GATE 2023 EE GA Q.10: solution and answer

GATE 2023 · EEGA Q.10 · 2 marks · MCQ

A square with sides of length 6 cm is given. The boundary of the shaded region is defined by two semi-circles whose diameters are the sides of the square, as shown.

The area of the shaded region is ______ $cm^{2}$ .

GATE 2023 EE GA question 10: A square of side 6 cm, with a double-headed arrow above its top edge labelled "6 cm". — Figure from the GATE 2023 EE paper. It is part of the question.

(A)
$6 π$
(B)
$18$
(C)
$20$
(D)
$9 π$

Answer and solution

ANSWER

The shaded area is 18 cm², option (B).

Official answer: BSource: Official answer key, GATE 2023 EE (IIT Kanpur)

Put the bottom-left corner of the square at the origin. The left semicircle has centre $(0, 3)$ and the bottom semicircle has centre $(3, 0)$ . Both have radius $r = 3 cm$ . The point $(3, 3)$ lies on both circles, because $(3 - 0)^{2} + (3 - 3)^{2} = 9$ and $(3 - 3)^{2} + (3 - 0)^{2} = 9$ . The point $(0, 0)$ also lies on both. So the arcs meet at $(0, 0)$ and at the square's centre $(3, 3)$ , and their common chord is the half-diagonal from $(0, 0)$ to $(3, 3)$ . Its length is $32$ .
Find the central angle from the chord length, using $c = 2 r sin \frac{θ}{2}$ : $32 = 2 (3) sin \frac{θ}{2} \Rightarrow sin \frac{θ}{2} = \frac{1}{2} \Rightarrow θ = 9 0^{\circ}$ for each circle.
Area of a $9 0^{\circ}$ circular segment $= sector - triangle = \frac{1}{4} π r^{2} - \frac{1}{2} r^{2} = \frac{9 π}{4} - \frac{9}{2}$ .
The white lens is made of two such segments, one from each circle, on opposite sides of the common chord: $Lens = 2 (\frac{9 π}{4} - \frac{9}{2}) = \frac{9 π}{2} - 9.$
The lens is the overlap, so it lies inside both semicircles. Each shaded piece is therefore a whole semicircle minus the whole lens: $\frac{π r ^{2}}{2} - Lens = \frac{9 π}{2} - (\frac{9 π}{2} - 9) = 9 cm^{2} .$
Shaded area $= 9 + 9 = 18 cm^{2}$ . The $π$ terms cancel, which rules out options (A) and (D).
Check by moving area. Compare the upper shaded piece with the triangle $(0, 0), (0, 6), (3, 3)$ . The piece has an extra $9 0^{\circ}$ segment outside the triangle, beyond the chord from $(0, 6)$ to $(3, 3)$ . It is missing an equal $9 0^{\circ}$ segment of the other circle, which lies inside the triangle and belongs to the lens. These cancel, so the piece has the triangle's area, $\frac{1}{2} \times 6 \times 3 = 9$ . By symmetry the lower piece is also 9, so the total is $2 \times 9 = 18$ . The answer is (B).

KEY IDEA

Only one region is left white: the lens where the two semicircles overlap. So each shaded piece is a semicircle minus the whole lens, where the lens is two 90° segments of radius 3. That gives 9π/2 − (9π/2 − 9) = 9 per piece, so the π terms cancel. Equivalently, each shaded piece has the same area as a triangle with base 6 and height 3.

WATCH OUT FOR

Adding the two semicircles, 2 × 9π/2 = 9π, and picking (D) without removing the white overlap. A second trap is subtracting the lens only once, which gives 9π/2 + 9 ≈ 23.1. The lens sits inside both semicircles, so it must come out of each one, two times in all. A third trap is taking the lens to be a quarter-circle, when it is two circular segments.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 22 Geometry questions, with notes

GATE 2023 EE GA Q.10: solution and answer

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