Question

find the area of an isosceles triangle whose base is 18 centimetre and one of its equal side is 10 centimetre​

Answer 1

Area of triangle =

\sqrt{s(s - a)(s - b)(s - c)}

S=a+b+c/2

a=18 b=15 c=15

S=48/2

s=24

\sqrt{24(24 - 18)(24 - 15) (24 - 15)}

\sqrt{24 \times 6 \times 9 \times 9}

\sqrt{6 \times 2 \times 2 \times 6 \times 9 \times 9}

=6*2*9

=

{108cm }^{2}

Answer 2

Hey mate

Here is your Answer

area of triangle= root over (s-a)(s-b)(s-c) .first we find s . s=a+b+c/2 ok .so,s=18+15+15/2=24.now area of triangle=rootover 24(24-18)(24-15)(24-15)=rootover 24(6)(9)(9) now we find the prime factor so, the prime factor is rootover 2×2×2×3×2×3×3×3×3×3=now we make the pair bcoz it is in the root ok.so,2×2×3×3×3=4×9×3=108. so, the area of triangle is 108m^2

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