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Angle Bisector Theorem

In ABC, AB ...

Question

In $△$ ABC, AB = AC = x(cm); BC = 10 cm and the area of the triangle is 60 $c m^{2}$ . Find x(cm).

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Solution

In $△ A B C$ ,
$A B = A C = x$ and $B C = 10$ cm
Draw a perpendicular from A on BC such that $A D ⊥ B C$
In an Isosceles triangle a perpendicular bisects the non- congruent side.
Hence, $B D = \frac{1}{2} (10) = 5$ cm
In $△ A D B$ ,
$A B^{2} = A D^{2} + B D^{2}$
$x^{2} = A D^{2} + 5^{2}$
$A D = \sqrt{x^{2} - 25}$
Area of triangle = $\frac{1}{2} A D \times B C$
$60 = \frac{1}{2} (\sqrt{x^{2} - 25}) (10)$
$12 = \sqrt{x^{2} - 25}$
$144 = x^{2} - 25$
$x^{2} = 169$
$x = 13$ cm

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