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ΔABC above is an isosceles triangle in which AB = AC. What i

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ΔABC above is an isosceles triangle in which AB = AC. What i [#permalink] New post   22 Sep 2020, 04:49
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\(ΔABC\) above is an isosceles triangle in which AB = AC. What is the area of \(ΔABC\) ?

A. 30

B. 24

C. 20

D. 15

E. 12
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E
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Re: ΔABC above is an isosceles triangle in which AB = AC. What i [#permalink] New post   29 Sep 2020, 10:31
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Re: ΔABC above is an isosceles triangle in which AB = AC. What i [#permalink] New post   29 Sep 2020, 20:17
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given isosceles triangle

so AC=AB=5

for finding height of triangle draw a peripendicular from point A to side BC

now BD=DC=3

height of triangle will be =\(\sqrt{(5^2 - 3^2 )}\)=4

area of triangle ABC=(1/2)*6*4 =12

option E is correct
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Re: ΔABC above is an isosceles triangle in which AB = AC. What i [#permalink] New post   30 Sep 2020, 04:23
Breaking down the triangle to 30:60:90 then we have our height^2=5^2-3^2=16
H=4
Area of the triangle is 1/2*4*6= 12
Ans E

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Re: ABC above is an isosceles triangle in which AB = AC. What i [#permalink] New post   10 Apr 2022, 19:02
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Just saying that height in a isosceles triangle always divides in two equal parts, such as BD=DC=3.
This also applies in a equilateral triangle.
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Re: ABC above is an isosceles triangle in which AB = AC. What i [#permalink]
10 Apr 2022, 19:02
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