In $△ A B C, ∠ B = 90^{o}$ , AB=4 cm, BC=3 cm. Perpendicular from B is drawn to line AC which meets AC at D. then ratio of $\frac{A D}{C D}$ is,

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Solution

The correct option is D

Ratio of $\frac{A D}{C D}$ is the square of $\frac{A B}{B C}$ which can be proved using the concept of similar triangles.

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In △ABC,∠B=90o, AB=4 cm, BC=3 cm. Perpendicular from B is drawn to line AC which meets AC at D. then ratio of ADCD is,

The correct option is D Ratio of ADCD is the square of ABBC which can be proved using the concept of similar triangles.

In $△ A B C, ∠ B = 90^{o}$ , AB=4 cm, BC=3 cm. Perpendicular from B is drawn to line AC which meets AC at D. then ratio of $\frac{A D}{C D}$ is,

The correct option is D

Ratio of $\frac{A D}{C D}$ is the square of $\frac{A B}{B C}$ which can be proved using the concept of similar triangles.