NCERT Solutions for Class 9 Maths Chapter 6 — End-of-Chapter Exercises#

The big square has side $a + b$, so its area is $(a + b)^2$. It is cut into a square $a^2$, a square $b^2$ and two rectangles $ab$. Adding the pieces gives $a^2 + 2ab + b^2$, so $(a + b)^2 = a^2 + 2ab + b^2$.

Start with a square of side $a$ and cut away a square of side $b$ from one corner. The shaded L-shape that remains has area $a^2 - b^2$.

Cut the L-shape into two rectangles: $P$, of size $a \times (a - b)$, and $Q$, of size $(a - b) \times b$. Turn $Q$ and place it at the end of $P$. Together they form a rectangle of length $a + b$ and width $a - b$:

So $(a + b)(a - b) = a^2 - b^2$.

Draw a square of side $a + b + c$, and divide each side into parts $a$, $b$, $c$. This gives 9 pieces:

There are three squares $a^2$, $b^2$, $c^2$, and the six rectangles come in pairs: two $ab$, two $bc$, two $ca$. Adding all the pieces:

$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

The base is:

$$\text{Base} = 40 - 15 - 15 = 10 \text{ cm}$$

The altitude to the base bisects it, so it forms a right triangle with hypotenuse 15 cm and base 5 cm:

$$h^2 = 15^2 - 5^2$$

$$h^2 = 200$$

$$\Rightarrow h = 10\sqrt{2} \text{ cm}$$

So:

$$\text{Area} = \dfrac{1}{2} \times 10 \times 10\sqrt{2}$$

$$\text{Area} = 50\sqrt{2}$$

∴ The area is $50\sqrt{2}$ cm² ≈ 70.71 cm². (Heron’s formula with $s = 20$ gives $\sqrt{20 \times 5 \times 5 \times 10} = \sqrt{5000} = 50\sqrt{2}$ as well.)

The area is 60 cm², so:

$$\dfrac{1}{2} \times 10 \times h = 60$$

$$\Rightarrow h = 12 \text{ cm}$$

The altitude bisects the base, so each equal side is the hypotenuse of a right triangle with legs 5 cm and 12 cm:

$$\text{Side}^2 = 5^2 + 12^2$$

$$\text{Side}^2 = 169$$

$$\Rightarrow \text{Side} = 13 \text{ cm}$$

∴ Each equal side is 13 cm.

The two legs are the base and height. The area is 54 cm², so:

$$\dfrac{1}{2} \times 12 \times b = 54$$

$$\Rightarrow b = 9 \text{ cm}$$

The hypotenuse is:

$$c^2 = 12^2 + 9^2$$

$$c^2 = 225$$

$$\Rightarrow c = 15 \text{ cm}$$

So:

$$\text{Perimeter} = 12 + 9 + 15$$

$$\text{Perimeter} = 36$$

∴ The perimeter is 36 cm.

Let the sides be $2k$, $3k$ and $4k$. The perimeter is 45 cm, so:

$$2k + 3k + 4k = 45$$

$$\Rightarrow 9k = 45$$

$$\Rightarrow k = 5$$

So the sides are 10 cm, 15 cm and 20 cm, and $s = 22.5$ cm. By Heron’s formula:

$$\text{Area} = \sqrt{22.5 \times 12.5 \times 7.5 \times 2.5}$$

$$\text{Area} = \sqrt{\dfrac{45 \times 25 \times 15 \times 5}{16}}$$

$$\text{Area} = \dfrac{\sqrt{84375}}{4}$$

$$\text{Area} = \dfrac{75\sqrt{15}}{4}$$

∴ The area is $\dfrac{75\sqrt{15}}{4}$ cm² ≈ 72.62 cm².

$$7^2 + 24^2 = 49 + 576 = 625 = 25^2$$

So the triangle is right-angled, with legs 7 cm and 24 cm (converse of the Baudhāyana–Pythagoras theorem):

$$\text{Area} = \dfrac{1}{2} \times 7 \times 24$$

$$\text{Area} = 84 \text{ cm}^2$$

$$s = \dfrac{7 + 24 + 25}{2} = 28$$$$\text{Area} = \sqrt{28(28 - 7)(28 - 24)(28 - 25)}$$

$$\text{Area} = \sqrt{28 \times 21 \times 4 \times 3}$$

$$\text{Area} = \sqrt{7056}$$

$$\text{Area} = 84 \text{ cm}^2$$

∴ Both methods give 84 cm².

