Question

an isosceles triangle has perimeter 30 cm and each of the equal sides is is is 12 cm find the area of the triangle​

Answer 1

Answer:

issoscles triangle has two side angle are same perimeter of isosceles triangle equal to number of angle side add

30+12 42 answer

area of the triangle 42 answer

Answer 2

Given : The Perimeter of an isosceles triangle is 30 cm and equal sides are of 12 cm .

Need To Find : Area of Triangle ?

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❍ Let's say that , unequal side of Triangle be c cm .

⠀▪︎⠀We know that Perimeter of Triangle is the sum of all sides of Triangle i.e, " Perimeter of ∆ = a + b + c " .

$\\\longrightarrow \pmb{\sf \:Perimeter_{\:(Triangle)\:}\:=\: a + b + c }$

$\longrightarrow \sf \:30\:=\: 12 + 12 + c$

$\longrightarrow \sf \:30\:=\: 24 + c$

$\longrightarrow \sf \:c\:=\: 30 - 24$

$\longrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:c\:\:=\:6\:cm\:}}}}}\:\bigstar \:$

⠀⠀⠀⠀⠀∴ Hence, Three Sides of an Isosceles Triangle are 12 cm , 12 cm & 6 cm .

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⠀⠀⠀⠀⠀¤ Area of Triangle :

$\qquad \:\star\:\underline{\boxed{\pmb{\sf{Area_{\;(Triangle)} = \sqrt{s\Bigg(s - a\Bigg)\Bigg(s - b\Bigg)\Bigg(s - c\Bigg)}}}}}$

$\frak{Where}\:\begin{cases}\:\quad \sf \pmb{\frak{ a \:,\:b\:\&\:c\:}}\: are\:sides \:of \:triangle \:.\:\\\\\:\quad \sf \pmb{\frak{ s\:}}\:=\:Semi-Perimeter \:=\:\cancel{\dfrac{30}{2}}\:=\: \pmb{\sf 15\:cm\:}\:\end{cases}$

$\qquad \dag\underline {\frak{ Substituting \:known\:Values \:in\:Formula \:\::\:}}$

$:\implies \sf Area_{\;(Triangle)} = \sqrt{s\Bigg(s - a\Bigg)\Bigg(s - b\Bigg)\Bigg(s - c\Bigg)} \:\\\\\\ :\implies \sf Area_{\;(Triangle)} = \sqrt{15\Bigg(15 - 12\Bigg)\Bigg(15 - 12\Bigg)\Bigg(15 - 6\Bigg)} \:\\\\\\ :\implies \sf Area_{\;(Triangle)} = \sqrt{15\times 3 \times 3 \times 9 } \:\\\\\\ :\implies \sf Area_{\;(Triangle)} = \sqrt{15\times 9 \times 9 } \:\\\\\\:\implies \sf Area_{\;(Triangle)} = \sqrt{15\times 9^2 } \:\\\\\\ :\implies \pmb {\underline {\boxed {\purple {\:\frak{ \:Area_{\;(Triangle)} = 9\sqrt{15 }\:cm^2\:}}}}}\:\bigstar \:$

$\:\therefore \:\underline {\sf Hence, \:Area \:of\:Triangle \:is\:\pmb{\sf{9\sqrt{15}\:cm^2\:}}.}$