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06 Sep 2017, 04:37
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D
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87% (01:17) correct
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In the figure above, the area of triangular region PQR is 36.What is the area of the triangular region SQR?
(A) 12
(B) 16
(C) 24
(D) 30
(E) 36
Attachment:
2017-09-06_1535.png [ 7.32 KiB | Viewed 2481 times ]
06 Sep 2017, 05:00
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Given figure triangle QPS is right triangle
so base PS = root(PS^2-PQ^2) = root(5^2-4^2)=3
Area PQS = 1/2 *4*3 = 6
Given area of PQR=36
So area SQR =Area PQR- Area PQS = 36-6 = 30
Answer: D
so base PS = root(PS^2-PQ^2) = root(5^2-4^2)=3
Area PQS = 1/2 *4*3 = 6
Given area of PQR=36
So area SQR =Area PQR- Area PQS = 36-6 = 30
Answer: D
06 Sep 2017, 05:13
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Quote:
Given figure triangle QPS is right triangle
so base PS = root(PS^2-PQ^2) = root(5^2-4^2)=3
so base PS = root(PS^2-PQ^2) = root(5^2-4^2)=3
You might also recognize that triangle QPS is a 3-4-5 right triangle to save some time.
I agree with D as the answer
