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
A) 24
B) 11
C) 15
D) 2*sqrt(30)
E) 12
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Right triangle problem
Source: Beat The GMAT — Problem Solving |
RS + ST = 12mattocks 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
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
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, 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)
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)
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.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)
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]
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]
Now that is an interesting approach and the takeaway is really helpful, thankskmittal82 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..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)
I am on a break !!
