Question

67) Two parallel sides of an isosceles trapezium are 16 cm and 24
cm respectively. If the lengths of each non-parallel side is 5 cm,
find the area of the trapezium.

please help​

Answer 1

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Answer 2

Answer:

60 cm sq.

Step-by-step explanation:

CF ll AD and AF ll CD

AFCD is a parallelogram.

AF = CD (opposite sides of parallelogram)

=16 cm

AD =CF (opposite sides of parallelogram)

=5 cm

BCF is an isosceles triangle.

FB = AB - AF

=24 cm -16 cm

=8 cm

FE = 8/2

= 4 cm

FB = FE

= 4 cm

In triangle CFE---

CF sq.= CE sq. - FE sq.

5 sq. = CE sq. - 4sq.

CE sq.= 5 sq. - 4 sq.

ce = \sqrt{25 - 16}ce=

25−16

ce = \sqrt{ 9}ce=

9

ce = 3 cmce=3cm

area of trapezium=1/2 × height × (sum of parallel sides)

=1/2 × 3 × (24 + 16)

=1/2 × 3 × 40

=60 cm sq.

This is the correct answer...