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

Big idea: if you cut a shape into pieces and put the pieces back together without gaps or overlaps, the total area never changes. Draw a square, cut it into labelled pieces, and the pieces' areas must add up to the whole square's area — that "must add up" is exactly what an algebraic identity says.First, let's see how the given example works, because both questions use the same trick.How \(\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\) becomes a picture. Draw a square whose side is \(\displaystyle (a+b)\). On the bottom side, mark a point at distance \(\displaystyle a\) from the left corner (so the rest of that side, length \(\displaystyle b\), is left over). Do the same on the left side: mark a point at distance \(\displaystyle a\) up from the bottom corner. Now draw one vertical line and one horizontal line through these two marks. This slices the big square into a \(\displaystyle 2\times2\) grid of $\displaystyle 4$ smaller rectangles:

a^{$\displaystyle 2$}	ab
ab	b^{$\displaystyle 2$}

The bottom-left piece is an \(\displaystyle a \times a\) square (area \(\displaystyle a^{2}\)), the top-right piece is a \(\displaystyle b\times b\) square (area \(\displaystyle b^{2}\)), and the other two pieces are both \(\displaystyle a\times b\) rectangles (area \(\displaystyle ab\) each). Since these $\displaystyle 4$ pieces exactly tile the big square with no gaps or overlaps, their areas must add to the square's total area \(\displaystyle (a+b)^{2}\): \[(a+b)^{2}=a^{2}+ab+ab+b^{2}=a^{2}+2ab+b^{2}. \] That's the whole trick: split each side into the two lengths being added, draw the grid, and read off the areas.Part $\displaystyle 1$ — the figure for \(\displaystyle (a+b)(a-b)=a^{2}-b^{2}\).Aside before you start: this identity needs \(\displaystyle a\) to be the bigger length, since you are about to cut a smaller square (side \(\displaystyle b\)) out of a bigger one (side \(\displaystyle a\)). If you mix up which letter is bigger, the construction won't make sense.1. Draw a square of side \(\displaystyle a\). 2. On the top side, mark a point at distance \(\displaystyle (a-b)\) from the left corner — so the leftover piece on that side, toward the right corner, has length \(\displaystyle b\). 3. Do the same on the right side: mark a point at distance \(\displaystyle (a-b)\) up from the bottom corner, leaving a length \(\displaystyle b\) at the top. 4. Through these two marks draw one horizontal line and one vertical line. This cuts the square into $\displaystyle 4$ pieces, and because you used lengths \(\displaystyle (a-b)\) and \(\displaystyle b\) instead of \(\displaystyle a\) and \(\displaystyle b\), the grid looks like this:

(a-b)^{$\displaystyle 2$}	(a-b)\,b
b\,(a-b)	b^{$\displaystyle 2$}

The four pieces sit at: bottom-left (a square of side \(\displaystyle (a-b)\)), bottom-right and top-left (two rectangles of size \(\displaystyle (a-b)\times b\)), and top-right, tucked right into the corner (a small square of side \(\displaystyle b\), area \(\displaystyle b^{2}\)).5. Now cut out that top-right corner square (area \(\displaystyle b^{2}\)) and throw it away. What's left is an L-shaped piece made of the other three regions, and its area is exactly \(\displaystyle a^{2}-b^{2}\) (the whole square minus the corner you removed). 6. This L-shape is really the bottom-left square plus the two matching rectangles glued to its right side and its top: \[a^{2}-b^{2} = (a-b)^{2} + (a-b)b + b(a-b) = (a-b)^{2} + 2b(a-b). \] 7. Here's the "rearranging" step that turns it into a rectangle: take the rectangle sitting on top of the bottom-left square — the one with width \(\displaystyle (a-b)\) and height \(\displaystyle b\) — and turn it on its side (rotate it a quarter turn). It now has width \(\displaystyle b\) and height \(\displaystyle (a-b)\), which is exactly the height of the big block below it. Slide this turned piece so it sits snugly against the right-hand side of the bottom-left-plus-bottom-right block.Because turning a piece over doesn't change its area, you haven't added or removed anything — you've only repositioned it. What you now have is one single rectangle: its height is \(\displaystyle (a-b)\) (unchanged), and its width is the old width \(\displaystyle a\) plus the extra \(\displaystyle b\) you just slid in, giving width \(\displaystyle (a+b)\). A single rectangle of width \(\displaystyle (a+b)\) and height \(\displaystyle (a-b)\) has area \[(a+b)(a-b). \] But this rectangle is made of the exact same L-shaped pieces as before, just rearranged — so its area must still equal \(\displaystyle a^{2}-b^{2}\) from step 6. That gives you the identity, straight from the picture: \[(a+b)(a-b)=a^{2}-b^{2}. \]Part $\displaystyle 2$ — the figure for \(\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\).This is the same grid idea as the very first example, just stretched from $\displaystyle 2$ pieces per side to 3.1. Draw a square of side \(\displaystyle (a+b+c)\). 2. On the bottom side, mark two points: one at distance \(\displaystyle a\) from the left corner, and a second at distance \(\displaystyle (a+b)\) from the left corner. This splits the bottom side into three pieces, in order: length \(\displaystyle a\), then length \(\displaystyle b\), then length \(\displaystyle c\). 3. On the left side, mark two points the same way — at distance \(\displaystyle a\) and at distance \(\displaystyle (a+b)\) from the bottom corner — so the left side is also split, in the same order, into \(\displaystyle a\), then \(\displaystyle b\), then \(\displaystyle c\).Aside: use the same order \(\displaystyle (a,b,c)\) on both sides. If you swap the order on one side, the picture still works out algebraically, but the three squares stop lining up along one diagonal and the pattern gets confusing to read — keeping the order matched is what makes the picture easy to check by eye.4. Through all four marked points, draw lines parallel to the sides of the square — two vertical lines from the bottom marks, two horizontal lines from the left-side marks. This carves the big square into a \(\displaystyle 3\times3\) grid of $\displaystyle 9$ rectangles. 5. Each cell's area is (its column length) times (its row length). Reading the grid out gives:

