Question 58

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:

Solution

Given that, $$AD\bot\ BC$$

$$BE=3$$

$$BC=5$$

Using Pythagoras' theorem,

$$AD^2+BD^2=AB^2$$ .....(1)

$$AD^2+DE^2=AE^2$$......(2)

$$AD^2+DC^2=AC^2$$......(3)

(1) - (2) gives

$$BD^2+DE^2=AB^2-AE^2$$

$$x^2-\left(3-x\right)^2=AB^2-AE^2$$

$$AB^2-AE^2=6x-9$$

$$AB^2-AE^2+6CD=6x-9+6\left(5-x\right)$$ ($$CD=\left(5-x\right)$$)

$$AB^2-AE^2+6CD=6x-9+30-6x$$

$$AB^2-AE^2+6CD=21$$

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