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Answers
Answer:
(i) ar(ΔBPQ) = 10 cm²
(ii) ar(ΔCDP) = 40 cm²
(iii) ar(ABCD) = 120 cm²
Step-by-step explanation:
Given:-
A Parallelogram ABCD, P being a point on BC, DP produced meets AB at Q.
BP : PC = 1 : 2
ΔCPQ = 20 cm²
To Find:-
(i) ar(ΔBPQ)
(ii) ar(ΔCDP)
(iii) ar(ABCD)
Proof:-
We have,
BP : PC = 1 : 2
So,
Let BP = 1x and PC = 2x ----- 1
Now,
In ΔCPQ,
ar(ΔCPQ) = 20 cm²
But
Area of Triangle = (1/2) × base × height
= (1/2)bh
So,
(1/2)bh = 20 cm²
Here,
Base = PC
From above eq.1,
b = x
In a triangle, its height must be perpendicular to its base.
So,
Let the height be some point on BC or a point produced from BC, where it is perpendicular, let's call it M.
Thus,
h = MQ
Hence,
(1/2)(PC)(MQ) = 20 cm²
From above eq.1,
(1/2)(2x)(MQ) = 20 cm²
x(MQ) = 20 cm² ----- 2
Similarly,
In ΔBPQ
b = BP
From above eq.1,
b = x
h = MQ
[This is because both the triangles have bases on the side BC, so that makes the perpendicular height same.]
So,
ar(ΔBPQ) = (1/2)bh
ar(ΔBPQ) = (1/2)(x)(MQ)
ar(ΔBPQ) = (1/2) × [x(MQ)]
From eq.2,
ar(ΔBPQ) = (1/2) × [(20)]
ar(ΔBPQ) = 10 cm² ----- 3
(ii)
In ΔBPQ and ΔCDP
∠BPQ = ∠CPD [Vertically Opposite Angles]
∠BQP = ∠CDP [Alternate Interior Angles]
∠PBQ = ∠PCD [Alternate Interior Angles]
∴ΔBPQ ~ ΔCDP (By A.A.A. Similarity)
Then,
(BP/PC) = (BQ/CD) = (PQ/PD)
[Corresponding Parts of Similar Triangles]
But we are given,
(BP/PC) = (1/2)
So,
(BP/PC) = (BQ/CD) = (PQ/PD) = (1/2) ----- 4
Now,
We know that,
If two triangles are similar, the ratio of their areas will be equal to the square of the ratios of their corresponding sides.
Hence,
[ar(ΔBPQ)]/[ar(ΔCDP)] = (BP/PC)²
From eq.3 and eq.4, we get,
[10 cm²]/[ar(ΔCDP)] = (1/2)²
[10 cm²]/[ar(ΔCDP)] = (1/4)
Cross multiplying we get,
ar(ΔCDP) × 1 = 4 × 10 cm²
ar(ΔCDP) = 40 cm² ----- 5
(iii)
In ΔBPQ and ΔADQ,
∠BQP = ∠AQD [Common Angle]
∠QAD = ∠QBP [Corresponding Angles]
∴ ΔBPQ ~ ΔADQ (By A.A. Similarity)
Then,
[ar(ΔBPQ)]/[ar(ΔADQ)] = (BP/AD)² ------ 6
Now,
We know that,
BC = BP + PC
From eq.1,
BC = x + 2x
BC = 3x
But we know that,
In a parallelogram, opposite sides are equal.
So,
BC = AD
∴ AD = 3x ------ 7
Then,
Continuing eq.6,
[ar(ΔBPQ)]/[ar(ΔADQ)] = (BP/AD)²
From eq.1 and eq.7,
[ar(ΔBPQ)]/[ar(ΔADQ)] = (1x/3x)²
[ar(ΔBPQ)]/[ar(ΔADQ)] = (1/3)²
[ar(ΔBPQ)]/[ar(ΔADQ)] = (1/9)
From eq.3,
[10 cm²]/[ar(ΔADQ)] = (1/9)
Cross multiplying we get,
[ar(ΔADQ)] × 1 = 9 × 10 cm²
ar(ΔADQ) = 90 cm² ------ 8
Then,
ar(ABPD) = ar(ΔAQD) - ar(ΔBPQ)
From eq.8 and eq.3,
ar(ABPD) = 90 cm² - 10 cm²
ar(ABPD) = 80 cm² ------ 9
Now,
ar(ABCD) = ar(ABPD) + ar(ΔCDP)
From eq.5 and eq.9,
ar(ABCD) = 80 cm² + 40 cm²
ar(ABCD) = 120 cm²
Hope it helped and believing you understood it........All the best
