These are the various types of solved problems on parallelogram.
1. Prove that any two adjacent angles of a parallelogram are supplementary.
Solution:
Let ABCD be a parallelogram
Then, AD ∥ BC and AB is a transversal.
Therefore, A + B = 180° [Since, sum of the interior angles on the same side of the transversal is 180°]
Similarly, ∠B + ∠C = 180°, ∠C + ∠D = 180° and ∠D + ∠A = 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 are as 2 : 3. Find the measure of each of its angles.
Solution:
Let ABCD be a given parallelogram
Then, ∠A and ∠B are its adjacent angles.
Let ∠A = (2x)° and ∠B = (3x)°.
Then, ∠A + ∠B = 180° [Since, sum of adjacent angles of a ∥gm is 180°]
⇒ 2x + 3x = 180
⇒ 5x = 180
⇒ x = 36.
Therefore, ∠A = (2 × 36)° = 72° and ∠B = (3 × 36°) = 108°.
Also, ∠B + ∠C = 180° [Since, ∠B and ∠C are adjacent angles]
= 108° + ∠C = 180° [Since, ∠B = 108°]
∠C = (180°  108°) = 72°.
Also, ∠C + ∠D = 180° [Since, ∠C and ∠D are adjacent angles]
⇒ 72° + ∠D = 180°
⇒ ∠D = (180°  72°) 108°.
Therefore, ∠A = 72°, ∠B = 108°, ∠C = 72°and ∠D = 108°.
3. In the adjoining figure, ABCD is a parallelogram in which ∠A = 75°. Find the measure of each of the angles ∠B, ∠C and ∠D.
Solution:
It is given that ABCD is a parallelogram in which ∠A = 75°.
Since the sum of any two adjacent angles of a parallelogram is 180°,
∠A + ∠B = 180°
⇒ 75° + ∠B = 180°
⇒∠B = (180°  75°) = 105°
Also, ∠B + ∠C = 180° [Since, ∠B and ∠C are adjacent angles]
⇒ 105° + ∠C = 180°
⇒ ∠C = (180°  105°) = 75°.
Further, ∠C + ∠D = 180° [Since, ∠C and ∠D are adjacent angles]
⇒ 75° + ∠D = 180°
⇒ ∠D = (180°  75°) = 105°.
Therefore, ∠B = 105°, ∠C = 75° and ∠D = 105°.
4. In the adjoining figure, ABCD is a parallelogram in which
∠BAD = 75° and ∠DBC = 60°. Calculate:
(i) ∠CDB and (ii) ∠ADB.
Solution:
We know that the opposite angles of a parallelogram are equal.
Therefore, ∠BCD = ∠BAD = 75°.
(i) Now, in ∆ BCD, we have
∠CDB + ∠DBC + ∠BCD = 180°
[Since, sum of the angles of a triangle is 180°]
⇒ ∠CDB + 60° + 75° = 180°
⇒ ∠CDB + 135° = 180°
⇒ ∠CDB = (180°  135°) = 45°.
(ii) AD ∥ BC and BD is the transversal.
Therefore, ∠ADB = ∠DBC = 60° [alternate interior angles]
Hence, ∠ADB = 60°.
5. In the adjoining figure, ABCD is a parallelogram in which
∠CAD = 40°, ∠BAC = 35° and ∠COD = 65°.
Calculate: (i) ∠ABD (ii) ∠BDC (iii) ∠ACB (iv) ∠CBD.
Solution:
(i) ∠AOB = ∠COD = 65° (vertically opposite angles)
Now, in ∆OAB, we have:
∠OAB + ∠ABO + ∠AOB =180° [Since, sum of the angles of a triangle is 180°]
⇒ 35°+ ∠ABO + 65° = 180°
⇒ ∠ABO + 100° = 180°
⇒ ∠ABO = (180°  100°) = 80°
⇒ ∠ABD = ∠ABO = 80°.
(ii) AB ∥ DC and BD is a transversal.
Therefore, ∠BDC = ∠ABD = 80° [alternate interior angles]
Hence, ∠BDC = 80°.
(iii) AD ∥ BC and AC is a transversal.
Therefore, ∠ACB = ∠CAD = 40° [alternate interior angles]
Hence, ∠ACB = 40°.
(iv) ∠BCD = ∠BAD = (35° + 40°) = 75° [opposite angles of a parallelogram]
Now, in ∆CBD, we have
∠BDC + ∠BCD + ∠CBD = 180° [sum of the angles of a triangle is 180°]
⇒ 80° + 75° + ∠CBD = 180°
⇒ 155° + ∠CBD = 180°
⇒ ∠CBD = (180°  155°) = 25°.
Hence, ∠CBD = 25°.
6. In the adjoining figure, ABCD is a parallelogram, AO and BO are the bisectors of ∠A and ∠B respectively. Prove that ∠AOB = 90°.
Solution:
We know that the sum of two adjacent angles of a parallelogram is 180°
Therefore, ∠A + ∠B = 180° ……………. (i)
Since AO and BO are the bisectors of ∠A and ∠B, respectively, we have
∠OAB = 1/2∠A and ∠ABO = 1/2∠B.
From ∆OAB, we have
∠OAB + ∠AOB + ∠ABO = 180° [Since, sum of the angles of a triangle is 180°]
⇒ ¹/₂∠A + ∠ABO + ¹/₂∠B = 180°
⇒ ¹/₂(∠A + ∠B) + ∠AOB = 180°
⇒ (¹/₂ × 180°) + ∠AOB = 180° [using (i)]
⇒ 90° + ∠AOB = 180°
⇒ ∠AOB = (180°  90°) = 90°.
Hence, ∠AOB = 90°.
7. The ratio of two sides of a parallelogram is 4 : 3. If its perimeter is 56 cm, find the
lengths of its sides.
Solution:
Let the lengths of two sides of the parallelogram be 4x cm and 3x cm respectively.
Then, its perimeter = 2(4x + 3x) cm = 8x + 6x = 14x cm.
Therefore, 14x = 56 ⇔ x = ⁵⁶/₁₄ = 4.
Therefore, one side = (4 × 4) cm = 16 cm and other side = (3 × 4) cm = 12 cm.
8. The length of a rectangle is 8 cm and each of its diagonals measures 10 cm. Find
its breadth.
Solution:
Let ABCD be the given rectangle in which length AB = 8 cm and diagonal AC = 10 cm.
Since each angle of a rectangle is a right angle, we have
∠ABC = 90°.
From right ∆ABC, we have
AB² + BC² = AC² [Pythagoras’ Theorem]
⇒ BC² = (AC²  AB²) = {(1O)²  (8)²} = (100  64) = 36
⇒ BC = √36 = 6cm.
Hence, breadth = 6 cm.
9. In the adjacent figure, ABCD is a rhombus whose diagonals AC and BD intersect at a point O. If side AB = 10cm and diagonal BD = 16 cm, find the length of diagonal AC.
Solution:
We know that the diagonals of a rhombus bisect each other at right angles
Therefore, BO = ¹/₂BD = (¹/₂ × 16) cm = 8 cm, AB = 10 cm and ∠AOB = 90°.
From right ∆OAB, we have
AB² = AO² + BO²
⇒ AO² = (AB² – BO²) = {(10) ²  (8)²} cm²
= (100  64) cm²
= 36 cm²
⇒ AO = √36 cm = 6 cm.
Therefore, AC = 2 × AO = (2 × 6) cm = 12 cm.
Parallelogram
Properties of a Rectangle Rhombus and Square
Practice Test on Parallelogram
Parallelogram  Worksheet
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