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Right triangle DEF has a hypotenuse of less than 7. What is the maximu

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

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New post 19 Jul 2017, 23:55
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Re: Right triangle DEF has a hypotenuse of less than 7. What is the maximu  [#permalink]

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New post 20 Jul 2017, 00:26
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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


Assume that a, b, c are 3 sides of that right triangle, where c is the hypotenuse.

According to Pythagorean theorem, we have \(a^2 + b^2 = c^2 < 7^2 = 49\)

Since we have \((a-b)^2 \geq 0 \iff a^2 + b^2 \geq 2ab\)
Hence \(49 > a^2 + b^2 \geq 2ab = 4S \implies S < \frac{49}{4} = 12.25\)

Since S is integer, we have \(S = 12\).

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

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

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New post 22 Jul 2017, 09:18
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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

Thus, max area=1/2 * 24 = 12
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Re: Right triangle DEF has a hypotenuse of less than 7. What is the maximu  [#permalink]

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New post 25 Jul 2017, 11:51
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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


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

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New post 08 Sep 2017, 06:21
QUITE SIMPLE LOGIC

let the sides be x y and 7
x^2 + y^2 = 49

for triangle to have maximum area x=y , since A= 0.5xy

2x^2 = 49
x = \(\frac{7}{√2}\)

A = 0.5 xy
A = \(\frac{49}{4}\)
since 49/4 is not an integer and for it to be integer the next lower value of Numerator is 48

A= 48/4 =12

A=12 OPTION D
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Re: Right triangle DEF has a hypotenuse of less than 7. What is the maximu  [#permalink]

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New post 08 Sep 2017, 06:28
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sahilvijay wrote:
QUITE SIMPLE LOGIC

let the sides be x y and 7
x^2 + y^2 = 49

for triangle to have maximum area x=y , since A= 0.5xy

2x^2 = 49
x = \(\frac{7}{√2}\)

A = 0.5 xy
A = \(\frac{49}{4}\)
since 49/4 is not an integer and for it to be integer the next lower value of Numerator is 48

A= 48/4 =12

A=12 OPTION D




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

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New post 08 Sep 2017, 07:26
One more method

On right triangle DEF keeping hypotenuse same / common make one more triangle

So figure must look like either a rectangle or a square

Now we know that if diogonal which is hypotenuse is same then square will have max area

A max = 0.5 D1D2
= 0.5 x 7x7
So 24.5
And area of triangle is half of 24.5
Which is 12.25

Amax triangle 12.25
But Area is integer so 12 option D

A=12 option D

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Re: Right triangle DEF has a hypotenuse of less than 7. What is the maximu &nbs [#permalink] 08 Sep 2017, 07:26
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