a^{$\displaystyle 2$}	ab	ac
ab	b^{$\displaystyle 2$}	bc
ac	bc	c^{$\displaystyle 2$}

Along the diagonal from bottom-left to top-right sit the three squares: \(\displaystyle a^{2}\), \(\displaystyle b^{2}\), \(\displaystyle c^{2}\). Every other cell is repeated exactly once more on the opposite side of that diagonal: \(\displaystyle ab\) appears twice, \(\displaystyle bc\) appears twice, and \(\displaystyle ac\) (the same as \(\displaystyle ca\)) appears twice.6. These $\displaystyle 9$ rectangles tile the whole big square with no gaps and no overlaps, so their areas must add up to the square's total area: \[(a+b+c)^{2} = a^{2}+b^{2}+c^{2} + (ab+ab) + (bc+bc) + (ac+ac) \] \[(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca. \]Notice \(\displaystyle 2ac\) and \(\displaystyle 2ca\) are the same term written two ways — it isn't a missing piece, it's just the pair of matching \(\displaystyle ac\)-rectangles counted once each.Answer: \(\displaystyle (a+b)(a-b)=a^{2}-b^{2}\) is shown by a square of side \(\displaystyle a\) with a corner square of side \(\displaystyle b\) cut out, whose remaining L-shaped piece (area \(\displaystyle a^{2}-b^{2}\)) can be re-cut and slid into a single rectangle of sides \(\displaystyle (a+b)\) and \(\displaystyle (a-b)\). \(\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\) is shown by a square of side \(\displaystyle (a+b+c)\) split by a \(\displaystyle 3\times3\) grid (segments \(\displaystyle a,b,c\) in the same order on both sides) into three squares \(\displaystyle a^{2},b^{2},c^{2}\) on the diagonal plus three matching pairs of rectangles \(\displaystyle ab,\,bc,\,ca\), each pair contributing \(\displaystyle 2ab\), \(\displaystyle 2bc\), \(\displaystyle 2ca\).

