Question
Easy
The distance between the parallel sides AB and CD of a trapezium ABCD is 12 cm. If AB<CD, CD=24 cm and AD=BC=13 cm, then area of the trapezium (in cm$^2$) is :
1
228
2
123.5
3
114
4
247
Question Details
Time to Solve: 12
Exam: CTET
Level/Paper: CTET_P2
Chapter: 2D Geometry
Topic: Trapeziums
Correct Answer
Option A
Explanation
To determine the area of the trapezium ABCD, we need to use the formula for the area of a trapezium, which is given by: \[ \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \] In this problem, the parallel sides (bases) of the trapezium are AB and CD. We are given the following information: - AB < CD - CD = 24 cm - The height (distance between theтАжRead More
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