ the picture shows: In Fig. $\displaystyle 6.46$ the picture shows exactly the same arrangement, but with $\displaystyle 4$ equal circles.No lengths are printed on either figure, so we will give each picture its own letter for the size of its circles.Step $\displaystyle 1$: Give the circles in Fig. $\displaystyle 6.45$ a letter, and write down the radius.Let the diameter of each circle in Fig. $\displaystyle 6.45$ be \(\displaystyle d \) cm. (Any unit works here — centimetres are just to keep the working tidy. The unit will cancel at the end.)The radius is half the diameter, so \[r = \frac{d}{2}\ \text{cm} \]Step $\displaystyle 2$: Read the breadth of the rectangle off the picture.Each circle touches the top edge and the bottom edge. The distance straight across a circle from top to bottom is its diameter, so the breadth of the rectangle is one whole diameter: \[b = d\ \text{cm} \]This is the diameter, not the radius: the rectangle's breadth reaches all the way across the circle, and that distance is \(\displaystyle d \) cm, which is \(\displaystyle 2r \) cm.Step $\displaystyle 3$: Read the length of the rectangle off the picture.The $\displaystyle 3$ circles sit in a row, each touching the next, and the two end circles touch the two ends of the rectangle. So the length is $\displaystyle 3$ diameters laid end to end: \[l = 3 \times d = 3d\ \text{cm} \]Step $\displaystyle 4$: Find the area of the rectangle in Fig. 6.45.Using area of a rectangle = length \(\displaystyle \times \) breadth, where \(\displaystyle l \) is the length and \(\displaystyle b \) is the breadth: \[\text{Area of rectangle} = l \times b = 3d \times d = 3d^{2}\ \text{cm}^{2} \]Step $\displaystyle 5$: Find the area covered by the $\displaystyle 3$ circles.Using area of a circle \(\displaystyle = \pi r^{2} \), where \(\displaystyle r \) is the radius and \(\displaystyle \pi \) is the number that connects a circle's circumference to its diameter: \[\text{Area of one circle} = \pi r^{2} = \pi \left( \frac{d}{2} \right)^{2} = \pi \times \frac{d^{2}}{4} = \frac{\pi d^{2}}{4}\ \text{cm}^{2} \]All $\displaystyle 3$ circles are the same size, so \[\text{Area of 3 circles} = 3 \times \frac{\pi d^{2}}{4} = \frac{3\pi d^{2}}{4}\ \text{cm}^{2} \]Step $\displaystyle 6$: Divide, to get the fraction of Fig. $\displaystyle 6.45$ that is covered.\[\text{Fraction covered} = \frac{\text{area covered by the circles}}{\text{area of the rectangle}} = \frac{\dfrac{3\pi d^{2}}{4}\ \text{cm}^{2}}{3d^{2}\ \text{cm}^{2}} \]The \(\displaystyle \text{cm}^{2} \) on the top and the \(\displaystyle \text{cm}^{2} \) on the bottom cancel, so the answer is a plain number with no unit — which is what a fraction of a shape should be. Cancelling the $\displaystyle 3$ and the \(\displaystyle d^{2} \) as well: \[\text{Fraction covered} = \frac{3\pi d^{2}}{4 \times 3d^{2}} = \frac{\pi}{4} \]Step $\displaystyle 7$: Now do Fig. $\displaystyle 6.46$ from scratch, with its own letter.Fig. $\displaystyle 6.46$ is a different rectangle with different circles, so give it a fresh letter. Let the diameter of each of its $\displaystyle 4$ circles be \(\displaystyle D \) cm. There is no reason for \(\displaystyle D \) to equal \(\displaystyle d \), and we will not need it to.Same three readings from the picture: \[\text{breadth} = D\ \text{cm}, \qquad \text{length} = 4 \times D = 4D\ \text{cm} \]\[\text{Area of rectangle} = 4D \times D = 4D^{2}\ \text{cm}^{2} \]\[\text{Area of 4 circles} = 4 \times \pi \left( \frac{D}{2} \right)^{2} = 4 \times \frac{\pi D^{2}}{4} = \pi D^{2}\ \text{cm}^{2} \]\[\text{Fraction covered} = \frac{\pi D^{2}\ \text{cm}^{2}}{4D^{2}\ \text{cm}^{2}} = \frac{\pi}{4} \]The same fraction as Fig. 6.45.Step $\displaystyle 8$: See why the two answers had to match.Take either picture and draw a vertical line straight down at every point where two circles touch. Each piece you cut