Question

find the area of an isosceles triangle whose perimeter is 60 cm and each of the equal side is 24 cm​

Answer 1

$\large{\boxed{\bf{AnswEr}}}$

• Area of the isosceles triangle is 36√15 cm.

Given :-

• Perimeter of an isosceles triangle is 60cm.

• Each of the equal sides is 24cm.

To Find :-

• Area of the isosceles triangle.

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$\large{\boxed{\bf{Solution}}}$

Here,

• Equal sides, S = 24

• Perimeter = 60cm

• Base = ?

★ We know that the Perimeter of an isosceles triangle is :-

• Perimeter = 2s + b

➪ 2 (24) + b = 60

➪ 48 + b = 60

➪ b = 60 - 48

➪ b = 12cm

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Here,

• Hypotenuse = 24

• Base = 6 [ when an altitude is drawn, it bisects the base ]

• Height = √{ (24)² - (6)² } ➪ √ 576 - 36 ➪ 6√15

★ Now, we'll find the area of the isosceles triangle :-

• Area = base × height /2

➪ 12 × 6√15 /2

➪ 6 × 6√15

➪ 36√15 cm

Hence, the area of the isosceles triangle is 36√15cm.

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Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow taking\:isosceles\:triangle\:as\: \triangle ABC$

$\sf\dashrightarrow perimeter\:of\: isosceles\:triangle\:= 60cm$

$\sf\dashrightarrow side\:of\:triangle\:(a)=24cm$

$\sf\dashrightarrow let\:the\:third\:side(base)\:be\:x\:cm$

$\red{\text{ NOTE:- ISOSCELES TRIANGLE CONTAINS, TWO EQUAL SIDES AND ONE UNEQUAL.}}$

$\large\underline\bold{TO\:FIND,}$

$\sf\large\dashrightarrow area\:of\: isosceles\:triangle$

TAKING TWO CASES,

$\sf\therefore in\:case\:one,\:we\:will\:find\:the\:base\:of\:an\:issosceles\:triangle.$

$\sf\therefore in\:case\:two,\:we\:will\:find\:the\:area\:of\:an\:issosceles\:triangle.$

✯FORMULA IN USE,

✯FOR CASE:-1,

$\bf{\boxed{\bf{ \star\:\: PERIMETER\:OF\: TRIANGLE [ISOSCELES]= SUM\:OF\:ALL\:SIDES\:\: \star}}}$

✯FOR CASE :-2

$\large{\boxed{\bf{ \star\:\: AREA\:OF\:TRIANGLE= \dfrac{1}{2} \times base\times \:height\:\: \star}}}$

$\large\underline\bold{SOLUTION,}$

$\large{\boxed{\bf{CASE:-1 }}}$

$\sf\therefore PERIMETER\:OF\: TRIANGLE [ISOSCELES]= SUM\:OF\:ALL\:SIDES$

$\sf\implies 60= 24+24+x$

$\sf\implies 60=48+x$

$\sf\implies 60-48= x$

$\sf\implies 12= x$

$\large{\boxed{\bf{ \star\:\: base (x)=12cm\:\: \star}}}$

$\sf\therefore note:-FOR\:FINDING\:AREA\:OF\:TRIANGLE\:WE\: DON'T\:KNOW\:HEIGHT,$

$\sf\dashrightarrow HEIGHT\:CAN\:BE\:FOUND\:USING\: Pythagoras\:theorem.$

$\red{\text{DRAWING A LINE PERPENDICULAR TO BC }}$

$\sf\therefore TWO\:TRIANGLES\:FORMED(\triangle ABD \:and \triangle ADC.)$

$\sf\therefore TAKING \triangle ADC.$

$\sf\dashrightarrow AC= 24cm$

$\sf\dashrightarrow DC= \dfrac{BC}{2}= \dfrac{12}{2} = 6cm$

$\sf\large\therefore by\: Pythagoras\:theorem,$

$\sf\therefore AC^2= AD^2+DC^2$

$\sf\dashrightarrow (24)^2= (h)^2+ (6)^2$

$\sf\implies 576= h^2+36$

$\sf\implies h^2= 576-36$

$\sf\implies h= \sqrt{ 540}$

$\sf\implies h= 23.23 cm$

$\large{\boxed{\bf{ \star\:\:height(h)= 23.23cm \:\: \star}}}$

$\large{\boxed{\bf{ CASE:-2}}}$

$\sf\therefore AREA\:OF\:TRIANGLE= \dfrac{1}{2} \times base\times \:height$

$\sf\implies \dfrac{1}{2} \times 12 \times (23.23)$

$\sf\implies \dfrac{1}{\cancel{2}} \times \cancel{12} \times (23.23)$

$\sf\implies 6 \times (23.23)$

$\sf\implies 139.38cm^2$

$\large{\boxed{\bf{ \star\:\: area\:of\:isosceles\:triangle= 139.38cm^2\:\: \star}}}$

$\large\underline\bold{AREA\:OF\:ISOSCLES\:TRIANGLE\:IS\:139.38cm^2}$

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