Question

find the area of an isosceles triangle whose base is 18 cm and one of its equal sides is 15 CM​

Answer 1

Area of triangle =
\sqrt{s(s - a)(s - b)(s - c)}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)}24(24−18)(24−15)(24−15)​ 
\sqrt{24 \times 6 \times 9 \times 9}24×6×9×9​ 
\sqrt{6 \times 2 \times 2 \times 6 \times 9 \times 9}6×2×2×6×9×9​ 
=6×2×9
=108 cm^2 

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