Question

the area of a Trapezium is 60 CM square the distance between its parallel sides is 6 CM if one of the parallel sides is 8 centimetre, find the other parallel side​

Answer 1

$\bold{\underline{\underline{\huge{\sf{AnsWer:}}}}}$

Other parallel side of the trapezium is 12 cm.

$\bold{\underline{\underline{\large{\sf{StEp\:by\:stEp\:explanation:}}}}}$

GiVeN :

• The area of a trapezium is 60 cm²
• Distance between its parallel sides is 6 cm
• One of the parallel sides is 8 centimetre

To FiNd :

• Other parallel side

SoLuTioN :

Let the other parallel side be x cm.

FoRmUlA :

$\bold{\large{\boxed{\tt{\red{Area\:of\:trapezium\:=\:{\dfrac{1}{2}\:\times\:(sum\:of\:parallel\:sides)\:\times\:height}}}}}}$

Where,

• Area of trapezium = 60 cm²
• One parallel side = 8 cm
• Other parallel side = x cm
• h = height = 6 cm

Block in the values in the formula,

$\hookrightarrow$ $\tt{60\:=\:{\dfrac{1}{2}\:\times\:(8+x)\:\times\:6}}$

$\hookrightarrow$ $\tt{60\:=\:{\dfrac{(8+x)}{2}\:\times\:6}}$

$\hookrightarrow$ $\tt{60\:=\:{\dfrac{6(8+x)}{2}}}$

$\hookrightarrow$ $\tt{60\:=\:{\dfrac{48+6x}{2}}}$

$\hookrightarrow$ $\tt{120\:=\:48\:+\:6x}$

$\hookrightarrow$ $\tt{120-48\:=\:6x}$

$\hookrightarrow$ $\tt{72\:=\:6x}$

$\hookrightarrow$$\tt{\dfrac{\cancel{72}}{\cancel{6}}\:=x}$

$\hookrightarrow$ $\tt{12=x}$

$\bold{\boxed{\red{\boxed{\therefore{\rm{Other\:parallel\:side\:=\:x\:=\:12\:cm}}}}}}$

$\bold{\huge{\underline{\tt{VeRiFiCaTiOn:}}}}$

To verify the answer, simply block in all the available data in the formula of area of the trapezium and check whether the LHS = RHS or vice versa. If yes, then our answer would be correct.

Data :

• Area of trapezium = 60 cm²
• One parallel side = 8 cm
• Other parallel side = 12 cm
• Height = 6 cm

Block in the values now,

$\hookrightarrow$ $\tt{60\:=\:{\dfrac{1}{2}\:\times\:(8+12)\:\times\:6}}$

$\hookrightarrow$ $\tt{60\:=\:{\dfrac{1}{2}\:\times\:(20)\:\times\:6}}$

$\hookrightarrow$ $\tt{60\:=\:{\dfrac{20\:\times\:6}{2}}}$

$\hookrightarrow$ $\tt{60\:=\:{\dfrac{\cancel{120}}{\cancel{2}}}}$

$\hookrightarrow$ $\tt{60\:=\:60}$

LHS = RHS.

Answer 2

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

Given ,

Area of trapezium = 60 cm²

Height of trapezium = 6 cm

Parallel side = 8 cm

Let ,

The other parallel side of trapezium be x

We know that , the area of trapezium is given by

$\large \mathtt{ \fbox{Area \: of \: trapezium = \frac{1}{2} (sum \: of \: parallel \: sides) \times height}}$

Substitute the values , we obtain

$\sf \hookrightarrow 60 = \frac{1}{2} (8 + x) \times 6 \\ \\\sf \hookrightarrow 120 = (8 + x)6 \\ \\\sf \hookrightarrow 20 = 8 + x \\ \\\sf \hookrightarrow x = 12$

Hence , the other parallel side of trapezium is 12 cm

$\large \sf \underline{ \: Verification : \: \: \: }$

Let's check our answer in the original problem by replacing x with 12

If our answer makes LHS = RHS in the equation then our answer is absolutely correct

$60 = \frac{1}{2} (8 + 12) \times 6 \\ \\ 60= \frac{1}{ \cancel2} ( \cancel{20}) \times 6 \\ \\ 60= 10 \times 6 \\ \\ 60 = 60$

$\star$ LHS = RHS

Hence , verified