Question

the parallel sides of a trapezium are 25cm and 10cm, while it's non parallel sides are 14cm and 13cm. find the area of the trapezium​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that ABCD is a trapezium with AB || CD and AD and BC are non parallel sides such that AB = 25 cm, CD = 10 cm, AD = 13 cm and BC = 14 cm.

Construction :- From C, draw a line CE || AD intersecting AB at E. And from C, draw CL perpendicular to AB.

Now,

AE || CD and CE || AD $\rm\implies \:$ AECD is a parallelogram.

So, AE = DC = 10 cm and AD = CE = 13 cm.

So, BE = AB - AE = 25 - 10 = 15 cm.

Now, In $\triangle$ BCE

We have, BC = 14 cm, BE = 15 cm and CE = 13 cm

So, Semi-perimeter, s = $\rm \: \dfrac{BE+EC+BC}{2}$

$\rm \: s = \dfrac{15 + 14 + 13}{2}$

$\rm \: s = \dfrac{42}{2}$

$\bf\implies \:s \: = \: 21 \: cm$

Now, Area of $\rm \: \triangle BCE$ using Heron's Formula is evaluated as

$\rm \: ar(\triangle BCE) = \sqrt{s(s - BE)(s - BC)(s - CE)}$

$\rm \: = \: \sqrt{21(21 - 15)(21 - 14)(21 - 13)}$

$\rm \: = \: \sqrt{21 \times 6 \times 7 \times 8}$

$\rm \: = \: \sqrt{7 \times 3 \times 3 \times 2 \times 7 \times 2 \times 2 \times 2}$

$\rm \: = \: 7 \times 3 \times 2 \times 2$

$\rm \: = \: 84 \: {cm}^{2}$

So,

$\bf\implies \:ar(\triangle BCE) = 84 \: {cm}^{2}$

Now, Also we know that

$\rm \: ar(\triangle BCE) = \dfrac{1}{2} \times BE \times CL$

$\rm \: 84 = \dfrac{1}{2} \times 15 \times CL$

$\bf\implies \:CL = \dfrac{56}{5} \: cm$

Now, we know that area of trapezium whose parallel sides are a units and b units and distance between parallel sides be h units is given by

$\boxed{ \rm{ \:Area_{(Trapezium)} = \frac{1}{2} \times (a + b) \times h \: }}$

So,

$\rm \: Area_{(Trapezium)} \: = \: \dfrac{1}{2} \times (AB + CD) \times CL$

$\rm \: = \: \dfrac{1}{2} \times (25 + 10) \times \dfrac{56}{5}$

$\rm \: = \: 35 \times \dfrac{28}{5}$

$\rm \: = \: \: 28 \times 7$

$\rm \: = \: \: 196 \: {cm}^{2}$

So,

$\rm\implies \:\boxed{ \bf{ \:Area_{(Trapezium)} = \: \: 196 \: {cm}^{2} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$