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Right triangle DEF has a hypotenuse of less than 7. What is the maximu [#permalink]

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20 Jul 2017, 09:40

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Maximum area of right angle triangle can be obtained only when its an isosceles triangle So, both the sides (Other than hypotenuse) will be equal : S^2 + S^2 = H^2 2S^2 = H^2 2S^2 < 7^2 2S^2 < 49 OR We can say 2S^2 = 48 S^2 = 24 or S = 2√6 Area of rt. Angle traiangle = ½ base x altitude In this case = ½ s^2 = ½ 2√6 x 2√6 = 12 D

Re: Right triangle DEF has a hypotenuse of less than 7. What is the maximu [#permalink]

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22 Jul 2017, 09:18

Bunuel wrote:

Right triangle DEF has a hypotenuse of less than 7. What is the maximum possible area of DEF, if the area of DEF is an integer?

A. 6 B. 9 C. 10 D. 12 E. 24

Answer: D

Theory: For a given polygon, the maximum are is achieved when all the sides are equal. So in a triangle max are is of an equilateral triangle followed by isosceles triangle.

Thus, 2x^2<49 x^2<24.5

Area of triangle = 1/2 bh = 1/2 x^2 Max integer area will be when x^2=24

Right triangle DEF has a hypotenuse of less than 7. What is the maximum possible area of DEF, if the area of DEF is an integer?

A. 6 B. 9 C. 10 D. 12 E. 24

We can let the legs of right triangle DEF be a and b. Thus, by the Pythagorean theorem, we have a^2 + b^2 = c^2, in which c is the hypotenuse. However, since the hypotenuse is less than 7, we have:

a^2 + b^2 < 7^2

a^2 + b^2 < 49

Recall that the legs of a right triangle are actually the base and height of the triangle, so the area of triangle DEF is A = 1/2ab.

To maximize the area of triangle DEF, we need a and b to be equal. So, we have:

(assume b = a)

a^2 + a^2 < 49

2a^2 < 49

a^2 < 49/2

a < 7/√2

Let’s say a = 7/√2, so we have the area of triangle DEF as:

(again assume b = a)

A = 1/2ab

A = 1/2(7/√2)(7/√2)

A = 49/4 = 12.25

However, since a is actually less than 7/√2, the area is less than 12.25. Since the area has to be an integer, the largest it can be is 12.

Answer: D
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