Split the triangle using its two equal sides — once you know all three sides, Heron's formula gives you the area directly, no height needed.Step $\displaystyle 1$: Find the third side.You're told the perimeter is $\displaystyle 40$ cm and two sides (the equal ones) are $\displaystyle 15$ cm each. The perimeter is just the sum of all three sides, so:\[\text{third side} = 40 - 15 - 15 = 10 \text{ cm} \]So the triangle has sides \(\displaystyle a = 15 \) cm, \(\displaystyle b = 15 \) cm, \(\displaystyle c = 10 \) cm.Common slip: don't assume the "third side" is also $\displaystyle 15$ cm just because the triangle is isosceles — isosceles means only two sides are equal, and here those are the two you were already given.Step $\displaystyle 2$: Set up Heron's formula.Heron's formula finds the area of a triangle from its three side lengths alone:\[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]where \(\displaystyle s \) is the semi-perimeter (half the perimeter) and \(\displaystyle a, b, c \) are the three side lengths.\[s = \frac{a+b+c}{2} = \frac{15+15+10}{2} = \frac{40}{2} = 20 \text{ cm} \]Step $\displaystyle 3$: Work out each bracket.\[s - a = 20 - 15 = 5 \text{ cm} \] \[s - b = 20 - 15 = 5 \text{ cm} \] \[s - c = 20 - 10 = 10 \text{ cm} \]Step $\displaystyle 4$: Plug into Heron's formula.\[\text{Area} = \sqrt{20 \times 5 \times 5 \times 10} = \sqrt{5000} \text{ cm}^2 \]Now simplify \(\displaystyle \sqrt{5000} \) instead of reaching for a decimal too early — break $\displaystyle 5000$ into a perfect square times a leftover factor:\[5000 = 2500 \times 2 \] \[\sqrt{5000} = \sqrt{2500} \times \sqrt{2} = 50\sqrt{2} \text{ cm}^2 \]Using \(\displaystyle \sqrt{2} \approx 1.414 \):\[\text{Area} \approx 50 \times 1.414 = 70.7 \text{ cm}^2 \]Note: this question doesn't involve \(\displaystyle \pi \) at all — the "use \(\displaystyle \frac{22}{7} \)" instruction in this chapter is only for parts with circles. Don't go looking for a circle here; this triangle's area comes purely from Heron's formula.Step $\displaystyle 5$: Quick check with a different method.Since the triangle is isosceles, dropping a perpendicular (height) from the top vertex to the base of $\displaystyle 10$ cm splits that base exactly in half — this is a property of isosceles triangles, the same perpendicular is also the line of symmetry. So it splits into two right triangles, each with: By the Pythagorean theorem, \(\displaystyle \text{height}^2 + \text{base}^2 = \text{hypotenuse}^2 \):\[h^2 = 15^2 - 5^2 = 225 - 25 = 200 \] \[h = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \text{ cm} \]Now use the triangle area formula, Area \(\displaystyle = \frac{1}{2} \times \text{base} \times \text{height} \), with the full base ($\displaystyle 10$ cm) and this height:\[\text{Area} = \frac{1}{2} \times 10 \times 10\sqrt{2} = 50\sqrt{2} \text{ cm}^2 \]Same answer both ways — that confirms it.Answer: The area of the triangle is \(\displaystyle 50\sqrt{2} \text{ cm}^2 \approx 70.7 \text{ cm}^2 \).

Split the isosceles triangle into two identical right triangles by drawing the altitude from the top vertex to the base — that altitude cuts the base exactly in half, and once you know the altitude, Pythagoras gives you the equal side.Step $\displaystyle 1$: Find the height using the area formula.Name the formula: Area of a triangle \(\displaystyle = \frac{1}{2} \times \text{base} \times \text{height} \).Here base \(\displaystyle = 10 \mathrm{~cm} \) and Area \(\displaystyle = 60 \mathrm{~cm}^{2} \). Let the height (from the top vertex straight down to the base) be \(\displaystyle h \) cm.\[60 = \frac{1}{2} \times 10 \times h \]\[60 = 5h \]\[h = 12 \mathrm{~cm} \]Step $\displaystyle 2$: Use the special property of an isosceles triangle's altitude.In an isosceles triangle, the altitude drawn from the vertex between the two equal sides down to the base does two things at once: it meets the base at a right angle, and it cuts the base into two exactly equal halves. This only works because the triangle is isosceles — don't use this shortcut on a scalene triangle.So each half of the base is:\[\frac{10}{2} = 5 \mathrm{~cm} \]This altitude also splits the whole triangle into two congruent right-angled triangles, each with: Careful here: the hypotenuse of this small right triangle is the slanted equal side, not the height. Don't accidentally use $\displaystyle 12$ cm as your final answer — that's only the vertical height, not the side length asked for.Step $\displaystyle 3$: Apply the Pythagoras theorem.Name the formula: in a right triangle, \(\displaystyle (\text{hypotenuse})^{2} = (\text{leg}_1)^{2} + (\text{leg}_2)^{2} \).Let the equal side be \(\displaystyle a \) cm.\[a^{2} = 12^{2} + 5^{2} \]\[a^{2} = 144 + 25 \]\[a^{2} = 169 \]\[a = \sqrt{169} = 13 \mathrm{~cm} \]Step $\displaystyle 4$: Check it makes sense.Both equal sides are $\displaystyle 13$ cm, base is $\displaystyle 10$ cm — this satisfies the triangle inequality ($\displaystyle 13$ + $\displaystyle 13$ > $\displaystyle 10$, and $\displaystyle 13$ + $\displaystyle 10$ > $\displaystyle 13$), so a triangle like this really can exist. You can also double-check the area: \(\displaystyle \frac{1}{2} \times 10 \times 12 = 60 \mathrm{~cm}^2 \), which matches what was given.Answer: Each of the equal sides is $\displaystyle 13$ cm long.

