Question

Find the area of an isosceles triangle whose Base is 16 cm and one of its equal side is 10 cm

Answer 1

if one of the side of the isosceles triangle is 10 cm
so it is forming a right angle triangle
so it base is 8cm and its hypotenuse is 10cm we have to find its height

l^2=h^2-b^2
l^2=10^2-8^2
l= √100-64
l= √36
l= 6cm

area of triangle is=1/2×b×h
=1/2×16×6
=8×6
=48cm^2
hope it will help you!!!!

Answer 2

AnswEr:

Let

• a = BC be the base of an isosceles triangle ABC = 16 cm

• b = AC be the side of an isosceles triangle ABC = 10 cm

• c = AB be the side of an isosceles triangle ABC = 10 cm

_________________________

• $\underline\mathfrak{Semi-perimeter\:of\:\triangle\:ABC:-}$

$\sf = s = \frac{a + b + c}{2} \\ \\ = \sf \: \frac{16 + 10 + 10}{2} \\ \\ \sf = \frac {36}{2} \\ \\ \sf = 18 \: cm$

Therefore,

• s - a = 18 - 16 = 2 cm
• s - b = 18 - 10 = 8 cm
• s - c = 18 - 10 = 8 cm

__________________________

• $\underline\mathfrak{Thus,\:area\:of\:a\:triangle\:ABC}$

$\sf = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \sf = \sqrt{18 \times 2 \times 8 \times 8} \: {cm}^{2} \\ \\ \sf = \sqrt{6 \times 3 \times 2 \times 8 \times 8} \: {cm}^{2} \\ \\ \sf = \sqrt{6 \times 6 \times 8 \times 8 } \: {cm}^{2} \\ \\ \sf = (6 \times 8) \: {cm}^{2} \\ \\ \sf = 48 \: {cm}^{2}$

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