off is \(\displaystyle d \) cm tall (that is the breadth of the rectangle) and \(\displaystyle d \) cm wide (that is one diameter of one circle). Tall and wide are equal, so each piece is a square of side \(\displaystyle d \) cm, and it holds exactly one circle, touching all four of its sides.Inside one such square: \[\frac{\text{area of the circle}}{\text{area of the square}} = \frac{\pi \left( \dfrac{d}{2} \right)^{2}\ \text{cm}^{2}}{d^{2}\ \text{cm}^{2}} = \frac{\dfrac{\pi d^{2}}{4}}{d^{2}} = \frac{\pi}{4} \]Two things drop out of that line, and both matter: One thing to be careful about: it is tempting to say Fig. $\displaystyle 6.46$'s circles are smaller "because four of them had to be squeezed into the same box". That is not what is happening. The two rectangles are separate drawings and neither is fixed by the other — the Fig. $\displaystyle 6.46$ rectangle is actually the longer of the two. The fractions agree for the square reason above, not because of any squeezing.Step $\displaystyle 9$: Write the fraction as a number.The chapter says to use the approximation \(\displaystyle \pi \approx \dfrac{22}{7} \). Putting that in: \[\frac{\pi}{4} \approx \frac{22}{7} \div 4 = \frac{22}{7 \times 4} = \frac{22}{28} = \frac{11}{14} \]As a decimal, \(\displaystyle \dfrac{11}{14} \approx 0.786 \), which is about \(\displaystyle 78.6\% \) of the rectangle.Because \(\displaystyle \dfrac{22}{7} \) is only an approximation of \(\displaystyle \pi \) and not equal to it, \(\displaystyle \dfrac{11}{14} \) is an approximate value too. The exact answer is \(\displaystyle \dfrac{\pi}{4} \).Answer: In Fig. $\displaystyle 6.45$ the $\displaystyle 3$ circles cover \(\displaystyle \dfrac{\pi}{4} \) of the rectangle, and in Fig. $\displaystyle 6.46$ the $\displaystyle 4$ circles cover \(\displaystyle \dfrac{\pi}{4} \) of the rectangle — the same fraction, because each circle sits in its own square and fills \(\displaystyle \dfrac{\pi}{4} \) of that square, so the number of squares cancels. Using \(\displaystyle \pi \approx \dfrac{22}{7} \), this is \(\displaystyle \dfrac{\pi}{4} \approx \dfrac{11}{14} \approx 0.786 \), or about \(\displaystyle 78.6\% \), in both figures.

One row of touching circles cuts the rectangle into equal squares, one square per circle — and a circle drawn inside a square always fills the same part of it, so the number of circles cannot change the answer.The arrangement meant here is the one from the previous question (Fig. $\displaystyle 6.45$ and Fig. $\displaystyle 6.46$): all the circles are the same size, they sit in a single row, each circle touches the top edge and the bottom edge of the rectangle, each circle touches the circles next to it, and the two end circles touch the left and right edges. There are no gaps and no overlaps.Step $\displaystyle 1$: Write down what the two earlier cases gave.In the previous question the answers were: The number of circles went from \(\displaystyle 3\) to \(\displaystyle 4\), and the fraction did not budge. That is the hint the question wants you to notice.Step $\displaystyle 2$: Make the conjecture.A conjecture is a careful guess that you have not proved yet. Here it is:Conjecture — circles fitted into a rectangle in this manner always cover \(\displaystyle \dfrac{\pi}{4}\) of the rectangle, no matter how many circles there are.Using \(\displaystyle \pi = \dfrac{22}{7}\), as the chapter instruction says: \[\frac{\pi}{4} = \frac{22}{7} \div 4 = \frac{22}{7 \times 4} = \frac{22}{28} = \frac{11}{14} \] So the conjecture predicts \(\displaystyle \dfrac{11}{14}\) every single time — which is exactly what both earlier pictures gave.Step $\displaystyle 3$: Test the conjecture for $\displaystyle 10$ circles, $\displaystyle 20$ circles and $\displaystyle 50$ circles.The conjecture says the circle size does not matter, so choose a size that keeps the arithmetic friendly. Take