Question

find the area of a trapezium if it's height 8cm and the lengths of parallel sides are 11cm and 17cm​

Answer 1

$\green{\large\underline{\sf{Given- }}}$

• The length of parallel sides of a trapezium are 11 cm and 17 cm.

• Distance between parallel lines is 8 cm

$\purple{\large\underline{\sf{To\:Find - }}}$

• Area of trapezium

$\pink{\begin{gathered}\large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}}$

The area of trapezium whose length of parallel sides are a units and b units respectively and distance between parallel lines be h units is

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

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

Given that,

• The length of parallel sides of a trapezium are 11 cm and 17 cm.

• Distance between parallel lines is 8 cm

So, here

• a = 11 cm

• b = 17 cm

• h = 8 cm

So,

$\rm :\longmapsto\:Area_{(trapezium)} = \dfrac{1}{2} \times (a + b) \times h$

$\rm :\longmapsto\:Area_{(trapezium)} = \dfrac{1}{2} \times (11 + 17) \times 8$

$\rm :\longmapsto\:Area_{(trapezium)} = (28) \times 4$

$\rm :\longmapsto\:Area_{(trapezium)} = 112 \: {cm}^{2}$

Hence,

$\\ \rm :\longmapsto\:\boxed{\tt{ \: \: \: Area_{(trapezium)} = 112 \: {cm}^{2} \: \: \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

Answer 2

$answers \: in \: the \: attachment$