Free Areas of Triangles and Parallelograms 01 Practice Test - 9th Grade 

Question 1

A parallelogram and a triangle are on equal base and between the same parallel lines. The ratio of their areas is ______.

A. 2:1
B. 3:1
C. 2:3
D. 1:2


Solution : A

 Consider the figure given below.

In this figure, ABCD is a parallelogram and ABE is a triangle. Both of them have the same base i.e., AB. The perpendicular EF is an altitude for both triangle and parallelogram. 
Now area of ΔABE=12AB×EF
And area of parallelogram is AB×EF.
Hence, Area of parallelogram is twice that of triangle and the ratio will be 2:1

Question 2

ABCD is a trapezium in which AB || DC. Diagonals AC and BD intersect each other at O. Find the triangle which is equal to the area of   BOC.



Solution : D

It can be observed that ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD.

∴ Area (ΔDAC) = Area (ΔDBC)
Subtracting Area (ΔDOC) on both the sides

⇒ Area (ΔDAC) − Area (ΔDOC) = Area (ΔDBC) − Area (ΔDOC)

⇒ Area (ΔAOD) = Area (ΔBOC)

Question 3

ABCD is a parallelogram in which CD is produced to P. DC = 6 cm and height BQ = 4 cm. Find the area of  APB 



A. cm2
B. 12 cm2
C. 24 cm2
D. 48 cm2


Solution : B

Theorem used:
Area of triangle is half the area of parallelogram, if both are on same base and between the same parallels.

 From the given figure:
ABCD    ( ABCD is  a parallelogram)
AB = CD = 6 cm (given)
Altitude BQ = 4 cm (given)

ABP and parallelogram ABCD are on same base and between the same parallels.

Area of ΔAPB=12×Area of ABCD

Area of ABCD=Base×Altitude

Area of ΔAPB=12×Area of ABCD


Area of ΔAPB=12 cm2 

Question 4

In the given figure, ABCDE is a pentagon with AC = 5 cm. A line through B parallel to AC meets DC produced at F. If the altitude of triangle ABC (perpendicular to AC) is 6 cm. Find the area of triangle ACF.

A. cm2
B. 10 cm2
C. 15 cm2
D. 30 cm2


Solution : C

Let BG be perpendicular to AC.
From the given figure:
AC BF                    (given)
AC = 5cm                 (given)
Length of the altitude BG, perpendicular to AC =  6 cm   (given)

ACB and ACF lie on the same base AC and are between the same parallels AC and BF. 

Area ACB = Area ACF 

Area ACB=12×base ×Height= 12×5×6=15 cm2

Area ACF =15 cm2 

Question 5

In the rectangle ABCD, O is any point inside the rectangle. If area( ΔAOD) = 30 cm2 and area( ΔBOC) = 60 cm2, area of the rectangle ABCD is


(in cm2)



Solution :


Draw a line through O parallel to ADas shown.Parallelogram APQD and triangle AOD are on same base AD and between same parallels.So, Area(AOD)=12Area (APQD)Area (APQD)=60cm2Parallelogram PBCQ and triangle BOC are on same base BC and between same parallels.So, Area (BOC)=12Area (PBCQ)Area (PBCQ)=120cm2Area (ABCD)=Area (APQD)+Area (PBCQ)=180cm2 

Question 6

ABC is a triangle in which D, E, F are the mid-points of BC, AC and AB respectively. If Area (ΔABC) = 32 cm2, then area of trapezium BFEC is ______        

A. cm2
B. 16 cm2
C. 24 cm2
D. 32 cm2


Solution : C

Given: In ABC,  D,E and F are midpoints of BC, CA and AB.
Area (ΔABC) = 32 cm2

To find: Area of trapezium BFEC

Consider ABC,
F and E are midpoints of AB and AC. (given)
  FE  BC      (Midpoint theorem)
  FE  BD     

Similarly ED AB and FD AC
FEDB, FDEC and FDEA are all parallelograms.