The two legs of a right-angled triangle are the base and the height, so the area plus one leg gives you the other leg — then the Pythagorean theorem gives the hypotenuse, and the three sides add up to the perimeter.You are told the area is \(\displaystyle 54\) sq. cm and that one leg is \(\displaystyle 12\) cm long. The legs are the two sides that meet at the right angle. Call the leg you do not know \(\displaystyle b\).Step $\displaystyle 1$: Use the area to find the second leg.Area of a triangle: \(\displaystyle \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\), where the height is measured at right angles to the base.In a right-angled triangle the two legs meet at \(\displaystyle 90^\circ\), so each leg is already at right angles to the other. That means you can take one leg as the base and the other leg as the height, with no extra line to draw:\[54 \text{ cm}^2 = \frac{1}{2} \times 12 \text{ cm} \times b \]\[54 \text{ cm}^2 = 6 \text{ cm} \times b \]\[b = \frac{54 \text{ cm}^2}{6 \text{ cm}} = 9 \text{ cm} \]So the two legs are \(\displaystyle 12\) cm and \(\displaystyle 9\) cm.Watch out: read the \(\displaystyle 12\) cm as a leg, not as the hypotenuse. If you take it to be the hypotenuse, you are left with no height to put into the area formula and the question looks impossible. The hypotenuse is the side opposite the right angle, and the question does not give it to you — you work it out in Step 2.Step $\displaystyle 2$: Find the hypotenuse with the Pythagorean theorem.Pythagorean theorem: \(\displaystyle a^2 + b^2 = c^2\), where \(\displaystyle a\) and \(\displaystyle b\) are the two legs and \(\displaystyle c\) is the hypotenuse, the longest side, lying opposite the right angle.\[c^2 = (12 \text{ cm})^2 + (9 \text{ cm})^2 = 144 \text{ cm}^2 + 81 \text{ cm}^2 = 225 \text{ cm}^2 \]\[c = \sqrt{225 \text{ cm}^2} = 15 \text{ cm} \]A quick check that \(\displaystyle 15\) is right and not a guess: \(\displaystyle 9, 12, 15\) is the familiar \(\displaystyle 3, 4, 5\) triangle with every side multiplied by \(\displaystyle 3\).A note on the aside above, so you are not left with a wrong rule: it is not that the hypotenuse can never be a base. You could use the hypotenuse as the base — here \(\displaystyle \frac{1}{2} \times 15 \text{ cm} \times 7.2 \text{ cm} = 54 \text{ cm}^2\), the same area — but only with the height drawn from the right angle down to the hypotenuse, which is \(\displaystyle 7.2\) cm. Nobody gives you that height in this question, so the legs are the sensible pair to use.Step $\displaystyle 3$: Add the three sides to get the perimeter.Perimeter of a triangle: \(\displaystyle P = \text{side}_1 + \text{side}_2 + \text{side}_3\), the total distance once around it.\[P = 12 \text{ cm} + 9 \text{ cm} + 15 \text{ cm} = 36 \text{ cm} \]Every number here came out whole, so there is nothing to round off. The chapter tells you to use \(\displaystyle \pi = \frac{22}{7}\), but this question has no circle in it, so \(\displaystyle \pi\) never appears in the working.Answer: The perimeter is \(\displaystyle 36\) cm (legs \(\displaystyle 12\) cm and \(\displaystyle 9\) cm, hypotenuse \(\displaystyle 15\) cm).

