ABCD is a parallelogram. P is any point on CD. If `ar(DeltaDPA)=15" cm"^(2)` and `ar(DeltaAPC)=20" cm"^(2)`, then `ar(DeltaAPB)=`

Options

A. `15" cm"^(2)`
B. `20" cm"^(2)`
C. `35" cm"^(2)`
D. `30" cm"^(2)`

Answer

To solve the problem, we need to find the area of triangle \( \Delta APB \) given that \( ar(\Delta DPA) = 15 \, \text{cm}^2 \) and \( ar(\Delta APC) = 20 \, \text{cm}^2 \).

### Step 1: Understand the relationship between the areas of the triangles
In the parallelogram \( ABCD \), we know that triangles \( DPA \) and \( APC \) share a common base \( AP \) and have heights that are perpendicular to this base.

### Step 2: Calculate the area of triangle \( \Delta ADC \)
The area of triangle \( \Delta ADC \) can be expressed as the sum of the areas of triangles \( DPA \) and \( APC \):
\[ ar(\Delta ADC) = ar(\Delta DPA) + ar(\Delta APC) = 15 \, \text{cm}^2 + 20 \, \text{cm}^2 = 35 \, \text{cm}^2 \]

### Step 3: Use the properties of parallelograms
In a parallelogram, opposite triangles formed by a diagonal have equal areas. Thus, the area of triangle \( \Delta APB \) is equal to the area of triangle \( \Delta ADC \):
\[ ar(\Delta APB) = ar(\Delta ADC) = 35 \, \text{cm}^2 \]

### Final Answer
Therefore, the area of triangle \( \Delta APB \) is:
\[ ar(\Delta APB) = 35 \, \text{cm}^2 \]

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