the radius to be \[r = 7 \text{ units}, \qquad \text{so the diameter is } d = 2r = 2 \times 7 = 14 \text{ units}. \] (The \(\displaystyle 7\) is chosen because \(\displaystyle \dfrac{22}{7}\) will cancel with it.)Area of one circle, using the formula \(\displaystyle A = \pi r^{2}\), where \(\displaystyle A\) is the area and \(\displaystyle r\) is the radius: \[A = \frac{22}{7} \times 7 \text{ units} \times 7 \text{ units} = 22 \times 7 = 154 \text{ square units} \]Two facts hold in all three cases: Case $\displaystyle 1$ — $\displaystyle 10$ circles. \[\text{length} = 10 \times 14 \text{ units} = 140 \text{ units} \] Area of the rectangle, using Area \(\displaystyle =\) length \(\displaystyle \times\) breadth: \[140 \text{ units} \times 14 \text{ units} = 1960 \text{ square units} \] Area of the $\displaystyle 10$ circles: \[10 \times 154 \text{ square units} = 1540 \text{ square units} \] Fraction covered, simplified one divide at a time: \[\frac{1540}{1960} = \frac{1540 \div 10}{1960 \div 10} = \frac{154}{196} = \frac{154 \div 14}{196 \div 14} = \frac{11}{14} \]Case $\displaystyle 2$ — $\displaystyle 20$ circles. \[\text{length} = 20 \times 14 \text{ units} = 280 \text{ units} \] \[\text{area of rectangle} = 280 \text{ units} \times 14 \text{ units} = 3920 \text{ square units} \] \[\text{area of the 20 circles} = 20 \times 154 \text{ square units} = 3080 \text{ square units} \] \[\frac{3080}{3920} = \frac{3080 \div 10}{3920 \div 10} = \frac{308}{392} = \frac{308 \div 14}{392 \div 14} = \frac{22}{28} = \frac{22 \div 2}{28 \div 2} = \frac{11}{14} \]Case $\displaystyle 3$ — $\displaystyle 50$ circles. \[\text{length} = 50 \times 14 \text{ units} = 700 \text{ units} \] \[\text{area of rectangle} = 700 \text{ units} \times 14 \text{ units} = 9800 \text{ square units} \] \[\text{area of the 50 circles} = 50 \times 154 \text{ square units} = 7700 \text{ square units} \] \[\frac{7700}{9800} = \frac{7700 \div 100}{9800 \div 100} = \frac{77}{98} = \frac{77 \div 7}{98 \div 7} = \frac{11}{14} \]Three tests, three times \(\displaystyle \dfrac{11}{14}\). The conjecture has survived, so now prove it.Step $\displaystyle 4$: Prove the conjecture for any number of circles.Testing three cases is not a proof — there are infinitely many values left untested. So do the whole thing with letters instead of numbers.Let there be \(\displaystyle n\) circles, each of radius \(\displaystyle r\) units, so each diameter is \[d = 2r \text{ units}. \]Breadth of the rectangle: every circle touches the top edge and the bottom edge, so the breadth is one whole diameter (the distance right across a circle), not one radius: \[b = d = 2r \text{ units} \]Length of the rectangle: the \(\displaystyle n\) circles lie end to end, each touching the next, with the end circles touching the sides, so the length is \(\displaystyle n\) diameters: \[l = n \times d = n \times 2r = 2nr \text{ units} \]Area of the rectangle, using Area \(\displaystyle =\) length \(\displaystyle \times\) breadth: \[A_{\text{rectangle}} = l \times b = (2nr) \times (2r) = 4nr^{2} \text{ square units} \]Area of all the circles, using Area of a circle \(\displaystyle = \pi r^{2}\) and there being \(\displaystyle n\) of them, all the same size: \[A_{\text{circles}} = n \times \pi r^{2} = n\pi r^{2} \text{ square units} \]Now divide to get the fraction covered: \[\frac{A_{\text{circles}}}{A_{\text{rectangle}}} = \frac{n\pi r^{2}}{4nr^{2}} = \frac{\pi}{4} \] The \(\displaystyle n\) on the top cancels the \(\displaystyle n\) on the bottom, and \(\displaystyle r^{2}\) on the top cancels \(\displaystyle r^{2}\) on the bottom. Both the number of circles and their size have vanished from the answer, so the fraction is \(\displaystyle \dfrac{\pi}{4}\) for every \(\displaystyle n\) and every circle size. The conjecture is proved.Here is the same proof in a picture, which is the idea in the