Question

Find the area of an isosceles triangle whose perimeter is 108 cm and base 48 CM

Answer 1

Length of base = 48 cm

length of 1 equal side = x cm

Total length of 2 equal sides = 2x cm

Perimeter = side1 + side2 + side3 = 48 cm + x cm + x cm = 48 cm + 2x cm = 108 

 = 48 cm + 2x cm = 108 cm

= 2x cm = 108 cm - 48 cm = 60 cm

= x = 60/2 cm = 30 cm

Therefore, all sides of triangle are 48 cm , 30 cm & 30 cm.

Area of isosceles triangle =  1/2 X base X sq. root of [ (ength of equal side2 - base2 / 4]

= 1/2 X 48 X sq. root of [ 30 X 30 - 48 X 48 / 4]

= 24 X sq. root of [ 900 - 576 ] = 24 X sq. root of [ 324 ]

= 24 X 18 = 432 cm2

Therefore, area of the triangle is 432 cm2

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Answer 2

$\pink{\underline{\huge{\ulcorner{\star\: Solution\: \star}\urcorner}}}$

$Let\: we\: consider \: the \colon \\ \triangle ABC whose \: one \: side \: BC (Base) 48cm$

$AB = AC = a \\ Perimeter = 108 \\ AB + BC + CA = 108 \\ a+a+48 = 108 \\ 2a= 108-48\\ 2a=60\\a=\frac{60}{2}\\ a=\frac{\cancel60}{\cancel2}\\a=30 \:cm$

$AB = AC = 30 \: cm$

$\red{\underline{BY\: Heron's\: formulae}}$

$S=\frac{AB+BC+CA}{2}\\\implies {108}{2}\\\implies 54$

$Area = \sqrt{S(S-AB)(S-BC)(S-CA)}\\\implies \sqrt{54(54-30)(54-30)(54-48)}\\\implies \sqrt{54\times24\times24\times6}\\\implies \sqrt{186624}\\\implies 432 \: cm^{2}$

$\green{\boxed{\bold{Area \: of \: \triangle \: ABC= 432\: cm^{2} }}}$