Areas of Parallelograms & Triangles — Competency Focused Questions

In which of the following, you find two polygons on the same base and between the same parallels?

Answer

In third figure,

Join BD and AC.

Triangle ABC and triangle ABD. They both share the same base AB. Crucially, the vertices C and D both lie on the same line, and that line is parallel to the base AB.

Hence, option 3 is the correct option.

In the figure, ABCD is a trapezium with parallel sides AB = x and CD = y. E and F are mid-points of the non-parallel sides AD and BC respectively. The ratio of ar (ABFE) and ar (EFCD) is :

1. x : y

2. (3x + y) : (x + 3y)

3. (x + 3y) : (3x + y)

4. (2x + y) : (3x + y)

Answer

Join BD which intersects EF at M.

In ∆ABD,

E is the midpoint of AD and EM || AB

By midpoint theorem,

M is the midpoint of BD

EM = $\dfrac{1}{2} AB$ ....(1)

In ∆CBD,

F is mid-point of BC and M is mid-point of BD so by mid-point theorem,

MF = $\dfrac{1}{2} CD$ ....(2)

So EF ∥ AB and EF ∥ CD

That means:

AB ∥ EF ∥ CD

Adding equations (1) and (2), we get:

EM + MF = $\dfrac{1}{2} AB + \dfrac{1}{2} CD = \dfrac{x + y}{2}$

Since:

AB ∥ EF ∥ CD

Let total height between AB and CD = H

EF lies exactly halfway between them,

∴ Height of trapezium EFCD = Height of trapezium ABEF = $\dfrac{H}{2}$ = h (let)

Area of trapezium = $\dfrac{1}{2} \times (\text{sum of parallel sides}) \times h$

Area of trapezium ABFE

$= \dfrac{1}{2} \times \Big(x + \dfrac{x + y}{2}\Big) \times h \\[1em] = \dfrac{1}{2} \times \Big(\dfrac{3x + y}{2}\Big) h \\[1em] = \dfrac{h}{4} \times (3x + y).$

Area of trapezium EFCD

$= \dfrac{1}{2} \times \Big(y + \dfrac{x + y}{2}\Big) \times h \\[1em] = \dfrac{1}{2} \times \Big(\dfrac{x + 3y}{2}\Big) h \\[1em] = \dfrac{h}{4} \times (x + 3y).$

Required ratio = Area of trapezium ABFE / Area of trapezium EFCD

By substituting the values,

$\text{Ratio} = \dfrac{\dfrac{h}{4}\times (3x + y)}{\dfrac{h}{4} \times (x + 3y)} \\[1em] \text{Ratio} = \dfrac{(3x + y)}{(x + 3y)} \\[1em] \text{Ratio} = (3x + y) : (x + 3y).$

The ratio of ar (ABFE) and ar (EFCD) is (3x + y) : (x + 3y).

Hence, option 2 is the correct option.

In the figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of ΔDFB is 3 cm², then area of the parallelogram ABCD is :

1. 9 cm²

2. 10 cm²

3. 12 cm²

4. 15 cm²

Answer

We know that,

Area of triangles on the same base and between the same parallel lines are equal.

△ ADF and △ DFB lie on same base DF and between same parallel lines AB and DC.

∴ Area of △ ADF = Area of △ DFB = 3 cm²

By converse of mid-point theorem,

If a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, it bisects the third side.

In △ ABE,

C is the mid-point of BE and CF || AB.

∴ F is the mid-point of AE. (By converse of mid-point theorem)

∴ EF = AF.

In △ ADF and △ EFC,

⇒ ∠AFD = ∠EFC (Vertically opposite angles are equal)

⇒ EF = AF (Proved above)

⇒ ∠DAF = ∠CEF (Alternate interior angles are equal)

∴ △ ADF ≅ △ ECF (By A.S.A. axiom)

We know that,

Area of congruent triangles are equal.

∴ Area of △ EFC = Area of △ ADF = 3 cm².

In △ BFE,

Since, C is the mid-point of BE.

∴ CF is the median of triangle BFE.

We know that,

Median of triangle divides it into two triangles of equal areas.

∴ Area of △ BFC = Area of △ EFC = 3 cm².

From figure,

⇒ Area of △ BDC = Area of △ BDF + Area of △ BFC

⇒ Area of △ BDC = 3 + 3 = 6 cm².

We know that,

The area of triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

||gm ABCD and △ BDC lies on same base DC and between same parallel lines AB and DC.

∴ Area of △ BDC = $\dfrac{1}{2}$ Area of ||gm ABCD

⇒ Area of ||gm ABCD = 2 × Area of △ BDC

⇒ Area of ||gm ABCD = 2 × 6 = 12 cm².

Hence, option 3 is the correct option.

Bansidhar is a farmer. He has a field in the form of a parallelogram ABCD. He took any point P on CD and joined it to points A and B. In how many parts the field is divided? What are the shapes of these parts? Bansidhar gave the three parts of the field to his two sons equally. How did he do it?

Answer

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

△ ABP and parallelogram ABCD both share the base AB and lie between the same parallel lines, AB and CD.

Area (△ ABP) = $\dfrac{1}{2}$ Area(∥gm ABCD).

Son 1: Received the area of △ ABP.

Since the total area of the field is the sum of the three triangles, and one triangle takes up exactly half, the other two triangles must together make up the remaining half.

Son 2: Received the combined area of △ ADP and △ BCP

Hence, son 1 received the area of △ ABP and son 2 received the area of △ ADP and △ BCP.

A parallelogram ABCD and a trapezium EFCD have the same base DC and are between the same parallels l and m. If the length of AB > length of EF, then compare the areas of ABCD and EFCD.

Answer

Let distance between lines l and m be h units.

We know that,

Area of parallelogram ABCD = base × height = DC × h

$\text{Area of trapezium} = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{ height} \\[1em] \text{Area of trapezium EFCD} = \dfrac{1}{2} \times (DC + EF) \times h.$

The length EF is less than AB (which is equal to DC). Since the average of DC and EF $\Big(\dfrac{DC + EF}{2}\Big)$ is smaller than DC itself, thus the parallelogram has a larger area.

ar(ABCD) > ar(EFCD).

Hence, ar(ABCD) > ar(EFCD).