Question

Question on photo is given

Answer 1

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

In trapezium PQRS, it is given that, Length of parallel sides are

PQ = 12 cm

RS = 8 cm

and

Distance between PQ and RS is 5.5 cm.

We know,

Area of trapezium whose parallel sides are a units and b units respectively and distance between them is h units is given by

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

Here, we have

$\rm \: a \: = \: 12 \: cm$

$\rm \: b \: = \: 8 \: cm$

$\rm \: h \: = \: 5.5 \: cm$

So, on substituting the values, we get

$\rm \: Area_{(PQRS)} = \dfrac{1}{2}(12 + 8) \times 5.5$

$\rm \: Area_{(PQRS)} = \dfrac{1}{2}(20) \times 5.5$

$\rm \: Area_{(PQRS)} = 10 \times 5.5$

$\rm\implies \:Area_{(PQRS)} \: = \: 55 \: {cm}^{2}$

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

In trapezium PQRS, it is given that, Length of parallel sides are

PQ = 21 m

RS = 12 m

and

Distance between PQ and RS is 14 m.

We know,

Area of trapezium whose parallel sides are a units and b units respectively and distance between them is h units is given by

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

Here, we have

$\rm \: a \: = \: 21 \: m$

$\rm \: b \: = \: 12 \: m$

$\rm \: h \: = \: 14 \: m$

So, on substituting the values, we get

$\rm \: Area_{(PQRS)} = \dfrac{1}{2}(21 + 12) \times 14$

$\rm \: Area_{(PQRS)} = 33 \times 7$

$\rm\implies \:Area_{(PQRS)} \: = \: 231 \: {m}^{2}$

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

Additional Information :-

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

Answer 2

Trapezium

A Trapezium is similar to a quadrilateral which has two parallel sides of unequal length and two non-parallel sides.

We know that, Area of a Trapezium,

${\boxed{\bullet\:{Area\:=\:{\dfrac{1}{2}}\:(a\:+\:b)\:\times\:h}}}$

where,

• $a$ and $b$ are two non-parallel sides
• $h$ is the height

As we went through the concept about Trapezium, Now, Let's move on finding the solution for our question.

A) In Trapezium PQRS, We have been given that,

• $\sf{a\:=\:8\:cm}$
• $\sf{b\:=\:12\:cm}$
• $\sf{h\:=\:5.5\:cm}$

Substituting values in Formula, we get,

$\sf\implies{{\dfrac{1}{2}}\:(a\:+\:b)\:\times\:h}$

$\sf\implies{{\dfrac{1}{2}}\:(8\:+\:12)\:\times\:5.5}$

$\sf\implies{{\dfrac{1}{2}}\:\times\:20\:\times\:5.5}$

$\sf\implies{{\dfrac{1}{\cancel{{2}}}}\:\times\:{\cancel{20}}\:\times\:5.5}$

$\sf\implies{10\:\times\:5.5}$

$\sf\implies{55\:cm^2}$

B) In Trapezium PQRS, We have been given that,

• $\sf{a\:=\:12\:cm}$
• $\sf{b\:=\:21\:cm}$
• $\sf{h\:=\:14\:cm}$

Substituting values in Formula, we get,

$\sf\implies{{\dfrac{1}{2}}\:(a\:+\:b)\:\times\:h}$

$\sf\implies{{\dfrac{1}{2}}\:(12\:+\:21)\:\times\:14}$

$\sf\implies{{\dfrac{1}{2}}\:\times\:33\:\times\:14}$

$\sf\implies{{\dfrac{1}{\cancel{{2}}}}\:\times\:33\:\times\:{\cancel{14}}}$

$\sf\implies{33\:\times\:7}$

$\sf\implies{231\:cm^2}$

Hence, The Area of the Trapezium A is 55 cm² and Trapezium B is 231 cm².