Question

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 AB²-AE²+6CD is:

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

Answer 1

Answer: (E)

According to the question,

 

Using the Pythagorean theorem:

\[ AB^2 = BD^2 + AD^2, \quad AE^2 = DE^2 + AD^2 \]

Also, since \(BD = 3 - DE\), we substitute this into the expression:

\[ AB^2 - AE^2 + 6CD = BD^2 + AD^2 - (DE^2 + AD^2) + 6CD \]

Simplifying:

\[ AB^2 - AE^2 + 6CD = BD^2 - DE^2 + 6CD \] 

Now substitute \(BD = 3 - DE\):

\[ AB^2 - AE^2 + 6CD = (3 - DE)^2 - DE^2 + 6(DE + 2) \]

Expand terms:

\[ (3 - DE)^2 = 9 + DE^2 - 6DE \]

So:

\[ AB^2 - AE^2 + 6CD = (9 + DE^2 - 6DE) - DE^2 + 6DE + 12 \]

Combine like terms:

\[ AB^2 - AE^2 + 6CD = 9 + 12 = 21 \]

Final Answer:

\[ \boxed{21} \] 
Hence, Option E is the correct answer.