Question

two parallel sides of an isosceles trapezium are 16 cm and 24 cm respectively and if the length of its non parallel side is 5 cm and the find the area of the trapezium​

Answer 1

SOLUTION

Draw CF||AD, intersecting AB in F &

CE perpendicular BF.

CF||AD and AF||CD

AFCD is a ||gm

AF= CD= 16cm (opposite sides of ||gm)

AD=CF= 5cm (opposite sides of ||gm)

BF=AB- AF = 24cm- 16cm= 8cm

∆BCF is an isosceles triangle

=) EF= EB

=) 8cm/2

=) 4cm

(In an isosceles ∆ altitude from the vertex bisects the base)

In ∆CFE

CE^2+ FE^2= CF^2

=) CE^2= CF^2- FE^2

=) CE^2= (5)^2- (4)^2

=) CE= √25-16

=)CE= √9

=) CE= 3cm

Area of trapezium

=) 1/2× sum of base× height

=) 1/2× (24+16)× 3

=) 1/2 × 40×3

=) 60cm^2

hope it helps ✔️

Answer 2

Answer:

100 cm² is the correct answer

Given:

two parallel sides of an isosceles trapezium are 16 cm and 24 cm

the length of its non-parallel side is 5 cm

To find:

the area of the trapezium

Solution:

area of trapezium = [(base 1 + base 2 ) / 2 x h]

let the area be Y.

So,

Y= [(16 + 24 ) / 2 x 5]

Y = 40/2 x 5

Y = 100 cm²

Hence, 100 cm² is the correct answer.

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