Question

A trapezium shaped field, its side along the river is parallel to and twice the side along the road. if the area of field is 10500 msq and perpendicular between the parallel sides is 100 msq .find the side along the river

Answer 1

$\huge\mathbb{\underline{\underline{SOLUTION:-}}}$

• A trapezium shaped field

• the area of trapezium and the height of trapezium

$\bold{Given}\begin{cases}\sf{Area\:of\: trapezium=10500m{}^{2}}\\ \sf{It's \: height=100m}\end{cases}$

$\mathbb{TO\: FIND:-}$

• We have to find the parallel sides of trapezium field
• which is given that one of its parallel side is twice the small one

let the parallel sides be x and 2x

$\sf\underline{\underline{\red{According\:to\: Question:-}}}$

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

$\sf 10500=\frac{1}{\cancel2}\times \cancel{100}\times (2x+x)\\ \\ \sf\implies 10500=50\times 3x\\ \\ \sf\implies \cancel{\frac{10500}{50}}=3x\\ \\ \sf\implies 210=3x\\ \\ \sf\implies \cancel{\frac{210}{3}}=x\\ \\ \sf\implies x=70$

• Now the value of x is 70cm means one of it's parallel side is 70cm and other is twice then
• 70×2=140cm other

• The parallel sides of trapezium shaped field is 70cm and 140cm

• Now let's verify our answer:-

$\tt Area\:of\: trapezium= \frac{1}{2}\times h\times (sum\:of\: parallel\:sides)\\ \\ \sf\leadsto 10500=\frac{1}{\cancel{2}}\times \cancel{100}\times (70+140)\\ \\ \sf\leadsto 10500 =50\times 210\\ \\ \sf\implies 10500=10500$

L.H.S=R.H.S

$\boxed{\sf{parallel\:sides=70cm\:and\:140cm}}$

#BAL

#answerwithquality

Answer 2

Hope it helps you.................