Class-8 Parallelogram
Introduction to Parallelogram
A quadrilateral having both pairs of opposite sides are parallel is known as parallelogram.
In the above figure, ABCD is a quadrilateral in which AB || DC and AD || BC. Hence ABCD is a parallelogram.
Properties of Parallelogram
- Opposite sides are parallel i.e., AB || DC and AD || BC
- Opposite sides are equal i.e., AB = CD and AD = BC
- Opposite angles are equal i.e., ∠A = ∠C and ∠B = ∠D
- Adjacent angles are supplementary i.e., ∠A + ∠B = 180°, ∠B + ∠C = 180°, ∠C + ∠D = 180°, ∠D + ∠A = 180°
- Each diagonal bisects the parallelogram into congruent triangles.
Example 1. In a parallelogram ABCD, if ∠A = 80°, then find ∠B, ∠C, and ∠D.
Solution. As we know adjacent angles are supplementary of a parallelogram
∠A + ∠B = 180°
=> 80° + ∠B = 180°
=> ∠B = 180° − 80°
=> ∠B = 100°
Since the opposite angles of a parallelogram is equal, we have
∠A = ∠C = 80° and ∠B = ∠D = 100°
Hence, ∠B = 100°, ∠C = 80° and ∠D = 100°.
Example 2. Two adjacent angles of a parallelogram are as 4 : 5. Find the measure of each of its angles.
Solution. Let ABCD be the given parallelogram.
Then ∠ and ∠ are its adjacent angles.
Let's assume ∠A = (4a)° and ∠B = (5a)°
As we know adjacent angles are supplementary of a parallelogram.
∠A + ∠B = 180°
=> 4a + 5a = 180°
=> 9a = 180°
=> a = 180°⁄9
=> a = 20°
∠A = 4 × 20° = 80°
∠B = 5 × 20° = 100°
Since the opposite angles of a parallelogram is equal, we have
∠A = ∠C = 80°
∠B = ∠D = 100°
Hence, ∠A = 80°, ∠B = 100°, ∠C = 80°, and ∠D = 100°
Example 3. In the below given figure, ABCD is a parallelogram in which ∠BAD = 65° and ∠CBD = 85°. Calculate ∠ADB and ∠CDB.
Solution. As we know opposite angles of a parallelogram is equal.
∠BAD = ∠BCD = 65°
Adjacent angles of a parallelogram are supplementary.
∠BAD + ∠ABC = 180°
=> ∠BAD + ∠ABD + ∠CBD = 180°
=> 65° + ∠ABD + 85° = 180°
=> ∠ABD + 150° = 180°
=> ∠ABD = 180° − 150°
=> ∠ABD = 30°
Addition of all the angles of a triangle is 180°.
∠BAD + ∠ABD + ∠ADB = 180°
=> 65° + 30° + ∠ADB = 180°
=> ∠ADB + 95° = 180°
=> ∠ADB = 180° − 95°
=> ∠ADB = 85°
∠ADC = ∠ABC = 85° + 30°
∠ADC = 115°
=> ∠ADB + ∠CDB = 115°
=> 85° + ∠CDB = 115°
=> ∠CDB = 115° − 85°
=> ∠CDB = 30°
Hence, ∠ADB = 85° and ∠CDB = 30°
Example 4. In the below given figure, ABCD is a parallelogram in which ∠CAD = 45°, ∠BAC = 30° and ∠COD = 120°. Find the value of ∠ABD, ∠BDC, ∠ACB and ∠CBD.
Solution. As we know vertical opposite angles are equal.
∠COD = ∠AOB = 120°
Sum of all angles of a triangle is equal to 180°.
∠AOB + ∠BAO + ∠ABO = 180°
=> 120° + 30° + ∠ABO = 180°
=> 150° + ∠ABO = 180°
=> ∠ABO = 180° − 150°
=> ∠ABO = 30°
∠ABO = ∠ABD = 30°
AB || DC and BD is a transversal.
Hence, ∠BDC = ∠ABD = 30°
AD || BC and AC is a transversal.
Hence, ∠ACB = CAD = 45°
Opposite angles of a parallelogram are equal.
Hence, ∠BCD = ∠BAD = 45° + 30° = 75°
Sum of all angles of a triangle is equal to 180°.
In triangle BCD, ∠CBD + ∠BDC + ∠BCD = 180°.
=> ∠CBD + 30° + 75° = 180°
=> ∠CBD + 105° = 180°
=> ∠CBD = 180° − 105°
=> ∠CBD = 75°
Hence, ∠ABD = 30°, ∠BDC = 30°, ∠ACB = 45° and ∠CBD = 75°.
Class-8 Parallelogram MCQ
Class-8 Parallelogram Worksheets
Answer Sheet
Parallelogram-AnswerDownload the pdf
Copyright © 2024 LetsPlayMaths.com. All Rights Reserved.
Email: feedback@letsplaymaths.com