Question

Q.9 The area of an isosceles triangle is 12cm². If one of its equal side is 5cm. Find its base.

Answer 1

Answer:

Solution:

$\sf{Let \ base \ be \ x} \\ \\ \sf{s=\dfrac{a+b+c}{2}} \\ \\ \sf{Here, \ s=\dfrac{5+5+x}{2}} \\ \\ \sf{\therefore{s=5+\dfrac{x}{2}}} \\ \\ \sf{By \ heron's \ formula} \\ \\ \boxed{\sf{A(\triangle)=\sqrt{s(s-a)(s-b)(s-c)}}} \\ \\ \sf{\therefore{12=\sqrt{(5+\dfrac{x}{2})(5+\dfrac{x}{2}-5)(5+\dfrac{x}{2}-5)(5+\dfrac{x}{2}-x)}}} \\ \\ \sf{On \ squaring \ both \ sides} \\ \\ \sf{\therefore{144=(5^{2}-\dfrac{x^{2}}{4})(\dfrac{x}{2})^{2}}} \\ \\ \sf{\therefore{144=(\dfrac{100-x^{2}}{4})(\dfrac{x^{2}}{4})}} \\ \\ \sf{\therefore{144=\dfrac{100x^{2}-x^{4}}{16}}} \\ \\ \sf{\therefore{100x^{2}-x^{4}=2304}} \\ \\ \sf{Substitute \ x^{2}=a} \\ \\ \sf{a^{2}-100a+2304=0}$

$\sf{By \ formula \ method} \\ \\ \sf{a=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}} \\ \\ \sf{\therefore{a=\dfrac{100\pm\sqrt{784}}{2}}} \\ \\ \sf{\therefore{a=50\pm14}} \\ \\ \sf{\therefore{a=64 \ or \ a=36}} \\ \\ \sf{\therefore{x^{2}=64 \ or \ 36}} \\ \\ \sf{Lengths \ are \ in \ positive} \\ \\ \sf{\therefore{x=8 \ cm \ or \ 6 \ cm}} \\ \\ \orange{\tt{\therefore{The \ base \ of \ the \ triangle \ is \ 8 \ cm \ or \ 6 \ cm}}}$