Question 60

∆ABC is an isosceles triangle with AB = AC = 15 cm and altitude from A on BC is 12 cm. Length of side BC is

Solution

Given : ABC is an isosceles triangle with AB = AC = 15 cm. AD is the altitude = 12 cm

To find : BC = ?

Solution : Altitude of an isosceles triangle bisects the opposite side, => BD = CD = $$\frac{BC}{2}$$

In $$\triangle$$ ADC,

=> $$(CD)^2=(AC)^2-(AD)^2$$

=> $$(CD)^2=(15)^2-(12)^2$$

=> $$(CD)^2=225-144=81$$

=> $$CD=\sqrt{81}=9$$ cm

$$\therefore$$ BC = $$2 \times $$ CD

= $$2 \times 9=18$$ cm

=> Ans - (C)

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