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

Split the quarter circle into three pieces you already know how to measure — the two semicircles and a leftover crescent — and watch the semicircles secretly add up to exactly the quarter circle's own area.Step $\displaystyle 1$ — Label the picture.Let the side of the square be \(\displaystyle s \). Call the vertex where the quarter circle is centred \(\displaystyle P \) (that's the corner both semicircles touch too), and call the two vertices next to it \(\displaystyle Q \) and \(\displaystyle S \) — the quarter circle's arc runs from \(\displaystyle Q \) to \(\displaystyle S \).Because the quarter circle passes through \(\displaystyle Q \) and \(\displaystyle S \), and \(\displaystyle PQ \), \(\displaystyle PS \) are sides of the square, the quarter circle's radius is the full side: \[\text{radius of quarter circle} = PQ = PS = s \]The two semicircles sit on \(\displaystyle PQ \) and \(\displaystyle PS \) — the two sides meeting at \(\displaystyle P \) — as diameters. So each semicircle's diameter is the side \(\displaystyle s \), which means its radius is only half of that: \[\text{radius of each semicircle} = \frac{s}{2} \]This is the step people get wrong: the side of the square is the semicircle's diameter, not its radius. Whenever you plug into the semicircle-area formula, you need half the side, not the side itself.The two semicircle arcs cross each other at \(\displaystyle P \) and at one more point — call it \(\displaystyle M \). Region A (the "football" shape) is the little patch between \(\displaystyle P \) and \(\displaystyle M \) that lies inside both semicircles at once. Region B is whatever is left of the quarter-circle's pie slice once you take both semicircles out of it.Step $\displaystyle 2$ — Area of the quarter circle.Formula: area of a full circle \(\displaystyle = \pi r^2 \), where \(\displaystyle r \) is the radius. A quarter circle is \(\displaystyle \tfrac14 \) of that: \[\text{Area of quarter circle} = \frac{1}{4}\pi s^2 \]Step $\displaystyle 3$ — Area of one semicircle.Formula: area of a semicircle \(\displaystyle = \dfrac{1}{2}\pi r^2 \), where \(\displaystyle r \) is its radius, \(\displaystyle \tfrac{s}{2} \) here: \[\text{Area of one semicircle} = \frac{1}{2}\pi\left(\frac{s}{2}\right)^2 = \frac{1}{2}\pi\cdot\frac{s^2}{4} = \frac{\pi s^2}{8} \]Both semicircles have the same radius \(\displaystyle \tfrac{s}{2} \), so both have this same area, \(\displaystyle \dfrac{\pi s^2}{8} \).Step $\displaystyle 4$ — Add the two semicircles together.\[\text{Semicircle}_1 + \text{Semicircle}_2 = \frac{\pi s^2}{8} + \frac{\pi s^2}{8} = \frac{\pi s^2}{4} \]Look at Step $\displaystyle 2$: this is the exact same number as the quarter circle's area, \(\displaystyle \tfrac14 \pi s^2 \).Why that isn't a coincidence: halving a radius divides a circle's area by $\displaystyle 4$ (since area depends on \(\displaystyle r^2 \)). So a circle of radius \(\displaystyle \tfrac{s}{2} \) has exactly \(\displaystyle \tfrac14 \) the area of a circle of radius \(\displaystyle s \). Two semicircles of radius \(\displaystyle \tfrac{s}{2} \) make up one full circle of radius \(\displaystyle \tfrac{s}{2} \) — so together they automatically weigh in at exactly one quarter of the big circle's area. That's precisely the quarter-circle sector in this figure.Step $\displaystyle 5$ — Use the picture itself.The quarter-circle pie slice (bounded by \(\displaystyle PQ \), \(\displaystyle PS \), and the arc \(\displaystyle QS \)) is completely covered, with no gaps, by: semicircle $\displaystyle 1$, plus semicircle $\displaystyle 2$, plus region B. The only overlap anywhere is region A, which sits inside both semicircles and so gets counted twice if you just add the two semicircle areas. So: \[\text{Area of quarter circle} = \text{Semicircle}_1 + \text{Semicircle}_2 - A + B \] (subtract \(\displaystyle A \) once, because adding the two semicircles counted it twice; add \(\displaystyle B \), the only bit of the sector neither semicircle reaches.)Step $\displaystyle 6$ — Substitute and cancel.From Steps $\displaystyle 2$ and $\displaystyle 4$, \(\displaystyle \text{Area of quarter circle} = \text{Semicircle}_1+\text{Semicircle}_2\) — they're the same number, \(\displaystyle \tfrac{\pi s^2}{4} \). Replace the left side of Step $\displaystyle 5$'s equation with that: \[\text{Semicircle}_1 + \text{Semicircle}_2 = \text{Semicircle}_1 + \text{Semicircle}_2 - A + B \]Subtract \(\displaystyle \text{Semicircle}_1 + \text{Semicircle}_2 \) from both sides: \[0 = -A + B \quad\Longrightarrow\quad A = B \]Notice \(\displaystyle s \) and \(\displaystyle \pi \) have both cancelled out completely — this didn't depend on how big the square is. It works for every square of every size.Answer: Area(A) = Area(B), because the two semicircles (radius = half the side) always add up to exactly the same area as the quarter circle (radius = the full side); once you remove the shared lens A from both sides of that equality, the two leftover pieces A and B are forced to be equal.

