Math, asked by samiksha6872, 10 months ago

Sports committee of Kaspa Municipal High School wants to construct a Kho-Kho court of dimension 29 m. × 16 m. This is to be a rectangular enclosure of area 558 m2. They want to leave space of equal width all around the court for the spectators. What would be the width of the space for spectators? Would it be enough?

Answers

Answered by AYUSH124544E
0

Step-by-step explanation:

area of court=29×19=551m²

total area of court = 558m²

area for spectators = 558-551=7m²

therefore it would be not enough

Answered by topwriters
1

Available width surrounding the court is only one metre along the perimeter for the spectators.

Step-by-step explanation:

Area of kho-kho court = 29 × 16 = 464m²

Total area of the field = 558 m²

Area available for spectators = 558 - 464 = 94 m²

Total length of the strip for spectators would be the perimeter of the court + the corner areas depending on the width of the spectator region.

Area of the spectator region of width one metre = twice (29 * 1) + twice ( 16 * 1) + corner areas four times (1*1)

= 29 + 29 + 16 + 16 + 4

= 94 m²

Sports committee of Kaspa Municipal High School can construct a Kho-Kho court of dimension 29 m × 16 m. with a space for spectators of width 1 metre surrounding it. area of kho-kho court = 29 × 16 = 464m²

total area of the field = 558 m²

area available for spectators = 558 - 464 = 94 m²

Total length of the strip for spectators would be the perimeter of the court + the corner areas depending on the width of the spectator region.

Area of the spectator region of width one metre = twice (29 * 1) + twice ( 16 * 1) + corner areas four times (1*1)

= 29 + 29 + 16 + 16 + 4

= 94 m²

Sports committee of Kaspa Municipal High School can construct a Kho-Kho court of dimension 29 m × 16 m. with a space for spectators of width 1 metre surrounding it. It would be a narrow area.

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