Find the actual side lengths from the ratio, then use Heron's formula to get the area — you don't know any angle here, so Heron's is the only tool that works.Step $\displaystyle 1$: Turn the ratio into real side lengths.The sides are in the ratio \(\displaystyle 2:3:4\), so call them \(\displaystyle 2x\), \(\displaystyle 3x\), \(\displaystyle 4x\) for some common length \(\displaystyle x\) (in cm). "Ratio $\displaystyle 2$:$\displaystyle 3$:$\displaystyle 4$" just means these three sides are $\displaystyle 2$ parts, $\displaystyle 3$ parts, and $\displaystyle 4$ parts of the same unknown part-length \(\displaystyle x\) — it does not mean the sides are $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$ cm.The perimeter is the sum of all three sides: \[2x + 3x + 4x = 45 \] \[9x = 45 \implies x = 5 \text{ cm} \]So the three sides are: \[a = 2x = 10 \text{ cm}, \quad b = 3x = 15 \text{ cm}, \quad c = 4x = 20 \text{ cm} \]Check: \(\displaystyle 10 + 15 + 20 = 45\) cm. Good, that matches the given perimeter.Step $\displaystyle 2$: Set up Heron's formula.Heron's formula finds the area of a triangle from its three sides alone: \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \] where \(\displaystyle s\) is the semi-perimeter — half the perimeter — and \(\displaystyle a, b, c\) are the three side lengths.\[s = \frac{a+b+c}{2} = \frac{10+15+20}{2} = \frac{45}{2} = 22.5 \text{ cm} \](A common slip here: people plug the full perimeter, $\displaystyle 45$, into the formula instead of half of it. Always divide by $\displaystyle 2$ first.)Step $\displaystyle 3$: Compute \(\displaystyle s-a\), \(\displaystyle s-b\), \(\displaystyle s-c\).\[s - a = 22.5 - 10 = 12.5 \text{ cm} \] \[s - b = 22.5 - 15 = 7.5 \text{ cm} \] \[s - c = 22.5 - 20 = 2.5 \text{ cm} \]Step $\displaystyle 4$: Multiply everything under the square root.To keep things exact (no rounding yet), write each decimal as a fraction: \[s = \frac{45}{2}, \quad s-a = \frac{25}{2}, \quad s-b = \frac{15}{2}, \quad s-c = \frac{5}{2} \]\[s(s-a)(s-b)(s-c) = \frac{45}{2} \times \frac{25}{2} \times \frac{15}{2} \times \frac{5}{2} = \frac{45 \times 25 \times 15 \times 5}{16} \]Multiply the top: \(\displaystyle 45 \times 25 = 1125\), then \(\displaystyle 1125 \times 15 = 16875\), then \(\displaystyle 16875 \times 5 = 84375\).\[s(s-a)(s-b)(s-c) = \frac{84375}{16} \]Step $\displaystyle 5$: Take the square root.\[\text{Area} = \sqrt{\frac{84375}{16}} = \frac{\sqrt{84375}}{4} \]Pull out the largest perfect square hiding inside 84375. Since \(\displaystyle 84375 = 5625 \times 15\) and \(\displaystyle 5625 = 75^2\): \[\sqrt{84375} = \sqrt{5625 \times 15} = 75\sqrt{15} \]So the exact area is: \[\text{Area} = \frac{75\sqrt{15}}{4} \text{ cm}^2 \]Step $\displaystyle 6$: Convert to a decimal.Using \(\displaystyle \sqrt{15} \approx 3.873\): \[\text{Area} \approx \frac{75 \times 3.873}{4} = \frac{290.475}{4} \approx 72.62 \text{ cm}^2 \]Answer: The sides are $\displaystyle 10$ cm, $\displaystyle 15$ cm, and $\displaystyle 20$ cm, and the area of the triangle is \(\displaystyle \dfrac{75\sqrt{15}}{4} \approx 72.62 \text{ cm}^2\).

Check first whether this is a right triangle — if it is, you get an instant shortcut.The sides given are \(\displaystyle 7\ \text{cm}\), \(\displaystyle 24\ \text{cm}\), \(\displaystyle 25\ \text{cm}\). Whenever you see a triangle's three sides, it is worth testing the Pythagoras relation on the two shorter sides against the longest one, because "$\displaystyle 3$-$\displaystyle 4$-$\displaystyle 5$-style" triples show up a lot in these problems.\[7^2 + 24^2 = 49 + 576 = 625 \] \[25^2 = 625 \]Both sides match: \(\displaystyle 7^2 + 24^2 = 25^2\). So this triangle is right-angled, and the right angle sits between the sides \(\displaystyle 7\ \text{cm}\) and \(\displaystyle 24\ \text{cm}\) (the two legs), with \(\displaystyle 25\ \text{cm}\) as the hypotenuse.Way $\displaystyle 1$: Use the right-angle shortcut — the two legs act as base and height.In a right triangle, once you know which two sides meet at the right angle, those two sides ARE the base and the height of the triangle — you don't need to draw or find any extra height, because they are already perpendicular to each other.Formula for the area of a triangle: \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]Take base \(\displaystyle = 24\ \text{cm}\) and height \(\displaystyle = 7\ \text{cm}\) (the two legs):\[\text{Area} = \frac{1}{2} \times 24\ \text{cm} \times 7\ \text{cm} = \frac{1}{2} \times 168\ \text{cm}^2 = 84\ \text{cm}^2 \]Common slip: don't pick the hypotenuse (\(\displaystyle 25\ \text{cm}\)) as one of the two multiplied sides here — the hypotenuse is never one of the two perpendicular sides in this shortcut. It's only used to identify which side is the hypotenuse in the first place.Way $\displaystyle 2$: Use Heron's formula — this works even without knowing it's a right triangle.Heron's formula finds the area of any triangle purely from its three side lengths \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\):\[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}, \qquad \text{where } s = \frac{a+b+c}{2} \]Here \(\displaystyle s\) is the semi-perimeter (half the perimeter), and \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\) are the three side lengths.Let \(\displaystyle a = 7\ \text{cm}\), \(\displaystyle b = 24\ \text{cm}\), \(\displaystyle c = 25\ \text{cm}\).Step $\displaystyle 1$ — find \(\displaystyle s\): \[s = \frac{a+b+c}{2} = \frac{7 + 24 + 25}{2} = \frac{56}{2} = 28\ \text{cm} \]Step $\displaystyle 2$ — find each bracket: \[s - a = 28 - 7 = 21\ \text{cm} \] \[s - b = 28 - 24 = 4\ \text{cm} \] \[s - c = 28 - 25 = 3\ \text{cm} \]Step $\displaystyle 3$ — multiply everything under the square root: \[s(s-a)(s-b)(s-c) = 28 \times 21 \times 4 \times 3 \]Multiply step by step (don't round anywhere here — keep exact whole numbers): \[28 \times 21 = 588 \] \[588 \times 4 = 2352 \] \[2352 \times 3 = 7056 \]Step $\displaystyle 4$ — take the square root: \[\text{Area} = \sqrt{7056}\ \text{cm}^2 = 84\ \text{cm}^2 \]Common slip: the units under the square root are \(\displaystyle \text{cm}^4\) (since you multiplied four lengths in cm), and \(\displaystyle \sqrt{\text{cm}^4} = \text{cm}^2\) — that's why the answer comes out in cm², matching an area, not cm⁴.Both methods land on exactly the same number, \(\displaystyle 84\ \text{cm}^2\) — which is a good sign you did each one correctly, since a right-triangle area shouldn't depend on which valid method you use to find it.Answer: The area of the triangle is \(\displaystyle 84\ \text{cm}^2\), found both by the right-angle shortcut \(\displaystyle \left(\frac{1}{2} \times 24 \times 7\right)\) and by Heron's formula \(\displaystyle \left(\sqrt{28 \times 21 \times 4 \times 3}\right)\).