Split the big square into $\displaystyle 4$ small squares, then study just ONE petal — the whole flower is $\displaystyle 4$ copies of it.Step $\displaystyle 1$ — Find the radius of each semicircleEach semicircle is drawn on one side of the square, so that side is the diameter of the semicircle, not its radius.Side of square = $\displaystyle 2$ units \(\displaystyle \Rightarrow\) diameter \(\displaystyle d = 2\) units \(\displaystyle \Rightarrow\) radius \(\displaystyle r = \dfrac{d}{2} = 1\) unit.Common trap: don't grab the "$\displaystyle 2$" from the figure and use it as the radius — the side is the diameter, and radius is always HALF the diameter.Step $\displaystyle 2$ — Perimeter: see what actually draws the outlineLook at Fig. $\displaystyle 6.50$ again: the wavy blue outline of the flower is made up of exactly the four curved arcs of the four semicircles — there is no straight edge anywhere on the boundary.Circumference of a full circle: \(\displaystyle C = 2\pi r\). A semicircular arc (just the curved part) is half of that: arc length \(\displaystyle = \pi r\).With \(\displaystyle r = 1\): \[\text{one arc length} = \pi (1) = \pi \text{ units} \]There are $\displaystyle 4$ such arcs (one on each side of the square), so: \[\text{Perimeter of flower} = 4 \times \pi = 4\pi \text{ units} \]Using \(\displaystyle \pi \approx \dfrac{22}{7}\): \[\text{Perimeter} = 4 \times \frac{22}{7} = \frac{88}{7} = 12\frac{4}{7} \approx 12.57 \text{ units} \]Step $\displaystyle 3$ — Area: zoom into one quarter of the squareDraw the horizontal and vertical line through the centre of the big square (joining midpoints of opposite sides). This cuts the \(\displaystyle 2 \times 2\) square into $\displaystyle 4$ small squares of side $\displaystyle 1$ unit each, and it cuts the flower into $\displaystyle 4$ identical petals — one petal sits inside each small square.Take the bottom-left small square (side $\displaystyle 1$ unit). Two of the original semicircles poke into it, and inside this small square each one looks like a quarter circle of radius $\displaystyle 1$, drawn from two opposite corners of the small square. These two quarter circles overlap in the middle, and that overlap is exactly one petal.Area of a quarter circle: \(\displaystyle \dfrac{1}{4}\pi r^2\). With \(\displaystyle r = 1\): \[\text{one quarter-circle area} = \frac{1}{4}\pi(1)^2 = \frac{\pi}{4} \text{ sq units} \]Since each quarter circle's radius equals the full side of the small square, together the two quarter circles cover the entire small square, with only the middle lens (the petal) counted twice when you just add the two areas: \[2 \times \frac{\pi}{4} = (\text{area of petal}) + (\text{area of small square}) \] \[\frac{\pi}{2} = (\text{area of petal}) + 1 \] \[\text{area of petal} = \frac{\pi}{2} - 1 \text{ sq units} \]Step $\displaystyle 4$ — Multiply by the $\displaystyle 4$ petals \[\text{Area of flower} = 4\left(\frac{\pi}{2} - 1\right) = 2\pi - 4 \text{ sq units} \]Using \(\displaystyle \pi \approx \dfrac{22}{7}\): \[\text{Area} = 2\times\frac{22}{7} - 4 = \frac{44}{7} - \frac{28}{7} = \frac{16}{7} = 2\frac{2}{7} \approx 2.29 \text{ sq units} \](No value of \(\displaystyle \pi\) was fixed in the question, so \(\displaystyle \pi \approx \frac{22}{7}\) was used above — that's the value this chapter uses by default.)Answer: Perimeter of the flower \(\displaystyle = 4\pi\) units \(\displaystyle = \dfrac{88}{7}\) units \(\displaystyle \approx 12.57\) units; Area of the flower \(\displaystyle = (2\pi - 4)\) sq units \(\displaystyle = \dfrac{16}{7}\) sq units \(\displaystyle \approx 2.29\) sq units.

