Question

Find the base of an isosceles triangle whose area is 60 sq.cm and length of equal side is 13 cm
(A) 24m or 10m (B) Only 10m
(C) Only 24m
(D) None of these​

Answer 1

◕ To Find :

The base of the Isosceles Triangle.

◕ We Know :

Area of a Isosceles Triangle :

$\purple{\sf{\underline{\boxed{A = \dfrac{b}{4}\left(\sqrt{4a^{2} - b^{2}}\right)}}}}$

Where ,

• A = Area of the Isosceles Triangle
• b = Base of the Isosceles Triangle
• a = Area of the triangle

◕ Solution :

→ Concept :

According to the question , we have to find the Base of the area , so by putting the given value in the Equation and solving will get the area of the triangle.

→ Calculation :

Given :

• Area of the Isosceles Triangle = 60 cm²

• Length of Equal side = 13 cm

Taken :

Let the base be b.

Using the formula and substituting the values in it , we get :

$\purple{\sf{A = \dfrac{b}{4}\left(\sqrt{4a^{2} - b^{2}}\right)}}$

$\sf{\Rightarrow 60 = \dfrac{b}{4}\left(\sqrt{4 \times 13^{2} - b^{2}}\right)}$

$\sf{\Rightarrow 60 = \dfrac{b}{4}\left(\sqrt{4 \times 169 - b^{2}}\right)}$

$\sf{\Rightarrow 60 = \dfrac{b}{4}\left(\sqrt{676 - b^{2}}\right)}$

$\sf{\Rightarrow 60 \times 4 = b\left(\sqrt{676 - b^{2}}\right)}$

$\sf{\Rightarrow 240 = b\left(\sqrt{676 - b^{2}}\right)}$

$\sf{\Rightarrow \dfrac{240}{b} = \left(\sqrt{676 - b^{2}}\right)}$

By Squaring on both the sides ,we get :

$\sf{\Rightarrow \left(\dfrac{240}{b}\right)^{2} = \left(\sqrt{676 - b^{2}}\right)^{2}}$

$\sf{\Rightarrow \dfrac{240^{2}}{b^{2}} = 676 - b^{2}}$

$\sf{\Rightarrow \dfrac{67600}{b^{2}} = 676 - b^{2}}$

By multiplying b² on both the sides , we get :

$\sf{\Rightarrow \dfrac{67600}{\cancel{b^{2}}} \times \cancel{b^{2}} = \big(676 - b^{2}\big) \times b^{2}}$

$\sf{\Rightarrow 67600 = 676b^{2} - b^{4}}$

$\sf{\Rightarrow b^{4} - 676b^{2} + 67600 = 0}$

Using the middle-splitting factor theorem :

$\sf{\Rightarrow b^{4} - (576 + 100)b^{2} + 67600 = 0}$

$\sf{\Rightarrow b^{4} - 576b^{2} - 100b^{2} + 67600 = 0}$

$\sf{\Rightarrow b^{2}(b^{2} - 576) - 100(b^{2} - 576) = 0}$

Taking the common i.e,(b² - 576) ,we get :

$\sf{\Rightarrow (b^{2} - 576)(b^{2} - 100) = 0}$

$\purple{\sf{\therefore (b^{2} - 576)(b^{2} - 100) = 0}}$

Hence ,the zeroes of the polynomial are ,

• $\sf{b^{2} - 576 = 0}$

$\sf{\Rightarrow b^{2} = 576}$

$\sf{\Rightarrow b = \sqrt{576}}$

$\sf{\Rightarrow b = 24}$

• $\sf{b^{2} - 100 = 0}$

$\sf{\Rightarrow b^{2} = 100}$

$\sf{\Rightarrow b = \sqrt{100}}$

$\sf{\Rightarrow b = 10}$

Hence , the base of the Isosceles Triangle is 24 cm and 10 cm.

» Additional information :

• Area of a Sector = lr/2

• Perimeter of a Sector = l + 2r

• Surface area of a Cylinder = 2πr(h + r)

• Lateral surface area of a Cylinder = 2πrh