FRANK Solutions for Class 9 Maths Chapter 21 - Areas Theorems on Parallelograms

Get quick access to online Frank Solutions for ICSE Class 9 Mathematics Chapter 21 Areas theorems on Parallelograms on TopperLearning. You will learn how to prove theorems in this chapter. You can use the textbook solutions as quick reference materials to revise the lessons.

Learn to use the accurate theorem to find the area of a parallelogram using the information of area of a triangle that’s part of the parallelogram. Revise similar area theorem problems effectively with our ICSE Class 9 Maths Frank solutions. For clarity on quadrilateral problems, you can get your doubts answered in our learning portal’s ‘UnDoubt’ section.

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Chapter 21 - Areas Theorems on Parallelograms Exercise Ex. 21.1

Question 1
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Solution 1
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Question 2
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Solution 2
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Question 3
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Solution 3
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Question 4

In the given figure area of gm PQRS is 30 cm2. Find the height of gm PQFE if PQ = 6 cm.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 4

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 5

In the given figure, PQRS is a gm. A straight line through P cuts SR at point T and QR produced at N. Prove that area of triangle QTR is equal to the area of triangle STN.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 5

ΔPQT and parallelogram PQRS are on the same base PQ and between the same parallel lines PQ and SR.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

ΔPSN and parallelogram PQRS are on the same base PS and between the same parallel lines PS and QN.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Adding equations (i) and (ii), we get

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Subtracting A(ΔRTN) from both the sides, we get

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 6

In the given figure, ST PR. Prove that: area of quadrilateral PQRS = area of ΔPQT.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 6

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 7

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 7

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 8

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 8

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 9

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 9

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 10

In the given figure, if AB DC FG and AE is a straight line. Also, AD FC. Prove that: area of gm ABCD = area of gm BFGE.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms 

Solution 10

Joining AC and FE, we get

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

ΔAFC and ΔAFE are on the same base AF and between the same parallels AF and CE.

A(ΔAFC) = A(ΔAFE)

A(ΔABF) + A(ΔABC) = A(ΔABF) + A(ΔBFE)

A(ΔABC) = A(ΔBFE)

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On ParallelogramsA(parallelogram ABCD) = Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On ParallelogramsA(parallelogram BFGE)  A(parallelogram ABCD) = A(parallelogram BFGE) 

Question 11

In the given figure, the perimeter of parallelogram PQRS is 42 cm. Find the lengths of PQ and PS.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 11

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 12

In the given figure, PT QR and QT RS. Show that: area of ΔPQR = area of ΔTQS.

 

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

 

*Question modified

Solution 12

Joining TR, we get

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

ΔPQR and ΔQTR are on the same base QR and between the same parallel lines QR and PT.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

ΔQTR and ΔTQS are on the same base QT and between the same parallel lines QT and RS.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

From (i) and (ii), we get

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 13

In the given figure, ΔPQR is right-angled at P. PABQ and QRST are squares on the side PQ and hypotenuse QR. If PN TS, show that:

(a) ΔQRB ΔPQT

(b) Area of square PABQ = area of rectangle QTNM.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 13

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 14

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 14

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 15

The diagonals of a parallelogram ABCD intersect at O. A line through O meets AB in P and CD in Q. Show that

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms 

 

Solution 15

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 16

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 16

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 17

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 17

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 18

In the given figure AF = BF and DCBF is a parallelogram. If the area of ΔABC is 30 square units, find the area of the parallelogram DCBF.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 18

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 19

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 19

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 20

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 20

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 21

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 21

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 22

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 22

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Question 23

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 23

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 24

In the given figure, BC DE.

(a) If area of ΔADC is 20 sq. units, find the area of ΔAEB.

(b) If the area of ΔBFD is 8 square units, find the area of ΔCEF

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 24

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 25

In the given figure, AB SQ DC and AD PR BC. If the area of quadrilateral ABCD is 24 square units, find the area of quadrilateral PQRS.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 25

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms 

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 26

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 26

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 27

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 27

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Question 28

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 28

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Question 29

In the given figure, PQ SR MN, PS QM and SM PN. Prove that:

ar. (SMNT) = ar. (PQRS).