Big idea: the chord BC is tangent to the small circle, so the radius \(\displaystyle OA\) is perpendicular to \(\displaystyle BC\) at the point of contact — and a perpendicular from the centre of a circle to a chord always bisects that chord. Those two facts turn this into a single right-triangle (Pythagoras) calculation.Setting up what the figure tells usLet the outer circle have radius \(\displaystyle R = OB = OC\) and the inner circle have radius \(\displaystyle r = OA\). (We are never told the actual values of \(\displaystyle R\) and \(\displaystyle r\) — and we won't need them separately. The whole point of the question is that they cancel out.)BC is a chord of the big circle, and it just touches the small circle at the single point A — that is what "touching" means here, i.e. BC is a tangent to the inner circle.Step $\displaystyle 1$ — Use the tangent–radius rule at point AFormula/fact used: the radius drawn to the point of contact of a tangent is perpendicular to the tangent.Here the radius is \(\displaystyle OA\) and the tangent is line \(\displaystyle BC\), touching at \(\displaystyle A\). So: \[OA \perp BC \]This is the detail everyone misses: A is not just "some point on BC" — it's the foot of the perpendicular from the centre \(\displaystyle O\) onto \(\displaystyle BC\).Step $\displaystyle 2$ — Use the perpendicular-from-centre-bisects-chord ruleFormula/fact used: the perpendicular from the centre of a circle to a chord bisects the chord.\(\displaystyle BC\) is a chord of the outer circle, and \(\displaystyle OA\) is perpendicular to it (from Step $\displaystyle 1$). So \(\displaystyle A\) is the midpoint of \(\displaystyle BC\): \[BA = AC = \frac{l}{2} \](Careful: \(\displaystyle l\) is the whole chord \(\displaystyle BC\); each half, \(\displaystyle BA\), is \(\displaystyle l/2\), not \(\displaystyle l\).)Step $\displaystyle 3$ — Apply Pythagoras in the right triangle \(\displaystyle OAB\)Since \(\displaystyle OA \perp BC\), triangle \(\displaystyle OAB\) has its right angle at \(\displaystyle A\). Its sides are: Pythagoras' theorem: \(\displaystyle (\text{hypotenuse})^2 = (\text{leg}_1)^2 + (\text{leg}_2)^2\)\[OB^{2} = OA^{2} + AB^{2} \] \[R^{2} = r^{2} + \left(\frac{l}{2}\right)^{2} = r^{2} + \frac{l^{2}}{4} \]Rearranging: \[R^{2} - r^{2} = \frac{l^{2}}{4} \qquad \text{...(★)} \]This is the key relationship — notice it connects \(\displaystyle R^2 - r^2\) to \(\displaystyle l\) directly, without needing \(\displaystyle R\) and \(\displaystyle r\) individually.Step $\displaystyle 4$ — Write the area of the green regionThe green region is the annulus — the area of the big circle minus the area of the small circle.Formula used: Area of a circle \(\displaystyle = \pi \times (\text{radius})^{2}\).\[\text{Area of green region} = \pi R^{2} - \pi r^{2} = \pi\left(R^{2} - r^{2}\right) \]Step $\displaystyle 5$ — Substitute (★)\[\text{Area of green region} = \pi \left(\frac{l^{2}}{4}\right) = \frac{1}{4}\pi l^{2} \]This is exactly what we were asked to show — and notice it didn't matter what \(\displaystyle R\) and \(\displaystyle r\) actually were; only the chord length \(\displaystyle l\) decides the green area.Answer: Area of the green (shaded) region \(\displaystyle = \dfrac{1}{4}\pi l^{2}\), shown using: tangent ⊥ radius at the point of contact, perpendicular from centre bisects the chord, and Pythagoras' theorem in \(\displaystyle \triangle OAB\).

The big semicircle on the hypotenuse is built from three pieces — the triangle plus two curved slivers — and those very same two slivers are the pieces you cut away from the two small semicircles to leave the crescents \(\displaystyle A\) and \(\displaystyle B\). Take the same two pieces off both sides, and what is left over must be equal.Step $\displaystyle 1$: Name every part of Fig. 6.52.Call the right-angle corner of the triangle \(\displaystyle P\) (bottom-left), the top corner \(\displaystyle Q\), and the bottom-right corner \(\displaystyle R\). Let\[PQ = a \ \text{units}, \qquad PR = b \ \text{units}, \qquad QR = c \ \text{units} \ \ (\text{the hypotenuse}). \]The figure shows three semicircles, each drawn on a side of the triangle as its diameter: The three shaded regions are: Step $\displaystyle 2$: Show that the big arc really does pass exactly through the corner \(\displaystyle P\).Let \(\displaystyle M\) be the midpoint of the hypotenuse \(\displaystyle QR\), so \(\displaystyle M\) is the centre of the semicircle on \(\displaystyle QR\) and its radius is \(\displaystyle \dfrac{c}{2}\) units. (This midpoint is the dot marked on \(\displaystyle QR\) in the figure.)Complete the rectangle: through \(\displaystyle Q\) draw a line parallel to \(\displaystyle