In the figure, ABCD is a rectangle. P is a point on the side CD and Q is a point on CD extended. AB = 5cm and BC = 3 cm. ? c) What is the area of the triangle ∆ ABP?
Answers
Given : ABCD is a rectangle. P is a point on the side CD and Q is a point on CD extended.
AB = 5cm and BC = 3 cm.
To Find :
a) What is the area of the rectangle ABCD?
b) What is the area of the triangle ∆ ABC?
c) What is the area of the triangle ∆ ABP?
d) What is the area of the triangle ∆ ABQ?
e) What is the specialty of the positions of the points C, P, and Q?
Solution:
area of the rectangle ABCD = AB * BC = 5 * 3 = 15 cm²
area of the triangle ∆ ABC = (1/2) * AB * BC = (1/2) * 5 * 3 = 7.5 cm²
Draw PM ⊥ AB and QN ⊥ AB
PM = QN = BC = 3 cm
area of the triangle ∆ ABP = (1/2) * AB * PM = (1/2) * 5 * 3 = 7.5 cm²
area of the triangle ∆ ABQ = (1/2) * AB * QN = (1/2) * 5 * 3 = 7.5 cm²
C, P, and Q are collinear points
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