Chapter 21 : Areas Theorems on Parallelograms - Frank Solutions for Class 9 Maths ICSE

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Chapter 21 - Areas Theorems on Parallelograms Excercise Ex. 21.1

Question 1
Solution 1
Question 2
Solution 2
Question 3

Solution 3

Question 4

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

Solution 4

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.

Solution 5

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

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

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

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

Question 6

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

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

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.

  

Solution 10

Joining AC and FE, we get

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)

 A(parallelogram ABCD) =  A(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.

Solution 11

Question 12

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

 

 

*Question modified

Solution 12

Joining TR, we get

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

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

From (i) and (ii), we get

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.

Solution 13

Question 14

Solution 14

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

  

 

Solution 15

Question 16

Solution 16

Question 17

Solution 17

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.

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

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

Solution 24

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.

Solution 25

  

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

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

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

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 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

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

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

Area of ΔABE = Area of ΔACE.

 

Solution 33

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

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

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

Area of ΔABC = 4 × Area of ΔABE

Solution 34

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.

In a parallelogram PQRS, SQ is the diagonal.

So, it bisects the parallelogram.

Area(DPSQ)

 

SM is the median of ΔPSQ.

Area(ΔPSM)

  

 

Again, MN is the median of ΔPSM.

Area(ΔPMN)   

  

  

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.

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

Solution 37

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