Question

the perimeter of an isosceles triangle is 42 cm and its base is 1 1/2 times each of the equal side find the length of each side of the triangle the area of the triangle and the height of the triangle .​

Answer 1

AnswEr :

$\bf{\purple{\large{\underline{\underline{\bf{Given\::}}}}}}$

The perimeter of an Isosceles triangle is 42 cm and it's base is 1 1/2 times each of the equal side.

$\bf{\red{\large{\underline{\underline{\bf{To\:find\::}}}}}}$

The length of each side of the triangle the area of the triangle and the height of the Δ.

$\bf{\orange{\large{\underline{\underline{\bf{Explanation\::}}}}}}$

$\bf{We\:have\:in\:Isosceles\: \triangle}\begin{cases}\sf{Perimeter\:(P)=42\:cm}\\ \sf{Two\:side\:are\:equal\:=\:r}\\ \sf{The\:base\:of\:\triangle = \dfrac{3r}{2} }\end{cases}}$

$\star$We know that formula of the perimeter of triangle :

$\longmapsto\sf{Perimeter\:of\:triangle=Side+Side+Side}\\\\\\\longmapsto\sf{42cm=r+r+\dfrac{3r}{2} }\\\\\\\longmapsto\sf{42cm=\dfrac{2r+2r+3r}{2} }\\\\\\\longmapsto\sf{84cm=7r}\\\\\\\longmapsto\sf{r=\cancel{\dfrac{84}{7} }}\\\\\\\longmapsto\sf{\red{r=12\:cm}}$

$\bf{\large{\underline{\underline{\bf{\blacksquare{Sides\:of\:the\:Isosceles\:\triangle\::}}}}}}$

$\bullet\sf{1st\:side=r=12\:cm}\\\bullet\sf{2nd\:side=r=12\:cm}\\\\\bullet{\sf{3rd\:side\:=\dfrac{3(12)}{2} =\dfrac{36}{2} =18\:cm}}$

$\bf{\large{\underline{\underline{\bf{\blacksquare{UsinG\:Heron's\:FormULA\::}}}}}}$

$\leadsto\tt{Semi-perimeter\:(S)=\dfrac{A+B+C}{2} }\\\\\\\leadsto\tt{Semi-perimeter=\dfrac{12cm+12cm+18cm}{2} }\\\\\\\leadsto\tt{Semi-perimeter=\cancel{\dfrac{42}{2}} cm}\\\\\\\leadsto\tt{\red{Semi-perimeter=21\:cm}}$

Now,

$\leadsto\bf{Area\:of\:\triangle=\sqrt{S(S-A)(S-B)(S-C)}\:cm^{2} }\\\\\\\leadsto\tt{Area\:of\:\triangle =\sqrt{21(21-12)(21-12)(21-18)}\: cm^{2} }\\\\\\\leadsto\tt{Area\:of\:\triangle=\sqrt{21(9)(9)(3)} \:cm^{2} }\\\\\\\leadsto\tt{Area\:of\: \triangle=\sqrt{3*7*3*3*3*3*3} \:cm^{2} }\\\\\\\leadsto\tt{\red{Area\:of\:\triangle=27\sqrt{7} \:cm^{2}}}$

$\bf{\large{\underline{\underline{\bf{\blacksquare{Height\:of\:the\:isosceles\:\triangle\::}}}}}}$

$\star$Formula use of the another area of the triangle :

$\Rightarrow\bf{Area\:of\:the\:triangle\:=\:\dfrac{1}{2} \times Base\times Height}\\\\\\\\\Rightarrow\sf{27\sqrt{7} =\dfrac{1}{\cancel{2}} \times \cancel{18}\times h}\\\\\\\\\Rightarrow\sf{27\sqrt{7} =9\times h}\\\\\\\\\Rightarrow\sf{h=\dfrac{\cancel{27}\sqrt{7} }{\cancel{9}} }\\\\\\\\\Rightarrow\sf{\red{h=3\sqrt{7} \:cm}}$

Thus,

$\underbrace{\bf{\green{The\:area\:of\:Isosceles\: \triangle=27\sqrt{7} \:cm^{2} }}}}}\\\\\underbrace{\bf{\green{The\:height\:of\: \triangle=3\sqrt{7} \:cm }}}}}$

Answer 2

||✪✪ QUESTION ✪✪||

The perimeter of an isosceles triangle is 42 cm and its base is 1 1/2 times each of the equal side find the length of each side of the triangle the area of the triangle and the height of the triangle . ?

|| ✰✰ ANSWER ✰✰ ||

Given That, Base of Isosceles ∆ is 1(1/2) Times each of The Equal sides.

So, Lets Assume That, Both Equal Sides of isosceles ∆ is X cm.

→ Than Base of ∆ = (3x/2)cm.

So,

→ perimeter of ∆ = (x + x + 3x/2)

→ (2x + 3x/2) = 42

→ (7x/2) = 42

→ 7x = 42*2

→ x = 12cm.

So, Equal sides of Isosceles ∆ are = 12 cm.

and, Base of ∆ is = 12*(3/2) = 18cm.

___________________________

Now Here Either we use Heron Formula as we have all sides , or we can use Pythagoras Theoram To Find Height of ∆ first.

we know That, Perpendicular Of a isosceles ∆ divide The base in Equal parts at Right angle .

So, using Pythagoras Theoram Now, we have :-

→ (18/2)² + (Perpendicular)² = (12)²

→ (Perpendicular)² = (12)² - (9)²

→ (Perpendicular)² = 144 - 81

→ (Perpendicular)² = 63

Square - root both sides we get,

→ Perpendicular = √(63) = √(7*3*3) = 3√7 cm.

___________________________

So,

→ Area of isosceles ∆ = (1/2) * Base * Height

→ Area = (1/2) * 18 * (3√7)

→ Area = 9 * (3√7)

→ Area = 27√7 cm².

Hence , Area of Required isosceles ∆ will be (27√7)cm² and Its Height on base 18cm will be 3√7cm.