PR\), through \(\displaystyle R\) draw a line parallel to \(\displaystyle PQ\), and let them meet at \(\displaystyle S\). Then \(\displaystyle PQSR\) is a rectangle, because \(\displaystyle \angle P = 90^\circ\) and the opposite sides are parallel. Its two diagonals are \(\displaystyle QR\) and \(\displaystyle PS\), and for a rectangle: So \(\displaystyle M\) is the midpoint of \(\displaystyle PS\) as well, which gives\[MP = \frac{PS}{2} = \frac{c}{2} \ \text{units}, \qquad MQ = MR = \frac{c}{2} \ \text{units}. \]\(\displaystyle P\) sits exactly \(\displaystyle \dfrac{c}{2}\) units from the centre \(\displaystyle M\) — the same as the radius — so \(\displaystyle P\) lies on the big semicircle's arc. That is why the arc runs through the corner \(\displaystyle P\) instead of missing it.Step $\displaystyle 3$: Cut the big semicircle into its three pieces.Because the arc passes through \(\displaystyle P\), the two straight legs \(\displaystyle PQ\) and \(\displaystyle PR\) cut the big semicircular region into three pieces that do not overlap: Areas of non-overlapping pieces add, so\[\text{Area(semicircle on } QR) \;=\; \text{Area}(C) + s_1 + s_2 \qquad \textbf{(1)} \]Note that \(\displaystyle s_1\) and \(\displaystyle s_2\) are not equal to each other in general; they are equal only in the special case \(\displaystyle a = b\). What matters below is only that the same two numbers \(\displaystyle s_1\) and \(\displaystyle s_2\) turn up again in the next step.Step $\displaystyle 4$: Say exactly what the crescents \(\displaystyle A\) and \(\displaystyle B\) are.Look again at crescent \(\displaystyle A\) in the figure: it is bounded by two arcs, not by an arc and a straight edge. Both arcs join the same two points \(\displaystyle P\) and \(\displaystyle Q\), and both lie on the same (left) side of \(\displaystyle PQ\): For the crescent to be "semicircle minus sliver", the sliver must lie inside the semicircle. Here is the check. Both regions sit on the same chord \(\displaystyle PQ\), so compare how far each arc bulges out from the midpoint \(\displaystyle N\) of \(\displaystyle PQ\): \[\frac{c}{2} - \frac{b}{2} = \frac{c-b}{2} \ \text{units}. \]In any triangle each side is shorter than the sum of the other two, so \(\displaystyle c < a + b\), which rearranges to \(\displaystyle c - b < a\) and hence \(\displaystyle \dfrac{c-b}{2} < \dfrac{a}{2}\). The big arc is the flatter of the two, so the sliver \(\displaystyle s_1\) lies completely inside the semicircle on \(\displaystyle PQ\), and\[\text{Area}(A) = \text{Area(semicircle on } PQ) - s_1 \qquad \textbf{(2)} \]The other leg works the same way: the centre \(\displaystyle M\) is \(\displaystyle \dfrac{a}{2}\) units from the line \(\displaystyle PR\), the big arc pokes past \(\displaystyle PR\) by \(\displaystyle \dfrac{c-a}{2}\) units, and \(\displaystyle c < a+b\) gives \(\displaystyle \dfrac{c-a}{2} < \dfrac{b}{2}\). So\[\text{Area}(B) = \text{Area(semicircle on } PR) - s_2 \qquad \textbf{(3)} \]Step $\displaystyle 5$: Write down the area of a semicircle.Area of a semicircle \(\displaystyle = \dfrac{1}{2}\pi r^2\), where \(\displaystyle r\) is the radius. The sides \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\) here are the diameters, so each radius is half a side. Writing the formula in terms of a diameter \(\displaystyle d\):\[\text{Area of semicircle} = \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{1}{2}\pi \cdot \frac{d^2}{4} = \frac{\pi d^2}{8} \]So, in square units,\[\text{Area(semicircle on } PQ) = \frac{\pi a^2}{8}, \qquad \text{Area(semicircle on } PR) = \frac{\pi b^2}{8}, \qquad \text{Area(semicircle on } QR) = \frac{\pi c^2}{8}. \]The chapter says to use \(\displaystyle \pi = \dfrac{22}{7}\) whenever a number is needed. In this proof \(\displaystyle \pi\) appears on both sides of every equation and cancels, so no number is needed to finish it. Step $\displaystyle 8$ puts \(\displaystyle \pi = \dfrac{22}{7}\) into a real example so you can watch it work.Step $\displaystyle 6$: Bring in Pythagoras' theorem.Since \(\displaystyle \angle P = 90^\circ\), Pythagoras' theorem ("in a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides") gives\[a^2 + b^2 = c^2. \]Multiply both sides by \(\displaystyle \dfrac{\pi}{8}\):\[\frac{\pi a^2}{8} + \frac{\pi b^2}{8} = \frac{\pi c^2}{8} \]which says, in words,\[\text{Area(semicircle on } PQ) + \text{Area(semicircle on } PR) = \text{Area(semicircle on } QR) \qquad \textbf{(4)} \]Step $\displaystyle 7$: Put the pieces together.Add equations ($\displaystyle 2$) and ($\displaystyle 3$):\[\text{Area}(A) + \text{Area}(B) = \Big[\text{Area(semicircle on } PQ) + \text{Area(semicircle on } PR)\Big] - (s_1 + s_2) \]By ($\displaystyle 