Free Areas of Triangles and Parallelograms 03 Practice Test - 9th Grade 

Question 1

Which of the following figures lie on the same base and between the same parallel lines?

A.

B.

C.

D.

SOLUTION

Solution : C

It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However, these do not lie between the same parallel lines.

It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC. However, these do not have any common base.

It can be observed that parallelogram PQRS and triangle TQR have a common base QR and they are lying between the same parallel lines PS and QR.

It can be observed that parallelogram PQRS and trapezium MNRS have a common base RS. However, their vertices, (i.e., opposite to the common base) P, Q of parallelogram and M, N of trapezium, are not lying on the same line.

Question 2

PQ is a line parallel to side BC and passing through vertex A of a triangle ABC. If BE || AC and CF || AB meet PQ at E and F respectively. If the base and altitude of ΔABC are 6 cm and 8 cm respectively, the area of ΔACF is ___.

A. 12 cm2
B. 24 cm2
C. 48 cm2
D. 48 cm2

SOLUTION

Solution : B

CF when drawn parallel to AB, forms parallelogram ABCF.

AC is the diagonal of the parallelogram that divides ABCF into 2 equal triangles ABC and ACF.

So, Area of ΔACF = Area of ΔABC = 12×6×8=24 cm2

Question 3

If A, B, C, D are mid-points of sides of parallelogram PQRS, area of PQRS is always ________ the area of shape formed by joining A,B,C,D.

A. twice
B. thrice
C. half
D. one-third

SOLUTION

Solution : A

Draw a line joining D and B.

See that parallelogram PDBQ and triangle ADB are on same base and between same parallels PQ and DB.

Area(ADB) = 12 Area(PDBQ) = 14 Area(PQRS)

Similarly, Area(DCB) = 12 Area(DBRS) = 14 Area(PQRS)

Area(DABC) = Area(ADB) + Area(DCB) = 12 Area(PQRS)

Area(PQRS) = 2 × Area(DABC)

Question 4

ABCD is a parallelogram and P,Q are the midpoints of DC and AB respectively. Then, area of parallelogram AQPD is equal to area of triangle ADB.

A.

True

B.

False

SOLUTION

Solution : A

Parallelogram ABCD and triangle ABD are on the same base and between same parallels .

Area(ADB) = 12 Area of parallelogram(ABCD)

As P and Q are the midpoints of DC and AB, Area of parallelogram(AQPD) = 12Area of parallelogram(ABCD)

So, Area(ADB) = Area of parallelogram(AQPD)

Question 5

In the given figure ABCD is a parallelogram. If area of parallelogram ABCD = 1728 square units, then area BEC = ____ square units.

A. 1728 square units   
B. 864 square units   
C. 432 square units         
D. 400 square units     

SOLUTION

Solution : B

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.

So, Area BEC =  12 × Area parallelogram ABCD

Area BEC = 12 × 1728

                      = 864 square units

Question 6

If 1, 2, 3 and 4 represent the areas of the four triangles (of a parallelogram) with different shades, then _______.

A. 2 = 3, 1 ≠ 4
B. 1 = 2 = 3 = 4
C. 1 ≠ 2, 3 = 4
D. 1 = 2, 3 ≠ 4

SOLUTION

Solution : B

Consider the figure below:

In DAO and OCB
DAO=OCB(Alternate angles)ADO=CBO(Alternate angles)AD=CB(Opposite sides of parallelogram)Hence, ΔADOΔCBO(AAS congruence)Similarly it can be proved thatArea(ΔDOC)=Area(Δ(BOA)
In ABD,  AO is the median.
A median divides a triangle into two equal parts. 
Hence 1 = 4
Similarly 2 = 3
1=2=3=4

Question 7

Which of the following figures lie on the same base and between the same parallels?

A.

Fig 1 

B.

Fig 2 

C.

Fig 3

D.

Fig 4

SOLUTION

Solution : A and D

In fig 1, BC is the base for quadrilateral ABCD and ΔBEC lies between two parallel lines (BC and AD). In fig 4, AB is the base for quadrilateral ABCD and ΔABD lies between two parallel lines. Hence, options A and D are correct.

Question 8

In the figure, ABCD is a parallelogram, AB and BC are equal to 15 cm and 6 cm respectively. If DF = 4 cm, what is the value of BE?

A.  6 cm
B. 8 cm
C. 10 cm  
D. 12 cm

SOLUTION

Solution : C

Given
BC = 6 cm
AD = BC (parallel sides of a parallelogram are equal)
Area (ABCD) = Base × Height
                        = AB × DF = AD × BE


AB × DF = 15 cm × 4 cm
                = 60 square cm
AD × BE = 60 square cm
BE × 6 cm = 60 square cm
BE = 10 cm

Question 9

If two figures have same area, they must be congruent.

A.

True

B.

False

SOLUTION

Solution : B

If two figures are congruent, they must have the same area but the reverse may not be true. Two figures having the same area need not be congruent.

 For example: Consider the figure below:

In this figure, both parallelograms have the same area because they have same base and height. But they aren't congruent.

Question 10

What is the area of ABC (in square cm) if AD is the median and area ADB= 18 square cm?


__

SOLUTION

Solution :

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

So, Area (ABC) = 2 × Area (ADB) = 2 × 18 = 36 square cm.