In one rotation the bicycle moves one circumference:

$$C = \pi d$$

$$C = \dfrac{22}{7} \times 60$$

$$C = \dfrac{1320}{7} \text{ cm}$$

In 100 rotations:

$$\text{Distance} = 100 \times \dfrac{1320}{7}$$

$$\text{Distance} = \dfrac{132000}{7}$$

$$\text{Distance} \approx 18857 \text{ cm}$$

∴ The cyclist travels about 18857 cm ≈ 188.57 m.

The circumference is 66 cm, so:

$$2 \times \dfrac{22}{7} \times r = 66$$

$$\Rightarrow r = \dfrac{66 \times 7}{44}$$

$$\Rightarrow r = 10.5 \text{ cm}$$

So:

$$\text{Quadrant} = \dfrac{1}{4} \times \dfrac{22}{7} \times 10.5^2$$

$$\text{Quadrant} = \dfrac{1}{4} \times 346.5$$

$$\text{Quadrant} = 86.625$$

∴ The area of the quadrant is 86.625 cm².

One turn moves the car one circumference:

$$C = 2 \times \dfrac{22}{7} \times 28$$

$$C = 176 \text{ cm}$$$$1 \text{ km} = 100000 \text{ cm}$$

So the number of turns is:

$$\text{Turns} = \dfrac{100000}{176}$$

$$\text{Turns} \approx 568.2$$

∴ The car travels 176 cm in one turn, and the wheel turns about 568 times in 1 km.

Yes. Let the common area be $A$ and the common perimeter be $P$, and let a rectangle have sides $x$ and $y$:

$$x + y = \dfrac{P}{2} \quad \cdots (1)$$

$$xy = A \quad \cdots (2)$$

From equation (1) and equation (2), we can find the difference of the sides:

$$(x - y)^2 = (x + y)^2 - 4xy$$

$$(x - y)^2 = \dfrac{P^2}{4} - 4A$$

So $x + y$ and the size of $x - y$ are fixed by $A$ and $P$, and therefore the two side lengths are fixed:

$$x = \dfrac{1}{2}\left(\dfrac{P}{2} + \sqrt{\dfrac{P^2}{4} - 4A}\right), \quad y = \dfrac{1}{2}\left(\dfrac{P}{2} - \sqrt{\dfrac{P^2}{4} - 4A}\right)$$

∴ Both rectangles have the same pair of sides, so they are congruent. (This is special to rectangles; for example, two different triangles can share both area and perimeter.)

In Fig. 6.42 the trapezium (parallel sides $a$ on top and $b$ at the bottom, height $h$) is split into a parallelogram (purple) with base $a$ and a triangle (pink) with base $b - a$, both of height $h$.

$$\text{Parallelogram} = a \times h$$

$$\text{Triangle} = \dfrac{1}{2}(b - a)h$$

Adding the two parts:

$$\text{Trapezium} = ah + \dfrac{1}{2}(b - a)h$$

$$\text{Trapezium} = \dfrac{1}{2}(2a + b - a)h$$

$$\text{Trapezium} = \dfrac{1}{2}(a + b)h$$

Hence shown.

Let $ABCD$ be a trapezium with $AB \parallel DC$, $AB = a$, $DC = b$ and height $h$. The diagonal $AC$ splits it into $\triangle ABC$ and $\triangle ACD$.

Both triangles have height $h$, because their bases lie on the two parallel sides:

$$\text{Area}(\triangle ABC) = \dfrac{1}{2}ah$$

$$\text{Area}(\triangle ACD) = \dfrac{1}{2}bh$$

Adding:

$$\text{Trapezium} = \dfrac{1}{2}ah + \dfrac{1}{2}bh$$

$$\text{Trapezium} = \dfrac{1}{2}(a + b)h$$

Hence shown.

Take a trapezium with parallel sides $a$ (top) and $b$ (bottom) and height $h$. Make a copy, turn it upside down (rotate it through 180°), and place it beside the first so that a slanting side of each matches.

The new shape has its top made of $a + b$ (top of the first plus top of the turned copy, which was its bottom $b$) and its bottom also $b + a$. Its opposite sides are parallel and equal, so it is a parallelogram with base $a + b$ and height $h$:

$$\text{Two trapeziums} = (a + b) \times h$$

Each trapezium is half of this:

$$\text{Trapezium} = \dfrac{1}{2}(a + b)h$$

Let the kite be $ABCD$ with $AB = AD$ and $CB = CD$. Its diagonal $AC$ (length $d_1$) is the line of symmetry, and it cuts the other diagonal $BD$ (length $d_2$) at right angles at its midpoint $O$, so $OB = OD = \dfrac{d_2}{2}$.