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 29

SM PN

SM TN

Also, SR MN

ST MN

Hence, SMNT is a parallelogram.

 

SM PN

SM PO

Also, PS QM

PS OM

Hence, SMOP is a parallelogram.

 

Now, parallelograms SMNT and SMOP are on the same base SM and between the same parallels SM and PN.

A(parallelogram SMNT) = A(parallelogram SMOP) ….(i)

 

Similarly, we can show that quadrilaterals PQRS is a parallelogram.

Now, parallelograms PQRS and SMOP are on the same base PS and between the same parallels PS and QM.

A(parallelogram PQRS) = A(parallelogram SMOP) ….(ii)

 

From (i) and (ii), we have

A(parallelogram SMNT) = A(parallelogram PQRS)

Question 31

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 31

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Question 32

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 32

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Question 33

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 33

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms
Question 34(a)

In the given figure, ABC is a triangle and AD is the median.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

If E is any point on the median AD. Show that:

Area of ΔABE = Area of ΔACE.

 

Solution 34(a)

AD is the median of ΔABC.

Therefore it will divide ΔABC into two triangles of equal areas.

Area(ΔABD) = Area(ΔACD) ….(i)

 

Similarly, ED is the median of ΔEBC.

Area(DEBD) = Area(DECD) ….(ii)

 

Subtracting equation (ii) from (i), we have

Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD) 

Area(ΔABE) = Area(ΔACE)

Question 34(b)

In the given figure, ABC is a triangle and AD is the median.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

If E is the midpoint of the median AD, prove that:

Area of ΔABC = 4 × Area of ΔABE

Solution 34(b)

AD is the median of ΔABC.

Therefore it will divide ΔABC into two triangles of equal areas.

Area(ΔABD) = Area(ΔACD) ….(i)

 

Similarly, ED is the median of ΔEBC.

Area(ΔEBD) = Area(ΔECD) ….(ii)

 

Subtracting equation (ii) from (i), we have

Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD) 

Area(ΔABE) = Area(ΔACE) ….(iii)

 

Since E is the mid-point of median AD,

AE = ED

 

Now,

ΔABE and ΔBED have equal bases and a common vertex B.

Area(ΔABE) = Area(ΔBED) ….(iv)

 

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

Area(ΔABE) = A(ΔBED) = Area(ΔACE) = Area(ΔEDC) ….(v)

Now,

Area(ΔABC) = Area(ΔABE) + A(ΔBED) + Area(ΔACE) + Area(ΔEDC)

= 4 × Area(ΔABE) [From (v)]

Question 35

In a parallelogram PQRS, M and N are the midpoints of the sides PQ and PS respectively. If area of ΔPMN is 20 square units, find the area of the parallelogram PQRS.

Solution 35

Construction: Join SM and SQ.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

In a parallelogram PQRS, SQ is the diagonal.

So, it bisects the parallelogram.

Area(DPSQ) Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

 

SM is the median of ΔPSQ.

Area(ΔPSM) Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms 

 

Again, MN is the median of ΔPSM.

Area(ΔPMN) Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms 

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms 

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms 

Question 36

In a parallelogram PQRS, T is any point on the diagonal PR. If the area of ΔPTQ is 18 square units find the area of ΔPTS.

Solution 36

Construction: Join QR. Let the diagonals PR and QS intersect each other at point O.

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Since diagonals of a parallelogram bisect each other, therefore O is the mid-point of both PR and QS.

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

 

In ΔPSQ, OP is the median.

Area(ΔPOS) = Area(ΔPOQ) ….(i)

 

Similarly, OT is the median of ΔTSQ.

Area(ΔTOS) = Area(ΔTOQ) ….(ii)

 

Subtracting equation (ii) from (i), we have

Area(ΔPOS) - Area(ΔTOS) = Area(ΔPOQ) - Area(ΔTOQ) 

Area(ΔPTQ) = Area(ΔPTS)

Area(ΔPTS) = 18 square units

Question 37

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Solution 37

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms

Frank Solutions Icse Class 9 Mathematics Chapter - Areas Theorems On Parallelograms