Since a diagonal divides a parallelogram into two congruent triangles, hence
=14Area(ΔABC)=8 cm2

=Area(ΔBFD)+Area(ΔEFD)+Area(ΔECD)=24 cm2

Question 7

ABCD is a parallelogram. P is any point on CD. If ar(DPA) = 35 cm2 and ar(APC) = 15 cm2, then area(APB) is

A. 15 cm2
B. 35 cm2
C. 50 cm2
D. 70 cm2


Solution : C

Given: ABCD is a parallelogram.
 area(DPA) = 35 cm2
 area(APC) = 15 cm2
To find: area(APB) 


Now, ABCD is a parallelogram.
AC is the diagonal of parallelogram ABCD.
A diagonal of a parallelogram divides it into two congruent triangles.

 Area(ACD)=Area(ACB)                               =50 cm2

ACB and APB are on same base AB and between same parallels AB and DC.
                        =50 cm2

Question 8

If AD is median of ΔABC and P is a point on AC such that ar(ΔADP) : ar(ΔABD) = 2 : 3, then ar(ΔPDC) : ar(ΔABC) is

A. 3 : 5
B. 2 : 5
C. 1 : 5
D. 1 : 6


Solution : D

Given : AD is median of Δ ABC
 ar(ΔADP) : ar(ΔABD) = 2 : 3
To find: ar(ΔPDC) : ar(ΔABC)
Construction:  Draw XY BC
Median divides the triangle into two equal areas and
Triangle ABD and ADC have equal base BD and CD and are within the same parallels XY and BC.

area Δ ABD = area Δ ADC...(i)
area Δ ABD : area Δ ABC = 1 : 2 ...(ii)

area Δ ADP : area Δ ABD = 2 : 3 … (iii)
area Δ ADC = area Δ ADP + area Δ PDC
area Δ ABD = area Δ ADP + area Δ PDC
area Δ PDC
= area Δ ABD - area Δ ADP
= area Δ ABD - 23area Δ ABD
=13 area Δ ABD

area Δ PDC : area Δ ABD = 1 : 3...(iv)

areaΔPDCareaΔABC=13×12 ….. (from equations (i) and (iv)

area Δ PDC : area Δ ABC = 1 : 6

Question 9

In parallelogram ABCD shown below, the vertical distance between the lines AD and BC is 5 cm and length of BC is 4 cm. P and Q are the midpoints of AB and AD respectively. Area of triangle AQP is ____.


A. 1.25 cm2
B. 2.5 cm2
C. cm2
D. Insufficient data


Solution : B

ABCD is a parallelogram.
Base BC = 4 cm
Height of a parallelogram = 5 cm

Area of parallelogram ABCD = Base x Height 
=5 cm×4 cm 
=20 cm2 

A diagonal divides a parallelogram in two congruent triangles which have equal areas.
 Area ΔABD = Area ΔBDC 
 Area ΔABD  =12 Area of  ABCD
= 10 cm2

Let R be the midpoint of the diagonal. Join P and Q to R.
P and Q are midpoints of AB and AD.
Therefore PQ is parallel to BD (Midpoint theorem)

Similarly, QR is parallel to AB and PR is parallel to AD
Therefore, APQR, DQPR and BPQR are all parallelograms
 Area ΔAQP = Area ΔRQP 
= Area ΔBPR = Area ΔQPR ...(i)

 Area ΔABD = Area ΔAQP + Area ΔRQP + Area ΔBPR + Area ΔQPR 

 Area ΔABD =4× Area ΔAQP = 14× Area ΔABD 
 Area ΔAQP = 14×10 

Question 10

In the adjoining figure, ΔQPR is right-angled at Q in which QR = 6 cm and PQ = 7 cm. Find the area of ΔQSR, given that PS is parallel to QR.


A. 21 cm2
B. 20 cm2
C. 10 cm2
D. 11 cm2


Solution : A

Area of right angled triangle PQR
= 12 x Base x Height
=12 x QR x PQ
=12 x 6 x 7
=21 cm2

ΔQPR  and ΔQSR lie on same base QR and are between same parallels hence, their areas are equal.
Area ΔQPR  = Area ΔQSR 
 Area of ΔQSR=21 cm2