In given figure, AC is perpendicular on BD such that P intersects BD in ratio 3:4 and the ratio CP:PD is 1:2. What is the ratio of areas of the triangles ABD and ACD?
Question
In given figure, AC is perpendicular on BD such that P intersects BD in ratio 3:4 and the ratio CP:PD is 1:2. What is the ratio of areas of the triangles ABD and ACD?
Solution 1
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given that AC is perpendicular to BD, triangles ABD and ACD are similar because they share an angle at A and have a right angle at C and D respectively.
The ratio of the sides CP:PD is given as 1:2. Since CP is a side in triangle ACD and PD is a side in triangle ABD, we can use this ratio for our calculation.
Therefore, the ratio of the areas of triangles ABD and ACD is (CP/PD)² = (1/2)² = 1/4.
So, the ratio of the areas of triangles ABD to ACD is 4:1.
Solution 2
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given that AC is perpendicular to BD, triangles ABD and ACD are similar.
The ratio of the corresponding sides CP and PD is given as 1:2.
Therefore, the ratio of the areas of triangles ABD and ACD is (1:2)^2 = 1:4.
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