# RD SHARMA Solutions for Class 9 Maths Chapter 13 - Quadrilaterals

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## Chapter 13 - Quadrilaterals Exercise Ex. 13.1

Question 1

Solution 1

Question 2

Solution 2

Question 3
The angles of quadrilateral are in the ratio 3: 5: 9: 13, Find all the angles of the quadrilateral.
Solution 3
Let the common ratio between the angles is x. So, the angles will be 3x, 5x, 9x and 13x respectively.
Since the sum of all interior angles of a quadrilateral is 360o.
3x + 5x + 9x + 13x = 360o

30x = 360o
x = 12o
Hence, the angles are
3x = 3  12 = 36o
5x = 5  12 = 60o
9x = 9  12 = 108o
13x = 13  12 = 156o

Question 4

Solution 4

## Chapter 13 - Quadrilaterals Exercise Ex. 13.2

Question 1

Two opposite angles of a parallelogram are (3x - 2)° and (50 - x)°. Find the measure of each of the parallelogram.

Solution 1

Question 2

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Solution 2

Question 3

Find the measure of all the angles of a parallelogram, if one angle is 24o less than twice the smallest angle.

Solution 3

Question 4

The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?

Solution 4

Question 5

In a parallelogram ABCD, D = 135°, determine the measures of A and B.

Solution 5

Question 6

ABCD is a parallelogram in which A = 70. Compute B, C and D.

Solution 6

Question 7

In fig., ABCD is a parallelogram in which DAB = 75° and DBC = 60°. Compute CDB and ADB.

Solution 7

Question 8

Solution 8

i. F

ii. T

iii. F

iv. F

v. T

vi. F

vii. F

viii. T

Question 9

In fig., ABCD is a parallelogram in which A = 60°. If the bisectors of A and b meet at P, prove that AD = DP, PC = BC and DC = 2AD.

Solution 9

Question 10

In fig., ABCD is a parallelogram and E is the mid-point of side BC. IF DE and AB when produced meet at F, prove that AF = 2AB.

Solution 10

## Chapter 13 - Quadrilaterals Exercise Ex. 13.3

Question 1

In a parallelogram ABCD, determine sum of angles C and D.

Solution 1

C and D are cosecutive interior angles on the same side of the transversal CD. Therefore,

C + D = 180o

Question 2

In a parallelogram ABCD, if B = 135°, determine the measures of its other angles.

Solution 2

Question 3

ABCD is a square. AC and BD intersect at O. State the measure of AOB.

Solution 3

Since, diagonals of a square bisect each other at right angle. Therefore, AOB = 90o

Question 4

Solution 4

Question 5

The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.

Solution 5

Question 6

P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.

Solution 6

Question 7

ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is square.

Solution 7

Question 8

ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.

Solution 8

Question 9

ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.

Solution 9

## Chapter 13 - Quadrilaterals Exercise Ex. 13.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

In fig., triangle ABC is right-angled at B. Give that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate

(i) The length of BC

Solution 7

Question 8

In fig., M, N, and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, Np = 3.5 cm and MP = 2.5 cm, Calculate BC, AB and AC.

Solution 8

Question 9

In fig., AB = AC and CP  BA and AP is the bisector of exterior CAD of ΔABC. Prove that (i) PAC = BCA (ii) ABCP is a parallelogram.

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Solution 12

Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.
Join PQ, QR, RS, SP and BD.
In ABD, S and P are mid points of AD and AB respectively.
So, By using mid-point theorem, we can say that
SP || BD and SP =   BD             ... (1)
Similarly in BCD
QR || BD and QR = BD               ... (2)
From equations (1) and (2), we have
SP || QR and SP = QR
As in quadrilateral SPQR one pair of opposite sides are equal and parallel to
each other.
So, SPQR is a parallelogram.
Since, diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Question 13

Fill in the blanks to make the following statements correct:

(i) The triangle formed by joining the mid-points of the sides of an isosceles traingle is ______.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is ______ .

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ______ .

Solution 13

(i) isosceles

(ii) right triangle

(iii) parallelogram

Question 14

Solution 14

Question 15

In fig., BE  AC. AD is any line from A to BC interesting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that PQR = 90°.

Solution 15

Question 16

Solution 16

Question 17

In fig., ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = (1/4)AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

Solution 17

Question 18

In fig., ABCD and PQRC are rectangle and Q is the mid-point of AC. Prove that

i. DP = PC ii. PR = (1/2) AC

Solution 18

Question 19

ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC and G, P and H respectively. Prove that GP = PH.