opening line: the rectangle is \(\displaystyle 2nr\) by \(\displaystyle 2r\), so it splits into \(\displaystyle n\) squares of side \(\displaystyle 2r\), each square holding exactly one circle that touches all four of its sides. One square has area \(\displaystyle (2r)^{2} = 4r^{2}\) square units and its circle has area \(\displaystyle \pi r^{2}\) square units, so each circle covers \(\displaystyle \dfrac{\pi r^{2}}{4r^{2}} = \dfrac{\pi}{4}\) of its own square. Every square is covered to the same extent, so the whole rectangle is too — and counting the squares never comes into it.Step $\displaystyle 5$: Put the fraction into numbers, and say how far it was rounded.Using \(\displaystyle \pi = \dfrac{22}{7}\): \[\frac{\pi}{4} = \frac{22}{28} = \frac{11}{14} \] \[\frac{11}{14} = 11 \div 14 = 0.785714\ldots \approx 0.786 \ \text{(rounded to 3 decimal places)} \] As a percentage: \[0.785714\ldots \times 100 = 78.5714\ldots\% \approx 78.6\% \ \text{(rounded to 1 decimal place, or } 78.57\% \text{ to 2 decimal places)} \]Step $\displaystyle 6$: Notice what the proof relied on.The proof used the arrangement in two places: the breadth being one diameter, and the length being \(\displaystyle n\) diameters. Both come from the circles being in one row, all the same size, touching the top and bottom edges, touching each other and touching the two ends. Change the packing — stack them in two rows, leave gaps, or use circles of different sizes — and those two lines stop being true, so \(\displaystyle \dfrac{\pi}{4}\) is no longer the answer.Answer: Conjecture — circles fitted into a rectangle in this manner always cover \(\displaystyle \dfrac{\pi}{4}\) of it, whatever their number; with \(\displaystyle \pi = \dfrac{22}{7}\) that is \(\displaystyle \dfrac{11}{14} \approx 0.786\), about \(\displaystyle 78.6\%\) (to $\displaystyle 1$ decimal place). It holds for $\displaystyle 10$, $\displaystyle 20$ and $\displaystyle 50$ circles — each gives \(\displaystyle \dfrac{1540}{1960} = \dfrac{3080}{3920} = \dfrac{7700}{9800} = \dfrac{11}{14}\) — and it is proved for every \(\displaystyle n\), since \(\displaystyle \dfrac{n\pi r^{2}}{4nr^{2}} = \dfrac{\pi}{4}\) with both \(\displaystyle n\) and \(\displaystyle r^{2}\) cancelling.

Split the big rectangle into its two rows, and notice that the small rectangles are turned sideways between the rows — that single fact hands you the equation you need.Look at the figure: the top row is $\displaystyle 4$ small rectangles standing side by side, and the bottom row is $\displaystyle 5$ small rectangles standing side by side, turned the other way. All $\displaystyle 9$ rectangles are identical (same length and same width) — the top row just uses them lying one way, and the bottom row uses them turned $\displaystyle 90$°.Let each small rectangle have: Step $\displaystyle 1$: Read the top row. In the top row, each rectangle lies with its length \(\displaystyle l \) running sideways (that's what makes the row wide) and its width \(\displaystyle w \) running up-down (that's the row's height). Four of them stand side by side: \[\text{width of top row} = 4l, \qquad \text{height of top row} = w \]Step $\displaystyle 2$: Read the bottom row. In the bottom row, each rectangle is turned the other way — its width \(\displaystyle w \) runs sideways and its length \(\displaystyle l \) runs up-down. Five of them stand side by side: \[\text{width of bottom row} = 5w, \qquad \text{height of bottom row} = l \]Step $\displaystyle 3$: The two rows sit on top of each other to make ONE rectangle, so their widths must be equal. This is the equation people usually miss — they look only at the area and forget the picture is telling them something too: \[4l = 5w \]Step $\displaystyle 4$: Use the area for a second equation. The $\displaystyle 9$ identical rectangles fit together edge to edge with no gaps and no overlaps, so the