Question

Find the area of Trapezium whose parallel sides are 25cm,13 cm and other sides are 15cm each​

Answer 1

Answer:

285cm^2

Step-by-step explanation:

area of trapezium

=1/2×h×(sum of parallel sides) =1/2×15(25+13)

=15/2×38

=15×19

=285cm^2

Answer 2

$\mathfrak{\huge{\underline{Answer:}}}$

Let ABCD be given Trapezium in which

AB = 25cm , DC = 13cm , BC = 15cm, AD = 15cm

Construction :

Through C, draw CF parallel to AD

Also, draw CE perpendicular to AB

Now,

FB = AB - AF

FB = 25 - 13

FB = 12 cm

In triangle FBC, FC = BC = 15cm

So, it's an isosceles triangle.

Also, CE is perpendicular to FB.

So, E is the mid-point of FB.

FE = 1/2 FB

FE = 1/2 × 12 cm

FE = 6 cm

In right angled triangle CEF,

[By Pythagoras Theorem]

$( {cf)}^{2} = ({fe})^{2} + (ce )^{2}$

${(15})^{2} = ( {6})^{2} + ({ce)}^{2}$

${(ce)}^{2} = 225 - 36$

${(ce)}^{2} = 189$

$ce = 3 \sqrt{21}$

Area of Trapezium ABCD

$\frac{1}{2} (ab + dc) \times ce$

$\frac{1}{2} (25 + 13) \times 3 \sqrt{21}$

$19 \times 3 \sqrt{21}$

$57 \sqrt{21}$