Question

find the altitude and area of an isosceles triangle whose perimeter is 64 cm and whose base is 24 cm​

Answer 1

Answer:

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 × h

2( 96 sq rt /2 ) = 24 × h

192 sq rt / 2 = 24 × h

192 sq rt /2 ÷ 24 = h

8 sq rt / 2 = h

11.31 cm = h or altitude

Answer 2

Answer:

height = 16 cm and area = 192 sq.cm

Step-by-step explanation:

let the unknown side be x and base = 24 cm

perimeter = 64 cm

24+x+x=64 cm

24+2x = 64 cm

2x =64-24

2x = 40

x = 20 cm

let the altitude of the triangle be AD

w.k.t altitude of an isosceles triangle divides the base into two equal half  and its perpendicular to the base

in triangle ABD,

$AD^{2}$ = $AB^{2}-BD^{2}$ ( by Pythagoras theorem )

AD = 16 cm

area of triangle = 1/2 x bx h

                           = 1/2 x24x16

                           = 192 sq. cm