# Problems on Parallelogram | Questions on Parallelogram with Solutions

Problems on Parallelogram are given in this article along with an explanation. It is easy to learn and understand the entire concept of a Parallelogram by solving every problem over here. There are various types of problems included according to the new updated syllabus. Get a good score in the exam and improve your preparation level immediately by working on your difficult topics.

1. Prove that any two adjacent angles of a parallelogram are supplementary?

Solution:
Let us take a parallelogram PQRS.

Then, PS ∥ QR and PQ is a transversal.
The sum of the interior angles on the same side of the transversal is 180°
Therefore, P + Q = 180°
Similarly, ∠R + ∠S = 180°, ∠Q + ∠R = 180°, and ∠S + ∠P = 180°.
Thus, the sum of any two adjacent angles of a parallelogram is 180°.

Hence, any two adjacent angles of a parallelogram are supplementary.

2. Two adjacent angles of a parallelogram PQRS are as 2 : 3. Find the measure of each of its angles?

Solution:
Let us take a parallelogram PQRS.

Then, ∠P and ∠Q are its adjacent angles.
Let ∠P = (2a)° and ∠Q = (3a)°.
The sum of adjacent angles of a parallelogram is 180°
Then, ∠P + ∠Q = 180°
⇒ 2a + 3a = 180
⇒ 5a = 180
⇒ a = 36.
Therefore, ∠P = (2 × 36)° = 72° and ∠Q = (3 × 36°) = 108°.
∠Q and ∠R are adjacent angles. By adding them, we get 180°
Also, ∠Q + ∠R = 180°
= 108° + ∠R = 180° [Since, ∠Q = 108°]
∠R = (180° – 108°) = 72°.
Also, ∠R + ∠S = 180°
⇒ 72° + ∠S = 180°
⇒ ∠S = (180° – 72°) 108°.

Therefore, ∠P = 72°, ∠Q = 108°, ∠R = 72°and ∠S = 108°.

3. In the adjoining figure, PQRS is a parallelogram in which ∠P = 75°. Find the measure of each of the angles ∠Q, ∠R, and ∠S.

Solution:
It is given that PQRS is a parallelogram in which ∠P = 75°.

Since the sum of any two adjacent angles of a parallelogram is 180°,
∠P + ∠Q = 180°
⇒ 75° + ∠Q = 180°
⇒∠Q = (180° – 75°) = 105°
Also, ∠Q + ∠R = 180°
⇒ 105° + ∠R = 180°
⇒ ∠R = (180° – 105°) = 75°.
∠R and ∠S are adjacent angles
Further, ∠R + ∠S = 180°
⇒ 75° + ∠S = 180°
⇒ ∠S = (180° – 75°) = 105°.

Therefore, ∠Q = 105°, ∠R = 75° and ∠S = 105°.

4. In the adjoining figure, PQRS is a parallelogram in which ∠QPS = 75° and ∠SQR = 60°. Calculate:
(i) ∠RSQ and (ii) ∠PSQ.

Solution:
Let us draw a parallelogram PQRS.

We know that the opposite angles of a parallelogram are equal.
Therefore, ∠QRS = ∠QPS = 75°.
(i) Now, in ∆ QRS, we have
The sum of the angles of a triangle is 180°
∠RSQ + ∠SQR + ∠QRS = 180°
⇒ ∠RSQ + 60° + 75° = 180°
⇒ ∠RSQ + 135° = 180°
⇒ ∠RSQ = (180° – 135°) = 45°.
(ii) PS ∥ QR and QS are the transversals.
Therefore, ∠PSQ = ∠SQR = 60° [alternate interior angles]

Hence, ∠PSQ = 60°.

5. In the adjoining figure, PQRS is a parallelogram in which ∠RPS = 40°, ∠QPR = 35°, and ∠ROS = 65°.
Calculate: (i) ∠PQS (ii) ∠QSR (iii) ∠PRQ (iv) ∠RQS.

