ML Aggarwal Solution Class 9 Chapter 14 Theorems on Area MCQs

 Multiple Choice Questions

Choose the correct answer from the given four options (1 to 8):

Question 1

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

(a) area of ||ABEF = area of rect. CDEF

(b) area of ||ABEF = area of quad. CBEF

(c) area of ||ABEF = 2 area of ∆ACF

(d) area of ||ABEF = 2 area of ∆EBD

Sol :

In the given figure,

⇒l ||m, AF || BE, FC ⊥ m and ED ⊥ m

∵ ||gm ABEF and rectangle CDEF are on the same base EF and between the same parallel

∴ area ||gm ABEF = area rect. CDEF (a)

Question 2

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

(a) 1 : 2

(b) 1 : 1

(c) 2 : 1

(d) 3 : 1

Sol :

⇒A triangle and a parallelogram are on the same base and between same parallel, then

∴ They are equal in area

∴ Their ratio 1:1 (b)

Question 3

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

(a) 1 : 3

(b) 1 : 2

(c) 3 : 1

(d) 1 : 4

Sol :

⇒A triangle and a parallelogram are on the same base and between same parallel, then area of triangle = $\frac{1}{2}$ area ||gm

∴ Their ratio 1 : 2 (b)

Question 4

A median of a triangle divides it into two

(a) triangles of equal area

(b) congruent triangles

(c) right triangles

(d) isosceles triangles

Sol :

⇒A median of a triangle divides it into two triangle equal in area. (a)

Question 5

In the given figure, area of parallelogram ABCD is

(a) AB×BM

(b) BC×BN

(c) DC×DL

(d) AD×DL

Sol :

In the given figure,

⇒Area of ||gm ABCD = AB×DL or DC×DL (∵ AB = DC)

Ans (c)

Question 6

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

(a) $\frac{1}{2}$ area of ∆ABC

(b) $\frac{1}{3}$ area of ∆ABC

(c) $\frac{1}{4}$ area of ∆ABC

(d) area of ∆ABC

Sol :

The mid point of the sides of a triangle along with any of vertices as the fourth point makes a parallelogram of area equal to $\frac{1}{2}$ the area of ∆ABC

i.e. area ||gm DEAF$=\frac{1}{2}$ area ∆ABC

Ans (a)

Question 7

In the given 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 ABEF and area of EFCD is

(a) a : b

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

(c) (a + 3b) : (3a + b)

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

Sol :

In the figure , ABCD is a trapezium in which 

AB||DC

AB=a, DC=b

E and F are mid points on DA and CB respectively.

Let h be the height  (∵EF||AB||DC)

∴EF$=\frac{1}{2}$(a+b)

Area of trapezium ABFE

$=\left[\frac{1}{2} \frac{(a+b)}{2} \times \frac{h}{2}\right]$

$=\frac{h}{4}\left(\frac{2 a+a+b}{2}\right)$

$=\frac{h}{8}(3 a+b)$

and area of trapezium EFCD

$=\frac{1}{2}[\mathrm{EF}+\mathrm{DC}] \times \frac{h}{2}$

$=\frac{h}{4}\left[\frac{a+b}{2}+b\right]$

$=\frac{h}{4}\left[\frac{a+b+2 b}{2}\right]$

$=\frac{h}{4}[a+3 b]$

∴Ratio$=\frac{h}{8}$(3a+b) : $\frac{h}{8}$(a+3b)

=(3a+b) : (a+3b) 

Ans (b)

Question 8

In the given 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?

(a) area of ∆ABC = area of ∆ABD

(b) area of ∆ACD = area of ∆BCD

(c) area of ∆OAB = area of ∆OCD

(d) area of ∆OAD = area of ∆OBC

Sol :

In the trapezium ABCD, AB||DC

⇒AB≠DC

The diagonals BD and AC intersect each other at O

⇒Only statement area of ΔOAB is not equal to area ΔCOD

⇒Other all statements are true 

⇒Only (b) is not true

Ans (b)