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23 Jul 2020, 04:43
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (01:53) correct
28%
(02:04)
wrong
based on 109
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The area of isosceles right triangle XYZ equals 9. Which of the following could be the perimeter?
I. 12
II. 13 + √145
III. 6√2 + 6
(A) I only
(B) I and III
(C) III only
(D) I and II
(E) All of the above
I. 12
II. 13 + √145
III. 6√2 + 6
(A) I only
(B) I and III
(C) III only
(D) I and II
(E) All of the above
23 Jul 2020, 05:13
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Bunuel
There can be only one Isosceles right ∆ with area 9
Let, sides of the isosceles triangle are x-x-x√2
ARea of this triangle = \((\frac{1}{2})*x*x = 9\)
i.e.\(x^2 = 18\)
i.e. \(x = 3√2\)
\(Perimeter = x+x+x√2 = 3√2+3√2+3*√2*√2 = 6+6√2\)
Answer: Option C
23 Jul 2020, 23:13
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Bunuel
Solution
- • XYZ is an isosceles right triangle.
- o ⟹ XYZ is a \(45° – 45° – 90°\) triangles.
o So, Let’s say if base is a, then
- Height = a
And, Hypotenuse =\(a\sqrt{2}\)
• Now, Area of X = \( \frac{1}{2}*base*height = \frac{1}{2}*a*a = 9⟹ a =\sqrt{18} = 3\sqrt{2}\)
• Therefore, perimeter of XYZ =\( 2a + a\sqrt{2} = 6\sqrt{2} + 3\sqrt{2} *\sqrt{2} = 6\sqrt{2} + 6\)
• So, out of the given statements only III is true.
Thus, the correct answer is Option C.
