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Question
In the following figure, triangle AEC is right-angled at E, B is a point on EC, BD is the altitude of triangle ABC, AC = 25 cm, BC = 7 cm and AE = 15 cm. Find the area of triangle ABC and the length of DB.
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Solution
Given, AC = 25 cm, BC = 7 cm, and AE = 15 cm
In ΔAEC, using Pythagoras theorem,
AC2 = AE2 + EC2
⇒ EC2 = AC2 – AE2
⇒ EC2 = (25)2 – (15)2 = 625 – 225 = 400
EC = `sqrt(400)` = 20 cm
And EB = EC – BC = 20 – 7 = 13 cm
Area of ΔAEC = `1/2` × AE × EC
= `1/2 xx 15 xx 20`
= 150 cm2
And Area of ΔAEB = `1/2` × AE × EB
= `1/2 xx 15 xx 13`
= 97.5 cm2
∴ Area of ΔABC = Area of ΔAEC – Area of ΔAEB
= 150 – 97.5
= 52.5 cm2
Again, Area of ΔABC = `1/2` × BD × AC
52.5 = `1/2` × BD × 25
⇒ BD = `(52.5 xx 2)/25` = 4.2 cm
Hence, the area of ΔABC is 52.5 cm2 and the length of DB is 4.2 cm.
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