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: Use a general property of diagonal-split quadrilaterals.
For any quadrilateral ABCD whose diagonals meet at P, the following identity always holds:
\[ \text{Area}(APB) \times \text{Area}(CPD) = \text{Area}(BPC) \times \text{Area}(APD) \]
This is true because angle APB and angle CPD are vertically opposite, so equal, and angle BPC and angle APD are also vertically opposite, so equal too. Writing each triangle's area as half the product of its two sides from P times the sine of the included angle, the sine terms cancel out between the two sides of the identity, leaving exactly this relationship between the segment lengths from P.

Step 2: Plug in the known areas.
We are given Area(BPC) = 12 and Area(APD) = 27, so their product is:
\[ \text{Area}(BPC) \times \text{Area}(APD) = 12 \times 27 = 324 \]

Step 3: Use the equal-area condition.
We are told Area(APB) = Area(CPD). Call this common area $x$. Then from Step 1's identity:
\[ x \times x = 324 \implies x^2 = 324 \implies x = \sqrt{324} = 18 \]
(taking the positive root, since an area cannot be negative).

Step 4: Sanity check against the given numbers.
An area of 18 sits between 12 and 27, which is exactly what we would expect: since Area(APB) = Area(CPD) = 18, the ratio 18/12 = 1.5 should match 27/18 = 1.5 as well, both being equal to $DP/BP$ found by the base-height method, and indeed both ratios come out to 1.5, confirming the answer.

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