ESSAY TYPE2. The ratio of the length and breadth of arectangular field

1.	ESSAY TYPE2. The ratio of the length and breadth of arectangular field is 3:2 and its area is 11094sq m. What will be the cost of fencing thefield at the rate of 2.50 per metre?TE
Answer» GIVEN The ratio of the length and breadth of a rectangular field is 3:2 and its area is 11094 sq m. TO FIND What will be the cost of fencing the field at the rate of 2.50 PER metre? SOLUTION Area of rectangle = 11094 m² ✒ According to the given condition ✞ Let the length be 3X and breadth be 2x Area of rectangle = 11094 ➡ length × breadth = 11094 ➡ 3x2x = 11094 ➡ 6x² = 11094 ➡ x² = 11094/6 ➡ x² = 1849 ➡ x = √1849 ➡ x = ± 43 ★ Length* and breadth never in negative Hence, Length = 3x = 3 × 43 = 129 m Breadth = 2x = 2 × 43 = 86 m _____________________ ✒Now, Perimeter of rectangle ➡ 2(Length + breadth) ➡ 2(129 + 86) ➡ 2 × 215 = 430 m Cost of fencing 1m field = Rs.2.50 Cost of fencing 430m field = Rs.2.50*430 = Rs.1075 _____________________

ESSAY TYPE2. The ratio of the length and breadth of arectangular field is 3:2 and its area is 11094sq m. What will be the cost of fencing thefield at the rate of 2.50 per metre?TE

Answer»

GIVEN

The ratio of the length and breadth of a

rectangular field is 3:2 and its area is 11094 sq m.

TO FIND

What will be the cost of fencing the

field at the rate of 2.50 PER metre?

SOLUTION

Area of rectangle = 11094 m²

✒ According to the given condition

✞ Let the length be 3X and breadth

be 2x

Area of rectangle = 11094

➡ length × breadth = 11094

➡ 3x*2x = 11094

➡ 6x² = 11094

➡ x² = 11094/6

➡ x² = 1849

➡ x = √1849

➡ x = ± 43

★ Length and breadth never in negative

Hence,

Length = 3x = 3 × 43 = 129 m

Breadth = 2x = 2 × 43 = 86 m

_____________________

✒Now, Perimeter of rectangle

➡ 2(Length + breadth)

➡ 2(129 + 86)

➡ 2 × 215 = 430 m

Cost of fencing 1m field = Rs.2.50

Cost of fencing 430m field

= Rs.2.50*430 = Rs.1075