Question

The inside perimeter of a running track shown in the figure is 400 m.

The length of each of the straight portions is 90 m, and the ends are

semicircles, if the track is 14 m wide everywhere, find the area of the

track. Also, find the length of the outer boundary of the track​

Answer 1

Answer:

Explanation:

Given :

• Inside perimeter of a running track is 400 m.
• The length of each of the straight portions is 90 m.
• The track is 14 m wide.

To Find :

• The area of track.
• The length of the outer boundary of the track.

Solution :

★The length of the remaining portion,

Inside perimeter of a running track - 2 × The length of each of the straight portion.

⇒400 - 2 × 90

⇒400 - 180

⇒220 m

★Inner Radius,

2πr = 220

⇒ 2 × 22/7 × r = 220

⇒ 44/7 × r = 220

⇒ r = 220 × 7/44

⇒ r = 5 × 7

⇒ r = 35 m

★Outer Radius,

R = Inner radius + wide of the track

⇒R = 35 + 14

⇒R = 49 m

★Area of the track,

Area of the track = Area of the two rectangles of dimensions + The area of the circular rings.

⇒ A = 2 × 90 × 14 + 22/7 × [(49)^2 - (35)^2]

⇒ A = 2520 + 22/7 × (2401 - 1225)

⇒ A = 2520 + 22/7 × 1176

⇒ A = 2520 + 22 × 168

⇒ A = 2520 + 3696

⇒ A = 6216 m^2

★The length of the outer boundary of the track.

Length of the outer running track = 2 × The length of each of the straight portions + 2πR

⇒ l = 2 × 90 + 2 × 22/7 × 49

⇒ l = 180 + 44 × 7

⇒ l = 180 + 308

⇒ I = 488 m

Answer 2

★Given:-

• Inside perimeter = 400m
• length of each of the straight portions is 90 m
• Ends are semicircles.
• Track is 14 m wide everywhere.

★To find:-

• Area of track
• length of the outer boundary of the track.

★Solution:-

The total length of the two straight portions = 90 + 90 = 180 m

Therefore the length of the remaining portion = 400 - 180 = 220 m

We have,

Circumference of a semi-circular = πr

Circumference of the two semi-circular portions =πr+πr=2πr.

Therefore,

$:\implies \sf 2\pi r=200\\\\:\implies \sf 2\times \frac{22}{7} \times r =200\\\\:\implies \sf 44r=200\times 7\\\\:\implies \underline{\sf r = 35m.}}$

Hence,the radius of inner circle = 35m.

Now,

Radius of outer track = Radius of inner track + width of the track Radius of outer track

⇒ 35 + 14 = 49m

Area of track:-

Area of the track = Area of the two rectangles of dimensions 90 × 14 + The area of the circular rings.

$:\implies \sf 2\times 90 \times 14 +\frac{22}{7} \times [49^2 -35^2 ]\\\\:\implies \sf 2520 + \frac{22}{7} \times (2401-1225)\\\\:\implies \sf 2520 + \frac{22\times 1176}{7}\\\\:\implies \sf 2520 + \frac{25872}{7}\\\\:\implies \sf 2520 + 3696 \\\\:\implies \underline{\boxed{\sf Area \ of\ track = 6216m^2.}}$

Length of outer boundary track:-

Length of outer track = 2× Length of rectangle + perimeter of two outer semi-circular ends.

$:\implies \sf Length\ of\ outer\ track = 2\times 90 + 2\pi r\\\\:\implies \sf 2\times 90 +(2\times \frac{22}{7} \times 49 ) \\\\:\implies \sf 180+308 = 488 \\\\:\implies \underline{\boxed{\sf Length = 488m.}}$

Hence,

The area of the  track is 6216m²

The length of the outer boundary of the track is 488m.

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