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 :
A. 5 : 8
B. 15 : 8
C. 8 : 5
D. 8 : 15
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & \frac{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,1}}{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,2}} = \frac{3}{2} \cr & \Rightarrow \frac{{\frac{1}{2} \times {B_1} \times 4}}{{\frac{1}{2} \times {B_2} \times 5}} = \frac{3}{2} \cr & \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{3}{2} \times \frac{5}{4} \cr & \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{{15}}{8} \cr & \therefore {B_1}:{B_2} = 15:8 \cr} $$Related Questions on Triangles
If ABC and PQR are similar triangles in which ∠A = 47° and ∠Q = 83°, then ∠C is:
A. 50°
B. 70°
C. 60°
D. 80°
In the following figure which of the following statements is true?
A. AB = BD
B. AC = CD
C. BC + CD
D. AD < Cd
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