4$) the square bracket is the big semicircle, so\[\text{Area}(A) + \text{Area}(B) = \text{Area(semicircle on } QR) - (s_1 + s_2) \]By ($\displaystyle 1$), rearranged, \(\displaystyle \text{Area(semicircle on } QR) - (s_1+s_2) = \text{Area}(C)\). Therefore\[\text{Area}(A) + \text{Area}(B) = \text{Area}(C), \]which is what had to be shown. In symbols, the common value of both sides is\[\frac{\pi}{8}\left(a^2+b^2\right) - (s_1+s_2) \ \text{square units}, \]and it is not \(\displaystyle \dfrac{\pi}{8}(a^2+b^2)\) on its own — that quantity is just the two small semicircles added together, before the slivers are removed.Step $\displaystyle 8$: The neat closed form, and a number check with \(\displaystyle \pi = \dfrac{22}{7}\).Region \(\displaystyle C\) is the right-angled triangle, and area of a triangle \(\displaystyle = \dfrac{1}{2} \times \text{base} \times \text{height}\), with the two legs serving as base and height:\[\text{Area}(C) = \frac{1}{2}ab \ \text{square units}. \]So the two curved crescents together are worth exactly \(\displaystyle \dfrac{1}{2}ab\) square units — a straight-sided answer with no \(\displaystyle \pi\) left in it at all.Test it on a \(\displaystyle 3\)-\(\displaystyle 4\)-\(\displaystyle 5\) triangle: \(\displaystyle a = 3\) units, \(\displaystyle b = 4\) units, \(\displaystyle c = 5\) units (and \(\displaystyle 3^2 + 4^2 = 9 + 16 = 25 = 5^2\), so it is right-angled). Using \(\displaystyle \pi = \dfrac{22}{7}\):\[\text{Area(semicircle on } PQ) = \frac{22}{7}\cdot\frac{3^2}{8} = \frac{22 \times 9}{56} = \frac{99}{28} \ \text{square units} \approx 3.54 \ \text{square units} \] \[\text{Area(semicircle on } PR) = \frac{22}{7}\cdot\frac{4^2}{8} = \frac{22 \times 16}{56} = \frac{176}{28} \ \text{square units} \approx 6.29 \ \text{square units} \] \[\text{Area(semicircle on } QR) = \frac{22}{7}\cdot\frac{5^2}{8} = \frac{22 \times 25}{56} = \frac{275}{28} \ \text{square units} \approx 9.82 \ \text{square units} \] \[\text{Area}(C) = \frac{1}{2}\times 3 \times 4 = 6 \ \text{square units} \]From ($\displaystyle 1$), the two slivers together come to\[s_1 + s_2 = \frac{275}{28} - 6 = \frac{275 - 168}{28} = \frac{107}{28} \ \text{square units} \approx 3.82 \ \text{square units}, \]and then\[\text{Area}(A) + \text{Area}(B) = \left(\frac{99}{28} + \frac{176}{28}\right) - \frac{107}{28} = \frac{275}{28} - \frac{107}{28} = \frac{168}{28} = 6 \ \text{square units} = \text{Area}(C). \ \checkmark \](The additions above are done with the exact fractions. The rounded decimals are only there to give you a feel for the sizes — adding \(\displaystyle 3.54\) and \(\displaystyle 6.29\) drifts by a hundredth from \(\displaystyle 9.82\), which is rounding, not an error.)Nothing in the argument used the numbers \(\displaystyle 3\), \(\displaystyle 4\), \(\displaystyle 5\): it used only \(\displaystyle a^2+b^2=c^2\), which Pythagoras guarantees for every right-angled triangle. So the result holds for all of them.Answer: Area(A) + Area(B) = Area(C) for every right-angled triangle. Both sides equal \(\displaystyle \dfrac{\pi}{8}(a^2+b^2) - (s_1+s_2)\), which is the same as \(\displaystyle \dfrac{\pi c^2}{8} - (s_1+s_2)\) by Pythagoras, and that is exactly the triangle \(\displaystyle C\), of area \(\displaystyle \dfrac{1}{2}ab\) square units — because the same two slivers \(\displaystyle s_1\) and \(\displaystyle s_2\) are subtracted on both sides.

Split the shaded lens into two matching circle-slivers — one bitten out of each circle by the line joining the two crossing points — and add them up.Look at the right-hand picture (Fig. $\displaystyle 6.53$). \(\displaystyle A\) and \(\displaystyle B\) are the two centres, each circle has radius \(\displaystyle r\), and \(\displaystyle C\), \(\displaystyle D\) are the two points where the circles cross (\(\displaystyle C\) on top, \(\displaystyle D\) on the bottom). The red shaded region — the one the question calls "the region enclosed by the two circles" — is only the overlap in the middle, not the whole area covered by both circles together. That is the trap in this question: read the picture, not just the words.Step $\displaystyle 1$ — Get the missing length: \(\displaystyle AB\).The question says each circle passes through the other one's centre. That means \(\displaystyle B\) (a point) lies ON circle \(\displaystyle A\), so the distance \(\displaystyle AB\) equals the radius of circle \(\displaystyle A\): \[AB = r \]Step $\displaystyle 2$ — Show \(\displaystyle \triangle ABC\) and \(\displaystyle \triangle ABD\) are equilateral.