$AC$ splits the kite into $\triangle ABC$ and $\triangle ADC$, each with base $AC = d_1$ and height $\dfrac{d_2}{2}$:

$$\text{Kite} = \dfrac{1}{2} \times d_1 \times \dfrac{d_2}{2} + \dfrac{1}{2} \times d_1 \times \dfrac{d_2}{2}$$

$$\text{Kite} = \dfrac{1}{2}d_1d_2$$

Draw a rectangle around the kite with sides parallel to the diagonals, so the rectangle measures $d_1 \times d_2$. The diagonals divide this rectangle into four smaller rectangles. Each side of the kite is a diagonal of one small rectangle, and cuts it into two equal triangles — one inside the kite and one outside.

So exactly half of the big rectangle is inside the kite:

$$\text{Kite} = \dfrac{1}{2} \times d_1 \times d_2$$

$$\text{Area}(PQRS) = 2a \times 2b = 4ab$$

$$\text{Area}(ABCD) = ab$$

So $PQRS$ has 4 times the area. Yes, 4 copies fit: arrange them in 2 rows of 2. Two copies side by side measure $2a \times b$, and two such rows stacked measure $2a \times 2b$.

For $\triangle PQR$, the semi-perimeter is $2s$, where $s$ is the semi-perimeter of $\triangle ABC$. By Heron’s formula:

$$\text{Area}(PQR) = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)}$$

$$\text{Area}(PQR) = \sqrt{16\,s(s - a)(s - b)(s - c)}$$

$$\text{Area}(PQR) = 4 \times \text{Area}(ABC)$$

Yes, 4 copies fit. Join the midpoints of the sides of $\triangle PQR$. This cuts it into 4 triangles, and each has sides $a$, $b$, $c$ (each side of a midpoint triangle is half of a side of $PQR$), so each is a copy of $\triangle ABC$ — three upright and one upside down in the middle.

The semi-perimeter of $\triangle PQR$ is $3s$:

$$\text{Area}(PQR) = \sqrt{3s(3s - 3a)(3s - 3b)(3s - 3c)}$$

$$\text{Area}(PQR) = \sqrt{81\,s(s - a)(s - b)(s - c)}$$

$$\text{Area}(PQR) = 9 \times \text{Area}(ABC)$$

Yes, 9 copies fit. Divide each side of $\triangle PQR$ into three equal parts and draw lines through the division points parallel to the sides. This makes a grid of 9 small triangles with sides $a$, $b$, $c$: 6 upright and 3 upside down (rows of 1, 3 and 5 triangles).

Let the triangle be $ABC$ with $A$ at the top. In the figure, $P$ is the midpoint of $AB$, and $Q$, $R$ divide $AC$ into three equal parts ($AQ = QR = RC$). The shaded region is $BPQR$.

$BPQR$ is $\triangle ABR$ with $\triangle APQ$ removed.

$\triangle ABR$ has the same height as $\triangle ABC$ (from $B$) and base $AR = \dfrac{2}{3}AC$:

$$\text{Area}(\triangle ABR) = \dfrac{2}{3}\text{Area}(\triangle ABC)$$

$\triangle APQ$ has $AP = \dfrac{1}{2}AB$ and $AQ = \dfrac{1}{3}AC$ with the same angle at $A$:

$$\text{Area}(\triangle APQ) = \dfrac{1}{2} \times \dfrac{1}{3} \times \text{Area}(\triangle ABC)$$

$$\text{Area}(\triangle APQ) = \dfrac{1}{6}\text{Area}(\triangle ABC)$$

So:

$$\text{Shaded} = \dfrac{2}{3} - \dfrac{1}{6}$$

$$\text{Shaded} = \dfrac{1}{2}$$

∴ Half of the triangle is shaded.

