Q122 of 131 Page 226

In Fig. 9.65, 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.

In ∆AEC, AE = 15 cm, AC = 25 cm


Applying Pythagoras theorem in ∆AEC,


(AE)2 + (EC)2 = (AC)2


(15)2 + (EC)2 = (25)2


(EC)2 = 625 - 225 = 400


EC =√400 = 20 cm


EB = EC – BC = 20 – 7 = 13 cm


Using the formula: Area of triangle = × b × h


Area of ΔAEC = × AE × EC = × 15 × 20 = 150 cm2


Area of ΔAEB = × AE × EB = × 15 × 13 = 97.5 cm2


Area of ΔABC = Area of ΔAEC - Area of ΔAEB


= 150 – 97.5 = 52.5 cm2


Also, area of ΔABC = × DB × AC


52.5 = × DB × 25


DB = = 4.2 cm


Hence, Area of ΔABC =52.5 cm2 and DB = 4.2 cm


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