In the given figure, $P$, $Q$, and $R$ are three points on a circle of radius 10 cm with $O$ as its center, $\overline{PQ} = \overline{RQ}$, and $\angle PQR = 45^{\circ}$. The figure is representative.
The area of the shaded region $PQRO$ is ______________ $\text{cm}^2$.

Question

On a certain principal, CI and SI at a certain rate of interest for 2 years is ₹ 16560 and ₹ 14400 respectively. Find the principal and rate of interest per annum.

Answer 1

The shaded patch PQRO is a kite-like quadrilateral made of two triangles glued along the line from the center O to the top point Q. Instead of jumping into a formula, let's build the picture piece by piece.

Because O is the circle's center, OP, OQ and OR are all radii, each 10 cm. We're also told the chords PQ and RQ are equal in length. Two triangles that share a side (OQ) and have their other two pairs of sides equal (OP=OR, QP=QR) must be mirror images of each other, congruent by SSS. That single fact does two things for us: it tells us OQ splits the quadrilateral exactly down the middle, and it tells us OQ cuts the given 45 degree angle at Q into two 22.5 degree pieces.

Find the apex angle at O: Look at just the triangle OPQ. Two of its sides, OP and OQ, are both radii of length 10, so the triangle is isosceles, and the base angles at P and Q must match. We already found the angle at Q inside this triangle is 22.5 degrees, so the angle at P is also 22.5 degrees. The angle sitting at O is whatever is left from 180 degrees: $180 - 22.5 - 22.5 = 135$ degrees.
Apply the two-sides-and-included-angle area rule: For any triangle where two sides $a$ and $b$ meet at a known angle $\theta$, the area is $\frac{1}{2}ab\sin\theta$. Plugging in $a=b=10$ and $\theta = 135^\circ$: $\frac{1}{2}(10)(10)\sin(135^\circ) = 50 \cdot \frac{\sqrt2}{2} = 25\sqrt2$ cm$^2$.
Double it for the mirror triangle: Since triangle OQR is the exact mirror of OPQ, it carries the identical area, $25\sqrt2$ cm$^2$. The full shaded region is simply these two matching triangles placed side by side along OQ.

Adding the two pieces: $25\sqrt2 + 25\sqrt2 = 50\sqrt2$ cm$^2$. This matches option (C), and it also makes sense as a sanity check, since the answer should scale with $r^2 = 100$, and $50\sqrt2 \approx 70.7$, a value smaller than a full quarter circle of area $25\pi \approx 78.5$, which is reasonable for a thin kite shape cut out of the circle.

\[ \boxed{50\sqrt{2} \text{ cm}^2} \]

Answer 2

The shaded patch PQRO is a kite-like quadrilateral made of two triangles glued along the line from the center O to the top point Q. Instead of jumping into a formula, let's build the picture piece by piece.

Because O is the circle's center, OP, OQ and OR are all radii, each 10 cm. We're also told the chords PQ and RQ are equal in length. Two triangles that share a side (OQ) and have their other two pairs of sides equal (OP=OR, QP=QR) must be mirror images of each other, congruent by SSS. That single fact does two things for us: it tells us OQ splits the quadrilateral exactly down the middle, and it tells us OQ cuts the given 45 degree angle at Q into two 22.5 degree pieces.

Find the apex angle at O: Look at just the triangle OPQ. Two of its sides, OP and OQ, are both radii of length 10, so the triangle is isosceles, and the base angles at P and Q must match. We already found the angle at Q inside this triangle is 22.5 degrees, so the angle at P is also 22.5 degrees. The angle sitting at O is whatever is left from 180 degrees: $180 - 22.5 - 22.5 = 135$ degrees.
Apply the two-sides-and-included-angle area rule: For any triangle where two sides $a$ and $b$ meet at a known angle $\theta$, the area is $\frac{1}{2}ab\sin\theta$. Plugging in $a=b=10$ and $\theta = 135^\circ$: $\frac{1}{2}(10)(10)\sin(135^\circ) = 50 \cdot \frac{\sqrt2}{2} = 25\sqrt2$ cm$^2$.
Double it for the mirror triangle: Since triangle OQR is the exact mirror of OPQ, it carries the identical area, $25\sqrt2$ cm$^2$. The full shaded region is simply these two matching triangles placed side by side along OQ.

Adding the two pieces: $25\sqrt2 + 25\sqrt2 = 50\sqrt2$ cm$^2$. This matches option (C), and it also makes sense as a sanity check, since the answer should scale with $r^2 = 100$, and $50\sqrt2 \approx 70.7$, a value smaller than a full quarter circle of area $25\pi \approx 78.5$, which is reasonable for a thin kite shape cut out of the circle.

\[ \boxed{50\sqrt{2} \text{ cm}^2} \]

Column 1		Column 2
(p)	First In First Out	(i)	Stacks
(q)	Lookup Operation	(ii)	Queues
(r)	Last In First Out	(iii)	Hash Tables

In the given figure, \(P\), \(Q\), and \(R\) are three points on a circle of radius 10 cm with \(O\) as its center, \(\overline{PQ} = \overline{RQ}\), and \(\angle PQR = 45^{\circ}\). The figure is representative.
The area of the shaded region \(PQRO\) is ______________ \(\text{cm}^2\).

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Quantitative Aptitude

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In the given figure, \(P\), \(Q\), and \(R\) are three points on a circle of radius 10 cm with \(O\) as its center, \(\overline{PQ} = \overline{RQ}\), and \(\angle PQR = 45^{\circ}\). The figure is representative. The area of the shaded region \(PQRO\) is ______________ \(\text{cm}^2\).