Each line joins a corner of the square to the midpoint of a side. Place the square with corners $(0, 0)$, $(2, 0)$, $(2, 2)$, $(0, 2)$, so its area is 4. Two of the lines are parallel:

$$\text{from } (0, 0) \text{ to } (2, 1): \quad x - 2y = 0$$

$$\text{from } (0, 1) \text{ to } (2, 2): \quad x - 2y + 2 = 0$$

The other two lines are perpendicular to these:

$$\text{from } (1, 0) \text{ to } (0, 2): \quad 2x + y - 2 = 0$$

$$\text{from } (2, 0) \text{ to } (1, 2): \quad 2x + y - 4 = 0$$

The shaded square’s side is the distance between each pair of parallel lines, which is $\dfrac{2}{\sqrt{5}}$ for both pairs:

$$\text{Shaded} = \left(\dfrac{2}{\sqrt{5}}\right)^2$$

$$\text{Shaded} = \dfrac{4}{5}$$$$\dfrac{\text{Shaded}}{\text{Square}} = \dfrac{4/5}{4}$$

$$\dfrac{\text{Shaded}}{\text{Square}} = \dfrac{1}{5}$$

∴ One-fifth of the square is shaded. (A cut-and-move view: the four corner pieces outside the shaded square can be rearranged into four more copies of it.)

Let each circle have diameter $d$, so radius $\dfrac{d}{2}$. The circles touch each other and the rectangle.

Fig. 6.45 (3 circles): the rectangle is $3d \times d$.

$$\text{Circles} = 3 \times \pi\left(\dfrac{d}{2}\right)^2 = \dfrac{3\pi d^2}{4}$$

$$\text{Fraction} = \dfrac{3\pi d^2/4}{3d^2} = \dfrac{\pi}{4}$$

Fig. 6.46 (4 circles): the rectangle is $4d \times d$.

$$\text{Circles} = 4 \times \pi\left(\dfrac{d}{2}\right)^2 = \pi d^2$$

$$\text{Fraction} = \dfrac{\pi d^2}{4d^2} = \dfrac{\pi}{4}$$

∴ In both cases the circles cover $\dfrac{\pi}{4} \approx \dfrac{11}{14} \approx 0.786$ of the rectangle.

Conjecture: however many circles are fitted in a row, they always cover $\dfrac{\pi}{4}$ of the rectangle.

Tests (diameter $d$):

Circles	Rectangle	Circles’ area	Fraction
10	$10d \times d = 10d^2$	$10 \times \dfrac{\pi d^2}{4}$	$\dfrac{\pi}{4}$
20	$20d^2$	$20 \times \dfrac{\pi d^2}{4}$	$\dfrac{\pi}{4}$
50	$50d^2$	$50 \times \dfrac{\pi d^2}{4}$	$\dfrac{\pi}{4}$

Proof: for $n$ circles of diameter $d$, the rectangle is $nd \times d$:

$$\text{Fraction} = \dfrac{n \times \pi d^2/4}{nd^2}$$

$$\text{Fraction} = \dfrac{\pi}{4}$$

The $n$ cancels, so the fraction does not depend on the number of circles. (The rectangle is really $n$ squares of side $d$, each holding one circle that covers $\dfrac{\pi}{4}$ of its square.)

Let each small rectangle have length $L$ and width $W$ ($L > W$). In Fig. 6.47 the top row has 4 rectangles lying flat and the bottom row has 5 standing upright, and both rows have the same total width, so:

$$4L = 5W$$

$$\Rightarrow L = \dfrac{5}{4}W \quad \cdots (1)$$

The nine rectangles make up 72 cm², so:

$$9LW = 72$$

$$\Rightarrow LW = 8 \quad \cdots (2)$$

Substituting $L = \dfrac{5}{4}W$ in equation (2):

$$\dfrac{5}{4}W^2 = 8$$

$$\Rightarrow W^2 = \dfrac{32}{5}$$

$$\Rightarrow W = \dfrac{4\sqrt{10}}{5} \approx 2.53 \text{ cm}$$

From equation (1), $L = \sqrt{10} \approx 3.16$ cm. So the perimeter of each small rectangle is:

$$\text{Perimeter} = 2(L + W)$$

$$\text{Perimeter} = 2 \times \dfrac{9}{4}W$$

$$\text{Perimeter} = \dfrac{9}{2} \times \dfrac{4\sqrt{10}}{5}$$

$$\text{Perimeter} = \dfrac{18\sqrt{10}}{5}$$

∴ Each small rectangle has perimeter $\dfrac{18\sqrt{10}}{5}$ cm ≈ 11.38 cm.

The opposite side is divided into three equal parts, so the blue and red triangles have equal bases. They share the same top vertex, so they have the same height (the distance from that vertex to the line of the base).

$$\text{Blue} = \dfrac{1}{2} \times \text{base} \times h = \text{Red}$$

∴ The two triangles have equal areas (each is one-third of the whole triangle).

