ABCD is a parallelogram. P is a point on DC such that area of triangle DAP= 25cm² and area of triangle BCP= 15cm². Find (a) area of parallelogram ABCD (ii) DP:PC
Pls give proper explanation and a diagram
Answers
DP/PC = (50/h)/(30/h) = 50/30=5/3
DP:PC = 5:3
Area of Parallelogram ABCD = 80 cm² & DP : PC = 5:3
Step-by-step explanation:
ABCD is a parallelogram
Area of Parallelogram = CD * ( height between AB & CD)
Area of ΔDAP = (1/2)DP * (height between AB & CD) = 25
Area of ΔBCP = (1/2)PC * (height between AB & CD) = 15
Area of ΔDAP + Area of ΔBCP = (1/2) (DP + PC)(height between AB & CD) = 25 + 15
=> (1/2) (CD)(height between AB & CD) = 40
=> (CD)(height between AB & CD) = 80
Area of Parallelogram ABCD = 80 cm²
Diagonal Bisect Parallelogram in two Equal Areas
=> Area of Δ ACD = 80/2 = 40 cm²
Area of ΔDAP = 25 cm²
Area of ΔCAP = 25 cm² = 40 - 25 = 15 cm²
Area of ΔDAP : Area of ΔCAP = 25 : 15
=> DP/PC = 25/15
=> DP/PC = 5/3
=> DP : PC = 5:3
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