Question

please answer these questions fast​

Answer 1

$\huge\boxed{\boxed{\blue{\sf{Solution-}}}}$

If we have to find the height of the isosceles trapezium

find height of ∆ADE

In trapezium ABCD

AE=FB

3 cm=3cm

AD= 5cm

By the Pythagoras theorm we find the height of ∆ADE

$\sf DE=\sqrt{AE{}^{2}-AD{}^{2}}\\ \sf DE =\sqrt{5{}^{2}-3{}^{2}}\\ \sf DE= \sqrt{25-9}\\ \sf DE=\sqrt{16}\\ \sf DE= 4cm$

The height of ∆ABE is equal to the height of trapezium ABCD

$\sf\implies 4cm=4cm$

The height of isosceles trapezium is 4cm

Check the attachment also !! !

$\boxed{\blue{\sf{Height=4cm}}}$