Question

the base of an isosceles triangle measures 80cm and its area is 360cm². find the perimeter of the triangle.

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Answer 1

Step-by-step explanation:

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Answer 2

A n s w e r

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G i v e n

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• Base of an isosceles triangle is 80 cm

• Area of the triangle is 360 sq. cm.

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F i n d

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• Perimeter of the triangle

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S o l u t i o n

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We know that in a isosceles triangle the sides except the base are equal

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• Let the equal sides be "z"

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$\underline{\bold{\texttt{Area of triangle :}}}$

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➠ $\sf \sqrt { s(s - a)(s - b)(s - c) }$ ⚊⚊⚊⚊ ⓵

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Where ,

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• s = Semi perimeter

• a = 1st Side

• b = 2nd Side

• c = 3rd Side

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➻ $\sf \dfrac { a + b + c } { 2 }$

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$\underline{\bold{\texttt{Area of given isosceles triangle :}}}$

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Let 1st and 3rd sides be equal

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So,

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➜ $\sf s = \dfrac { z + 80 + z } { 2 }$

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➜ $\sf s = \dfrac { 2(z + 40) } { 2 }$

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• s = z + 40

• a = z

• b = 80

• c = z

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⟮ Putting the above values in ⓵ ⟯

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

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➜ $\sf \sqrt { (z + 40)(z + 40 - z)(z + 40 - 80)(z + 40 - z) }$

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➜ $\sf \sqrt { (z + 40)(40)(z - 40)(40 ) }$

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➜ $\sf 40\sqrt { (z + 40)(z - 40) }$

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➜ $\sf 40\sqrt {z ^2 - 40 ^2 }$

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Given that the area is 360 sq. cm.

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So,

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➜ $\sf 40\sqrt {z ^2 - 40 ^2 } = 360$

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➜ $\sf \sqrt {z ^2 - 40 ^2 } = 9$

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⟮ Squaring both side ⟯

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➜ $\sf (\sqrt {z ^2 - 40 ^2 } )^2 = 9 ^2$

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➜ $\sf z ^2 - 40 ^2 = 81$

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➜ $\sf z ^2 = 81 + 40 ^2$

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➜ $\sf z ^2 = 81 + 1600$

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➜ $\sf z ^2 = 1681$

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➜ $\sf z = \sqrt { 1681 }$

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➨ z = 41 cm

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• Hence the other two sides are of 41 cm each

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$\underline{\bold{\texttt{Perimeter of triangle :}}}$

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➠ a + b + c ⚊⚊⚊⚊ ⓶

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Where ,

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• a = 1st side

• b = 2nd side

• c = 3rd side

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$\underline{\bold{\texttt{Perimeter of given isosceles triangle :}}}$

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• a = 41

• b = 80

• c = 41

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⟮ Putting the above values in ⓶ ⟯

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➜ a + b + c

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➜ 41 + 80 + 41

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➨ 162 cm

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• Hence the perimeter of the given triangle is 162 cm

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