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Question
ΔABC is an isosceles triangle with AB = AC = 13cm. The length of altitude from A on BC is 5cm. Find BC.
Solution
It is given that Δ ABC is an isosceles triangle.
Also, AB = AC = 13 cm
Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.
AD =5 cm
Δ 𝐴𝐷𝐵 𝑎𝑛𝑑 Δ 𝐴𝐷𝐶 are right-angled triangles.
Applying Pythagoras theorem, we have;
`AB^2=AD^2+BD^2`
`BD^2=AB^2-AD^2=13^2-5^2`
`BD^2=169-25=144`
`BD=sqrt144=12`
Hence,
`BC=2(BD)=2xx12=24 cm`
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