One full turn of the wheel carries the cycle forward by exactly one circumference — so find the circumference, then multiply by 100.Step $\displaystyle 1$: Get the circumference from the diameter.The formula for the circumference of a circle in terms of its diameter is\[C = \pi d \]where \(\displaystyle C\) is the circumference and \(\displaystyle d\) is the diameter (the full width straight across the circle, through the centre).You're given \(\displaystyle d = 60\) cm — this is already the diameter, not the radius, so don't halve it here. (If a problem instead gives you the radius \(\displaystyle r\), remember \(\displaystyle d = 2r\), and the formula becomes \(\displaystyle C = 2\pi r\). Mixing up radius and diameter is the single most common slip in wheel problems.)Using \(\displaystyle \pi = \dfrac{22}{7}\) as the instructions say:\[C = \frac{22}{7} \times 60 \text{ cm} = \frac{1320}{7} \text{ cm} \]As a decimal, \(\displaystyle \dfrac{1320}{7} = 188.5714...\), so\[C \approx 188.57 \text{ cm (one full rotation)} \]Step $\displaystyle 2$: One rotation = one circumference, so $\displaystyle 100$ rotations = $\displaystyle 100$ × circumference.Every time the wheel turns once, the bicycle moves forward by exactly \(\displaystyle C\) — the wheel is rolling, not slipping, so the ground distance covered equals the length that unrolls off the rim, which is the circumference.\[\text{Distance} = 100 \times C = 100 \times \frac{1320}{7} \text{ cm} = \frac{132000}{7} \text{ cm} \]\[\frac{132000}{7} = 18857.142857... \]So the distance is about \(\displaystyle 18857.14\) cm.Step $\displaystyle 3$: Convert to a sensible unit (metres).A distance a cyclist travels is usually read in metres, not centimetres. Since \(\displaystyle 1\) m \(\displaystyle = 100\) cm, divide by $\displaystyle 100$:\[\frac{132000}{7} \text{ cm} \div 100 = \frac{1320}{7} \text{ m} \approx 188.57 \text{ m} \](Notice this is the same fraction as the circumference in Step $\displaystyle 1$, just with the units shifted from cm to m — because we divided both the "$\displaystyle 100$ rotations" multiplication and the final unit conversion consistently. That's a good self-check: the number $\displaystyle 1320$/$\displaystyle 7$ showing up twice, once in cm and once in m, isn't a coincidence — it's $\displaystyle 100$ cm-circumferences becoming $\displaystyle 100$/$\displaystyle 100$ = $\displaystyle 1$ m-circumference-worth of digits.)Rounding to two decimal places, the cyclist travels approximately \(\displaystyle 188.57\) m after the wheel has rotated $\displaystyle 100$ times.Answer: The cyclist travels \(\displaystyle \dfrac{1320}{7}\) m \(\displaystyle \approx 188.57\) m (that is, about \(\displaystyle 18857.14\) cm) after the wheel rotates $\displaystyle 100$ times.

