In a parallelogram ABCD, the mid points of adjacent sides AB , BC are p and Q respectively, then the area of parallelogram ABCD : area of triangle DPQ=?
Answers
Given : In a parallelogram ABCD, the mid points of adjacent sides AB , BC are p and Q respectively,
To Find : the area of parallelogram ABCD : area of triangle DPQ
Solution:
Diagonal of parallelogram divided it in 2 Equal areas
Let say area of parallelogram ABCD = 8A sq units
Then area of ΔABD and ΔCBD = 8A/2 = 4A sq units
Then area of ΔABD and ΔCBD = 8A/2 = 4A sq units
area of ΔABC and ΔADC = 8A/2 = 4A sq units
in Δ ADB
P is mid point of AB
Hence Area of ΔAPD = 4A/2 = 2A Sq units
in Δ CBD
Q is mid point of BC
Hence Area of ΔDQC = 4A/2 = 2A Sq units
Now in ΔABC
PQ || BC as P & Q are mid points of AB & BC
=> ΔPBQ ≈ ΔABC
and PQ = BC/2
=> Area of ΔPBQ = (1/2)² ΔABC
=> Area of ΔPBQ = (1/2)² 4A
=> Area of ΔPBQ = A sq units
Area of ΔAPD + Area of ΔDQC + Area of ΔPBQ + Area of ΔDPQ = area of parallelogram
=> 2A + 2A + A + Area of ΔDPQ = 8A
=> Area of ΔDPQ = 3A
area of parallelogram ABCD : area of triangle DPQ
= 8A : 3A
= 8 : 3
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