area of the big rectangle is exactly $\displaystyle 9$ times the area of one small rectangle: \[9 \times (l \times w) = 72\ \mathrm{cm}^2 \quad\Rightarrow\quad l \times w = 8\ \mathrm{cm}^2 \]Step $\displaystyle 5$: Solve the two equations together. From Step $\displaystyle 3$: \(\displaystyle w = \dfrac{4l}{5} \). Put this into \(\displaystyle lw = 8 \): \[l \times \frac{4l}{5} = 8 \quad\Rightarrow\quad \frac{4l^2}{5} = 8 \quad\Rightarrow\quad l^2 = 10 \quad\Rightarrow\quad l = \sqrt{10}\ \mathrm{cm} \](Watch out: \(\displaystyle l^2 = 10 \) is not a perfect square — and that's fine. This is a starred, harder question, and \(\displaystyle \sqrt{10} \) is a perfectly good exact length. Don't try to round it here; keep it as \(\displaystyle \sqrt{10} \) and round only in the very last step.)Now find \(\displaystyle w \): \[w = \frac{4l}{5} = \frac{4\sqrt{10}}{5}\ \mathrm{cm} \]Step $\displaystyle 6$: Check the picture actually closes up to area 72. Width of the big rectangle: \(\displaystyle 4l = 4\sqrt{10} \), and also \(\displaystyle 5w = 5 \times \dfrac{4\sqrt{10}}{5} = 4\sqrt{10} \) — the two rows agree on the width, so the shape really is a rectangle. Height of the big rectangle: \(\displaystyle w + l = \dfrac{4\sqrt{10}}{5} + \sqrt{10} = \dfrac{4\sqrt{10}+5\sqrt{10}}{5} = \dfrac{9\sqrt{10}}{5} \). Area: \(\displaystyle 4\sqrt{10} \times \dfrac{9\sqrt{10}}{5} = \dfrac{36\times10}{5} = \dfrac{360}{5} = 72\ \mathrm{cm}^2 \) — matches the question exactly.Step $\displaystyle 7$: Find the perimeter of one small rectangle. Perimeter of a rectangle \(\displaystyle = 2(l+w) \), where \(\displaystyle l \) and \(\displaystyle w \) are its length and width. \[2(l+w) = 2\left(\sqrt{10} + \frac{4\sqrt{10}}{5}\right) = 2 \times \frac{9\sqrt{10}}{5} = \frac{18\sqrt{10}}{5}\ \mathrm{cm} \]Round only at the end: \(\displaystyle \sqrt{10} \approx 3.1623 \), so \[\frac{18\sqrt{10}}{5} \approx \frac{18\times 3.1623}{5} \approx \frac{56.92}{5} \approx 11.38\ \mathrm{cm} \]Answer: Perimeter of each small rectangle \(\displaystyle = \dfrac{18\sqrt{10}}{5}\ \mathrm{cm} = 3.6\sqrt{10}\ \mathrm{cm} \approx 11.38\ \mathrm{cm}. \)

Equal bases plus one shared apex means one shared height, so both shaded triangles have area \(\displaystyle \tfrac12 bh \) sq. units — and because the trisection puts the two apexes exactly two base-lengths apart, blue can be slid onto red's base and then rearranged onto red in exactly five pieces.Naming the points in Fig. 6.48. The apex is \(\displaystyle A \). Going along the base from left to right, the four points are \(\displaystyle B \), \(\displaystyle P \), \(\displaystyle Q \), \(\displaystyle C \), where \(\displaystyle P \) and \(\displaystyle Q \) are the two points of trisection, so\[BP \;=\; PQ \;=\; QC \;=\; \tfrac{1}{3}\,BC \]The two segments \(\displaystyle AP \) and \(\displaystyle AQ \) split the big triangle into three smaller ones: (If your copy has the colours the other way round, swap the words "blue" and "red" everywhere below and read the picture right-to-left; every step works the same.)Give the two repeated lengths short names: \[b \;=\; \tfrac13\,BC \ \text{units} \qquad\text{and}\qquad h \;=\; \text{the perpendicular distance from } A \text{ to the line } BC,\ \text{in units} \]Part $\displaystyle 1$ — showing the two areas are equal.Formula (area of a triangle): \[\text{Area} \;=\; \tfrac12 \times \text{base} \times \text{height} \ \ \text{sq. units} \] Here "base" is the length in units of the side you choose to stand the triangle on, and "height" is the perpendicular distance in units from the opposite vertex to the line that base lies on. Height is not the length of a slanted side such as \(\displaystyle AB \), \(\displaystyle AP \), \(\displaystyle AQ \) or \(\displaystyle AC \).This is the one place to go slowly. The three small triangles are tilted differently, so it is