|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
24 Dec 2021, 02:30
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
65% (01:56) correct
35%
(03:12)
wrong
based on 43
sessions
History
Date
Time
Result
Not Attempted Yet
A right triangle is drawn. The opposite side of right angle is 10 and sum of the sides next to right angle is 12. What is the area of the triangle?
A. 7
B. 14
C. 10
D. 11
E. 13
A. 7
B. 14
C. 10
D. 11
E. 13
24 Dec 2021, 03:46
Kudos
Bookmarks
In the figure give below we know that
\(10^{2} = X^{2} + Y^{2}\) ------- (1)
Also, X + Y = 12 ------- (2)
Equation 1 -
\(10^{2} = X^{2} + Y^{2} + 2XY - 2XY\)
\(10^{2} = (X+Y)^{2} - 2XY\)
\(12^{2} - 10^{2} = 2XY\)
XY = 22 ------- (3)
As we are given the length of the sides, we can use Hero's formula to get the area.
Area = \(\sqrt {(S)*(S-X)*(S-Y)*(S-Z)}\)
S = Semi - Perimeter = \(\frac{(X + Y + Z) }{ 2}\)
X, Y , Z = Sides of the triangle
Here Z= 10
S = \(\frac{(12 + 10) }{ 2}\) = 11
Area = \(\sqrt {(11)*(11-10)*(S-Y)*(S-X)}\)
Area = \(\sqrt {(11)*(11-10)*(S^2 - S(X + Y) + XY)}\)
Substituting the values
Area = \(\sqrt {(11)*(11-10)*(11^2 - 11(12) + 22)}\)
Area = \(\sqrt {(11)*(1)*(121 - 132 + 22)}\)
Area = \(\sqrt {(11)*(1)*(11)}\)
Area = 11
IMO D
\(10^{2} = X^{2} + Y^{2}\) ------- (1)
Also, X + Y = 12 ------- (2)
Equation 1 -
\(10^{2} = X^{2} + Y^{2} + 2XY - 2XY\)
\(10^{2} = (X+Y)^{2} - 2XY\)
\(12^{2} - 10^{2} = 2XY\)
XY = 22 ------- (3)
As we are given the length of the sides, we can use Hero's formula to get the area.
Area = \(\sqrt {(S)*(S-X)*(S-Y)*(S-Z)}\)
S = Semi - Perimeter = \(\frac{(X + Y + Z) }{ 2}\)
X, Y , Z = Sides of the triangle
Here Z= 10
S = \(\frac{(12 + 10) }{ 2}\) = 11
Area = \(\sqrt {(11)*(11-10)*(S-Y)*(S-X)}\)
Area = \(\sqrt {(11)*(11-10)*(S^2 - S(X + Y) + XY)}\)
Substituting the values
Area = \(\sqrt {(11)*(11-10)*(11^2 - 11(12) + 22)}\)
Area = \(\sqrt {(11)*(1)*(121 - 132 + 22)}\)
Area = \(\sqrt {(11)*(1)*(11)}\)
Area = 11
IMO D
Attachments
Triangle.png [ 7 KiB | Viewed 1504 times ]
24 Dec 2021, 17:17
Kudos
Bookmarks
Bunuel
Breaking Down the Info:
The opposite side of the right angle is 10: The hypotenuse is 10.
sum of the sides next to the right angle is 12: The two legs add up to 12. We can declare \(A + B = 12\), and from above \(A^2 + B^2 = 10^2\).
Our goal is to find the area which is expressed as \(\frac{1}{2}A*B\). We don't need to solve for A or B, we can square \(A + B = 12\) to get:
\(A^2 + 2AB + B^2 = 144\), then subtract \(A^2 + B^2 = 10^2\) to get \(2AB = 44\) and \(\frac{1}{2}AB = 11\).
Answer: D