Split it into two steps: first find the radius using the circumference, then find one-quarter of the circle's area.Step $\displaystyle 1$ — find the radius from the circumferenceThe formula for the circumference of a circle is \[C = 2\pi r \] where \(\displaystyle C\) is the circumference and \(\displaystyle r\) is the radius (not the diameter — a common slip is to plug the circumference in and get the diameter instead).We're told \(\displaystyle C = 66\) cm, and the instructions say to use \(\displaystyle \pi = \dfrac{22}{7}\). Substitute: \[66 = 2 \times \frac{22}{7} \times r \] \[66 = \frac{44}{7} \times r \]Multiply both sides by \(\displaystyle \dfrac{7}{44}\) to isolate \(\displaystyle r\): \[r = 66 \times \frac{7}{44} = \frac{462}{44} = \frac{21}{2} = 10.5 \text{ cm} \]Step $\displaystyle 2$ — find the area of the full circleThe formula for the area of a circle is \[A = \pi r^2 \] where \(\displaystyle r\) is the same radius as above. Using \(\displaystyle r = \dfrac{21}{2}\) cm keeps the numbers as clean fractions (squaring $\displaystyle 10.5$ by decimal works too, but the fraction avoids rounding): \[A = \frac{22}{7} \times \left(\frac{21}{2}\right)^2 = \frac{22}{7} \times \frac{441}{4} \]Cancel the $\displaystyle 7$ into $\displaystyle 441$ first (\(\displaystyle 441 \div 7 = 63\)) before multiplying anything out: \[A = \frac{22 \times 63}{4} = \frac{1386}{4} = 346.5 \text{ cm}^2 \]That is the area of the whole circle.Step $\displaystyle 3$ — take one quadrantA quadrant is a quarter of a circle — the region swept out by \(\displaystyle 90°\) out of the full \(\displaystyle 360°\) around the centre. So its area is exactly \(\displaystyle \dfrac{1}{4}\) of the full circle's area: \[\text{Area of quadrant} = \frac{1}{4} \times A = \frac{1}{4} \times 346.5 = 86.625 \text{ cm}^2 \]As an exact fraction this is \(\displaystyle \dfrac{693}{8}\) cm², which equals $\displaystyle 86.625$ cm² — nothing was rounded along the way, only written as a decimal at the very end.Answer: The area of the quadrant is \(\displaystyle 86.625 \text{ cm}^2\) (i.e. \(\displaystyle \dfrac{693}{8}\) cm²).

Each full turn of the wheel moves the car forward by exactly the wheel's circumference — so find the circumference first, then divide the total journey distance by it.Step $\displaystyle 1$: Write down what is given.The outer radius of the wheel is \(\displaystyle r = 28 \) cm. This is already the radius (centre to rim) — not the diameter — so nothing here needs to be doubled or halved. We're told to use \(\displaystyle \pi = \dfrac{22}{7} \) throughout.Step $\displaystyle 2$: Find the distance covered in one turn (the circumference).The circumference of a circle is \[C = 2\pi r \] where \(\displaystyle r \) is the radius.Substitute \(\displaystyle r = 28 \) cm and \(\displaystyle \pi = \dfrac{22}{7} \): \[C = 2 \times \dfrac{22}{7} \times 28 \text{ cm} \]Cancel the $\displaystyle 7$ into $\displaystyle 28$ first (\(\displaystyle 28 \div 7 = 4 \)) so the answer comes out exact, with no rounding: \[C = 2 \times 22 \times 4 \text{ cm} = 176 \text{ cm} \]So every single turn of the wheel carries the car forward by $\displaystyle 176$ cm.Step $\displaystyle 3$: Convert the journey length to the same unit.The journey is $\displaystyle 1$ km, but the circumference is in centimetres — these must match before dividing. Convert in two clean hops so it's easy to check: \[1 \text{ km} = 1000 \text{ m} = 1000 \times 100 \text{ cm} = 100000 \text{ cm} \](This is where people go wrong — writing \(\displaystyle 1 \text{ km} = 1000 \text{ cm} \) and skipping the metres-to-centimetres step.)Step $\displaystyle 4$: Divide the total distance by the distance covered in one turn.\[\text{Number of turns} = \dfrac{\text{total distance}}{\text{distance per turn}} = \dfrac{100000}{176} \]Simplify the fraction (divide top and bottom by $\displaystyle 16$): \[\dfrac{100000}{176} = \dfrac{6250}{11} \]Now convert to a decimal: \(\displaystyle 6250 \div 11 = 568.1818\ldots \), which is \(\displaystyle 568\tfrac{2}{11} \) turns.Check this makes sense: after $\displaystyle 568$ complete turns the wheel has covered \[568 \times 176 \text{ cm} = 99968 \text{ cm} = 999.68 \text{ m} \] That's $\displaystyle 32$ cm short of $\displaystyle 1$ km, so the wheel needs to turn a little further (about \(\displaystyle \tfrac{2}{11}\) of one more turn) to finish the kilometre — which is exactly why the count isn't a whole number.Answer: The car travels $\displaystyle 176$ cm in one complete turn of the wheel, and the wheel makes about $\displaystyle 568.18$ turns (\(\displaystyle 568\tfrac{2}{11}\) turns) to cover a $\displaystyle 1$ km journey.

