2021-11-13

Dare2solve | Area of isosceles trapezium where length of diagonal is given

This question was uploaded on 12/11/21 on social media accounts.



Solution:

This area can be divided into two parts (Upper and lower triangle):
Area of the upper triangle = 6a/2 = 3a,
Area of the upper triangle = 6b/2 = 3b,
Total area = 3(a+b),
In the above figure:
\[(a+b)^2+1^2=6^2\]
\[\Rightarrow (a+b)=\sqrt{35}\]
\[\Rightarrow Area=3\sqrt{35}\]

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