How to find the length of diagonal of an Isosceles Trapezium?
In SSC CGL Tier-2 2015, A question based on Trapezium was asked where it was required to find the Diagonal of an Isosceles Trapezium. Many students mailed us the same question and requested to provide the concept behind that and also the solution. In this post, We are providing you the concept along with few questions based on that. At the end, We are providing the same question. I hope you would be able to solve that question.
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Anyway, let's move ahead. Have a look on below given image. It's an Isosceles Trapezium having equal sides AC & BD and parallel sides AB & CD. Let the length of AB be X and the length of CD be Y where Y>X. BE is the height of the trapezium. You need to find the length of BD. So, How will you find?
According to formula,
- ED = (Y - X)/2 .
- CE = CD- ED or (Y - ED)
Triangle CEB is a Right Angle traingle where BE is the height and CB is the hypotenius. By Applying pythagores theorem, we can find the length of diagonal BC i.e
- BC^{2} = BE^{2} + CE^{2}
or
- BC = √ BE^{2} + CE^{2}
Sample Question with Solution
Q1. The length of parallel sides of an Isosceles trapezium are 12 cm and 6 cm and the height is 4 cm. The length of the diagonal (in cm) is:
- √95
- √97
- 28
- 26
Sol - It's given in the question that the length of parallel sides are 12 and 6 cm. It means,
- AB = 6 cm and
- CD = 12 cm
Now, Applying the formula, we get
Applying pythagores theorem, we get
- BC^{2} = BE^{2} + CE^{2}
- BC = √ BE^{2} + CE^{2}
- BC = √ 9^{2} + 4^{2} = √97
Questions for Practice- Can you solve this?
Q1. The area of an isosceles trapezium is 176 square cm and the height is 2/11th of the sum of its parallel sides. If the ratio of the length of the parallel sides is 4:7, then the length of a diagonal (in cm) is (Asked in SSC CGL Tier-2 2015)
- √137
- 24
- 2 √137
- 28
What's your answer? Comment! .
If you are still not able to solve this question, Inform us via comments. We will provide you the solution as well.
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