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# The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh

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The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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23 Jan 2015, 06:27
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81% (01:30) correct 19% (01:41) wrong based on 160 sessions

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The perimeter of an isosceles right triangle is equal to $$6\sqrt{2}(\sqrt{2}+1)$$. What is the area of that triangle?

A. 12
B. 18
C. 24
D. 30
E. 36

Kudos for a correct solution.

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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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23 Jan 2015, 06:29
1
B 18...
side is 6 and hyp 6*(2)^1/2...
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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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23 Jan 2015, 06:36
1
Let the side of the triangle be S.

Then 2S + S(2)^1/2 = 6(2)^1/2((2)^1/2 + 1)

(2)^1/2(S)( (2)^1/2+1 ) = 6(2)^1/2((2)^1/2 + 1)

Comparing both equations Hyp = 6(2)^1/2.

Side will be 6

Area = 1/2.6.6 = 18
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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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23 Jan 2015, 06:39
1
Bunuel wrote:
The perimeter of an isosceles right triangle is equal to $$6\sqrt{2}(\sqrt{2}+1)$$. What is the area of that triangle?

A. 12
B. 18
C. 24
D. 30
E. 36

Kudos for a correct solution.

in an isosceles right triangle two sides are equal and the hyp in \sqrt{2} of any one side .
there lets take one of the legs as x .
therefore permeter =x+x+x\sqrt{2}=6\sqrt{2}(\sqrt{2}+1)
2x+x\sqrt{2}12+6\sqrt{2}
x(2+\sqrt{2})=6(2+\sqrt{2}
x=6

area of triangle = 1/2*6*6=18 ans B
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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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23 Jan 2015, 17:10
1
Distribute. We get that the area is really 6*rad2 + 12. If we take 6*rad2 as the hypotenuse, then each leg is 6. We double-check that this satisfies our 1:1:rad2 ratio for a 45-45-90 triangle. So now we have each leg = 6. Our legs serve as base and height. A = bh/2 = 6*6/2 = 18. The answer is B.
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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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23 Jan 2015, 23:52
1
Isoceles rt angle has sides in ratio 1:1:sqrt{2}

Perimeter = sum of the sides of the triangle = X+X+sqrt{2}X = 2X+sqrt{2}X

X(2+sqrt{2}) = 6sqrt{2} (sqrt{2}+1)

Solving equation for X will be 6.
So sides are 6:6:6sqrt{2}
Area = 0.5* b*h = 0.5*6*6 = 18

Ans B
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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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26 Jan 2015, 03:24
Bunuel wrote:
The perimeter of an isosceles right triangle is equal to $$6\sqrt{2}(\sqrt{2}+1)$$. What is the area of that triangle?

A. 12
B. 18
C. 24
D. 30
E. 36

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Distributing the multiplication for the given perimeter information, you go from 6√2(√2+1) to 12+6√2. To get that into the isosceles right triangle side ratio of x:x:x√2, you'll find that this makes the sides 6:6:6√2. Because the two sides of 6 will form a right angle, you can use those two sides as the base and height in the area formula Area = 1/2(base)(height). 1/2(6)(6)=1/2*36=18.

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of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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03 Feb 2015, 00:02
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Let "a" is a side of isosceles right triangle. Its hypotenuse $$= \sqrt{a^2 + a^2} = a \sqrt{2}$$

Attachment:

root.png [ 2.31 KiB | Viewed 10788 times ]

Perimeter $$= 2a + a\sqrt{2} = 6\sqrt{2}(\sqrt{2}+1)$$

a = 6

Area of the triangle $$= \frac{1}{2} * 6^2 = 18$$
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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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28 Feb 2015, 21:31
perimeter is

12+6*sqrt2

in 1:1:sqrt2 ratio it is only 6:6:6*sqrt2

so area=6*6/2=18

B
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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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29 Jul 2016, 10:46
Here the easiest way to solve this to use the identity => in an IS(right) triangle => with hypotenuse y and the two sides x => y= √2*x=> x=6
hence area = 1/2 *base *height = 36/2=18
Smash that B
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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh  [#permalink]

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23 Aug 2018, 16:01
Bunuel wrote:
The perimeter of an isosceles right triangle is equal to $$6\sqrt{2}(\sqrt{2}+1)$$. What is the area of that triangle?

A. 12
B. 18
C. 24
D. 30
E. 36

We are given that the perimeter of an isosceles right triangle is 6√2(√2 + 1) = 12 + 6√2. Recall that the sides of an isosceles right triangle are in a ratio of x : x : x√2. Therefore, the sides of this isosceles right triangle must be 6, 6 and 6√2. Lastly, the area of an isosceles right triangle is s^2/2 where s is a leg of the triangle; thus, the area of this isosceles right triangle is 6^2/2 = 36/2 = 18.

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Re: The perimeter of an isosceles right triangle is equal to 62√(2√+1). Wh &nbs [#permalink] 23 Aug 2018, 16:01
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