\(\displaystyle AC = r\) (radius of circle \(\displaystyle A\)), \(\displaystyle BC = r\) (radius of circle \(\displaystyle B\), since \(\displaystyle C\) is also on circle \(\displaystyle B\)), and \(\displaystyle AB = r\) (Step $\displaystyle 1$). All three sides equal \(\displaystyle r\), so \[\triangle ABC \text{ is equilateral} \implies \angle CAB = 60^\circ \] The same argument with \(\displaystyle D\) gives \(\displaystyle AD = BD = AB = r\), so \[\triangle ABD \text{ is equilateral} \implies \angle DAB = 60^\circ \]Step $\displaystyle 3$ — Find the angle \(\displaystyle \angle CAD\) (this is the one people get wrong).\(\displaystyle C\) is above line \(\displaystyle AB\) and \(\displaystyle D\) is below it, so the two \(\displaystyle 60^\circ\) angles sit on opposite sides of \(\displaystyle AB\) and add up: \[\angle CAD = \angle CAB + \angle BAD = 60^\circ + 60^\circ = 120^\circ \] A very common slip here is to stop at \(\displaystyle 60^\circ\) and forget there are two equilateral triangles stacked together, one above \(\displaystyle AB\) and one below.Step $\displaystyle 4$ — Area of the sector \(\displaystyle ACD\) of circle \(\displaystyle A\).Sector area formula: \(\displaystyle \text{Area of sector} = \dfrac{\theta}{360^\circ} \times \pi r^2\), where \(\displaystyle \theta\) is the central angle and \(\displaystyle r\) is the radius. \[\text{Area of sector } ACD = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3}\pi r^2 \]This sector is a "pie-slice" bounded by the two straight radii \(\displaystyle AC\), \(\displaystyle AD\), and the curved arc \(\displaystyle CD\). To get just the curved sliver (the part that's actually in the shaded lens), we must cut off the straight-edged triangle \(\displaystyle ACD\) sitting inside it.Step $\displaystyle 5$ — Area of triangle \(\displaystyle ACD\), using the rhombus \(\displaystyle ACBD\).Since \(\displaystyle AC = CB = BD = DA = r\), the four points \(\displaystyle A, C, B, D\) form a rhombus (all four sides equal \(\displaystyle r\)).Split this rhombus along diagonal \(\displaystyle AB\): that gives exactly the two equilateral triangles from Step $\displaystyle 2$, so \[\text{Area of rhombus } ACBD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ABD) = \frac{\sqrt3}{4}r^2 + \frac{\sqrt3}{4}r^2 = \frac{\sqrt3}{2}r^2 \] (using the equilateral-triangle-area rule \(\displaystyle \text{Area} = \frac{\sqrt3}{4}(\text{side})^2\), which comes from Pythagoras: height \(\displaystyle =\frac{\sqrt3}{2}\times\text{side}\)).Now split the same rhombus the other way, along diagonal \(\displaystyle CD\): this gives triangles \(\displaystyle ACD\) and \(\displaystyle BCD\). Because all four sides of the rhombus are equal to \(\displaystyle r\), these two triangles are congruent (SSS: \(\displaystyle AC=BC\), \(\displaystyle AD=BD\), \(\displaystyle CD\) shared), so they share the rhombus's area equally: \[\text{Area}(\triangle ACD) = \frac{1}{2}\times\frac{\sqrt3}{2}r^2 = \frac{\sqrt3}{4}r^2 \]Don't reach for a harder formula (like two-sides-and-included-angle) for this \(\displaystyle 120^\circ\)-angled triangle — the rhombus trick gives it for free, using only the equilateral-triangle formula you already know.Step $\displaystyle 6$ — The curved sliver (segment), and the full lens.\[\text{Segment} = \text{Sector } ACD - \triangle ACD = \frac{1}{3}\pi r^2 - \frac{\sqrt3}{4}r^2 \]The shaded lens is made of exactly two such slivers — this one bitten out of circle \(\displaystyle A\) on the side near \(\displaystyle B\), and its mirror image bitten out of circle \(\displaystyle B\) on the side near \(\displaystyle A\) — meeting exactly along the line \(\displaystyle CD\). So: \[\text{Area of lens} = 2\left(\frac{1}{3}\pi r^2 - \frac{\sqrt3}{4}r^2\right) = \frac{2}{3}\pi r^2 - \frac{\sqrt3}{2}r^2 \]Step $\displaystyle 7$ — Put in \(\displaystyle \pi = \dfrac{22}{7}\) (as the chapter tells us to).\[\frac{2}{3}\times\frac{22}{7} = \frac{44}{21} \] \[\text{Area} = \frac{44}{21}r^2 - \frac{\sqrt3}{2}r^2 \]Writing both terms over one denominator (\(\displaystyle 42\)): \[\text{Area} = \frac{88}{42}r^2 - \frac{21\sqrt3}{42}r^2 = \frac{(88-21\sqrt3)}{42}\,r^2 \]Step $\displaystyle 8$ — A decimal, just to see the size of the number.Using \(\displaystyle \sqrt3 \approx 1.732\) (only now, at the very end): \[21 \times 1.732 = 36.372, \qquad 88 - 36.372 = 51.628, \qquad \frac{51.628}{42} \approx 1.23 \]So the lens is a little more than one square of side \(\displaystyle r\) in area — that's a sanity check, not a substitute for the exact answer, since the question asks for the area "in terms of \(\displaystyle r\)".Answer: The common area enclosed by the two circles is \(\displaystyle \left(\dfrac{2}{3}\pi - \dfrac{\sqrt3}{2}\right)r^2 = \dfrac{(88-21\sqrt3)}{42}\,r^2 \approx 1.23\,r^2\) square units (with \(\displaystyle \pi=\frac{22}{7}\)).

