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In the figure shown, if E and F are midpoints of the respective edges

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Bunuel
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In the figure shown, if E and F are midpoints of the respective edges [#permalink] In the figure shown, if E and F are midpoints of the respective edges   20 Oct 2022, 06:34
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A
B
C
D
E

Difficulty:

35% (medium)

Question Stats:

66% (02:24) correct 34% (02:41) wrong based on 59 sessions
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In the figure shown, if E and F are midpoints of the respective edges of the square ABCD, what percentage of the area of the square is common to the area of the triangle AEF?

(A) 33%
(B) 38%
(C) 48%
(D) 52%
(E) 62%






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divyadna
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Re: In the figure shown, if E and F are midpoints of the respective edges [#permalink] In the figure shown, if E and F are midpoints of the respective edges   20 Oct 2022, 07:01
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Re: In the figure shown, if E and F are midpoints of the respective edges [#permalink] In the figure shown, if E and F are midpoints of the respective edges   20 Oct 2022, 07:15
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Bunuel wrote:

In the figure shown, if E and F are midpoints of the respective edges of the square ABCD, what percentage of the area of the square is common to the area of the triangle AEF?

(A) 33%
(B) 38%
(C) 48%
(D) 52%
(E) 62%


Show Spoiler
Attachment:
EG.jpeg


Let, side of square = 4

Area of Square \(= 4^2 = 16\)

EF \(= \sqrt{2^2+2^2} = √8 = 2√2\)
AE = AF \(= \sqrt{4^2+2^2} = √20\)
i.e. ∆AEF has sides 2√2, √20, √20
Height drawn from point A on EF \(= \sqrt{20-2} = √18\)

ARea of ∆AEF \(= \frac{1}{2}*2√2*√18 = 6\)

Common area as percentage of area of square \(= \frac{6}{16}*100 = 37.5\)

Answer: Option B

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Re: In the figure shown, if E and F are midpoints of the respective edges [#permalink] In the figure shown, if E and F are midpoints of the respective edges   20 Oct 2022, 13:04
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Bunuel wrote:

In the figure shown, if E and F are midpoints of the respective edges of the square ABCD, what percentage of the area of the square is common to the area of the triangle AEF?

(A) 33%
(B) 38%
(C) 48%
(D) 52%
(E) 62%


Show Spoiler
Attachment:
EG.jpeg


Set a side equal to 2.

ADF and ABE are congruent.

Area of each is 1/2 * 1* 2 = 1.

So both together = 2.

EFC has half the height but same base as the others, so its area is

1/2

Area of these 3 triangles is then

2 1/2.

Subtracted from the area of the square, 4, gives the area of the central triangle

1 1/2

This area as a percentage of the square's is

100*3/2*1/4 = 38%

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Re: In the figure shown, if E and F are midpoints of the respective edges [#permalink]
20 Oct 2022, 13:04
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Bunuel
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