Question

ABCD is a quadrilateral whose diagonals intersect at point P. The area of triangle APD is 27 and the area of triangle BPC is 12. If the areas of triangles APB and CPD are equal, then the area of triangle APB is

• A. 21
• B. 18
• C. 16
• D. 15

Hint:

Recall that for any quadrilateral with diagonals meeting at P, area(APB) times area(CPD) always equals area(BPC) times area(APD).

Answer 1

The Correct Option is B

Step 1: Set up heights and bases from P.
Let the diagonals be AC and BD meeting at P. Triangles APD and BPC share the same pair of vertical angles at P (angle APD = angle BPC). Let \(h_1\) be the height from A to line BD, and \(h_2\) be the height from C to line BD.

Step 2: Write areas as base times height at P.
\[ \text{Area}(APD) = \frac{1}{2} h_1 \cdot DP, \qquad \text{Area}(BPC) = \frac{1}{2} h_2 \cdot BP \]
Dividing these two areas:
\[ \frac{\text{Area}(APD)}{\text{Area}(BPC)} = \frac{h_1 \cdot DP}{h_2 \cdot BP} = \frac{27}{12} \quad \text{...(ii)} \]

Step 3: Bring in the equal-area condition.
\[ \text{Area}(APB) = \frac{1}{2} h_1 \cdot BP, \qquad \text{Area}(CPD) = \frac{1}{2} h_2 \cdot DP \]
Since these are given equal: \(h_1 \cdot BP = h_2 \cdot DP\), which rearranges to \(\dfrac{h_1}{h_2} = \dfrac{DP}{BP}\).

Step 4: Combine with equation (ii).
Substituting \(\dfrac{h_1}{h_2} = \dfrac{DP}{BP}\) into (ii):
\[ \frac{DP}{BP}\cdot\frac{DP}{BP} = \frac{27}{12} \implies \left(\frac{DP}{BP}\right)^2 = \frac{9}{4} \implies \frac{DP}{BP} = \frac{3}{2} \]

Step 5: Use this ratio to find the area of APB.
\[ \frac{\text{Area}(APD)}{\text{Area}(APB)} = \frac{\frac{1}{2}h_1\cdot DP}{\frac{1}{2}h_1\cdot BP} = \frac{DP}{BP} = \frac{3}{2} \]
So:
\[ \text{Area}(APB) = \text{Area}(APD) \times \frac{2}{3} = 27 \times \frac{2}{3} = 18 \]

Final Answer:
The area of triangle APB is 18.\[ \boxed{18} \]