Math Olympiad Test: Area of Parallelograms and Triangles- 2 - Class 9 MCQ

Area of ∆ = 1/2 x b x h = A
Area of parallelogram = bh = 2A.

Area of parallelogram
= AM × CD = AD × CN
⇒ 7 × 10 = AD × 8 
⇒ 

Joining QS, it can be clearly seen that,
QS ∥ DC ∥ AB, 

[Area between QS and DC]

[Area between QS and AB]
Adding both,
ar (PQS) + ar (SRQ) = + ar (ASQB)]
ar (PQRS) = 1/2 ar (ABCD) 

In ∆ABC, AD is the median

Let AE ⊥ BC, then,
ar(∆ABD) = 1/2 × AE × BD,
ar(∆ADC) = 1/2 × AE × DC = 1/2 × AE × BD
∴  ar(∆ABD) = ar(∆ADC) 

Area of trapezium RQBC = ar(∆RBP) + ar(∆PQR) + ar(∆QPC)
= 3 × ar(∆PQR)

∵ AD median of ∆ABC

∴ ar (ABD) = ar(ADC) = 1/2 ar(∆ABC) …(i)


∴ Required ratio = = 1 : 6 

∵ XY ∥ AB ∥ DC
[Let the length of XY be x cm]
∵ ar (DCYX) + ar (XYBA) = ar (ABCD)
∴ 
…(i)
∵ DN = NM = 
∴ The equation (i) reduces to:

⇒ 100 + 2x =  200
⇒ 2x = 100
⇒ x = 50 cm.

Required area:

In ∆ADB,

ar(∆ADB) = 1/2 × 12 × 9 = 54 cm²
ar(∆DBC) = 1/2 × 15 × 8 = 60 cm²
∴ Total area = 54 + 60 = 114 cm²

ar(∆AFB) = 1/2 ar (parallelogram ABCD)
{areas between same parallels and same base}
⇒ ar (parallelogram ABCD) = 2 × ar(∆AFB) = 2 × ar(∆DCB) …(i) 
In ∆ADF and ∆ECF
∠AFD = ∠EFC {vertically opposite angles}
∠ACF = ∠ADF {alternate angles}
BC = CE = AD
∴ ∆ADF ≅ ∆ECF {by AAS congruency} 
∴ DF = CF
ar(parallelogram ABCD) = 2 × ar(∆DCB)
= 2 × 2 × (ar(∆DFB)) 
= 2 × 2 × 3 
= 12 cm²

ar(∆AQE) = 1/2ar (parallelogram AQED)
= 1/2 x 1/2 (ar(parallelogram ABCD)) 
= 1/4 × (ar(parallelogram FECG)) 
12 cm² = 1/4 × ar(parallelogram FECG) 
∴ ar(FECG) = 48 cm²
⇒ ar(FGQB) = 1/2 ar(FECG) = 48/2 = 24 cm²

ar(∆PQB) + ar(∆DQC) = 1/2 ar(parallelogram ABCD)
⇒ ar(parallelogram ABCD)
= 2 [ar(∆PQB) + ar(∆DQC)] 
= 2 [10 cm² + ar(∆DQC)]
ar(∆DQC) = (3 + 3 + 1) × ar(∆PQB) 
= 7 × 10 = 70 cm²
∴ ar(parallelogram ABCD) = 2 × 80
= 160 cm²

Q ar(ΔXMR) = 1/2 ar(∆PQR) 

∴ ar(∆XMR) = 1/2 × 12cm² = 6 cm² 

Let AB = CD = x, and BC = AD = y, AM = h₁, BN = h₂
ar (∆ADQ) = 1/2 × AM × DQ = 
ar (∆APB) =  
ar (∆PQC) = 
ar (parallelogram ABCD) = AM × DC = BN × AD = h₁x = h₂y
∴ ar(∆AQC) = h₁x – 

OC = 6.5 cm,
∴ AC = BC = 6.5
[By similarly of ∆AOC and ∆BOC] 
∴ AB = AC + BC = 2AC = 2 × 6.5 = 13 cm. 

∴ ar(∆AOB) = 1/2 × 5 × 12 = 30 cm²