find the altitude and area of an isosceles triangle whose perimeter is 64 cm
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Step-by-step explanation:
let the equal sides be x and x
perimeter of a triangle = sum of all sides
64 = x + x+ 24
64 = 2x +24
64-24 =2x
40 =2x
40÷2 = x
20 = x
the sides are 20cm , 20cm ,24 cm
by heron's formula,
semi perimeter = perimeter ÷2
s =64÷2
=32 cm
area of triangle = sq rt / s ( s-a) ( s-b) (s-c)
= sq rt / 32 ( 32 - 24 ) ( 32 - 20 ) (32- 20)
= sq rt / 2 ×2×2×2 ( 8 ) ( 12 ) ( 12 )
= 4 sq rt / 2×2 ×2×2 ×2×2 × 3×3 ×2
= 4×8×3 sq rt / 2
= 32×3 sq rt / 2
= 96 sq rt / 2 cm²
area of triangle = 1/2 × base × height
96 sq rt /2 = 1/2 × 24 × hieght
2( 96 sq rt /2 ) = 24 × hieght
192 sq rt / 2 = 24 × hieght
192 sq rt /2 ÷ 24 = hieght
8 sq rt / 2 = hieght
11.31 cm = hieght
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Answer:
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