Theorems on Area — Multiple Choice Questions

In the adjoining figure, if l || m, AF || BE, FC ⊥ m and ED ⊥ m, then the correct statement is

1. area of || ABEF = area of rect. CDEF

2. area of || ABEF = area of quad. CBEF

3. area of || ABEF = 2 area of △ACF

4. area of || ABEF = 2 area of △EBD

Answer

We know that,

A parallelogram and a rectangle on the same base and between the same parallel lines are equal in area.

Since, || ABEF and rectangle CDEF are on same base EF and between same parallel lines l and m.

∴ area of || ABEF = area of rect. CDEF.

Hence, Option 1 is the correct option.

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is

1. 1 : 2

2. 1 : 1

3. 2 : 1

4. 3 : 1

Answer

We know that,

Two parallelograms on equal bases and between the same parallel lines have equal areas.

So, ratio of their areas = 1 : 1.

Hence, Option 2 is the correct option.

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

1. 1 : 3

2. 1 : 2

3. 3 : 1

4. 1 : 4

Answer

We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallel lines.

Let area of parallelogram be x so area of triangle = $\dfrac{x}{2}$.

Ratio of area of the triangle to the area of the parallelogram = $\dfrac{\dfrac{x}{2}}{x} = \dfrac{1}{2} = 1 : 2.$

Hence, Option 2 is the correct option.

A median of a triangle divides it into two

1. triangles of equal area

2. congruent triangles

3. right triangles

4. isosceles triangles

Answer

A median of a triangle divides it into two triangles of equal area.

Hence, Option 1 is the correct option.

In the adjoining figure, area of parallelogram ABCD is

1. AB × BM

2. BC × BN

3. DC × DL

4. AD × DL

Answer

Area of || gm = base × height.

From figure,

Base = DC and Height = DL.

∴ Area of || gm = DC × DL.

Hence, Option 3 is the correct option.

The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to

1. $\dfrac{1}{2}$ area of △ABC

2. $\dfrac{1}{3}$ area of △ABC

3. $\dfrac{1}{4}$ area of △ABC

4. area of △ABC

Answer

Let CDEF be a parallelogram.

So, diagonal DF divides it into two triangles of equal area.

∴ area of △CDF = area of △EDF ........(i)

E and F are mid-points of side AB and AC respectively.

By mid-point theorem,

EF = $\dfrac{1}{2}$ BC = BD (As D is mid-point of BC) and EF || BC.

Since, FE || BC so from figure,

EF || BD and EF = BD.

So, EBDF is a paralleogram with diagonal ED dividing it into two triangles of equal area.

∴ area of △EBD = area of △EDF ........(ii)

E and D are mid-points of side AB and BC respectively.

By mid-point theorem,

ED = $\dfrac{1}{2}$ AC = AF (As F is mid-point of AC) and ED || AC.

Since, ED || AC so from figure,

ED || AF.

So, AEDF is a paralleogram with diagonal EF dividing it into two triangles of equal area.

∴ area of △AEF = area of △EDF ........(iii)

From (i), (ii) and (iii) we get,

area of △EDF = area of △CDF = area of △EBD = area of △AEF = x

From figure,

area of △ABC = area of △EDF + area of △CDF + area of △EBD + area of △AEF = 4x.

area of || gm EDCF = area of △EDF + area of △CDF = 2x.

So,

$\Rightarrow \dfrac{\text{area of || gm EDCF}}{\text{area of △ABC}} = \dfrac{2x}{4x} \\[1em] \Rightarrow \dfrac{\text{area of || gm EDCF}}{\text{area of △ABC}} = \dfrac{1}{2} \\[1em]$

area of || gm EDCF = $\dfrac{1}{2}$ area of △ABC.

Hence, Option 1 is the correct option.

In the adjoining figure, ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are mid-points of the non-parallel sides. The ratio of area of ABFE and area of EFCD is

1. a : b

2. (3a + b) : (a + 3b)

3. (a + 3b) : (3a + b)

4. (2a + b) : (3a + b)

Answer

It is given that

AB = a cm

DC = b cm

AB || DC

E and F are the mid-points of AD and BC

Consider h as the distance between AB, CD and EF

Now 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

and

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)

Adding equations (1) and (2)

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

EF = $\dfrac{1}{2}$ (AB + CD)

EF = $\dfrac{1}{2}$ (a + b)

Here,

Area of trapezium ABFE = $\dfrac{1}{2}$ [sum of parallel sides] × [distance between parallel sides]

Substituting the values,

$\text{Area of trap. ABFE} = \dfrac{1}{2} \Big[a + \dfrac{1}{2} (a + b)\Big] × h \\[1em] = \dfrac{1}{2}\Big[\dfrac{2a + a + b}{2}\Big]h \\[1em] = \dfrac{h}{4}[3a + b].$

Similarly,

$\text{Area of trap. EFCD } = \dfrac{1}{2}\Big[b + \dfrac{1}{2}(a + b)\Big] × h \\[1em] = \dfrac{1}{2}\Big[\dfrac{2b + a + b}{2}\Big]h \\[1em] = \dfrac{h}{4}[3b + a].$

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

By substituting the values,

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

Hence, Option 2 is the correct option.

In the adjoining figure, AB || DC and AB ≠ DC. If the diagonals AC and BD of the trapezium ABCD intersect at O, then which of the following statements is not true ?

1. area of △ABC = area of △ABD

2. area of △ACD = area of △BCD

3. area of △OAB = area of △OCD

4. area of △OAD = area of △OBC

Answer

We know that,

Triangles on the same base and between same parallel lines are equal in area.

Hence,

⇒ area of △ABC = area of △ABD ........ (1)

⇒ area of △ACD = area of △BCD

From figure and eqn. (1),

area of (△AOB + △OAD) = area of (△AOB + △OBC)

⇒ area of △OAD = area of △OBC.

Hence, Option 3 is the correct option.

Consider the following two statements:

Statement 1: The line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.

Statement 2: Diagonals of a parallelogram divide it into four triangles of equal area.

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

Let ABCD be a parallelogram in which E and F are mid-points of AB and CD respectively. Join EF.

Let us construct DG ⊥ AB and let DG = h, where h is the altitude on side AB.

Area of || gm ABCD = base × height = AB × h

Area of ||gm AEFD = AE × h = $\dfrac{AB}{2}$ × h ...................(1) [Since E is the mid-point of AB]

Area of ||gm EBCF = EB × h = $\dfrac{AB}{2}$ × h ...................(2) [Since E is the mid-point of AB]

From (1) and (2),

Area of || gm AEFD = Area of || gm EBCF.

Thus, the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.

∴ Statement 1 is true.

From figure,

The diagonals AC and BD cut at point O.

In parallelogram, the diagonals bisect each other.

∴ AO = OC

In ∆ACD, O is the mid-point of AC.

∴ OD is the median.

Area of ∆AOD = Area of ∆COD ................. (3) [Median of ∆ divides it into two triangles of equal areas.]

Similarly, in ∆ABC

O is the mid-point of AC.

∴ OB is the median.

Area of ∆AOB = Area of ∆COB ................. (4) [Median of ∆ divides it into two triangles of equal areas.]

In ∆ADB,

O is the mid-point of BD.

∴ OA is the median.

Area of ∆AOD = Area of ∆AOB ................. (5)

From (3), (4) and (5) we get,

Area of ∆AOB = Area of ∆COB = Area of ∆COD = Area of ∆AOD

So proved, that the diagonals of a parallelogram divide it into four triangles of equal area.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.