tempting to give each of them its own separate height. They do not need one. All four points \(\displaystyle B, P, Q, C \) lie on a single straight line, and \(\displaystyle A \) is one single fixed point, so there is only one perpendicular distance from \(\displaystyle A \) down to that line — the \(\displaystyle h \) units defined above — and it serves as the height of all three triangles at once.So, standing each triangle on its piece of the base line: \[\text{ar}(\triangle ABP) \;=\; \tfrac12 \times BP \times h \;=\; \tfrac12 \times b \times h \ \ \text{sq. units} \] \[\text{ar}(\triangle AQC) \;=\; \tfrac12 \times QC \times h \;=\; \tfrac12 \times b \times h \ \ \text{sq. units} \] Both bases equal \(\displaystyle b \) units, because \(\displaystyle P \) and \(\displaystyle Q \) trisect \(\displaystyle BC \). Both heights equal \(\displaystyle h \) units. Therefore \[\text{ar}(\triangle ABP) \;=\; \tfrac12\,b\,h \;=\; \text{ar}(\triangle AQC) \ \ \text{sq. units} \]In this particular figure the two triangles are not congruent: the apex \(\displaystyle A \) does not sit above the midpoint of \(\displaystyle BC \), so blue and red really are different shapes carrying the same area. (Had \(\displaystyle A \) been placed above the midpoint of \(\displaystyle BC \), the two outer triangles would have been mirror images and so congruent — which is why the book, when it says the same thing about a median, adds the words "in general".)Part $\displaystyle 2$ — cutting the blue triangle up to cover the red one.One honest word before starting. The book says outright that it will not tell you the least number of pieces. The route below is one route that definitely works, and it finishes with five pieces. Five is not a claim about the smallest possible number.Two moves are used, and each one is checked as it is used: Step $\displaystyle 1$ — slide the blue triangle across. No scissors yet.Pick up the whole blue triangle \(\displaystyle \triangle ABP \) and slide it to the right, without turning it, by the distance \[BQ \;=\; BP + PQ \;=\; b + b \;=\; 2b \ \text{units} \] Since \(\displaystyle BP = b \) units, \(\displaystyle B \) lands exactly on \(\displaystyle Q \) and \(\displaystyle P \) lands exactly on \(\displaystyle C \). Blue's base now lies exactly on red's base \(\displaystyle QC \). The apex \(\displaystyle A \) travels to a new point \(\displaystyle A_1 \); a slide moves every point the same distance in the same direction, so \(\displaystyle A_1 \) is still \(\displaystyle h \) units above the base line, just \(\displaystyle 2b \) units further right.This step is the whole idea of the dissection. Blue's copy \(\displaystyle \triangle QCA_1 \) and the red triangle \(\displaystyle \triangle QCA \) now stand on the same base \(\displaystyle QC \), with their apexes \(\displaystyle A_1 \) and \(\displaystyle A \) on one and the same line parallel to \(\displaystyle BC \), \(\displaystyle h \) units up. Nothing has been cut. All that is left to do is move the apex from \(\displaystyle A_1 \) across to \(\displaystyle A \), and that gap is \[A_1A \;=\; 2b \ \text{units} \;=\; \text{exactly two base lengths} \] Hold on to that number — it is what makes the next two steps so short.Step $\displaystyle 2$ — one cut and one half-turn make blue into a parallelogram.Let \(\displaystyle M \) be the midpoint of \(\displaystyle QA_1 \) and \(\displaystyle N \) the midpoint of \(\displaystyle CA_1 \). Cut along \(\displaystyle MN \). That gives two pieces: the small triangle \(\displaystyle A_1MN \) on top, and the trapezium \(\displaystyle QCNM \) underneath. By the midpoint theorem (the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long), \[MN \parallel QC \qquad\text{and}\qquad MN \;=\; \tfrac12 \, QC \;=\; \tfrac{b}{2} \ \text{units} \] and \(\displaystyle MN \) sits at height \(\displaystyle \dfrac{h}{2} \) units.Now