Question

\(ABCD\) is a rectangle with \(AD = 10\). \(P\) is a point on \(BC\) such that \(\angle APD = 90^{\circ}\). If \(DP = 8\), find the length of \(BP\).

• A. 6.4
• B. 5.2
• C. 4.8
• D. 3.6

Hint:

First use the right triangle APD to find AP, then set up two more Pythagoras equations for triangles ABP and DPC using AB = DC = x and BP = y, PC = 10 - y.

Answer 1

The Correct Option is D

There is a shortcut here that avoids setting up two separate Pythagoras equations: since $\angle APD = 90^{\circ}$, the two smaller right triangles ABP and DCP formed on either side of P turn out to be similar to each other. Once that is spotted, one clean ratio gives the answer directly.

1. Find AP first: in right triangle APD, the hypotenuse is $AD = 10$ and one leg is $DP = 8$. By the Pythagoras theorem, $AP = \sqrt{AD^2 - DP^2} = \sqrt{100 - 64} = \sqrt{36} = 6$.
2. Spot the similar triangles: angle $ABP$ and angle $DCP$ are both $90^{\circ}$, corners of the rectangle. Also, angle $APB$ plus angle $DPC$ must add up to $90^{\circ}$, because angle $APD$ itself takes up $90^{\circ}$ out of the straight line $BC$ at point P. Comparing complementary angles in the two right triangles shows angle $BAP$ equals angle $DPC$, and angle $APB$ equals angle $CDP$. So triangle $ABP$ is similar to triangle $PCD$ by the angle-angle rule.
3. Write the similarity ratio: matching corresponding sides of triangle $ABP \sim$ triangle $PCD$ gives $\dfrac{AP}{PD} = \dfrac{BP}{CD} = \dfrac{AB}{PC}$.
4. Use the first two ratios: since $CD = AB$, opposite sides of the rectangle, call this common length $x$. So $\dfrac{6}{8} = \dfrac{BP}{x}$, which gives $BP = \dfrac{6x}{8} = \dfrac{3x}{4}$.
5. Bring in the leftover side lengths: $BP + PC = BC = AD = 10$, opposite sides of the rectangle are equal. From $\dfrac{AP}{PD} = \dfrac{AB}{PC}$: $\dfrac{6}{8} = \dfrac{x}{PC}$, so $PC = \dfrac{8x}{6} = \dfrac{4x}{3}$.
6. Solve using $BP + PC = 10$: $\dfrac{3x}{4} + \dfrac{4x}{3} = 10$. Multiply throughout by $12$ to clear denominators: $9x + 16x = 120$, so $25x = 120$, giving $x = 4.8$.
7. Find BP: $BP = \dfrac{3x}{4} = \dfrac{3(4.8)}{4} = \dfrac{14.4}{4} = 3.6$.

Let's summarize:

• Once AP is found using the Pythagoras theorem in triangle APD, spotting that triangle ABP is similar to triangle PCD avoids setting up two separate equations from scratch.
• Solving the resulting ratio alongside $BP + PC = 10$ gives $x = AB = 4.8$ and $BP = 3.6$, matching the direct algebraic approach.

So $BP = 3.6$.