Any triangle can be cut into 2 pieces that form a parallelogram: cut along the line joining the midpoints of the two slanting sides, and swing the small top triangle down (a half-turn about one of those midpoints) beside the lower piece. The result is a parallelogram with the same base and half the height.

Both triangles become parallelograms with the same base and the same height $\dfrac{h}{2}$. These two parallelograms differ only in their slant, and one can be turned into the other by cutting off a triangle from one end and moving it to the other end.

So: cut the blue triangle into pieces that form this parallelogram, reshape the parallelogram, then undo the red triangle’s own two-piece cut. In this way the blue triangle is cut into a few pieces that, rearranged, cover the red triangle exactly.

Let the square have side $s$.

$$\text{Quarter circle} = \dfrac{1}{4}\pi s^2$$

Each semicircle has diameter $s$:

$$\text{One semicircle} = \dfrac{1}{2}\pi\left(\dfrac{s}{2}\right)^2 = \dfrac{1}{8}\pi s^2$$

Both semicircles lie inside the quarter circle and overlap in region $A$. Region $B$ is the part of the quarter circle outside both semicircles. Adding the two semicircles counts $A$ twice, so the region covered by the semicircles is:

$$\text{Covered} = \dfrac{1}{8}\pi s^2 + \dfrac{1}{8}\pi s^2 - A$$

$$\text{Covered} = \dfrac{1}{4}\pi s^2 - A$$

So:

$$B = \text{Quarter circle} - \text{Covered}$$

$$B = \dfrac{1}{4}\pi s^2 - \left(\dfrac{1}{4}\pi s^2 - A\right)$$

$$B = A$$

Hence $A$ and $B$ have equal area.

Each semicircle has radius 1 unit. Each petal is the overlap of two neighbouring semicircles and is bounded by two quarter circles of radius 1, from a corner of the square to its centre (as in Exercise Set 6.1, Q7).

Perimeter: 4 petals × 2 quarter circles = 8 quarter circles = 2 full circles of radius 1:

$$\text{Perimeter} = 2 \times 2\pi(1)$$

$$\text{Perimeter} = 4\pi$$

$$\text{Perimeter} \approx 12.57 \text{ units}$$

Area: each half of a petal is a quarter circle with a right triangle (legs 1 and 1) removed:

$$\text{Half petal} = \dfrac{1}{4}\pi(1)^2 - \dfrac{1}{2}(1)(1)$$

$$\text{Half petal} = \dfrac{\pi}{4} - \dfrac{1}{2}$$

The flower has 8 half-petals:

$$\text{Area} = 8\left(\dfrac{\pi}{4} - \dfrac{1}{2}\right)$$

$$\text{Area} = 2\pi - 4$$

$$\text{Area} \approx 2.29 \text{ square units}$$

∴ The flower has perimeter $4\pi$ ≈ 12.57 units and area $2\pi - 4$ ≈ 2.29 square units.

Let the radii be $R$ (large) and $r$ (small). $OA = r$, and $OA \perp BC$ (the radius to the point where a line touches a circle is perpendicular to that line). By Theorem 5 of Chapter 5, the perpendicular from the centre bisects the chord, so $AC = \dfrac{l}{2}$.

In right $\triangle OAC$:

$$OC^2 = OA^2 + AC^2$$

$$R^2 = r^2 + \dfrac{l^2}{4}$$

$$\Rightarrow R^2 - r^2 = \dfrac{l^2}{4}$$

The green region is the large circle minus the small one:

$$\text{Green} = \pi R^2 - \pi r^2$$

$$\text{Green} = \pi(R^2 - r^2)$$

$$\text{Green} = \dfrac{1}{4}\pi l^2$$

Hence shown. (The area depends only on the chord length, not on the size of the circles.)

Let the legs be $a$ and $b$ and the hypotenuse $c$, so $a^2 + b^2 = c^2$. In Fig. 6.52, $C$ is the triangle, and $A$ and $B$ are the crescent-shaped regions (“lunes”) on the two legs: each is the semicircle on a leg, minus the part covered by the semicircle on the hypotenuse.

The semicircle on the hypotenuse passes through the right-angle vertex (angle in a semicircle). It is made of the triangle $C$ and the two small segments $S_a$ and $S_b$ cut off by the legs:

$$\dfrac{1}{8}\pi c^2 = C + S_a + S_b \quad \cdots (1)$$

Each lune is a semicircle on a leg minus the segment on that leg:

$$A = \dfrac{1}{8}\pi a^2 - S_a$$

$$B = \dfrac{1}{8}\pi b^2 - S_b$$

Adding:

$$A + B = \dfrac{1}{8}\pi(a^2 + b^2) - (S_a + S_b)$$

$$A + B = \dfrac{1}{8}\pi c^2 - (S_a + S_b)$$

Using equation (1):

$$A + B = C$$

Hence shown. (These are the lunes of Hippocrates.)

