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Question 72
In a triangle ABC, PQ is a straight line parallel to AC, such that Area ABC : Area PBQ = 3 : 1 Then CB : CQ is equal to:
Solution
$$\frac{\left(AB\right)^2}{\left(PQ\right)^2}\ =\ \frac{3}{1}$$
$$\frac{AB}{PQ}\ =\ \frac{\sqrt{\ 3}}{1}$$
Area ABC : Area PBQ = 3 : 1
Here in triangle ABC, AB = AC = BC
Here in triangle PBQ, PB = BQ = PQ
$$\frac{Area\ ABC}{Area\ PBQ}\ =\ \frac{3}{1}$$
$$\frac{\frac{\sqrt{\ 3}}{4}\times\ \left(AB\right)^2}{\frac{\sqrt{\ 3}}{4}\times\ \left(PQ\right)^2}\ =\ \frac{3}{1}$$$$\frac{\left(AB\right)^2}{\left(PQ\right)^2}\ =\ \frac{3}{1}$$
$$\frac{AB}{PQ}\ =\ \frac{\sqrt{\ 3}}{1}$$
AB = AC = BC =Â $$\sqrt{\ 3}$$
PB = BQ = PQ = 1
CQ = BC-BQ
CQ =Â $$\sqrt{\ 3}-1$$
$$\frac{CB}{CQ}\ =\ \frac{\sqrt{\ 3}}{\sqrt{\ 3}-1}$$
= $$\frac{\sqrt{\ 3}}{\sqrt{\ 3}-1}\times\ \frac{\sqrt{\ 3}+1}{\sqrt{\ 3}+1}$$
= $$\frac{\sqrt{\ 3}\left(\sqrt{\ 3}+1\right)}{3-1}$$
= $$\frac{\sqrt{\ 3}\left(\sqrt{\ 3}+1\right)}{2}$$
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