Question

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

Answer 1

Answer:

Refer to the attached pic...

Hope it helps.....

Answer 2

Answer:

$\orange { Perimeter \: of \: \triangle }\green {=64\:cm }$

$\blue {Area \: of \triangle ABC }\green{=192\: cm^{2}}$

Step-by-step explanation:

Given:

∆ABC is an isosceles triangle.

AD is perpendicular to BC.

Let altitude AD = x cm,

AB = AC = (x + 4) cm,

BC = 24 cm ,

solution:

∆ADC is a right angled triangle .

AC² = AD² + DC²

( Phythagorean theorem )

=> ( x + 4 )² = x² + 12²

=> x² + 8x + 16 = x² + 144

=> 8x = 144 - 16

=> 8x = 128

=> x = 128/8 = 16 cm

Therefore.,

AD = x = 16 cm

AB = AC = x + 4 = 16 + 4 = 20 cm,

$\orange { Perimeter \: of \: \triangle }\\= AB + BC + AC\\</p><p>= 20 + 24 + 20\\= 64 \: cm$

$\blue {Area \: of \triangle ABC }\\= \frac{1}{2} \times BC \times AD$

$= \frac{1}{2} \times 24 \times 16$

$= 12 \times 16\\= 192\: cm^{2}$

Therefore.,

$\orange { Perimeter \: of \: \triangle }\green {=64\:cm }$

$\blue {Area \: of \triangle ABC }\green{=192\: cm^{2}}$

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