Question

find the area of a trapezium when the sum of the length if whose base is 60cm and whose area is 600cm

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Answer 1

Given:-

Sum of length of bases=60 cm

Area of trapezium=600 cm

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Find:-

The height of the trapezium

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Solve by using formula given below⬇️

Area of trapezium=$\dfrac{1}{2}$×Sum of parallel sides×height

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Solution:-

600=$\dfrac{1}{2}$×60×h

600×2=60 h

∴1200=60 h

∴h=$\dfrac{1200}{60}$

∴h=$\dfrac{120}{6}$

∴h=$\dfrac{20}{1}$

∴h=20 cm..

Hence height of trapezium is 20 cm ✔️

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$\sf \pink{Learn \ More}$

Remember that sum of two adjecent sides is equal to 180°

In trapezium only opposite sides are parallel and equal..

Answer 2

Answer:

Correct Question :

Find the height of a trapezium when the sum of the length if whose base is 60cm and whose area is 600 cm².

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

Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf y\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 60\ cm  }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 20\ cm  }\end{picture}$

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

Given :

• ↠ Base of trapezium = 60cm
• ↠ Area of trapezium = 600 cm²

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

To Find :

• ↠ Height of trapezium

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

Concept :

★ Here the concept of Area of Trapezium has been used. We are given that base of trapezium is 60 cm and area of trapezium is 600 cm.We need to find the height of trapezium.

★ So,We'll find the height of trapezium by insert the values in the required  formula.

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

Using Formula :

$\bigstar{\underline{\boxed{\bf{\red{Area_{(Trapezium)}\: =\dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times height}}}}}$

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

Solution :

$\red\bigstar$ Here

• ↠ Area of trapezium = 600 cm
• ↠ Sum of parallel side (base) = 60 cm

$\red\bigstar$ Finding the height of trapezium.

${\dashrightarrow{\pmb{\sf{Area_{(Trapezium)}\: =\dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times height}}}}$

• Substuting the values

${\dashrightarrow{\sf{600 =\dfrac{1}{2} \times60 \times height}}}$

${\dashrightarrow{\sf{600 =\dfrac{1 \times 60}{2} \times height}}}$

${\dashrightarrow{\sf{600 =\dfrac{60}{2} \times height}}}$

${\dashrightarrow{\sf{600 = \cancel\dfrac{60}{2} \times height}}}$

${\dashrightarrow{\sf{600 =30 \times height}}}$

${\dashrightarrow{\sf{Height_{(Trapezium)} = \dfrac{600}{30} }}}$

${\dashrightarrow{\sf{{Height_{(Trapezium)}=\cancel\dfrac{600}{30}}}}}$

${\dashrightarrow{\sf{Height_{(Trapezium)}\: = 20 \: cm }}}$

${\bigstar{\underline{\boxed{\bf{\purple{Height_{(Trapezium)}= 20 \: cm }}}}}}$

∴ Height of trapezium is 20 cm.

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

Verification :

$\red\bigstar$ Let's check our answer

${\dashrightarrow{\pmb{\sf{Area_{(Trapezium)}\: =\dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times height}}}}$

• Substuting the values

${\dashrightarrow{\sf{600 \: {cm}^{2} =\dfrac{1}{2} \times (60) \times 20}}}$

${\dashrightarrow{\sf{600 \: {cm}^{2} =\dfrac{1 \times 60 \times 20}{2}}}}$

${\dashrightarrow{\sf{600 \: {cm}^{2} =\dfrac{1200}{2}}}}$

${\dashrightarrow{\sf{600 \: {cm}^{2} = \cancel\dfrac{1200}{2}}}}$

${\dashrightarrow{\sf{600 \: {cm}^{2} =600 \: {cm}^{2} }}}$

${\bigstar{\underline{\boxed{\bf{\purple{LHS = RHS }}}}}}$

∴ Hence Verified!.

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

Learn More :

$\red\bigstar$ Formulas of area

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

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

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