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  • Find the area of an isosceles triangle whose equal sides are of length 15 cm each and the third side is 12 cm.

Find the area of an isosceles triangle whose equal sides are of length 15 cm each and the third side is 12 cm.

Answers (2)

$ Side length a = b =15 cm, c= 12 cm \\ s =(a+b+c)/2 = 42/2 =21 cm $ \\ \text { Area }=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{21 \times (21-15)(21-15)(21-12)} \\ =& \sqrt{21 \times 6 \times 6 \times 9} \\ &=6 \times 3\sqrt{21 } \\ &=18\sqrt{21} \ \mathrm{cm}^{2} \\ &=82.48 \ \mathrm{cm}^{2}

Posted by

Ravindra Pindel

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√s(s-a) (s-b) (s-c) = √21*(21-15) (21-15) (21-12)

√21*6*6*9)= (3*7*2*3*2*3*3*3) = √3*3*2√7

=18√7

Posted by

Cl choudhary

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