CBSE Class 9 Mathematics Quadrilaterals Assignment Set 03

Case I. Read the following passage and answer the questions from 1 to 5.
Anjali and Meena were trying to prove mid point theorem.
They draw a triangle ABC, where D and E are found to be the midpoints of AB and AC respectively. DE was joined and extended to F such that DE = EF and FC is also joined. 

1. ΔADE and ΔCFE are congruent by which criterion?
(a) SSS
(b) SAS
(c) RHS
(d) ASA
Answer : B

2. ∠EFC is equal to which angle?
(a) ∠DAE
(b) ∠EDA
(c) ∠AED
(d) ∠DBC
Answer : B

3. ∠ECF is equal to which angle?
(a) ∠EAD
(b) ∠ADE
(c) ∠AED
(d) ∠B
Answer : A

4. CF is equal to
(a) EC
(b) BE
(c) BC
(d) AD
Answer : D

5. CF is parallel to
(a) AE
(b) CE
(c) BD
(d) AC
Answer : C

Very Short Answer Type Questions

Question. Two consecutive angles of a parallelogram are (x + 60°) and (2x + 30°). What special name can you give to this parallelogram?
Answer : We know that consecutive interior angles of a parallelogram are supplementary.
∴ (x + 60°) + (2x + 30°) = 180°
∴ 3x + 90° = 180° ⇒ 3x = 90° ⇒ x = 30°
Thus, two consecutive angles are (30° + 60°), (2 × 30° + 30°) i.e., 90° and 90°.
Hence, the special name of the given parallelogram is rectangle.

Question. In the given figure, PQRS is a parallelogram in which ∠PSR = 125°. Find the measure of ∠RQT.

Answer : ∠PQR = ∠PSR = 125°
(∵ Opposite angles of a parallelogram are equal)
Now, ∠PQR + ∠RQT = 180° (Linear pair)
⇒ 125° + ∠RQT = 180° ⇒ ∠RQT = 55°

Question. ABCD is a parallelogram in which ∠A = 78°. Compute ∠B, ∠C and ∠D.
Answer : Since, ∠A + ∠B = 180°
[Co-interior angles]
⇒ ∠B = 180° – 78° = 102°
Now, ∠B = ∠D = 102°
and, ∠A = ∠C = 78°
[∵ opposite angles of a parallelogram are equal]

Question. In the given figure, ABCD is a parallelogram in which ∠DAB = 60° and ∠DBC = 55°. Compute ∠CDB and ∠ADB.

Answer : We have, ∠A + ∠B = 180°          [Co-interior angles]
⇒ 60° + ∠ABD + 55° = 180° ⇒ ∠ABD = 65°
Also, ∠ABD = ∠CDB
           [Alternate interior angles are equal]
∴ ∠CDB = ∠ABD = 65°
We have, ∠ADB = ∠DBC
           [Alternate interior angles are equal]
⇒ ∠ADB = 55°

Question. Can the angles 110°, 80°, 70° and 95° be the angles of a quadrilateral ? Why or why not?
Answer : No.
Sum of the angles = 110° + 80° + 70° + 95° = 355° ≠ 360°
Thus, the given angles cannot be the angles of a quadrilateral.

Question. In the figure, it is given that QLMN and NLRM are parallelograms. Can you say that QL = LR? Why or why not?

Answer : Yes, QL = LR
As, opposite sides of a parallelogram are equal.
∴ In parallelogram QLMN, QL = NM …(i)
In parallelogram NLRM, NM = LR …(ii)
From (i) and (ii), QL = LR

 Question. In a rhombus ABCD, if ∠A = 60°, then find the sum of ∠A and ∠C.
Answer : Since a rhombus is a parallelogram.
∴ Its opposite angles are equal.
⇒ ∠A = ∠C
∴ ∠C = 60° [ ∠A = 60° (Given)]
Now, required sum = ∠A + ∠C
= 60° + 60° = 120°

Question. ABCD is a trapezium in which AB || DC and ∠A = ∠B = 45°. Find angles C and D of the trapezium.
Answer : We have given, a trapezium
ABCD, whose parallel sides are AB and DC.
Since, AB | | CD and AD is a transversal.
∴ ∠A + ∠D = 180° [Angles on same side of transversal]
⇒ ∠D = 180° – ∠A = 180° – 45° = 135°
Similarly, ∠C = 135°

Question. In the given figure, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC. Prove that ∠PAC = ∠BCA and ABCP is a parallelogram.

Answer : We have, AB = AC ⇒ ∠BCA = ∠B
Now, ∠CAD = ∠B + ∠BCA      [Exterior angle property]
⇒ 2∠CAP = 2∠BCA               [∵ AP is the bisector of ∠CAD]
⇒ ∠CAP = ∠BCA ⇒ AP || BC
Also, AB || CP                                [Given]
Hence, ABCP is a parallelogram.

Question. If one angle of a rhombus is a right angle, then it is necessarily a _______ .
Answer : If one angle of a rhombus is a right angle, then it is necessarily a square

Question. Diagonals of a quadrilateral ABCD bisect each other. ∠A = 45° and ∠B = 135°. Is it true?
Justify your answer.
Answer : True. Given, ABCD is a quadrilateral whose diagonals bisect each other. Then, it should be a parallelogram.
Also, ∠A and ∠B are adjacent angles of parallelogram
ABCD. So, their sum should be 180°.
Now, ∠A + ∠B = 45° + 135° = 180°

Quadrilaterals

Q 1 Name a quadrilateral whose each pair of opposite sides is equal.

Q 2 What is the sum of two consecutive angles in a parallelogram?

Q 3 The angles of quadrilateral are respectively 100 , 30 , 92 and x. Find the value of x.

Q 4 The angles of quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.

Q 5 Thee sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.

Q 6 In a triangle ABC, P,Q and R are the mid – points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.

Q 7 Find the four angles P, Q, R and S in the parallelogram PQRS as shown below.

Q 8 Two opposite angles of a parallelogram are (5x + 1)° and (49 – 3x)°. Find the measure of these opposite angles of the parallelogram.

Q 9 Prove that each of the four sides of a rhombus is of the same length.

Q 10 ABCD is a rhombus. Show that diagonals AC bisects angle A as well as angle C.

Click on link below to download CBSE Class 9 Mathematics Quadrilaterals Assignment Set C