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Right triangle problem

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by GMAT_Gladiator » Sun Jul 18, 2010 10:33 am
Is it B
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by aloneontheedge » Sun Jul 18, 2010 10:52 am
mattocks wrote:For right triangle RST, RS and RT are legs and RT+RS=12. ST is hypotenuse with length 10. What is the area of the right triangle?

A) 24
B) 11
C) 15
D) 2*sqrt(30)
E) 12
RS + ST = 12
using pythagoras theorem
100 = x^2 + (x-12)^2

solving we get X as 6+sqrt(14) and 6-sqrt(14)

Area = 1/2(6+sqrt(14) and 6-sqrt(14))
= 11 Hence B
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by kmittal82 » Sun Jul 18, 2010 11:34 am
Area of triangle = 1/2 x RS x RT

Now, ST^2 = RS^2 + RT^2

RS^2 + RT^2 = (RS+RT)^2 - (2xRTxRS)

Thus,

ST^2 = (12)^2 - 2xRTxRS

=> 2RTxRS = 100 - 144
=> RT x RS = 22
=> 1/2 x RT x RS = 11

This, answer is (B)

In other words, (hypotenuse)^2 = (sum of sides)^2 - (4x area of triangle)
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by mattocks » Mon Jul 19, 2010 4:09 pm
A circle is circumscribed about an equilateral triangle of side 'a'. The diameter of the circle is?

A. a*root3
B. a/root3
C. 2a/root3
D. 2*root3(3a)
E. 3root3(a)
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by Vipulvp » Tue Jul 20, 2010 5:00 am
mattocks wrote:A circle is circumscribed about an equilateral triangle of side 'a'. The diameter of the circle is?

A. a*root3
B. a/root3
C. 2a/root3
D. 2*root3(3a)
E. 3root3(a)
[spoiler]Option (C)[/spoiler]. Circumradius = (2/3 * length of altitude). We can remember this as a rule.
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by VS2013 » Wed Jul 21, 2010 5:07 pm
Hey there, my answer is C.

Diameter of circumscribed circle of any triangle is abc/(2*area of triangle), where a, b, c are the sides of the triangle.

1. We know that all the sides are equal and length = a
2. To find the area of the triangle, use the 30 60 90 triangle rule, whereby the height will be (root3*a)/2
Therefore, the area of the triangle = 1/2 base* height
= 1/2*a*(root3*a)/2)
= (a*a*root3)/4
3. Diameter= (a*a*a)/((a*a*root3)/4)
= 2a/root3



[/u]
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by surajgarg » Wed Jul 21, 2010 8:49 pm
kmittal82 wrote:Area of triangle = 1/2 x RS x RT

Now, ST^2 = RS^2 + RT^2

RS^2 + RT^2 = (RS+RT)^2 - (2xRTxRS)

Thus,

ST^2 = (12)^2 - 2xRTxRS

=> 2RTxRS = 100 - 144
=> RT x RS = 22
=> 1/2 x RT x RS = 11

This, answer is (B)

In other words, (hypotenuse)^2 = (sum of sides)^2 - (4x area of triangle)
Now that is an interesting approach and the takeaway is really helpful, thanks :)
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by sumanr84 » Wed Jul 21, 2010 9:51 pm
kmittal82 wrote:Area of triangle = 1/2 x RS x RT

Now, ST^2 = RS^2 + RT^2

RS^2 + RT^2 = (RS+RT)^2 - (2xRTxRS)

Thus,

ST^2 = (12)^2 - 2xRTxRS

=> 2RTxRS = 100 - 144
=> RT x RS = 22
=> 1/2 x RT x RS = 11

This, answer is (B)

In other words, (hypotenuse)^2 = (sum of sides)^2 - (4x area of triangle)
Nice solution without any approximation..
I am on a break !!
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