The trick: split one big triangle into three, using a right angle hiding in the picture.Look at the figure. Three straight lines come out of the bottom-left corner of the rectangle — call this corner \(\displaystyle L \). One line goes to a point on the top side (call it \(\displaystyle P \)), one goes to a point on the right side (call it \(\displaystyle R \)), and one goes to a point inside the rectangle (call it \(\displaystyle Q \)). Now look carefully at \(\displaystyle Q \): it sits exactly below \(\displaystyle P \) and exactly to the left of \(\displaystyle R \). So \(\displaystyle PQ \) is a vertical segment, \(\displaystyle QR \) is a horizontal segment, and they meet at a right angle at \(\displaystyle Q \). This is the one fact the picture gives you that the words don't — without that right angle, the three areas don't split up this cleanly.So the big triangle \(\displaystyle LPR \) (bottom-left corner, the top point, the side point) has been cut into exactly the three coloured triangles: \[A = \text{area of } \triangle LPQ, \qquad B = \text{area of } \triangle LQR, \qquad C = \text{area of } \triangle PQR. \]Name the four lengths we'll need (the figure fixes these for us): Because \(\displaystyle P \) sits on the top side and \(\displaystyle R \) sits on the right side, adding these pieces up gives you the whole rectangle: \[\text{width of rectangle} = p + w, \qquad \text{height of rectangle} = h + q. \]Aside — this is the step people rush past: \(\displaystyle p+w \) is the width only because \(\displaystyle P \) is on the top side and \(\displaystyle R \) is on the right side, so the horizontal strip from the line through \(\displaystyle P,Q \) to the right wall has total length \(\displaystyle p+w \). This only works because \(\displaystyle Q \) lines up directly under \(\displaystyle P \) and directly left of \(\displaystyle R \) — exactly what the picture showed you.Now find A, B, C in terms of \(\displaystyle p, q, h, w \).Formula used every time: area of a triangle \(\displaystyle = \dfrac{1}{2} \times \text{base} \times \text{height} \), where "height" means the perpendicular distance from the base to the opposite vertex.For \(\displaystyle A = \triangle LPQ \): take \(\displaystyle PQ \) (length \(\displaystyle h \)) as the base. It is vertical, so the perpendicular distance from \(\displaystyle L \) to this line is just the horizontal gap, \(\displaystyle p \). \[A = \frac{1}{2} \times h \times p \]For \(\displaystyle C = \triangle PQR \): this triangle has its right angle at \(\displaystyle Q \), so its two sides \(\displaystyle PQ \) and \(\displaystyle QR \) are already a base and a height, perpendicular to each other. \[C = \frac{1}{2} \times h \times w \]For \(\displaystyle B = \triangle LQR \): take \(\displaystyle QR \) (length \(\displaystyle w \)) as the base. It is horizontal, so the perpendicular distance from \(\displaystyle L \) up to this line is the vertical gap, \(\displaystyle q \). \[B = \frac{1}{2} \times w \times q \]Aside — it's tempting to use \(\displaystyle LP \) or \(\displaystyle LR \) as the base for \(\displaystyle A \) or \(\displaystyle B \) instead. Don't: those sides are slanted, and you'd need a perpendicular height onto a slanted line, which the figure doesn't hand you. \(\displaystyle PQ \) and \(\displaystyle QR \) are the only two segments that are exactly horizontal or vertical, so they are the only safe bases to pick.Add A and C — the pieces \(\displaystyle p \) and \(\displaystyle w \) rebuild the whole width.\[A + C = \frac{1}{2}hp + \frac{1}{2}hw = \frac{1}{2}h(p+w) = \frac{1}{2}h \times (\text{width of rectangle}) \]So: \[\text{width of rectangle} = \frac{2(A+C)}{h} \qquad \text{...(1)} \]Add B and C the same way — this time \(\displaystyle q \) and \(\displaystyle h \) rebuild the whole height.\[B + C = \frac{1}{2}wq + \frac{1}{2}hw = \frac{1}{2}w(q+h) = \frac{1}{2}w \times (\text{height of rectangle}) \]So: \[\text{height of rectangle} = \frac{2(B+C)}{w} \qquad \text{...(2)} \]Multiply ($\displaystyle 1$) and ($\displaystyle 2$) to get the area of the rectangle.\[\text{Area of rectangle} = \text{width} \times \text{height} = \frac{2(A+C)}{h} \times \frac{2(B+C)}{w} = \frac{4(A+C)(B+C)}{hw} \]Now bring back the formula for \(\displaystyle C \): we found \(\displaystyle C = \frac{1}{2}hw \), so \(\displaystyle hw = 2C \). Substitute that in:\[\text{Area of rectangle} = \frac{4(A+C)(B+C)}{2C} = \frac{2(A+C)(B+C)}{C} \]which is exactly the formula we were asked to show.Aside — notice \(\displaystyle C \) does two jobs in this proof: first it's just one of the three given areas, but at the very end it comes back as \(\displaystyle hw = 2C \), the "missing link" that turns the width-fact and the height-fact into one product. That's only possible because \(\displaystyle C \) is the one triangle whose two sides are already a clean horizontal length and a clean vertical length.Answer: Area of the rectangle \(\displaystyle = \dfrac{2(A+C)(B+C)}{C} \) — shown by cutting triangle \(\displaystyle LPR \) into \(\displaystyle A \), \(\displaystyle B \), \(\displaystyle C \) using the right angle at \(\displaystyle Q \), writing each area as \(\displaystyle \tfrac12 \times \text{base} \times \text{height} \), and combining \(\displaystyle A+C \) and \(\displaystyle B+C \) to rebuild the rectangle's width and height.

