RD SHARMA Solutions for Class 9 Maths Chapter 13 - Quadrilaterals
Chapter 13 - Quadrilaterals Exercise Ex. 13.1
Since the sum of all interior angles of a quadrilateral is 360o.
30x = 360o
x = 12o
Hence, the angles are
3x = 3

5x = 5

9x = 9

13x = 13

Chapter 13 - Quadrilaterals Exercise Ex. 13.2
Two opposite angles of a parallelogram are (3x - 2)° and (50 - x)°. Find the measure of each of the parallelogram.
If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.
Find the measure of all the angles of a parallelogram, if one angle is 24o less than twice the smallest angle.
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?
In a parallelogram ABCD, ∠D = 135°, determine the measures of ∠A and ∠B.
ABCD is a parallelogram in which ∠A = 70. Compute ∠B, ∠C and ∠D.
In fig., ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.
i. F
ii. T
iii. F
iv. F
v. T
vi. F
vii. F
viii. T
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.
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.
Chapter 13 - Quadrilaterals Exercise Ex. 13.3
In a parallelogram ABCD, determine sum of angles ∠C and ∠D.
C and
D are cosecutive interior angles on the same side of the transversal CD. Therefore,
C +
D = 180o
In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.
ABCD is a square. AC and BD intersect at O. State the measure of AOB.
Since, diagonals of a square bisect each other at right angle. Therefore, AOB = 90o
The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.
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.
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.
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.
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.
Chapter 13 - Quadrilaterals Exercise Ex. 13.4
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.
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.
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.
Join PQ, QR, RS, SP and BD.
In

So, By using mid-point theorem, we can say that
SP || BD and SP =

Similarly in

QR || BD and QR =

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.
Hence, PR and QS bisect each other.
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 ______ .
(i) isosceles
(ii) right triangle
(iii) parallelogram
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°.
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.
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
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.
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.
Chapter 13 - Quadrilaterals Exercise 13.70
The opposite sides of a quadrilateral have
(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points
ABCD is a Quadrilateral.
The opposite sides AB and DC, AD and BC have no common point.
Hence, correct option is (a).
The consecutive sides of a quadrilateral have
(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points
Consecutive sides of a Quadrilateral ABCD are
AB and BC,
BC and CD,
CD and AD,
AD and AB,
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
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°
For a rhombus, the angle between the diagonals is 90° and not 60°.
Hence, correct option is (d).
Diagonals necessarily bisect opposite angles in a
(a) rectangle
(b) parallelogram
(c) isosceles trapezium
(d) square
Diagonals necessarily bisect opposite angles in a square.
Hence, correct option is (d).
The two diagonals are equal in a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) trapezium
The two diagonals are equal in a rectangle (property).
Hence, correct option is (c).
We get a rhombus by joining the mid-points of the sides of a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) triangle
The bisectors of any two adjacent angles of a parallelogram intersect at
(a) 30°
(b) 45°
(c) 60°
(d) 90°
The bisectors of the angle of a parallelogram enclose a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) square
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).
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
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
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
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
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
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°
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
Chapter 13 - Quadrilaterals Exercise 13.72
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°
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°
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).
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
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
Chapter 13 - Quadrilaterals Exercise 13.73
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