Question

find the area of an isocles triangle whose base is 24 cm and one of its equal sides is 13cm

Answer 1

${\huge {\underline {\blue {\bf {Question}}}}}$

• Find the area of an isocles triangle whose base is 24 cm and one of its equal sides is 13cm.

${\huge {\underline {\blue {\bf {Answer}}}}}$

Let us consider three sides of a triangle as

• a = 12 cm

• b = 13 cm

• c = 24 cm.

Now,

• Semi-perimeter (s) of triangle is given by

$\underline{\boxed{\sf Semi Perimeter \: (s) \ of \ a \ triangle=1/2(a+b+c)}}$

$\rm :\implies\:s \: = \dfrac{13 + 13 + 24}{2}$

$\rm :\implies\:s \: = \: \dfrac{50}{2}$

$\rm :\implies\:\boxed{ \red{ \bf \:s = 25}}$

Now,

• Area of triangle is given by

$\underline{\boxed{\bf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

$\rm :\implies\: \sqrt{25(25 - 13)(25 - 13)(25 - 24)}$

$\rm :\implies\: = \sqrt{25 \times 12 \times 12 \times 1}$

$\rm :\implies\: = 5 \times 12$

$\rm :\implies\: = 60 \: {cm}^{2}$

$\rm :\implies\:\underline{ \pink{\boxed{\bf Area \ of \ triangle \: = \: 60 \: {cm}^{2} }}}$