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

Step 1: Find \(AP\) using triangle \(APD\).
Triangle \(APD\) has a right angle at \(P\), since \(\angle APD = 90^{\circ}\), with hypotenuse \(AD = 10\) and one leg \(DP = 8\).
By the Pythagoras theorem, \(AD^2 = AP^2 + DP^2\), so \(AP^2 = 10^2 - 8^2 = 100 - 64 = 36\), giving \(AP = 6\).

Step 2: Name the unknown lengths.
Let \(AB = DC = x\), the two equal widths of the rectangle, and let \(BP = y\), the length we want to find.
Since \(P\) lies on \(BC\) and \(BC = AD = 10\), opposite sides of a rectangle are equal, the rest of \(BC\) is \(PC = 10 - y\).

Step 3: Write the Pythagoras equation for triangle \(ABP\).
Triangle \(ABP\) has a right angle at \(B\), a corner of the rectangle, so \(AB^2 + BP^2 = AP^2\).
Using the variables: \(x^2 + y^2 = 36\). Call this equation (ii).

Step 4: Write the Pythagoras equation for triangle \(DPC\).
Triangle \(DPC\) has a right angle at \(C\), another corner of the rectangle, so \(DC^2 + PC^2 = DP^2\).
Using the variables: \(x^2 + (10-y)^2 = 64\). Call this equation (ii).

Step 5: Expand equation (ii) and subtract equation (ii).
Expanding: \(x^2 + 100 - 20y + y^2 = 64\).
From equation (ii), \(x^2 + y^2 = 36\), so substitute this in: \(36 + 100 - 20y = 64\).
This gives \(136 - 20y = 64\), so \(20y = 72\), and \(y = 3.6\).

Step 6: Check the wrong options.
6.4 would actually be \(10 - 3.6\), which is \(PC\), not \(BP\), so it is a mix-up of the two segments.
5.2 and 4.8 do not satisfy equation (ii) when checked back; plugging \(y=4.8\), for example, gives \(x^2 = 36 - 23.04 = 12.96\) and then \(x^2+(10-4.8)^2 = 12.96+27.04=40\), which does not equal \(64\), so it fails the check.
Only \(y = 3.6\) satisfies both equations (ii) and (ii) at once.

Final Answer:
\(BP = 3.6\).
\[ \boxed{BP = 3.6} \]