Question

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

Answer 1

Answer:

area = 192cm^2

perimeter = 64 cm

Step-by-step explanation:

let height be x

side of isosceles triangle will be x + 4

base is 24

In ∆ABC area = 1/2 × base × height

the height will divide the triangle in two right angle triangle

side will become the hypotenuse

base will be bisected

base = 12 hyp = x + 4 other side = x

${(x + 4)}^{2} = {x}^{2} + {12}^{2} \\ {x}^{2} + 16 + 8x = {x}^{2} + 144 \\ 16 + 8x = 144 \\ 8x = 144 - 16 \\ 8x = 128 \\ x = \frac{128}{8} = 16$

height of the isosceles triangle is 16 cm

base is 24 cm

area of triangle ABC

$\frac{1}{2} \times 24 \times 16 = 12 \times 16 = 192 {cm}^{2} \\ \\ perimeter \: \\ = (x + 4) + (x + 4) + 24 \\ = 16 + 4 + 16 + 4 + 24 \\ = 64cm$

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 = $192cm^{2}$