Question

# If a ∆ ABC, AD is the bisector of ∠A, meeting side BC at D. (i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC. (ii) If BD = 2 cm, AB = 5 cm and DC = 3 cm, find AC. (iii) If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD. (iv) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC. (v) If AC = 4.2 cm, DC = 6 cm and BC = 10 cm. find AB. (vi) If AB = 5.6 cm, AC = 6 cm and DC = 3 cm, find BC. (vii) If AD = 5.6 cm, BC = 6 cm and BD = 3.2 cm, find AC. (viii) If AB = 10 cm, AC = 6 cm and BC = 12 cm, find BD and DC.

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Solution

## (i) It is given that , and . In , is the bisector of, meeting side at . We have to find . Since is bisector Then Hence (ii) It is given that , and. In , is the bisector of, meeting side at . We have to find . Since is bisector So (is bisector of and side) Then Hence (iii) It is given that , and . In , is the bisector of, meeting side at . We have to find BD. Since is bisector So (is bisector of and side) Then Hence (iv) It is given that , and . In , is the bisector of, meeting side at . We have to find BD and . Since is bisector So (is bisector of and side) Let BD = x cm. Then CD = (6 − x) cm Then, $\frac{10}{14}=\frac{x}{6-x}\phantom{\rule{0ex}{0ex}}⇒14x=60-10x\phantom{\rule{0ex}{0ex}}⇒24x=60\phantom{\rule{0ex}{0ex}}⇒x=\frac{60}{24}=2.5$ Hence, BD = 2.5 cm and DC = 6 − 2.5 = 3.5 cm. (v) It is given that , and . In , is the bisector of, meeting side at . We have to find . Since is bisector So Then Hence (vi) It is given that , and . In , is the bisector of, meeting side at . We have to find . Since is bisector So Then So Hence (vii) If it is given that AB = 5.6 cm, and . In, is the bisector of, meeting side at (viii) It is given that , and . In , is the bisector of, meeting side at . We have to find BD and . Since is bisector So Let BD = x cm Then ⇒ Now Hence and

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