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

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 $Δ A B E = \frac{1}{2} A B \times E F$
And area of parallelogram is $A B \times E F$ .
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.

A. ΔADC

B. ΔAOB

C. ΔDOC

D. ΔAOD

SOLUTION

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. 6

c m^{2}

B. 12

c m^{2}

C. 24

c m^{2}

D. 48

c m^{2}

SOLUTION

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:
$A B ∥ C D$ ( $∵$ ABCD is a parallelogram)
AB = CD = 6 cm (given)
Altitude BQ = 4 cm (given)

$△ A B P$ and parallelogram ABCD are on same base and between the same parallels.

$∴ Area of Δ A P B = \frac{1}{2} \times Area of A B C D$

$Area of A B C D = Base \times Altitude$

$= 6 \times 4 = 24 c m^{2}$

$Area of Δ A P B = \frac{1}{2} \times Area of A B C D$

$= \frac{1}{2} \times 24$

$Area of Δ A P B = 12 c m^{2}$

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. 5

c m^{2}

B. 10

c m^{2}

C. 15

c m^{2}

D. 30

c m^{2}

SOLUTION

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)

$△ A C B and △ A C F lie on the same base AC and are between the same parallels AC and BF.$

$∴ Area △ A C B = Area △ A C F$

$Area △ A C B = \frac{1}{2} \times base \times Height = \frac{1}{2} \times 5 \times 6 = 15 c m^{2}$

$∴ Area △ A C F = 15 c m^{2}$

Question 5

In the rectangle ABCD, O is any point inside the rectangle. If area( ΔAOD) = 30 $c m^{2}$ and area( ΔBOC) = 60 $c m^{2}$ , area of the rectangle ABCD is

___

(in $c m^{2}$ )

SOLUTION

Solution :

$Draw a line through O parallel to A D as shown. Parallelogram A P Q D and triangle A O D are on same base A D and between same parallels. So, Area (A O D) = \frac{1}{2} Area (A P Q D) Area (A P Q D) = 60 c m^{2} Parallelogram P B C Q and triangle B O C are on same base BC and between same parallels. So, Area (B O C) = \frac{1}{2} Area (P B C Q) Area (P B C Q) = 120 c m^{2} Area (A B C D) = Area (A P Q D) + Area (P B C Q) = 180 c m^{2}$

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. 8

c m^{2}

B. 16

c m^{2}

C. 24

c m^{2}

D. 32

c m^{2}

SOLUTION

Solution : C

Given: In $△ A B C,$ D,E and F are midpoints of BC, CA and AB.
Area (ΔABC) = 32 cm²

To find: Area of trapezium BFEC

Consider $△ A B C,$
F and E are midpoints of AB and AC. (given)
$∴$ FE $∥ B C$ (Midpoint theorem)
$∴$ FE $∥ B D$

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

Since a diagonal divides a parallelogram into two congruent triangles, hence
$A r e a (Δ B F D) = A r e a (Δ E F D) = A r e a (Δ E C D) = A r e a (Δ E F A)$
$= \frac{1}{4} A r e a (Δ A B C) = 8 c m^{2}$

Area(BFEC)
$= A r e a (Δ B F D) + A r e a (Δ E F D) + A r e a (Δ E C D) = 24 c m^{2}$

Question 7

ABCD is a parallelogram. P is any point on CD. If ar( $△$ DPA) = 35 $c m^{2}$ and ar( $△$ APC) = 15 $c m^{2}$ , then area( $△$ APB) is

A. 15

c m^{2}

B. 35

c m^{2}

C. 50

c m^{2}

D. 70

c m^{2}

SOLUTION

Solution : C

Given: ABCD is a parallelogram.
area( $△$ DPA) = 35 $c m^{2}$
area( $△$ APC) = 15 $c m^{2}$
To find: area( $△$ APB)

$A r e a (Δ A C D) = A r e a (Δ D P A) + A r e a (Δ A P C)$
$= 35 c m^{2} + 15 c m^{2}$
$= 50 c m^{2}$

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

$\Rightarrow A r e a (△ A C D) = A r e a (△ A C B) = 50 c m^{2}$

$△ A C B$ and $△ A P B$ are on same base AB and between same parallels AB and DC.
Hence,
$A r e a (Δ A P B) = A r e a (Δ A C B)$
$= 50 c m^{2}$

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

Solution : D

Given : AD is median of $Δ$ ABC
$∴ B D = D C$
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 - $\frac{2}{3}$ area $Δ$ ABD
= $\frac{1}{3}$ area $Δ$ ABD

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

$\frac{a r e a Δ P D C}{a r e a Δ A B C} = \frac{1}{3} \times \frac{1}{2}$ ….. (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

c m^{2}

B. 2.5

c m^{2}

C. 5

c m^{2}

D. Insufficient data

SOLUTION

Solution : B

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

Area of parallelogram ABCD = Base x Height
$= 5 c m \times 4 c m$
$= 20 c m^{2}$

A diagonal divides a parallelogram in two congruent triangles which have equal areas.
$∴ A r e a Δ A B D = A r e a Δ B D C$
$∴ A r e a Δ A B D$ = $\frac{1}{2}$ Area of ABCD
$= 10 c m^{2}$

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
$∴ A r e a Δ A Q P = A r e a Δ R Q P$
$= A r e a Δ B P R = A r e a Δ Q P R$ ...(i)

$A r e a Δ A B D = A r e a Δ A Q P + A r e a Δ R Q P + A r e a Δ B P R + A r e a Δ Q P R$

$A r e a Δ A B D = 4 \times A r e a Δ A Q P$ $= \frac{1}{4} \times A r e a Δ A B D$
$∴ A r e a Δ A Q P = \frac{1}{4} \times 10$
$= 2.5 c m^{2}$

Question 10

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

21 c m^{2}

20 c m^{2}

10 c m^{2}

11 c m^{2}

SOLUTION

Solution : A

Area of right angled triangle PQR
= $\frac{1}{2}$ x Base x Height
= $\frac{1}{2}$ x QR x PQ
= $\frac{1}{2}$ x 6 x 7
$= 21 c m^{2}$

$Δ Q P R$ and $Δ Q S R$ lie on same base QR and are between same parallels hence, their areas are equal.
Area $Δ Q P R$ = Area $Δ Q S R$
$∴ A r e a o f Δ Q S R = 21 c m^{2}$

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

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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Related Exams
Introduction to Euclid's Geometry 01
Triangles 02
Statistics 02