turn the small triangle \(\displaystyle A_1MN \) half a turn (\(\displaystyle 180^\circ \)) about the point \(\displaystyle N \). A half-turn about \(\displaystyle N \) swaps the two ends of any segment whose midpoint is \(\displaystyle N \). Since \(\displaystyle N \) is the midpoint of \(\displaystyle A_1C \), the apex \(\displaystyle A_1 \) lands exactly on \(\displaystyle C \). And \(\displaystyle M \) travels along the line \(\displaystyle MN \), through \(\displaystyle N \), to a point \(\displaystyle M'' \) with \[NM'' \;=\; NM \;=\; \tfrac{b}{2} \ \text{units} \] The turned piece is now the triangle \(\displaystyle C\,M''\,N \), and it meets the trapezium along the shared edge \(\displaystyle CN \) — same edge, same length, no gap and no overlap.Put the two pieces together and read off the outline: \(\displaystyle Q \to C \) along the bottom, up to \(\displaystyle M'' \), along the top through \(\displaystyle N \) to \(\displaystyle M \), back down to \(\displaystyle Q \). The bottom side \(\displaystyle QC \) is \(\displaystyle b \) units long. The top side is \[MM'' \;=\; MN + NM'' \;=\; \tfrac{b}{2} + \tfrac{b}{2} \;=\; b \ \text{units} \] and it is parallel to \(\displaystyle QC \), because \(\displaystyle MN \parallel QC \) and \(\displaystyle M'' \) lies on that same line. A quadrilateral with one pair of opposite sides both equal in length and parallel is a parallelogram. So blue is now a parallelogram on base \(\displaystyle QC \), of height \(\displaystyle \dfrac{h}{2} \) units, and its area is \[\text{base} \times \text{height} \;=\; b \times \tfrac{h}{2} \;=\; \tfrac12\,b\,h \ \ \text{sq. units} \] the same as the triangle it came from, as it has to be, since only a cut and a turn were used. Pieces so far: $\displaystyle 2$.Step $\displaystyle 3$ — one more cut turns blue's parallelogram into red's.Now do the same fold to the red triangle, but only in pencil — red is the target and is never cut. Let \(\displaystyle M_r \) be the midpoint of \(\displaystyle QA \) and \(\displaystyle N_r \) the midpoint of \(\displaystyle CA \). Folding red the same way would give a parallelogram on the same base \(\displaystyle QC \), of the same height \(\displaystyle \dfrac{h}{2} \) units, with top side running from \(\displaystyle M_r \) one base length to the right.Where do the two top sides sit relative to each other? The fold always puts the top-left corner at the midpoint of the segment from the base's left end \(\displaystyle Q \) to the apex. Blue's apex is \(\displaystyle A_1 \), red's is \(\displaystyle A \), and \(\displaystyle A_1 \) is \(\displaystyle 2b \) units right of \(\displaystyle A \). Taking midpoints halves that gap: \[M\,M_r \;=\; \tfrac12 \times A_1A \;=\; \tfrac12 \times 2b \;=\; b \ \text{units} \] So red's parallelogram is blue's parallelogram with the top side pushed exactly one base length to the left, the base staying put. That is precisely the extreme case allowed by the slide move — and at exactly one base length the required cut is simply the parallelogram's own diagonal:Cut blue's parallelogram along the diagonal from \(\displaystyle C \) (bottom-right corner) to \(\displaystyle M \) (top-left corner). Take the right-hand piece, triangle \(\displaystyle C\,M''\,M \), and slide it \(\displaystyle b \) units to the left. Under that slide \(\displaystyle C \to Q \), \(\displaystyle M'' \to M \) and \(\displaystyle M \to M_r \), so the piece lands as triangle \(\displaystyle Q\,M\,M_r \), fitting against the piece left behind, triangle \(\displaystyle Q\,C\,M \). The two together have bottom side \(\displaystyle QC \) and top side running from \(\displaystyle M_r \) to \(\displaystyle M \) — which is red's folded parallelogram exactly.How many pieces is that? \(\displaystyle C \) and \(\displaystyle M \) are both corners of the trapezium \(\displaystyle QCNM \), and a trapezium