Solution:
(i) ∠POQ = ∠ROS = 65° (vertically opposite angles)
Now, from ∆OPQ, we can write as:
The sum of the angles of a triangle is 180°
∠OPQ + ∠PQO + ∠POQ =180°
⇒ 35°+ ∠PQO + 65° = 180°
⇒ ∠PQO + 100° = 180°
⇒ ∠PQO = (180° – 100°) = 80°
⇒ ∠PQS = ∠PQO = 80°.
(ii) PQ ∥ SR and QS is a transversal.
Therefore, ∠QSR = ∠PQS = 80° [alternate interior angles]
Hence, ∠QSR = 80°.
(iii) PS ∥ QR and PR is a transversal.
Therefore, ∠PRQ = ∠RPS = 40° [alternate interior angles]
Hence, ∠PRQ = 40°.
(iv) ∠QRS = ∠QPS = (35° + 40°) = 75° [opposite angles of a parallelogram]
Now, in ∆RQS, we have
The sum of the angles of a triangle is 180°.
∠QSR + ∠QRS + ∠RQS = 180°
⇒ 80° + 75° + ∠RQS = 180°
⇒ 155° + ∠RQS = 180°
⇒ ∠RQS = (180° – 155°) = 25°.
Hence, ∠RQS = 25°.

6. In the adjoining figure, PQRS is a parallelogram, PO and QO are the bisectors of ∠P and ∠Q respectively. Prove that ∠POQ = 90°.

Solution:
We know that the sum of two adjacent angles of a parallelogram is 180°
Therefore, ∠P + ∠Q = 180° ……………. (i)
Since PO and QO are the bisectors of ∠P and ∠Q, respectively, we have
∠OPQ = 1/2∠P and ∠PQO = 1/2∠Q.
From ∆OPQ, we have
The sum of the angles of a triangle is 180°
∠OPQ + ∠POQ + ∠PQO = 180°
⇒ ¹/₂∠P + ∠PQO + ¹/₂∠Q = 180°
⇒ ¹/₂(∠P + ∠Q) + ∠POQ = 180°
⇒ (¹/₂ × 180°) + ∠POQ = 180° [from equation (i)]
⇒ 90° + ∠POQ = 180°
⇒ ∠POQ = (180° – 90°) = 90°.

Hence, ∠POQ = 90°.

7. The ratio of two sides of a parallelogram is 5: 4. If its perimeter is 54 cm, find the lengths of its sides?

Solution:
Let the lengths of two sides of the parallelogram be 5a cm and 4a cm respectively.
Find the perimeter using given values.
Then, its perimeter = 2(5a + 4a) cm = 2 (9a) cm = 18a cm.
Therefore, 18a = 54 ⇔ a = 54/18 = 3.

Therefore, one side = (5 × 3) cm = 15 cm and other side = (4 × 3) cm = 12 cm.

8. The length of a rectangle is 16 cm and each of its diagonals measures 20 cm. Find its breadth?

Solution:
Let PQRS be the given rectangle in which length PQ = 16 cm and diagonal PR = 20 cm.

Since each angle of a rectangle is a right angle, we have
∠PQR = 90°.
From the right ∆PQR, we have
PQ² + QR² = PR² [From Pythagoras’ Theorem]
⇒ QR² = (PR² – PQ²) = {(20)² – (16)²} = (400 – 256) = 144

⇒ QR = √144 = 12 cm.

9. In the below figure, PQRS is a rhombus whose diagonals PR and QS intersect at a point O. If side PQ = 20 cm and diagonal QS = 32 cm, find the length of diagonal PR.

Solution:
We know that the diagonals of a rhombus bisect each other at right angles.
Therefore, QO = ¹/₂QS = (¹/₂ × 32) cm = 16 cm, PQ = 20 cm and ∠POQ = 90°.
From right ∆OPQ, we have PQ² = PO² + QO²
⇒ PO² = (PQ² – QO²) = {(20) ² – (16)²} cm²
= (400 – 256) cm²
= 144 cm²
⇒ PO = √144 cm = 12 cm.

Therefore, PR = 2 × PO = (2 × 12) cm = 24 cm.