Question

find the area of isosceles trapezium whose parallel sides are 25 cm and 19 cm and each of the non-parallel sides is 5 cm.​

Answer 1

parallel sides measures:-

25 cm

19cm

Non parallel side measures:-

5cm

To find:-

area of trapezium

SOLUTION:-

height:-

$a^2=b^2+c^2$

$5^2=3^2+c^2$

$25=9+c^2$

$c^2=25-9$

$c=4cm$

height=4cm

_______________

area of trapezium:-

1/2(sum of parallel sides)height

1/2(25+19)4

1/2*44*4

88 sq.cm

Answer 2

Make the trapezium a parallelogram by 1 pair of sides as 19cm and other 5cm.

the rest of the trapezium becomes an isosceles triangle whose equal sides(a)=5cm and base(b) = 25cm-19cm=6cm.

area of this triangle = 1/4 *b (√4a²-b²)

= 1/4 * 6*(√4*5² - 6²)

= 3/2*(√100-36)

=3/2*(√64)

=3/2*8

=12cm²

area = 1/2* base * height

12=(1/2)*6*h

(12*2)/6=height

4 cm = height

height of the triangle=height of the trapezium= 4cm

area of trapezium =1/2*h*(sumof parallel sides)

=1/2*4cm *(19+25)

=2cm*44cm

88cm2