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# What is the area of the above right triangle, in terms of x?

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Math Expert
Joined: 02 Sep 2009
Posts: 49303
What is the area of the above right triangle, in terms of x?  [#permalink]

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09 Sep 2018, 09:03
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Difficulty:

25% (medium)

Question Stats:

86% (00:53) correct 14% (00:49) wrong based on 44 sessions

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What is the area of the above right triangle, in terms of x?

A. x^2/8

B. x^2/4

C. $$\frac{x^2\sqrt{2}}{2}$$

D. $$x^2\sqrt{2}$$

E. $$\frac{x^2\sqrt{2}}{4}$$

Attachment:

image015.jpg [ 1.5 KiB | Viewed 434 times ]

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What is the area of the above right triangle, in terms of x?  [#permalink]

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09 Sep 2018, 09:19
Bunuel wrote:

What is the area of the above right triangle, in terms of x?

A. x^2/8

B. x^2/4

C. $$\frac{x^2\sqrt{2}}{2}$$

D. $$x^2\sqrt{2}$$

E. $$\frac{x^2\sqrt{2}}{4}$$

Attachment:
image015.jpg

The measure of the angles of the given triangle is surely 45:45:90
the ratio of sides of such triangle is 1:1:$$\sqrt{2}$$

now it is given that
$$\sqrt{2}$$*p = x (where p is the measure of each side)
p = x/$$\sqrt{2}$$

area =$$\frac{b*h}{2}$$ = x/$$\sqrt{2}$$* x/$$\sqrt{2}$$ * 1/2 = $$x^2$$/4
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Joined: 28 Feb 2018
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Re: What is the area of the above right triangle, in terms of x?  [#permalink]

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09 Sep 2018, 09:53
as both sides of right angle triangle are equal
2*side^2=x^2
side^2=x^2/2

area of triangle=(1/2)*(side^2)
=(1/2)*((x^2)/2)=(1/4)*(x^2)
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Location: India
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Re: What is the area of the above right triangle, in terms of x?  [#permalink]

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09 Sep 2018, 19:02
Angle 45 indicates that it is a isosceles right angle triangle in which the sides are in the ratio 1:1:√2
√2a= x
a= x/√2
area= 1/2bh
½ *(x/√2)* (x/√2)
Comes to x2/4
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Re: What is the area of the above right triangle, in terms of x?  [#permalink]

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10 Sep 2018, 07:50
Bunuel wrote:

What is the area of the above right triangle, in terms of x?

A. x^2/8

B. x^2/4

C. $$\frac{x^2\sqrt{2}}{2}$$

D. $$x^2\sqrt{2}$$

E. $$\frac{x^2\sqrt{2}}{4}$$

Attachment:
image015.jpg

The given triangle is an isosceles right-angled triangle...

So, $$a^2 + a^2 = x^2$$

Or, $$2a^2 = x^2$$

Or, $$a^2 = \frac{x^2}{2}$$

Since area of a triangle is $$\frac{1}{2}*base*Altitude$$ = $$\frac{x^2}{4}$$, Answer must be (B)
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Thanks and Regards

Abhishek....

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Re: What is the area of the above right triangle, in terms of x? &nbs [#permalink] 10 Sep 2018, 07:50
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