Please enter the following details
Free Daily Targets & Mocks
Study Material & Formulas PDFs
1 on 1 Mentorship
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
$$BD^2+DE^2=AB^2-AE^2$$
$$x^2-\left(3-x\right)^2=AB^2-AE^2$$
$$AB^2-AE^2=6x-9$$
$$AB^2-AE^2+6CD=6x-9+6\left(5-x\right)$$ ($$CD=\left(5-x\right)$$)
$$AB^2-AE^2+6CD=6x-9+30-6x$$
$$AB^2-AE^2+6CD=21$$
Video Solution
Create a FREE account and get:
- All Quant Formulas and shortcuts PDF
- 15 XAT previous papers with solutions PDF
- XAT Trial Classes for FREE
XAT Quant Questions | XAT Quantitative Ability
XAT Venn Diagrams QuestionsXAT Functions, Graphs and Statistics QuestionsXAT Geometry QuestionsXAT Averages, Ratio and Proportion QuestionsXAT Profit, Loss and Interest Questions
XAT DILR Questions | LRDI Questions For XAT
XAT Scheduling QuestionsXAT Truth Lie Concept QuestionsXAT Charts QuestionsXAT Quant Based DI QuestionsXAT Arrangement Questions
