Sides AB and CD of a trapezium ABCD are parallel. Area of ABCD is 25 sq. cm. P is any point on CD. Find area of triangle ABP when length of AB and CD are 6 cm and 4 cm respectively
Answers
Answer:
ou know that the area of a parallelogram is calculated as one side (taken as base) times the height with respect to that base. That height is the distance from the base to a vertex not on the base. Of course that distance is measured along a perpendicular to the base (extending that base as needed). With all that, you see that if you take side CD as the base, the length of AP is the height. As a consequence 9AP=72, and AP=72/9=8.
P could be between C and D, or it could be outside the parallelogram, but It is on the line CD. So, if AP is perpendicular to CD, it is perpendicular to PD. That means that APD is a right triangle, with sides AP and PD being its legs, and side AD being the hypotenuse. You can calculate the area of triangle APD as base times height divided by 2, as
(AP) × (PD) ÷ 2 , because if you take one of the legs as base of the triangle, the other leg is the height worth respect to that base.
You can find PD using the Pythagorean theorem as square root of AD squared minus the square of AP. You have to fond out the square root of
256-64=192=64×3.
The exact value of that square root is 8 times square root of 3. If you are expected to give the answer as a decimal, you could approximate that as PD=13.8564 for now. I world type the exact value as 8sqrt(3).
The area of APD is
(AP) × (PD) ÷ 2 = 8 × 8sqrt(3) ÷ 2 = 32sqrt(3).
The approximate value is 55.4, and can be calculated as
(AP) × (PD) ÷ 2 = 8 × ÷ 2 = 15.4256 , and then rounded as needed.