Question

The area of a trapezium is 84cm'. The length of one of the parallel sides is 15cm
and the distance between parallel sides is 8cm. Find the length of the other parallel
side.​

Answer 1

$\Huge{\mathfrak{\blue{\underline{\red{Solution}}}}}$

Given:-

• Area of trapezium = 84$\sf{cm}^{2}$
• The length of one of the parallel side is 15cm.
• The distance between parallel sides is 8cm.

To Find:-

• Find the length of the other parallel side =?

Solution:-

Let's understand

Here, Area of trapezium is 84sqcm and length of one of the parallel side is 15cm and distance between parallel sides is 8cm and we have to find other parallel side.

We know that,

$\large\sf\blue{Area \:of\:trapezium={\frac{1}{2}(Sum\:of\:parallel\:sides)×Height}}$

Let the other parallel side be x.

Now, Put the value

$\sf \implies \: 84 {cm}^{2} = \frac{1}{2} (15 + x) \times 8 \\ \\ \sf \implies \: 84 \times 2 = (15 + x) \times 8 \\ \\ \sf \implies \: 84 \times 2 = 120 + 8x \\ \\ \sf \implies \: 168 = 120 + 8x \\ \\ \sf \implies \: 168 - 120 = 8x \\ \\ \sf \implies \: 48 = 8x \\ \\ \sf \implies \therefore \: x = 6$

Hence, Other parallel side is 6cm.

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Answer 2

AnswEr-:

Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 8\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 15\ cm  }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf x\ cm  }\end{picture}$

$\frak{\pink{\bf {Given \:\: -:}}} \begin{cases} \sf{The\:Area \:of\:the\:Trapezium \: \:is\:= \frak{84\:cm^{2}}} & \\\\ \sf{Height \:or\:Distance \:Between \:Parallel\:Sides\:is \:=\:\frak{8cm}}& \\\\ \sf{Length \:of\:one \:Parallel\:Side\:is \:=\:\frak{15cm}}\end{cases}$

$\frak{\bf{To\:Find\:-:}}\sf{The\:Length \:of\:other\:Parallel \:side\:}$

$\dag{\sf{\bf{Solution \:of\:Question \::}}}\sf{ \:\:Let's \:Assume \:the\:length\:other\:Parallel \:side\:be\:\bf{x\:cm}}$

$\boxed{\sf{\pink{Area_{(Trapezium)} = \dfrac{1}{2} \times Height \times ( a + b ) \:Parallel \:Sides\:}}}$

$\frak{\pink{\bf {Here \:\: -:}}} \begin{cases} \sf{The\:Area \:of\:the\:Trapezium \: \:is\:= \frak{84\:cm^{2}}} & \\\\ \sf{Height \:or\:Distance \:Between \:Parallel\:Sides\:is \:=\:\frak{8cm}}& \\\\ \sf{Length \:of\:one \:Parallel\:Sides\:is \:=\:\frak{15cm}}& \\\\ \sf{Length \:of\:other \:Parallel\:Side\:is \:=\:\frak{x\:cm}}\end{cases}$

$\qquad \quad \sf{\underline {\star{ Now \:By\:Putting \:the \:Given \:Values\:-:}}}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \:\dfrac{1}{2} \times 8 \times (15 + \:x ) = 84cm^{2} }}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \:\dfrac{1}{\cancel {2}} \times \cancel {8} \times (15 + \:x ) = 84cm^{2} }}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \: 4 \times (15 + \:x ) = 84cm^{2} }}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \: 15 + \:x =\dfrac{ 84}{4} }}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \: 15 + \:x = \dfrac{\cancel {84}}{\cancel {4}} }}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \: 15 + \:x = 21 }}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \: \:x = 21-15 }}$

$\qquad \quad \qquad \quad \underline {\boxed{\pink{\mathrm { \: \:x = 6 }}}}$

Hence ,

• $\dag\:\:{\underline {\pink{\mathrm {The\:Length \:of\:other\:Parallel \:side\:is\:6\:cm\:}}}}$

_________________________________________

$\large { \boxed{\red {\mathrm |\:\:{\underline {Verification \::}}\:\:|}}}$

$\boxed{\sf{\pink{Area_{(Trapezium)} = \dfrac{1}{2} \times Height \times ( a + b ) \:\:Parallel \:Sides\:}}}$

$\frak{\pink{\bf {Here \:\: -:}}} \begin{cases} \sf{The\:Area \:of\:the\:Trapezium \: \:is\:= \frak{84\:cm^{2}}} & \\\\ \sf{Height \:or\:Distance \:Between \:Parallel\:Sides\:is \:=\:\frak{8cm}}& \\\\ \sf{Length \:of\:one \:Parallel\:Sides\:is \:=\:\frak{15cm}}& \\\\ \sf{Length \:of\:other \:Parallel\:Side\:is \:=\:\frak{6cm}}\end{cases}$

$\qquad \quad \sf{\underline {\star{ Now \:By\:Putting \:the \:Known \:Values\:-:}}}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \:\dfrac{1}{2} \times 8 \times (15 + \:6 ) = 84cm^{2} }}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \:\dfrac{1}{\cancel {2}} \times \cancel {8} \times (15 + \:6 ) = 84cm^{2} }}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \: 4 \times (15 + \:6 ) = 84cm^{2} }}$

$\qquad \quad \qquad \quad \longmapsto {\mathrm { \: 4 \times 21 =84cm^{2} }}$

$\qquad \quad \qquad \quad \underline {\boxed{\pink{\mathrm { \: \:84\:cm^{2} = 84cm^{2} }}}}$

$\qquad \qquad \qquad \dag\:\:\:\: {\sf{\bf{ Hence\:Verified \:}}}$

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