Question : In ΔABC with sides 8 cm, 9 cm and 12 cm, the angle bisector of the largest angle divides the opposite sides into two segments. What is the length of the shorter segment?
Option 1: $5 \frac{11}{17} {~cm}$
Option 2: $4\frac{11}{17} {~cm}$
Option 3: $6\frac{13}{17} {~cm}$
Option 4: $3\frac{9}{17} {~cm}$
Correct Answer: $5 \frac{11}{17} {~cm}$
Solution : Suppose AB = 8 cm, BC = 9 cm and AC = 12 cm AD is the angle bisector of $\angle BAC$ We know that, An angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle ⇒ $\frac{DC}{BD} = \frac{12}{8}$ ⇒ $\frac{DC}{BD}+1 = \frac{3}{2}+1$ ⇒ $\frac{DC + BD}{BD} = \frac52$ ⇒ $\frac{9}{BD} = \frac52$ ⇒ $BD = \frac{18}{5}$ cm = $3\frac{3}{5}$ Hence, the correct answer is $3\frac{3}{5}$ cm.
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Question : In $\triangle$ ABC, $\angle$ C = 90$^{\circ}$. M and N are the midpoints of sides AB and AC, respectively. CM and BN intersect each other at D and $\angle$ BDC = 90$^{\circ}$. If BC = 8 cm, then the length of BN is:
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Question : The value of $\frac{2}{3} \div \frac{3}{10}$ of $\frac{4}{9}-\frac{4}{5} \times 1 \frac{1}{9} \div \frac{8}{15}+\frac{3}{4} \div \frac{1}{2}$ is:
Question : The value of $\frac{2}{3} \div \frac{3}{10}$ of $\frac{4}{9}-\frac{4}{5} \times 1 \frac{1}{9} \div \frac{8}{15}-\frac{3}{4}+\frac{3}{4} \div \frac{1}{2}$ is:
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