Solution 19

Question 20

BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.

Solution 20

## Chapter 13 - Quadrilaterals Exercise 13.70

Question 1

The opposite sides of a quadrilateral have

(a) no common point

(b) one common point

(c) two common points

(d) infinitely many common points

Solution 1

The opposite sides AB and DC, AD and BC have no common point.

Hence, correct option is (a).

Question 2

The consecutive sides of a quadrilateral have

(a) no common point

(b) one common point

(c) two common points

(d) infinitely many common points

Solution 2

Consecutive sides of a Quadrilateral ABCD are

AB and BC,

BC and CD,

which have only one point in common

i.e the joint point of their ends.

Hence, correct option is (b).

## Chapter 13 - Quadrilaterals Exercise 13.71

Question 3

Solution 3

Question 4

Which of the following quadrilateral is not a rhombus?

(a) All four sides are equal

(b) Diagonals bisect each other

(c) Diagonals bisect opposite angles

(d) One angle between the diagonals is 60°

Solution 4

For a rhombus, the angle between the diagonals is 90° and not 60°.

Hence, correct option is (d).

Question 5

Diagonals necessarily bisect opposite angles in a

(a) rectangle

(b) parallelogram

(c) isosceles trapezium

(d) square

Solution 5

Diagonals necessarily bisect opposite angles in a square.

Hence, correct option is (d).

Question 6

The two diagonals are equal in a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) trapezium

Solution 6

The two diagonals are equal in a rectangle (property).

Hence, correct option is (c).

Question 7

We get a rhombus by joining the mid-points of the sides of a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) triangle

Solution 7

Question 8

The bisectors of any two adjacent angles of a parallelogram intersect at

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Solution 8

Question 9

The bisectors of the angle of a parallelogram enclose a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) square

Solution 9

AR, BR, CP, DP are the bisectors of angles of parallelogram.

Because two bisectors of adjacent angles make 90° between them So  PQRS is a Rectangle

Because DP and BR are acute angle bisectors so the distance between them PQ < PS (The distance between other two bisectors)

So PQ ≠ PS (So PQRS  is not a square, but only a rectangle)

Hence, correct option is (c).

Question 10

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a

(a) parallelogram

(b) rectangle

(c) square

(d) rhombus

Solution 10

Question 11

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a

(a) square

(b) rhombus

(c) trapezium

(d) none of these

Solution 11

Question 12

The figure formed by joining the mid-points of the adjacent sides of a rhombus is a

(a) square

(b) rectangle

(c) trapezium

(d) none of these

Solution 12

Question 13

The figure formed by joining the mid-points of the adjacent sides of a square is a

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

Solution 13

Question 14

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a

(a) rectangle

(b) parallelogram

(c) rhombus

(d) square

Solution 14

Question 15

If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is

(a) 176°

(b) 68°

(c) 112°

(d) 102°

Solution 15

Question 16

Solution 16

Question 17

ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =

(a) AE

(b) BE

(c) CE

(d) DE

Solution 17

## Chapter 13 - Quadrilaterals Exercise 13.72

Question 18

Solution 18

Question 19

If an angle of  a parallelogram is is two-third of its adjacent angle, the smallest angle of the parallelogram is

(a) 108°

(b) 54°

(c) 72°

(d) 81°

Solution 19

Question 20

If the degree measures of the angles of quadrilateral are 4x, 7x, 9x and 10x, what is the sum of the measures of the smallest angle and largest angle?

(a) 140°

(b) 150°

(c) 168°

(d) 180°

Solution 20

Sum of all angles of a Quadrilateral = 360°

4x + 7x + 9x + 10x = 360°

30x = 360°

x = 12°

So, sum of smallest and largest angle,

i.e. 4x + 10x = 14x = 14 × 12 = 168°

Hence, correct option is (c).

Question 21

If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to

(a) 16 cm

(b) 15 cm

(c) 20 cm

(d) 17 cm

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 30

Solution 30

Question 27

Solution 27

Question 28

Solution 28

Question 29

In ΔABC, E is the mid-point of median AD such that BE produced meets AC at F. If AC = 10.5 cm, then AF =

(a) 3 cm

(b) 3.5 cm

(c) 2.5 cm

(d) 5 cm

Solution 29

Question 31

Solution 31

Question 32

Solution 32