is convex, so the diagonal \(\displaystyle CM \) lies entirely inside the trapezium and never touches the turned triangle. It splits the trapezium in two and leaves the turned triangle whole. Pieces so far: $\displaystyle 3$.Step $\displaystyle 4$ — unfold, and red is covered.Red's parallelogram is red's triangle with its top folded down, so undo that fold. First locate the fold point: \(\displaystyle M_r \) is halfway across from \(\displaystyle Q \) to \(\displaystyle A \), and \(\displaystyle N_r \) is halfway across from \(\displaystyle C \) to \(\displaystyle A \), so the sideways gap between them is half of \(\displaystyle QC \), that is \(\displaystyle \dfrac{b}{2} \) units — exactly half of the top side. In other words, \(\displaystyle N_r \) is the midpoint of the top side of the parallelogram you are now holding.Cut from \(\displaystyle C \) (the bottom-right corner) to \(\displaystyle N_r \) (that midpoint). Take the piece to the right of this cut, triangle \(\displaystyle C\,M\,N_r \), and turn it half a turn about \(\displaystyle N_r \). Because \(\displaystyle N_r \) is the midpoint of \(\displaystyle CA \), the corner \(\displaystyle C \) lands on \(\displaystyle A \); because \(\displaystyle N_r \) is the midpoint of the top side \(\displaystyle M_rM \), the corner \(\displaystyle M \) lands on \(\displaystyle M_r \). That piece therefore lands as the triangle \(\displaystyle A\,M_r\,N_r \), which is the top of the red triangle, and what stays behind is the trapezium \(\displaystyle Q\,C\,N_r\,M_r \), which is the bottom of the red triangle. The red triangle \(\displaystyle \triangle AQC \) is now completely covered by paper that started life as the blue triangle.Counting the pieces. The seam left over from Step $\displaystyle 3$ runs from \(\displaystyle Q \) up to \(\displaystyle M \). The new cut runs from \(\displaystyle C \) up to \(\displaystyle N_r \). Since \(\displaystyle C \) is to the right of \(\displaystyle Q \) along the bottom while \(\displaystyle N_r \) is to the left of \(\displaystyle M \) along the top, those two segments cross each other, at one point — call it \(\displaystyle X \). Crossing that seam means the new cut slices through two of the three pieces, turning each of them into two. So the finished dissection has \[3 + 2 \;=\; 5 \ \text{pieces} \] Three of them settle into red's bottom trapezium — \(\displaystyle QCX \), \(\displaystyle QXN_r \) and \(\displaystyle QN_rM_r \) — and two of them, \(\displaystyle CMX \) and \(\displaystyle XMN_r \), make the half-turn together to become red's top triangle.Every single move used was a straight cut, a slide, or a half-turn. None of these stretches or shrinks anything, so the five pieces keep their exact areas and shapes throughout, and the five of them fit inside \(\displaystyle \triangle AQC \) with no gap and no overlap. Their total area is \(\displaystyle \tfrac12 bh \) sq. units, matching Part $\displaystyle 1$ as expected.Answer: The two shaded triangles have equal area, \(\displaystyle \text{ar}(\triangle ABP) = \tfrac12\,b\,h = \text{ar}(\triangle AQC) \) sq. units where \(\displaystyle b = \tfrac13 BC \) units and \(\displaystyle h \) units is the height of \(\displaystyle A \) above line \(\displaystyle BC \), because the trisection makes their bases equal and the single apex \(\displaystyle A \) gives them one common height; and the blue triangle can be cut into $\displaystyle 5$ pieces that cover the red one — slide blue \(\displaystyle 2b \) units right onto base \(\displaystyle QC \), cut along the midline and half-turn the top piece to get a parallelogram, cut that parallelogram's diagonal from \(\displaystyle C \) to the top-left corner and slide that half one base length left, then cut from \(\displaystyle C \) to the midpoint of the top side and half-turn that piece about the midpoint of \(\displaystyle CA \).