Question

A trapezium with parallel sides of length as 7:3 is cut from a rectangle 30 DM by 4 DM so as to have an area of one-third the latter. find the lengths of the parallel sides.
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Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

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• A trapezium with parallel sides of length as 7 : 3 is cut from a rectangle 30 dm by 4 dm so as to have an area of one-third the latter. Find the lengths of the parallel sides.

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$\huge{AηsωeR} ✍$

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$\large\underline\blue{\bold{Given:-  }}$

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$\begin{gathered}\begin{gathered}\bf given : - \begin{cases} &\sf{length \: of \: rectangle = 30 \: dm} \\ &\sf{breadth \: of \: rectangle = 4 \: dm}\\ &\sf{parallel \: sides \: of \: trapezium \: in \: 7 : 3} \\ &\sf{area \: of \: trap. \: = \frac{1}{3} \: area \: of \: rectangle }\end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: find :- \begin{cases} &\sf{the \: parallel \: sides \: of \: trapezium} \end{cases}\end{gathered}\end{gathered}$

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$\text{\large\underline{\orange{Formula:-}}}$

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$\begin{gathered}\begin{gathered}\bf \begin{cases} &\sf{Area_{(rectangle)} = length × breadth} \\ &\sf{Area_{(trapezium)} = \frac{1}{2} \times (sum \: of \: parallel \: sides)×height} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

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☆ Dimensions of rectangle

$\begin{gathered}\begin{gathered}\bf \begin{cases} &\sf{length \: of \: rectangle = 30 \: dm} \\ &\sf{breadth \: of \: rectangle \: = 4 \: dm} \end{cases}\end{gathered}\end{gathered}$

$\bf\implies \:Area_{(rectangle)} = length × breadth$

$\sf \:  ⟼Area_{(rectangle)} = 30 \times 4 = 120 \: {dm}^{2}$

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☆ Dimensions of trapezium.

$\bf\red{According\:to\:the\:question,}$

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☆ Parallel sides of trapezium are in the ratio 7 : 3

$\begin{gathered}\begin{gathered}\bf Let : - \begin{cases} &\sf{one \: parallel \: side \: = 7x \: dm} \\ &\sf{other \: parallel \: side \: = 3x \: dm} \end{cases}\end{gathered}\end{gathered}$

☆ Height of trapezium = breadth of rectangle = 4 dm

$\bf \:Area_{(trapezium)} = \frac{1}{2} (sum \: of \: parallel \: sides)×height$

$\sf \:  ⟼Area_{(trapezium)} = \dfrac{1}{2} \times (7x + 3x) \times 4$

$\sf \:  ⟼Area_{(trapezium)} = 2 \times 10x = 20x \: {dm}^{2}$

$\bf\green{According\:to\:the\:question,}$

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$\bf \:  ⟼Area_{(trapezium)} = \dfrac{1}{3} \times Area_{rectangle}$

☆ On substituting the values, we get

$\sf \:  20x = \dfrac{1}{3} \times 120$

$\sf \:  ⟼20x = 40$

$\bf\implies \:x = 2$

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$\begin{gathered}\begin{gathered}\bf Hence : - \begin{cases} &\sf{one \: parallel \: side \: = 7 \times 2 = 14\: dm} \\ &\sf{other \: parallel \: side \: = 3 \times 2 = 6 \: dm} \end{cases}\end{gathered}\end{gathered}$

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Answer 2

Step-by-step explanation:

Given Question:−

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A trapezium with parallel sides of length as 7 : 3 is cut from a rectangle 30 dm by 4 dm so as to have an area of one-third the latter. Find the lengths of the parallel sides.

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\huge{AηsωeR} ✍AηsωeR ✍

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\large\underline\blue{\bold{Given:- }}

Given:−

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\begin{gathered}\begin{gathered}\begin{gathered}\bf given : - \begin{cases} &\sf{length \: of \: rectangle = 30 \: dm} \\ &\sf{breadth \: of \: rectangle = 4 \: dm}\\ &\sf{parallel \: sides \: of \: trapezium \: in \: 7 : 3} \\ &\sf{area \: of \: trap. \: = \frac{1}{3} \: area \: of \: rectangle }\end{cases}\end{gathered}\end{gathered}\end{gathered}

given:−

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lengthofrectangle=30dm

breadthofrectangle=4dm

parallelsidesoftrapeziumin7:3

areaoftrap.=

3

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areaofrectangle

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\begin{gathered}\begin{gathered}\bf To \: find :- \begin{cases} &\sf{the \: parallel \: sides \: of \: trapezium} \end{cases}\end{gathered}\end{gathered}