Triangle Perpendicular Side Relation

XAT 2023 · Quantitative Ability & Data Interpretation · Medium · Geometry

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ABC is a triangle with $BC=5$ . D is the foot of the perpendicular from A on BC. E is a point on CD such that $BE=3$ . The value of $AB^2 - AE^2 + 6CD$ is:

A.
5
B.
10
C.
14
D.
18
E.
21

Answer

Explanation

Let $BD = x \implies DE = 3 - x$ and $CD = 5 - x$ . In right $\triangle ABD$ , $AB^2 = AD^2 + x^2$ . In right $\triangle ADE$ , $AE^2 = AD^2 + (3-x)^2$ . $AB^2 - AE^2 = x^2 - (3-x)^2 = 6x - 9$ . $AB^2 - AE^2 + 6CD = 6x - 9 + 6(5-x) = 6x - 9 + 30 - 6x = 21$ .

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Triangle Perpendicular Side Relation

Explanation

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