Question

the area of a trapezium is 475sq.cm. parallel sides are in the ratio 2:3 and height is 10cm. the length of the longer side is
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Answer 1

• The length of the longer side of the trapezium = 57 cm.

Given :

• The area of a trapezium = 475 cm².
• The ratio of the parallel sides of the trapezium = 2 : 3.
• The height of the trapezium = 10 cm.

To Find :

• The length of the longer side.

Solution :

Let,

The length of the smaller side of the trapezium be 2x.

The length of the longer side of the trapezium be 3x.

Given,

Area of the trapezium = 475 cm²

We know that,

$\huge\blue\star \: \: \large\boxed{\large{\orange{\sf{Area \: of \: trapezium= \dfrac{1}{2} \times (Sum \: of \: parallel \: sides) \times h}}}}$

So,

$\bf \implies \dfrac{1}{2} \times (Sum \: of \: parallel \: sides) \times h = 475 \: {cm}^{2}$

We have,

• Sum of parallel sides = 2x + 3x.
• h = height = 10 cm.

First, we need to find the value of x.

Now, substitute both the given values in the formula of area of the trapezium.

$\bf \implies \dfrac{1}{\not2} \times (2x + 3x) \times \not10 \: cm = {475 \: cm}^{2}\\\\\\\bf \implies 1 \times (5x) \times 5 ={475 \: cm}^{2}\\\\\\\bf \implies 5x= \frac{ {475 \not{cm}}^{2} }{5 \not cm}\\\\\\\bf \implies 5x= \dfrac{\not475}{\not5} \: cm\\\\\\\bf \implies 5x = 95 \: cm\\\\\\\bf \implies x = \frac{ \not95}{ \not5} \: cm\\\\\\\bf \implies x = 19 \: cm\\\\\\ \: \: \: \bf \therefore \: \: x = 19 \: cm$

So, the length of the parallel sides of the trapezium are :-

★ Length of the smaller side of the trapezium = 2x = 2 × 19 cm = 38 cm.

★ Length of the longer side of the trapezium = 3x = 3 × 19 cm = 57 cm.

Hence,

The length of the longer side of the trapezium is 57 cm.

Answer 2

Given

• Area of a Trapezium is 475cm².
• Parallel sides are in ratio 2:3.
• Height of the Trapezium is 10cm.

⠀

To find

• Length of the longer side.

⠀

Solution

• Let the ratio be x.

⠀

❍ Shorter side = 2x

❍ Longer side = 3x

⠀

• As it is given in the question that the area of trapezium is 475cm².

⠀

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 10\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 3x }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 2x }\end{picture}$

⠀⠀⠀⠀⠀⠀⠀⠀Required Diagram

⠀

★ We know that

$\large{\boxed{\boxed{\sf{Area_{(Trapezium)} = \dfrac{Sum\: of\: parallel\: lines}{2} \times height}}}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{2x + 3x}{\cancel{2}} \times \cancel{10} = 475}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {5x \times 5 = 475}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {25x = 475}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {x = \dfrac{475}{25}}$

⠀

$\bf:\implies\: \: \: \: \: \: \: \: {x = 19}$

⠀

• We have, x = 19

⠀

★ Now

$\mathcal\longrightarrow{Smaller\: side = 2x = 38cm}$

$\mathcal\longrightarrow{Longer\: side = 3x = 57cm}$

⠀

Hence,

• The longer side of the Trapezium is 57cm.

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