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If the area of an isosceles right triangle is 9/8 square inches, what
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17 Feb 2022, 07:15
17 Feb 2022, 07:15
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If the area of an isosceles right triangle is 9/8 square inches, what is the perimeter of the triangle?
(A) 3 + √3/4
(B) 3 + √3/2
(D) 3 + 3√2/2
(E) 7√3/2
(C) 4 + 3√3/2
(A) 3 + √3/4
(B) 3 + √3/2
(D) 3 + 3√2/2
(E) 7√3/2
(C) 4 + 3√3/2
If the area of an isosceles right triangle is 9/8 square inches, what
[#permalink]
18 Feb 2022, 03:31
18 Feb 2022, 03:31
1
Area of an isosceles right triangle with side \(a = \frac{1}{2} a^2\)
\(\frac{1}{2}a^2 = \frac{9}{8}\)
\(a = \frac{3}{2}\)
So two sides = \(\frac{3}{2}\) & \(\frac{3}{2}\)
It's a right triangle so the third side can be found by the Pythagoras theorem.
Third side = \(\sqrt{\frac{9}{4} + \frac{9}{4}}\)
\(= \frac{3}{\sqrt{2}}\)
Perimeter \(= \frac{3}{2} + \frac{3}{2} + \frac{3}{\sqrt{2}}\)
\(= 3 + \frac{3}{\sqrt{2}} = 3 + \frac{3\sqrt{2}}{2}\)
Answer D
\(\frac{1}{2}a^2 = \frac{9}{8}\)
\(a = \frac{3}{2}\)
So two sides = \(\frac{3}{2}\) & \(\frac{3}{2}\)
It's a right triangle so the third side can be found by the Pythagoras theorem.
Third side = \(\sqrt{\frac{9}{4} + \frac{9}{4}}\)
\(= \frac{3}{\sqrt{2}}\)
Perimeter \(= \frac{3}{2} + \frac{3}{2} + \frac{3}{\sqrt{2}}\)
\(= 3 + \frac{3}{\sqrt{2}} = 3 + \frac{3\sqrt{2}}{2}\)
Answer D
gmatclubot
If the area of an isosceles right triangle is 9/8 square inches, what [#permalink]
18 Feb 2022, 03:31
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