Question

en would get 3 oranges less Find the
14 Mr Dinesh owns a agricultural format village Talver The length of the formi
10 meter more than twice the breadte in order to harvest roin water, he dog
oore shaped pond inside the farm. The side of pond is of the breadth of the
form. The area of the form is 20 times the area of the pond Fd the length an
breadth of the form and of the rond​

Answer 1

Answer:

The length of the field is 100 metre.

The breadth of the field is 45 metre.

The side of the pond is 15 metre.

Step-by-step-explanation:

NOTE: Kindly refer to the attachment first.

Let the breadth of the field be x metre.

And the length of the field be ( 2x + 10 ) metre.

The field is rectangular in shape.

We know that,

$\large\boxed{\red{\sf\:Area\:of\:field\:=\:length\:\times\:breadth}}$

$\therefore\sf\:Area\:of\:field\:=\:(\:2x\:+\:10\:)\:\times\:x\:m^{2}$

Now, the side of the pond = $\sf\frac{1}{3}\:x$ metre.

The pond is square in shape.

We know that,

$\large\boxed{\red{\sf\:Area\:of\:square\:=\:(\:side\:)^{2}}}$

$\therefore\sf\:Area\:of\:pond\:=\:(\frac{1}{3}\:x)^{2}\:metre$

From the given condition,

$\sf\:Area\:of\:rectangular\:field\:=\:20\:\times\:\sf\:Area\:of\:pond\\\\\therefore\sf\:(\:2x\:+\:10\:)\:\times\:x\:=\:20\:\times\:(\:\frac{1}{3}\:x\:)^{2}\\\\\implies\sf\:2x^{2}\:+\:10x\:=\:20\:\times\:(\:\frac{1}{9}\:x^{2}\:)\\\\\implies\sf\:18x^{2}\:+\:90x\:=\:20x^{2}\:\:\:...\:[\:Multiplying\:both\:sides\:by\:9\:]$

$\implies\sf\:{20x}^{2}\:=\:{18x}^{2}\:+\:90x\\\\\implies\sf\:{20x}^{2}\:-\:{18x}^{2}\:-\:90x\:=\:0\\\\\implies\sf\:{2x}^{2}\:-\:90x\:=\:0\\\\\implies\sf\:x^{2}\:-\:45x\:=\:0\:\:\:\:...[\:Dividing\:both\:sides\:by\:2\:]$

$\implies\sf\:x\:(\:x\:-\:45\:)\:=\:0\\\\\implies\sf\:x\:=\:0\:\:or\:\:x\:-\:45\:=\:0\\\\\implies\boxed{\red{\sf\:x\:=\:0}}\:\:\:\sf\:or\:\:\:\boxed{\pink{\sf\:x\:=\:45}}$

But, the length can't be zero.

$\therefore\sf\:x\:=\:0\:is\:unacceptable\\\\\therefore\boxed{\red{\sf\:x\:=\:45}}$

$\therefore$ The breadth of the field = ( x ) = 45 metre.

$\therefore$ The length of the field

= ( 2x + 10 )

= ( 2 × 45 ) + 10

= 90 + 10

= 100 metre.

$\therefore\sf\:The\:side\:of\:pond\:=\:\sf(\:\frac{1}{3}\:x\:)\:=\:(\:\frac{1}{\cancel3}\:\times\:\cancel{45}\:)\:=\:15\:metre$