CTET 2 Math and Science Sep 2014 Paper

Question : 45 of 120

Marks: +1, -0

The base of an isosceles Δ ABC is 48 cm and its area is

168 cm^2

. The length of one of its equal sides is

Solution:

Area of isoceles triangle

⇒168=\;\;{1}/{4}×a√{4b^{2}-a^{2}}

⇒168=\;\;{1}/{4}×48×√{4b^{2}-(48)^{2}}

⇒\;\;{168×4}/{48}=√{4b^{2}-(48)^{2}}

⇒14=√{4b^{2}-(48)^{2}}

Squaring both sides. We get

⇒196=4b^{2}-(48)^{2}

⇒4b^{2}=196+(48)^{2}

⇒b^{2}=\;\;{196}/{4}+\;\;{48×48}/{4}

⇒b^{2}=49+12×48

⇒b^{2}=49+576=625

∴b=√{625}=25

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