ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CO:OP = 3:1. If ar(ΔPBO)=10CM². Find the area of parallelogram ABCD.
Answers
Step-by-step explanation:
take angle similarity theorem and u will get a answer
Given : ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CQ: QP = 3:1 and ar(ΔPBQ) = 10 cm².
To find : Area of parallelogram ABCD.
Proof :
Let, CQ = 3x & PQ = x
ar (∆PBQ) = 10 cm²……….(1)
We know that, Area of ∆ = ½ × Base × Height
ar (∆PBQ) = ½ × PQ × BQ
10 = ½ × x × h
[Let BQ = h ]
10 × 2 = x h
xh = 20 ……..(2)
ar (∆BQC) = ½ × QC × BQ
ar (∆BQC) = ½ × 3(x × h)
ar (∆BQC) = ½ × 3 × 20
[From eq 2]
ar (∆BQC) = 3 × 10
ar (∆BQC) = 30 cm² ………….(3)
Now,
ar (∆PCB) = ar (∆PBQ) + ar (∆BQC)
ar (∆PCB) = 10 + 30
[From eq 1 & 3]
ar (∆PCB) = 40 cm²
½ × PB × BC = 40 cm²
PB × BC = (40 × 2 )cm²
PB × BC = 80 cm² …………(4)
Now, area of parallelogram = Base × Height
ar (|| gm ABCD) = AB × BC
ar (|| gm ABCD) = 2 PB × BC
[AB = 2 BP , P is the mid point of AB]
Area (ABCD) = 2 × 80
[From eq 4]
ar (ABCD) = 160 cm²
Hence, the area of parallelogram ABCD is 160 cm²
HOPE THIS ANSWER WILL HELP YOU…..
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