If a rectangle's perimeter tells you the sum of its two sides, and its area tells you their product, then knowing BOTH numbers pins down the two side lengths exactly — so yes, the two rectangles must be congruent. Here is the full proof.Set up the two rectangles.Let the first rectangle have length \(\displaystyle a\) and breadth \(\displaystyle b\), where \(\displaystyle a \ge b\) (just call the longer side "length" — that's a labeling choice, not a restriction). Let the second rectangle have length \(\displaystyle c\) and breadth \(\displaystyle d\), where \(\displaystyle c \ge d\).We are told:Same perimeter — using Perimeter \(\displaystyle =2(\text{length}+\text{breadth})\): \[2(a+b) = 2(c+d) \quad\Rightarrow\quad a+b = c+d \]Same area — using Area \(\displaystyle =\text{length}\times\text{breadth}\): \[ab = cd \]The tool: connect sum, product and difference.You already know this identity from algebra: \[(x-y)^2 = (x+y)^2 - 4xy \]Apply it to each rectangle's pair of sides: \[(a-b)^2 = (a+b)^2 - 4ab \] \[(c-d)^2 = (c+d)^2 - 4cd \]Since \(\displaystyle a+b=c+d\) and \(\displaystyle ab=cd\), the right-hand sides above are built from the exact same numbers, so they are equal — which forces the left-hand sides to be equal too: \[(a-b)^2 = (c-d)^2 \]Because \(\displaystyle a\ge b\) and \(\displaystyle c\ge d\), both \(\displaystyle a-b\) and \(\displaystyle c-d\) are zero or positive, so taking the square root of both sides is safe here: \[a - b = c - d \]Aside — the step people skip: squaring hides the sign, so \(\displaystyle (a-b)^2=(c-d)^2\) only lets you conclude \(\displaystyle a-b=c-d\) once you know both differences are \(\displaystyle \ge 0\). That is exactly why "length" was defined as the longer side for each rectangle at the start.Solve for the sides.Now there are two equations: \[a + b = c + d \qquad \text{...(1)} \] \[a - b = c - d \qquad \text{...(2)} \]Add ($\displaystyle 1$) and ($\displaystyle 2$): \[2a = 2c \quad\Rightarrow\quad a = c \]Subtract ($\displaystyle 2$) from ($\displaystyle 1$): \[2b = 2d \quad\Rightarrow\quad b = d \]So the two rectangles have the same length AND the same breadth. That is exactly what "congruent" means for rectangles — one can be placed exactly on top of the other.Check it with numbers.Take a rectangle \(\displaystyle 8~\text{cm} \times 6~\text{cm}\): perimeter \(\displaystyle =2(8+6)=28~\text{cm}\), area \(\displaystyle =8\times 6 = 48~\text{cm}^2\).Is there a different pair of positive numbers that also adds to \(\displaystyle 14\) (half the perimeter) and multiplies to \(\displaystyle 48\)? Using the same identity: \[(c-d)^2 = (c+d)^2 - 4cd = 14^2 - 4(48) = 196-192 = 4 \quad\Rightarrow\quad c-d = 2 \] Solving \(\displaystyle c+d=14\) and \(\displaystyle c-d=2\) together gives \(\displaystyle c=8,\ d=6\) — the same rectangle again. There is no second, differently-shaped rectangle hiding in there.Aside — where the confusion usually comes from: same area alone does NOT force the same shape. A \(\displaystyle 2~\text{cm}\times 18~\text{cm}\) rectangle and a \(\displaystyle 6~\text{cm}\times 6~\text{cm}\) rectangle both have area \(\displaystyle 36~\text{cm}^2\), but their perimeters are \(\displaystyle 40~\text{cm}\) and \(\displaystyle 24~\text{cm}\) — different. It is only once you also demand the same perimeter that both the sum and the product of the sides get fixed, and the working above shows that is enough to force the sides to match exactly.Answer: Yes — two rectangles with both the same area and the same perimeter must be congruent (same length and same breadth). Matching the area alone is not enough — e.g. $\displaystyle 2$ cm × $\displaystyle 18$ cm and $\displaystyle 6$ cm × $\displaystyle 6$ cm both have area $\displaystyle 36$ cm² but different perimeters ($\displaystyle 40$ cm vs $\displaystyle 24$ cm).