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In the given 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,

AC^{2} = AE^{2} + EC^{2}

⇒ EC^{2} = AC^{2} – AE^{2}

⇒ EC^{2} = (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 cm^{2}

and Area of ΔAEB = `1/2` × AE × EB

= `1/2 xx 15 xx 13`

= 97.5 cm^{2}

∴ Area of ΔABC = Area of ΔAEC – Area of ΔAEB

= 150 – 97.5

= 52.5 cm^{2}

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 cm^{2} and the length of DB is 4.2 cm.

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