RD SHARMA SOLUTION CHAPTER -13 Quadrilaterals| CLASS 9TH MATHEMATICS-EDUGROWN

Chapter 13 – Quadrilaterals Exercise Ex. 13.1

Question 1

Solution 1

Question 2

Solution 2

Question 3The angles of quadrilateral are in the ratio 3: 5: 9: 13, Find all the angles of the quadrilateral.Solution 3Let 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 360^o.
 3x + 5x + 9x + 13x = 360^o
30x = 360^o
    x = 12^o
Hence, the angles are
3x = 3  12 = 36^o
5x = 5  12 = 60^o
9x = 9  12 = 108^o
13x = 13  12 = 156^o

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 24^o 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. TQuestion 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 = 180^oQuestion 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 = 90^oQuestion 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

(ii) The area of ADE.

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 12Show 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) parallelogramQuestion 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