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Two sides of a triangle are each 8 units long, and the third
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23 Mar 2020, 23:21
23 Mar 2020, 23:21
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Two sides of a triangle are each 8 units long, and the third side is 6 units long. What is the area of the triangle, expressed in square units?
(A) 14
(B) 12\(\sqrt{3}\)
(C) 18
(D) 22
(E) 3\(\sqrt{55}\)
(A) 14
(B) 12\(\sqrt{3}\)
(C) 18
(D) 22
(E) 3\(\sqrt{55}\)
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Two sides of a triangle are each 8 units long, and the third
[#permalink]
12 Feb 2021, 06:53
12 Feb 2021, 06:53
1
Approach 1:
Let AB = AC = 8 and BC = 6
Drop a perpendicular from point A to BC and name it is as E
Perpendicular bisects the base in Isosceles Triangle.
Therefore, BE = CE = \(\frac{6}{2}\) = 3
Now apply Pythagoras Theorem in triangle ABE or ACE to get AE = \(\sqrt{55}\)
So, Area of triangle ABC = \(\frac{1}{2}\) x 6 x \(\sqrt{55}\) = \(3\sqrt{55}\)
Approach 2: Use HERON's Formula;
Area of triangle = \(\sqrt{s (s-a) (s-b) (s-c)}\)
Where, s = semi-perimeter = \(\frac{(a+b+c)}{2}\)
a, b, and c = sides of the triangle
Calculate s = \(\frac{(8+8+6)}{2}\) = 11
Area = \(\sqrt{11 (11-8) (11-8) (11-6)}\) = \(\sqrt{11 (3) (3) (5)}\) = \(3\sqrt{55}\)
Hence, option E
Let AB = AC = 8 and BC = 6
Drop a perpendicular from point A to BC and name it is as E
Perpendicular bisects the base in Isosceles Triangle.
Therefore, BE = CE = \(\frac{6}{2}\) = 3
Now apply Pythagoras Theorem in triangle ABE or ACE to get AE = \(\sqrt{55}\)
So, Area of triangle ABC = \(\frac{1}{2}\) x 6 x \(\sqrt{55}\) = \(3\sqrt{55}\)
Approach 2: Use HERON's Formula;
Area of triangle = \(\sqrt{s (s-a) (s-b) (s-c)}\)
Where, s = semi-perimeter = \(\frac{(a+b+c)}{2}\)
a, b, and c = sides of the triangle
Calculate s = \(\frac{(8+8+6)}{2}\) = 11
Area = \(\sqrt{11 (11-8) (11-8) (11-6)}\) = \(\sqrt{11 (3) (3) (5)}\) = \(3\sqrt{55}\)
Hence, option E
gmatclubot
Two sides of a triangle are each 8 units long, and the third [#permalink]
12 Feb 2021, 06:53
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