Question

The length of 2 parallel sides of a trapezium are 10 cm and 15 cm and the distance between the parallel lines is 16 cm.then its area is

Answer 1

$\huge\bf{\underline{Solution:-}}$

Given :-

•Parallel sides of trapezium 10cm and 15cm

•Distance b/w them or (Height)= 16cm

To find

• The area of trapezium

Now

A.T.Q :-

$\boxed{\sf{Area\:of\: trapezium=\dfrac{1}{2}\times (sum\:of\: parallel\:sides)\times height}}$

$\sf\implies Area=\dfrac{1}{2}\times (10+15)\times 16\\ \\ \sf\implies Area=\dfrac{1}{\cancel2}\times 25\times \cancel{16}\\ \\ \sf\implies Area=25\times 8\\ \\ \sf\implies Area\:of\: trapezium=200cm{}^{2}$

$\sf{\underline{\dag{\:\:Area\:of\: trapezium=200cm{}^{2}}}}$

Answer 2

AnswEr :

200cm².

$\bf{\green{\underline{\underline{\bf{Given\::}}}}}$

The length of two parallel sides of a trapezium are 10 cm and 15 cm and the distance between the parallel lines is 16 cm.

$\bf{\green{\underline{\underline{\bf{To\:find\::}}}}}$

The area of trapezium.

$\bf{\green{\underline{\underline{\bf{Explanation\::}}}}}$

We know that the formula of the area of trapezium :

$\bf{\boxed{\bf{Area\:of\:trap.=\frac{1}{2} \times (sum\:of\:base)\times height}}}}}$

$\bf{We\:have}\begin{cases}\sf{1st\:side\:of\:the\:parallel\:(a_{1})=10cm}\\ \sf{2nd\:side\:of\:the\:parallel\:(b_{2})=15cm}\\ \sf{Height\:of\:trap.(h)=16cm}\end{cases}}$

A/q

$\mapsto\sf{Area\:of\:trap.=\dfrac{1}{\cancel{2}} \times (10+15)\times \cancel{16}}\\\\\\\mapsto\sf{Area\:of\:trap.=(25\times 8)\:cm^{2} }\\\\\\\mapsto\sf{\red{Area\:of\:trap.=200\:cm^{2} }}$

Thus;

$\underbrace{\sf{The\:area\:of\:the\:trapezium\:=\:200\:cm^{2} }}}}}$