Math Olympiad Test: Area of Parallelograms and Triangles- 1 - Class 9 MCQ

If the parallelograms lie on the same base and between the same parallel lines, then their areas are equal.

Base = b, Distance between parallels = h
Then, Area of rectangle = area of parallelogram = bh

∵ O lies in mid of the parallelogram. 
∴ PQ divides parallelogram in two parts of equal areas.
∴ ar(PQDA) = 1/2 ar(ABCD) 

∵ ABCD and ∆PDC lie on same base, i.e., CD and between the same parallels, i.e., AB and CD. 
∴ 
Similarly, 

⇒ ar(∆PDC) = ar(∆AQD) = 1/2 ar(quad. ABCD) 

Using midpoint theorem,
BC = 2 QR, AB = 2 PQ, AC = 2 PR.
Let the area of ∆PQR be A, then, ABC is a ∆ resulted by doubling the length of every side of ∆PQR.
∴ ar (∆ABC) = 4 ar(∆PQR) [using Heron’s formula] 

ar (parallelogram PQRS)

Let AB = x, AC = y, then,
BC =  (Pythagoras’ theorem) 
Area (ABMN) = x²
Area (ACFG) = y²
ar(BCED) = 
∴ ar(BCED) = ar(ABMN) + ar(ACFG) 

According to question


⇒ (40 + 50) = K (60 + 50)
⇒ k = 9/11

∵ ∆ADC and ∆BDC lie on same base DC and between the same parallels, i.e., AB and DC. 
∴ ar(∆ADC) = ar(∆BDC) ar(AOD) + ar(DOC) = ar(BOC) + ar(DOC) ⇒ ar(AOD) = ar(BOC). 

Draw XY ∥ AD ∥ BC and PQ ∥ AB ∥ DC

Now ar(∆AOB) = 

⇒ ar(∆AOB) + ar(∆DOC) = 
⇒ ar(∆AOD) + ar(∆BOC) = 
⇒ ar(ABCD) = 2 × (3 + 6) = 18 cm²