Question

Find the area of a trapezium which has a height 20 cm and its parallel sides are 7cm and 8 cm long

Answer 1

Answer:

150cmsq

Step-by-step explanation:

area of trapezium is

$\frac{1}{2} \times height \times (sum \: of \: parallel \: sides)$

$\frac{1}{2} \times 20(7 + 8)$

10*15=150

therefore area of trapezium is 150cmsquare

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Answer 2

Answer:

Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf }\put(3,2.4){ \bf }\put(-0.3,-0.3){ \bf }\put(4,-0.3){ \bf }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 20\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 8\ cm }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 7\ cm  }\end{picture}$

$\begin{gathered}\end{gathered}$

Given :

• » Parallel sides to trapezium = 7 cm and 8 cm
• » Height of trapezium = 20 cm

$\begin{gathered}\end{gathered}$

To Find :

• » Area of Trapezium

$\begin{gathered}\end{gathered}$

Concept :

$\small\bigstar$ Here the concept of area of trapezium has been used. We are given that parallel sides of trapezium are 7cm and 8cm, height of trapezium is 20cm. We need to find the Area of trapezium.

$\small\bigstar$ So, we'll find the area of trapezium by insert the values in the required  formula.

Using Formula :

$\normalsize\bigstar{\underline{\boxed{\sf{\pink{Area  \: of \:  trapezium =  \dfrac{1}{2}\bigg(a + b \bigg)h}}}}}$

$\begin{gathered}\end{gathered}$

Solution :

$\small\bigstar$ Here, Let the :-

• a = 7cm
• b = 8cm
• h = 20cm

$\rule{200}2$

$\small\bigstar$ Now, finding the area of trapezium by substituting the values in the formula :-

$\begin{gathered}\small{\dashrightarrow{\sf{Area_{(Trapezium)}=  \dfrac{1}{2}\bigg(a + b \bigg)h}}} \\ \end{gathered}$

$\begin{gathered}\small{\dashrightarrow{\sf{Area_{(Trapezium)}=  \dfrac{1}{2}\bigg( \: 7 + 8 \: \bigg)20}}} \\ \end{gathered}$

$\begin{gathered}\small{\dashrightarrow{\sf{Area_{(Trapezium)}=  \dfrac{1}{2}\bigg( \: \: 15 \: \: \bigg)20}}} \\ \end{gathered}$

$\begin{gathered}\small{\dashrightarrow{\sf{Area_{(Trapezium)}=  \dfrac{1}{2} \times 15 \times 20}}} \\ \end{gathered}$

$\begin{gathered}\small{\dashrightarrow{\sf{Area_{(Trapezium)}=  \dfrac{1}{\cancel{2}} \times 15 \times \cancel{20}}}}\\ \end{gathered}$

$\begin{gathered}\small{\dashrightarrow{\sf{Area_{(Trapezium)}=  1\times 15 \times 10}}}\\ \end{gathered}$

$\begin{gathered} \small{\dashrightarrow{\sf{Area_{(Trapezium)}=  15 \times 10}}}\\ \end{gathered}$

$\begin{gathered}\small{\dashrightarrow{\underline{\underline{\sf{Area_{(Trapezium)}= 150 \: {cm}^{2}}}}}}\\ \end{gathered}$

$\begin{gathered} \small\bigstar{\underline{\boxed{\sf{\pink{Area_{(Trapezium)}=  150 \: {cm}^{2}}}}}} \end{gathered}$

$\small\bigstar$ Hence, The area of trapezium is 150 cm².

$\begin{gathered}\end{gathered}$

Learn More :

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

$\begin{gathered}\underline{\rule{200pt}{3pt}} \end{gathered}$