Question

find the area of a trianglr whose perimeter is 132 cm one side is 60 cm and the difference of the lenghth of the other two sides is 50 cm​

Answer 1

$\underline{\underline{ \blacksquare \: \bold{correct \: question : }}}$

★ Find the area of a triangle whose perimeter is 132 cm one of it's side is 60 cm and the difference of the length of the other two side is 50 cm .

$\odot \: \underline{\underline \bold{given : }}$

$\leadsto \sf \red{ \: perimeter = 132 \: cm}$

$\leadsto \: \sf \red{measure \: of \: one \: side = 60 \: cm}$

$\sf \leadsto \red{ \: difference \: of \: the \: length \: of \: two \: other \: sides = 50 \: cm \: }$

$\odot \: \underline{\underline \bold{find : }}$

$\leadsto \sf \purple{area \: of \: the \: \triangle}$

_______________________________

$\sf \star \: Let \: the \: sides \: be \: - a \: ,b \: , \: c$

As one is side is 60 cm

Perimeter of triangle = 2s = 132 cm

a+b+c= 132

b+ c=132-60

b+c= 72 .. (1)

Difference between two sides is 50 cm

b - c = 50 ..(2)

equation ( 1 ) + ( 2 )

b + c + b - c

72 + 50

= 122

2b = 122

b = 122/2

b = 61

Substituting b = 61 in equation (1)

61 + c = 72

c = 72 - 61

c = 11

As we know , the formula of area of triangle with sides a, b , c

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

$\sf \: 2s = 132 \\ \sf s = \cancel \frac{132}{2} = 66$

$\sqrt{66(66 - 60)(66 - 61)(66 - 11)}$

$\sqrt{66 \times 6 \times 5 \times 55 }$

$\sqrt{108900}$

$\overline{ \underline{ \fbox{ = 330}}}$

$\bold{ \blacktriangleright \: the \: area \: of \: \triangle = 330 \: cm^{2} }$