The quarter circle and the semicircle secretly have the exact same area — so if you peel the same in-between sliver off each one, what's left over on both sides has to match too.First, put letters on the figure so every step has something to point at. Aside: the radius of the quarter circle is a LEG of the triangle, not the hypotenuse — mixing that up is the easiest way to wreck this whole question.Now look at what the hypotenuse \(\displaystyle AB \) does to each curved shape: So the two shaded regions in the figure are the triangle \(\displaystyle OAB \) and the crescent \(\displaystyle L \). We need to show \(\displaystyle \text{area}(OAB) = \text{area}(L) \).Step $\displaystyle 1$: Find the hypotenuse. \(\displaystyle OA = OB = r \) and the angle at \(\displaystyle O \) is \(\displaystyle 90^\circ \), so by Pythagoras' theorem \(\displaystyle AB^2 = OA^2 + OB^2 \): \[AB^2 = r^2 + r^2 = 2r^2 \quad\implies\quad AB = r\sqrt{2} \]Step $\displaystyle 2$: Area of the quarter circle. Formula for a sector: area \(\displaystyle = \dfrac{\theta}{360^\circ}\times \pi r^2 \), where \(\displaystyle \theta \) is the angle at the centre and \(\displaystyle r \) is the radius. Here \(\displaystyle \theta = 90^\circ \): \[\text{Area(quarter circle)} = \frac{90^\circ}{360^\circ}\times \pi r^2 = \frac{\pi r^2}{4} \]Step $\displaystyle 3$: Area of the semicircle on \(\displaystyle AB \). Formula for a circle: area \(\displaystyle = \pi \rho^2 \), where \(\displaystyle \rho \) is the RADIUS — half the diameter. This is the exact spot people slip: you must halve \(\displaystyle AB \) before you square it, not use \(\displaystyle AB \) itself. \[\rho = \frac{AB}{2} = \frac{r\sqrt2}{2} \] \[\text{Area(semicircle)} = \frac12 \times \pi \rho^2 = \frac12 \times \pi \left(\frac{r\sqrt2}{2}\right)^2 = \frac12 \times \pi \times \frac{2r^2}{4} = \frac{\pi r^2}{4} \]Step $\displaystyle 4$: The two curved areas are exactly equal. \[\text{Area(quarter circle)} = \text{Area(semicircle)} = \frac{\pi r^2}{4} \] This is the whole trick of the problem, and it's worth pausing on: this equality doesn't depend on which value you use for \(\displaystyle \pi \) at all — \(\displaystyle \pi \) sits on both sides, so it would cancel out whether you used \(\displaystyle \frac{22}{7} \) or \(\displaystyle 3.14 \). It's also true for every value of \(\displaystyle r \), not just one particular size of triangle. That's why we can leave \(\displaystyle \pi \) as a symbol and never need to plug in \(\displaystyle \frac{22}{7} \) here.Step $\displaystyle 5$: Split each curved area into (triangle-or-sliver) + (leftover), and cancel.The quarter circle splits as triangle plus sliver: \[\text{Area(quarter circle)} = \text{area}(OAB) + \text{area}(S) \] The semicircle splits as sliver plus crescent (since the semicircle fully covers \(\displaystyle S \) before it runs out): \[\text{Area(semicircle)} = \text{area}(S) + \text{area}(L) \] But the left-hand sides are equal (Step $\displaystyle 4$), so the right-hand sides must be equal too: \[\text{area}(OAB) + \text{area}(S) = \text{area}(S) + \text{area}(L) \] Cancelling the sliver \(\displaystyle S \) — the piece both curves share — from both sides: \[\text{area}(OAB) = \text{area}(L) \] That is exactly what we were asked to show: the two shaded regions are equal.Step $\displaystyle 6$: Check it with actual numbers, as a confirmation. Area of the right triangle, using area \(\displaystyle = \frac12 \times \text{base} \times \text{height} \) with the two legs as base and height: \[\text{area}(OAB) = \frac12 \times r \times r = \frac{r^2}{2} \] Area of the sliver, from Step $\displaystyle 5$'s first equation: \[\text{area}(S) = \frac{\pi r^2}{4} - \frac{r^2}{2} \] Area of the crescent, from Step $\displaystyle 5$'s second equation: \[\text{area}(L) = \frac{\pi r^2}{4} - \text{area}(S) = \frac{\pi r^2}{4} - \left(\frac{\pi r^2}{4} - \frac{r^2}{2}\right) = \frac{r^2}{2} \] Both come out to \(\displaystyle \dfrac{r^2}{2} \) — matching, exactly as Step $\displaystyle 5$ already promised.Answer: The two shaded regions are equal. Each has area \(\displaystyle \dfrac{r^2}{2} \) (where \(\displaystyle r = OA = OB \) is a leg of the triangle), because the quarter circle and the semicircle on the hypotenuse have the same area \(\displaystyle \dfrac{\pi r^2}{4} \), and taking away the identical middle sliver \(\displaystyle S \) from each one leaves identical leftovers — the triangle on one side, the crescent on the other.