Question

Each of equal sides of an isosceles triangle is 4cm greater than its height.If base of the triangle is 24cm; calculate the perimeter and tge area of the triangle

Answer 1

Here is your answer,

Let there be a triangle  ABC with base BC and AB & AC and the height be AD touching BC at  D.

Now,  AD will bisect BC {because it is isosceles triangle}

Hence in trag. ADC we have,

(AD)^2 + (DC)^2 = (AC)^2

(X)^2 + (12)^2 = (X+4)^2              {SAY AD = AC = X cm}

X^2 + 144= X^2 +16+ 8X

144= 16+ 8X

8X = 128

X= 16 cm

Hence,

Height = 20 cm

Base = 24 cm

Equal  side = 16 cm

Perimeter =  (16+16+24) cm = 56 cm

Area = (1/2 * 24 *20) cm^2  = 240 cm^2

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 = $192cm^{2}$