# Isosceles triangle 9

Given an isosceles triangle ABC where AB= AC. The perimeter is 64cm and altitude is 24cm. Find the area of the isosceles triangle

Result

A =  168 cm2

#### Solution:

$p = 64 \ cm \ \\ h = 24 \ cm \ \\ \ \\ p = 2a + b \ \\ \ \\ a^2 = (b/2)^2 + h^2 \ \\ a^2 = ((p-2a)/2)^2 + h^2 \ \\ a^2 = (p/2-a)^2 + h^2 \ \\ a^2 = p^2/4-pa + a^2 + h^2 \ \\ pa = p^2/4 + h^2 \ \\ \ \\ a = p/4 + h^2/p = 64/4 + 24^2/64 = 25 \ cm \ \\ \ \\ b = p - 2 \cdot \ a = 64 - 2 \cdot \ 25 = 14 \ cm \ \\ \ \\ p_{ 1 } = 2 \cdot \ a + b = 2 \cdot \ 25 + 14 = 64 \ cm \ \\ p_{ 1 } = p \ \\ \ \\ A = \dfrac{ b \cdot \ h }{ 2 } = \dfrac{ 14 \cdot \ 24 }{ 2 } = 168 = 168 \ cm^2$

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Showing 2 comments:
Steve
No, your solution starts with h=25cm, but the question says that h=24cm.

So, looking at one-half of the isosceles triangle, x2 + 242 = (32-x)2    solves to x=7cm,   which gives the area of the isosceles triangle as 7x24 = 168cm2

Dr Math
thank you, we corrected 25 to 24 as altitude....

#### Following knowledge from mathematics are needed to solve this word math problem:

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