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The length of one of the sides of an acute angled triangle

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by BTGmoderatorDC » Mon Aug 12, 2019 8:05 pm
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The length of one of the sides of an acute angled triangle is 13 units. If the area of the triangle is 90 units^2 and the length of the another side of the triangle is 15 units. Find the length of the third side.

A. √124
B. √134
C. √224
D. √234
E. √244

OA E

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by Jay@ManhattanReview » Mon Aug 12, 2019 10:33 pm
BTGmoderatorDC wrote:The length of one of the sides of an acute angled triangle is 13 units. If the area of the triangle is 90 units^2 and the length of the another side of the triangle is 15 units. Find the length of the third side.

A. √124
B. √134
C. √224
D. √234
E. √244

OA E

Source: e-GMAT
Image

See the image above.

∆ABC is an acute-angled triangle. We are given that AB = 13 and BC = 15. Also, the area of ∆ABC = 90 unit^2. We have to find out AC.

Since ∆ABC is an acute-angled triangle and its area = 90, assuming BC as the base, we can have a perpendicular dropped from vertex A to AC. Thus, AA' would be called ∆ABC's height.

Area of ∆ ABC = 90 = 1/2 * AA' * BC => 180 = AA' * 15 => AA' = 12

Now, since ∆ABA' is a right-angled triangle, we have 13^2 = 12^ + A'B^2 => A'B = 5 => A'C = 15 – 5 = 10

Again, since ∆AA'C is a right-angled triangle, we have AC^2 = AA'^2 + A'C^2 => AC^2 = 12^2 + 10^2 => AC = √244

The correct answer: E

Hope this helps!

-Jay
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