Question

answer these please
this is a case based maths question and is carrying 5m​

Answer 1

Given:

• The length of each straight portion of the track is 90m

• Both the ends of the race track are semi circles

To Find:

• The answer of the following sub questions

Answers :

${ \blue{ \boxed{ \rm{Solution \: 1}}}}$

Given :

• Perimeter of the inner track is 400 m

To Find :

• The inner radius of the trak

Solution:

✦ Now, we know that the diagram is a fusion of a rectangle and 2 semi circles.

↷ According to the question

◕ The perimeter of the inner track is 400m

Now,

↦ Let's firstly find the lenght boundary of the inner perimeter excluding that of the circles

➼ 90m + 90m

➼ 180m

∴ The perimeter of two semicircles will be 400m - 180m

which is equal to 220m

We know,

↷ Perimeter of a semicircle = πr

So,

↷ Perimeter of 2 semicircles = 2πr

Now,

↦ Perimeter = 2πr

↦ 220m = 2 × 22 / 7 × r

↦ r = 220 × 7 / 22 × 2

↦ r = 1540/ 2 × 22

↦ r = 35m

• Henceforth the inner radius of the track is 35m

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${ \blue{ \boxed{ \rm{Solution \: 2}}}}$

Given:

• The outer radius of the track is 488m

To Find :

• The outer radius of the circle

Solution :

↷ According to the question :

 ✦ The perimeter of the outer track is 488m

Now,

↦ Let's firstly find the lenght boundary of the inner perimeter excluding that of the circles.

So,

• 90m + 90m = 180

∴ The perimeter of two semicircles will be 488m - 180m

which is equal to 308m

We know,

↷ Perimeter of a semicircle = πr

So,

↷ Perimeter of 2 semicircles = 2πr

Now,

↦ Perimeter = 2πR

↦ 308 = 2 × 22/7 × R

↦ R = 308 × 7 / 2 × 22

↦ R = 2156 / 44

↦ R = 49m

• Hence forth the outer radius of the track is 49m

${ \blue{ \boxed{ \rm{Solution \: 3}}}}$

Given :

• Outer radius = 49
• Inner radius = 35cm

To Find :

• The width of the track

Solution :

↷ Now,

• Let's find the width

◕ We know,

Width = Outer radius - Inner radius

↷ So,

Width = R - r

Width = 49 - 35

Width = 14m

• Henceforth the width of the track is 14m

${ \blue{ \boxed{ \rm{Solution \: 4}}}}$

We have,

• Width of the track = 14m
• Inner radius = 35m
• Outer radius = 49cm

To Find:

• The area of the track

Solution:

↷ Now, area is

↦ 2 [ 14 × 90 ] + [ {22 ÷ 7 × 49 × 49} - { 22 ÷ 7 × 35 ×35}]

↦ 2520m² + [ 7546m² - 3850m²]

↦ 2520m² + 3936m²

↦ 6216m²

• Henceforth the area of the track is 6216m²

${ \blue{ \boxed{ \rm{Solution \: 5}}}}$

We have,

• Area 6216m²
• Cost = 12.5 at m²

To Find,

• Total cost

Solution,

↷ Total cost

↦ Area × cost per m²

↦6216 × 12.50

↦77700

• Henceforth the cost of developing the track is 77700 Rupees

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Answer 2

We have,

Area 6216m²

Cost = 12.5 at m²

To Find,

Total cost

Solution,

↷ Total cost

↦ Area × cost per m²

↦6216 × 12.50

↦77700

Henceforth the cost of developing the track is 77700 Rupees

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