Math, asked by 2807apam22, 8 months ago

the perimeter of an isosceles triangle is 32cm.the ratio of the equal side to its base is 3:2. find the area of the triangle.​

Answers

Answered by bagkakali
2

Step-by-step explanation:

let base is x cm and one equal side is y cm

x+y+y=32

x+2y=32

y:x=3:2

y/x=3/2

2y=3x

x+2y=32

x+3x=32

4x=32

x=32/4=8

so base is 8 cm and equal sides are (32-8)/2cm

12 cm each

height of the triangle is√(12^2-4^2)cm

=√128cm=8√2cm

Area of the triangle is 1/2.8.8√2sqcm

=32√2sqcm

Answered by sfsgzdfgxggmailcom
0

The perimeter of an issosceles ∆-32 cm.

given that the ratio of the Equal sides to its

base is 3:2

so the ratio of the three sides of the issosceles

∆=3:3:2

sum the ratio 3+3+2=8

length of the Equal sides=3/8×32= 3×4=12

length of the base =2/8×32=2×4=8

according to heron's fromula.area of an

issosceles ∆ with Equal sides a & base b

= 1/4×b×√ (4a2 - B2)

Area of the given issosceles triangle where

a =12 & b =8

1/4×8×√(4×12)-8²

2×√(4×144)-64=2×√576-64

2×√512-2×√2×16×16

2×16×√2=32√2

** m=2

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