Let the centres be $A$ and $B$, with $AB = r$, and let the circles meet at $C$ and $D$. Then $AC = AD = BC = BD = AB = r$, so $\triangle ABC$ and $\triangle ABD$ are equilateral and $\angle CAD = 120°$.

The common region is two equal segments, one from each circle, cut off by the chord $CD$. For the circle with centre $A$:

$$\text{Sector } ACD = \dfrac{120}{360}\pi r^2 = \dfrac{1}{3}\pi r^2$$

$\triangle ACD$ has two sides $r$ with $120°$ between them; it is made of the two halves of equilateral triangles, so:

$$\text{Area}(\triangle ACD) = \dfrac{\sqrt{3}}{4}r^2$$

So one segment is:

$$\text{Segment} = \dfrac{1}{3}\pi r^2 - \dfrac{\sqrt{3}}{4}r^2$$

The region has two such segments:

$$\text{Region} = 2\left(\dfrac{1}{3}\pi r^2 - \dfrac{\sqrt{3}}{4}r^2\right)$$

$$\text{Region} = \left(\dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}\right)r^2$$

∴ The area is $\left(\dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}\right)r^2 \approx 1.23\,r^2$.

Place the rectangle with corners $O(0, 0)$, $(W, 0)$, $(W, H)$, $(0, H)$. In the figure, the triangles meet at $O$, at $P(p, H)$ on the top side, at $Q(W, q)$ on the right side, and at $R(p, q)$, where $PR$ is vertical and $RQ$ is horizontal.

• $C = \triangle PRQ$, right-angled at $R$: $C = \dfrac{1}{2}(W - p)(H - q)$
• $A = \triangle OPR$: base $PR = H - q$, height $p$ (distance from $O$): $A = \dfrac{1}{2}p(H - q)$
• $B = \triangle ORQ$: base $RQ = W - p$, height $q$: $B = \dfrac{1}{2}q(W - p)$

Then:

$$A + C = \dfrac{1}{2}(H - q)(p + W - p) = \dfrac{1}{2}W(H - q)$$

$$B + C = \dfrac{1}{2}(W - p)(q + H - q) = \dfrac{1}{2}H(W - p)$$

Multiplying:

$$(A + C)(B + C) = \dfrac{1}{4}WH(W - p)(H - q)$$

$$(A + C)(B + C) = \dfrac{1}{2}WH \times C$$

So:

$$\dfrac{2(A + C)(B + C)}{C} = WH$$

which is the area of the rectangle. Hence shown.

In Fig. 6.55, $O$ is the centre of the large circle of radius $r$, with $OA = OB = OC = r$ and $\angle AOB = 90°$. The two shaded regions are:

• the lune $AEBF$, between the semicircle drawn on $AB$ (centre $D$) and the arc $AFB$ of the large circle;
• the triangle $AOB$.

In right $\triangle AOB$:

$$AB^2 = r^2 + r^2 = 2r^2$$

The semicircle on $AB$ has radius $\dfrac{AB}{2}$:

$$\text{Semicircle} = \dfrac{1}{2}\pi\left(\dfrac{AB}{2}\right)^2$$

$$\text{Semicircle} = \dfrac{1}{8}\pi(2r^2)$$

$$\text{Semicircle} = \dfrac{1}{4}\pi r^2$$

The segment $AFB$ of the large circle is the quarter circle $OAB$ minus the triangle:

$$\text{Segment} = \dfrac{1}{4}\pi r^2 - \dfrac{1}{2}r^2$$

The lune is the semicircle minus this segment:

$$\text{Lune} = \dfrac{1}{4}\pi r^2 - \left(\dfrac{1}{4}\pi r^2 - \dfrac{1}{2}r^2\right)$$

$$\text{Lune} = \dfrac{1}{2}r^2$$

And:

$$\text{Area}(\triangle AOB) = \dfrac{1}{2} \times r \times r = \dfrac{1}{2}r^2$$

∴ The two shaded regions have equal areas, $\dfrac{1}{2}r^2$ each. (Another lune of Hippocrates — a curved region with exactly the area of a triangle.)