Question

P is a point on the segment joining the feet of two vertical poles of heights a and b. The angles of elevation of the tops of the poles from P are $45^{\circ$ each. Then, the square of the distance between the tops of the poles is

• A. $\frac{a^{2}+b^{2}}{2}$
• B. $a^{2}+b^{2}$
• C. $2(a^2+b^2)$
• D. $4(a^2+b^2)$

Hint:

$(x+y)^2 + (x-y)^2 = 2(x^2+y^2)$.

Answer 1

The Correct Option is C

Step 1: Geometry
In $\Delta APD$, $\tan 45^\circ = a/AP \Rightarrow AP = a$. Similarly, in $\Delta BPC$, $BP = b$.
Step 2: Coordinate geometry
The tops are $D(-a, a)$ and $C(b, b)$ relative to P. $DE = a+b$ (horizontal distance) and $CE = b-a$ (vertical difference).
Step 3: Calculate distance
$DC^2 = (a+b)^2 + (b-a)^2$.
Step 4: Final calculation
$= a^2+b^2+2ab + a^2+b^2-2ab = 2(a^2+b^2)$.
Final Answer: (C)