Question

Q.4) The parallel sides of a trapezium are 30cm and 20cm. The distance between
them is 15cm. Find the area of the trapezium.​

Answer 1

Question:

The parallel sides of a trapezium are 30cm and 20cm. The distance between

them is 15cm. Find the area of the trapezium.

Answer:

1st parallel side = 30 cm

2nd parallel side = 20 cm

Height = 15 cm

$area \: of \: trapezium \: = \frac{1}{2} \times height \times sum \: of \: parallel \: sides$

Substitute the values…

$= \frac{1}{2} \times 15 \times (30 + 20)$

$= \frac{1}{2} \times 15 \times 50$

$= \frac{750}{2}$

$= 375$

Hence area of the trapezium is 375 cm²…

Answer 2

$\boxed{\huge{\bf{\star{Correct\:question \:-:}}}}$

• The parallel sides of a trapezium are 30cm and 20cm. The distance between
• them is 15cm. Find the area of the trapezium.

AnswEr-:

• $\underline{\boxed{\star{\sf{\blue{ \:\:Area\:of\:Trapezium \:=375cm^{2} }}}}}$

EXPLANATION-:

• $\frak{Given \:\: -:} \begin{cases} \sf{The\:parallel\:sides\:of\:Trapezium \:\:is\:= \frak{30cm\;and\:20cm.}} & \\\\ \sf{Height\:or\:Distance \:Between\:Parallel \:Sides\:of\:Trapezium \:is \:=\:\frak{15cm}}\end{cases}$

• $\frak{To \:Find\: -:} \begin{cases} \sf{The\:Area\:\:of\:Trapezium \:\:}\end{cases}$

$\dag{\sf{\large { Solution:\:}}}$

• $\underline{\boxed{\star{\sf{\blue{ \:\:Area\:of\:Trapezium \:= \left( \dfrac{1}{2}\times Height \times (a+b) \:or \: Sum\: Parallel \:Sides \: \right) square \:units }}}}}$

• $\frak{Here \:\: -:} \begin{cases} \sf{(a \:and\:b) = The\:parallel\:sides\:of\:Trapezium \:\:is\:= \frak{30cm\;and\:20cm.}} & \\\\ \sf{H =Height\:or\:Distance \:Between\:Parallel \:Sides\:of\:Trapezium \:is \:=\:\frak{15cm}}\end{cases}$

$\dag{\sf{\large { Now -:\:}}}$

• $\implies{\sf{\large { \:\:\dfrac{1}{2} \times 15 \times (30 + 20) \: }}}$
• $\implies{\sf{\large { \:\:\dfrac{1}{2} \times 15 \times (50) \: }}}$
• $\implies{\sf{\large { \:\: 15 \times 25 \: }}}$
• $\implies{\sf{\large { \:\:375cm^{2} \: }}}$

$\dag{\sf{\large { Hence -:\:}}}$

• $\underline{\boxed{\star{\sf{\blue{ \:\:Area\:of\:Trapezium \:=375cm^{2} }}}}}$

♡ Figure of Trapezium-:

• $\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 15\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 30\ cm  }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 20\ cm  }\end{picture}$

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♡ More To Know -:

• Formulas of area :

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

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