Question
Given that the ratio of altitudes of two triangles is 4 : 5, ratio of their areas is 3 : 2, the ratio of their corresponding bases is :
Answer: Option B
$$\eqalign{
& \frac{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,1}}{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,2}} = \frac{3}{2} \cr
& \Rightarrow \frac{{\frac{1}{2} \times {B_1} \times 4}}{{\frac{1}{2} \times {B_2} \times 5}} = \frac{3}{2} \cr
& \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{3}{2} \times \frac{5}{4} \cr
& \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{{15}}{8} \cr
& \therefore {B_1}:{B_2} = 15:8 \cr} $$
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$$\eqalign{
& \frac{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,1}}{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,2}} = \frac{3}{2} \cr
& \Rightarrow \frac{{\frac{1}{2} \times {B_1} \times 4}}{{\frac{1}{2} \times {B_2} \times 5}} = \frac{3}{2} \cr
& \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{3}{2} \times \frac{5}{4} \cr
& \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{{15}}{8} \cr
& \therefore {B_1}:{B_2} = 15:8 \cr} $$
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