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In a right isosceles triangle, the lengths of the two nonhypotenuse si

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Bunuel
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In a right isosceles triangle, the lengths of the two nonhypotenuse si [#permalink] New post   30 Oct 2019, 02:56
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In a right isosceles triangle, the lengths of the two nonhypotenuse sides are designated a. What is the area of the triangle in terms of a?

A. 1/3*a^2

B. \(\frac{\sqrt{2}}{3}*a^2\)

C. 1/2*a^2

D. \(\frac{\sqrt{3}}{2}*a^2\)

E. \(\sqrt{2}a^2\)
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Re: In a right isosceles triangle, the lengths of the two nonhypotenuse si [#permalink] New post   30 Oct 2019, 03:10
Let ABC be an Isosceles Right triangle with Angle(B) =90 deg
Then by the given data AB = BC = non-hypotenuse sides = 'a' and AC= Hypotenuse

Area of a right triangle = 1/2*base*height

Here the base = height = 'a'
Therefore, Area = (1/2)*a*a = (1/2)a^2

Ans: C
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Re: In a right isosceles triangle, the lengths of the two nonhypotenuse si [#permalink] New post   30 Oct 2019, 09:38
Given is Base(a) and Height(a) of a right angled isosceles triangle.

\(Area = \frac{1}{2}*base*height => \frac{1}{2}* a * a\)

IMO Answer is Option C!
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Re: In a right isosceles triangle, the lengths of the two nonhypotenuse si [#permalink] New post   30 Oct 2019, 11:18
Bunuel wrote:
In a right isosceles triangle, the lengths of the two nonhypotenuse sides are designated a. What is the area of the triangle in terms of a?

A. 1/3*a^2

B. \(\frac{\sqrt{2}}{3}*a^2\)

C. 1/2*a^2

D. \(\frac{\sqrt{3}}{2}*a^2\)

E. \(\sqrt{2}a^2\)


Hypotenuse of a right isosceles triangle is > The non hypotenuse sides

Now, We have altitude = base of the Triangle = a

So, Area of the triangle is \(\frac{1}{2}*a^2\), Answer must be (C)
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Re: In a right isosceles triangle, the lengths of the two nonhypotenuse si [#permalink] New post   31 Oct 2019, 05:32
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Solution


Given:
    • In a right isosceles triangle, the lengths of the two non-hypotenuse sides are designated a

To find:
    • The area of the triangle

Approach and Working Out:
    • Area of triangle = \(\frac{1}{2}\) * base * height = \(\frac{1}{2}\) * a * a = \(\frac{a^2}{2}\)

Hence, the correct answer is Option C.

Answer: C
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Re: In a right isosceles triangle, the lengths of the two nonhypotenuse si [#permalink] New post   01 Nov 2019, 01:48
Bunuel wrote:
In a right isosceles triangle, the lengths of the two nonhypotenuse sides are designated a. What is the area of the triangle in terms of a?

A. 1/3*a^2

B. \(\frac{\sqrt{2}}{3}*a^2\)

C. 1/2*a^2

D. \(\frac{\sqrt{3}}{2}*a^2\)

E. \(\sqrt{2}a^2\)


Area of the triangle = 1/2 * a * a * sin 90 = 1/2 a^2

IMO C
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Re: In a right isosceles triangle, the lengths of the two nonhypotenuse si [#permalink] New post   01 Nov 2019, 19:43
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Bunuel wrote:
In a right isosceles triangle, the lengths of the two nonhypotenuse sides are designated a. What is the area of the triangle in terms of a?

A. 1/3*a^2

B. \(\frac{\sqrt{2}}{3}*a^2\)

C. 1/2*a^2

D. \(\frac{\sqrt{3}}{2}*a^2\)

E. \(\sqrt{2}a^2\)



Since the area of a right triangle is half the product of its legs and the two legs of the right triangle here are equal to a, then the area is 1/2 * a^2.

Answer: C
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Re: In a right isosceles triangle, the lengths of the two nonhypotenuse si [#permalink]
01 Nov 2019, 19:43
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