Class-8 Trapezium
Properties of Isosceles Trapezium
Introduction to Trapezium
A quadrilateral in which one pair of opposite sides is parallel is known as trapezium.
In the above figure, ABCD is a quadrilateral in which AB || CD. Hence ABCD is a trapezium. Here AD and BC are not parallel to each other and known as oblique sides
Properties of Trapezium
- One pair of opposite sides is parallel to each other
- Two pair of adjacent angles are supplementary, i.e., ∠A + ∠D = 180° and ∠B + ∠C = 180°
Isosceles Trapezium
If the two non-parallel sides of a trapezium are equal, then it is known as isosceles trapezium.
Properties of an Isosceles Trapezium
In an isosceles trapezium ABCD,
- One pair of opposite sides is parallel i.e., AB || CD.
- Two pair of adjacent angles are supplementary, i.e., ∠A + ∠D = 180° and ∠B + ∠C = 180°.
- Angles on the same base are equal, i.e., ∠A = ∠B and ∠C = ∠D.
- Diagonals are equal in length, i.e., AC = BD.
Example 1. Find the perimeter of the below given isosceles trapezium.
Solution. In an isosceles trapezium two non-parallel sides are equal i.e., AD = BC = 4 cm
Perimeter of the ABCD isosceles trapezium = AB + BC + CD + AD
= 7 cm + 4 cm + 5 cm + 4 cm
= 20 cm.
Hence, perimeter of ABCD isosceles trapezium is 20 cm.
Example 2. ABCD is an isosceles trapezium, find the value of ∠A and ∠C.
Solution. Isosceles trapezium angles on the same base are equal, i.e., ∠A = ∠B = 70° and ∠C = ∠D = 110°.
Hence, ∠A = 70° and ∠C = 110°.
Example 3. ABCD is an isosceles trapezium, find the value of ∠B, ∠C and ∠D.
Solution. In an isosceles trapezium two pair of adjacent angles are supplementary, i.e., ∠A + ∠D = 180° and ∠B + ∠C = 180°.
∠A + ∠D = 180°
=> 75° + ∠D = 180°
=> ∠D = 180° − 75°
=> ∠D = 105°
As we know isosceles trapezium angles on the same base are equal, i.e., ∠A = ∠B and ∠C = ∠D.
∠A = ∠B = 75° ∠C = ∠D = 105°
Hence, ∠B = 75°
Example 4. Find the value of p in the below given trapezium ABCD.
Solution. In the given ABCD trapezium ∠A = (p − 20)° and ∠D = (p + 40)°.
As we know, in a trapezium, the angles on either sides of the base are supplementary.
∠A + ∠D = 180°
=> (p − 20) + (p + 40) = 180
=> 2p + 20 = 180
=> 2p = 180 − 20
=> 2p = 160
=> p = 160⁄2
=> p = 80
Hence, the value of p is equal to 80.
Example 5. ABCD is a trapezium such that AB || CD, ∠A : ∠D = 2 : 1 and ∠B : ∠C = 7 : 5. Find the measure of all the angles of the trapezium.
Solution. Let's assume ∠A = 2p and ∠D = p.
As we know ∠A and ∠D are supplementary.
∠A + ∠D = 180°
=> 2p + p = 180°
=> 3p = 180°
=> p = 180°⁄3
=> p = 60°
∠A = 2p = 2 × 60° = 120°
∠D = 60°
Let's assume ∠B = 7q and ∠C = 5q are supplementary.
Similarly, ∠B and ∠C are supplementary.
∠B + ∠C = 180°
=> 7q + 5q = 180°
=> 12q = 180°
=> q = 180°⁄12
=> q = 15°
∠B = 7q = 7 × 15° = 105°
∠C = 5q = 5 × 15° = 75°
Hence, ∠A = 120°, ∠B = 105°, ∠C = 75° and ∠D = 60°.
Example 6. In trapezium ABCD, DO and CO are bisectors of ∠D and ∠C respectively. Find ∠A and ∠B.
Solution. ∠ODC = ∠ODA = 30° and ∠OCD = ∠OCB = 30°
So, ∠D = 60° and ∠C = 60°.
Since, ABCD is a trapezium,
∠A + ∠D = 180° and ∠B + ∠C = 180°
=> ∠A + 60° = 180° => ∠B + 60° = 180°
=> ∠A = 180° − 60° => ∠B = 180° − 60°
=> ∠A = 120° => ∠B = 120°
Hence, ∠A = 120° and ∠B = 120°.
Class-8 Trapezium MCQ
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