Question

ABCD is a trapezium in which AB parallel to DC,AB is 29 cm , DC is 20m ,AD is 15,BC is 12cm find the area of trapezium

Answer 1

Hope this will help you

Answer :- 294 cm^2

Answer 2

Area of trapezium is $294\ cm^{2}$

Step-by-step explanation:

Given,

$ABCD$ is a trapezium in which $AB \parallel DC$,$AB=29\ cm,DC=20\ cm,AD=15\ cm\ and\ BC=12\ cm$

Draw a parallel line with $BC$ through point $D$ as shown in figure.

In $\triangle ADE$,

Area of $\triangle ADE=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{(a+b+c)}{2}$

∴ $s=\frac{(15+12+9)}{2}$

⇒$s=\frac{(36)}{2}$

⇒$s=18$

∴ Area of $\triangle ADE=\sqrt{18(18-15)(18-12)(18-9)}$

                                      $=\sqrt{18\times 3\times 6\times 9}$

                                      $=\sqrt{2916}$

                                      $=54\ cm^{2}$

Also known Area of $\triangle ADE=\frac{1}{2}\times b\times h$ where $b=Base\ and\ h=Height$

In figure $b=Base=AE=9\ cm$

∴ $54=\frac{1}{2}\times 9\times h$

⇒$9\times h=108$

⇒$h=12\ cm$

∴ Height of the trapezium is $12 \ cm$

∴ Area of trapezium $ABCD=\frac{AB+DC}{2}\times h$

                                            $=\frac{29+20}{2}\times 12=294\ cm^{2}$  

So, Area of trapezium is $294\ cm^{2}$