Question

Calculate the area and perimeter of a triangle in which base is 24 cm and each of equal sides of an isosceles triangle is 4cm greater than its height

Answer 1

perimeter of the given triangles = 24+4+4=32cm ( s1+s2+s3)

area of triangle=1/2*24*4=48 cm (1/2*base*height)

please make this as barinliest answer

Answer 2

Answer:

Perimeter = 64cm

Area = 192cm^2

Step-by-step explanation:

Base = 24cm

Height = xcm

(x+4)^2 = x^2 + (12)^2

x^2 + 8x + 16 = x^2 + (12)^2

8x = x^2 - x^2 + 144 - 16

8x = 128

x = 128/8

x = 16

Perimeter = 24 + 16+ 4 + 16 + 4

Perimeter = 64cm

Area = $\sqrt{s(s-a)(s-b)(s-c)}$

{S = 64/2 = 32 ; A = 24 ; B = 20 ; C = 20}

Area = $\sqrt{32(32-24)(32-20)(32-20)}$

Area = $\sqrt{32*8*12*12}$

Area = $\sqrt{8*4*8*12*12}$

Area = 8*2*12

Area = $192cm^{2}$