Question

In parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. What is the area, in sq cm, of triangle APD? a) 36 b) 48 c) 54 d) 72

Answer 1

In this case, the area of triangle APD is 32√3 square centimeters.

The answer is option ⇒1

Step-by-step explanation:

To find the area of triangle APD, we first need to find the length of AP. Since AP is perpendicular to CD, triangle APD is a right triangle. We can use the Pythagorean theorem to find the length of AP:

AP2 = AD2 - DP2

AP2 = 162 - 92

AP2 = 256 - 81

AP2 = 175

AP = √175 = 5√7

Now that we have the length of AP, we can find the area of triangle APD:

Area = 1/2 * base * height

Area = 1/2 * AP * DP

Area = 1/2 * 5√7 * 9

Area = 45/2 * √7

Area = 32√3

So, the area of triangle APD is 32√3 square centimeters.

The answer is option ⇒1

Your question is incomplete, but most probably the full question was:

In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

1. 32√3

2. 18√3

3. 24√3

4 12√3