Question

the parallel sides of a trapezium are 25cm and 11cm, while its non parallel sides are 15 cm and 13 cm . Find the area of the trapezium

Answer 1

$\textbf{ \huge{ Hello Mate! }}$

Thanks for your question!

Area of triangle by heron's formula which is;

$= \sqrt{s(s - a)(s - b)(s - c)} \\ s = \frac{a + b + c}{2} \\ s = \frac{15 + 14 + 13}{2} = 21 \: cm \\ = \sqrt{21(21 - 15)(21 - 14)(21 - 13)} \\ = \sqrt{3 \times 7 \times 2 \times 3 \times 7 \times {2}^{3} } \\ = 3 \times 7 \times 2 \times 2 \: {cm}^{2} = 84 \: {cm}^{2}$

Hence, height of triangle will be

$area = \frac{1}{2} \times base \times height \\ 84 \: {cm}^{2} = \frac{1}{2} \times 14 \: cm \times h \\ 12 \: cm = height$

Now, area of trapezium.

$area = \frac{1}{2} (sum \: of || sides)h \\ area = \frac{1}{2} \times (11 + 25)cm \times 12cm \\ = 6cm \times 36cm = 216 \: {cm}^{2}$

$\boxed{ \textsf{ \red{ Area\:of\:triangle\:is\:216\:sq.\:cm }}}$

Problem related to my solution?

1. The line I draw in inside of trapezium is not diagonal but I have drawn it parallel to 15 cm line such that we get one parallelogram and one triangle.

2. Any more question? No problem, leave link in inbox to ask questions or comment below.

Have great future ahead!

Answer 2

HEY MATE HERE IS YOUR ANSWER

Sides are = 13,14,15

Heron's formula =
$\frac{ a+ b + c}{2}$

$= \frac{13 + 14 + 15}{2}$
$= \frac{42}{2}$
$= 21$
To find

$area \: of \: triangle = \sqrt{s(s - a)( s-b )(s -c )}$
$\sqrt{21(21 - 13)(21 - 14)(21 - 15)}$
$21 \times 8 \times 7 \times 6$
Make factors

$3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 3 \times 2$
$3 \times 7 \times 2 \times 2$
$21 \times 4$
$84 {cm}^{2}$

Area of triangle =
$\frac{1}{2} \times base \times height$
$84 = \frac{1}{2} \times 15 \times h$
$h = \frac{84 \times 2}{15}$
$h = \frac{56}{2}$
Area of trapezium =
$\frac{1}{2} (sum \: of \: parallel \: sides)h$
$\frac{1}{2} (36)( \frac{56}{5} )$
Area of trapezium =
$216 {cm}^{2}$

Hope it will help you
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