ABCD Is a rectangle that represents a park.
The lines show all the paths in the park.
The circular path is in the centre of the rectangle and has a diameter of 15m.
Calculate the shortest distance from A to C across the park,using only the lines shown.

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ABCD Is a rectangle that represents a park The lines show all the paths in the park The circular path is in the centre of the rectangle and has a diameter of 15 class=

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Answer:

Shortest distance from A to C = 102.9005 m.

Step-by-step explanation:

It is given that, ABCD is a rectangular park.

The length of the park is 80 m.

The breadth of the park is 50 m.

The diameter of the circle = 15 m.

We have to calculate the shortest distance from A to C across the park.

The distance AC = [tex]\sqrt{50^{2}+80^{2} }[/tex] = 94.339 m.

As one has to pass only through the lines shown, he cannot pass through the circle.

So, we have to subtract the diameter of 15 m from AC

=> 94.339 m - 15 m = 79.339 m.

One must pass through either half of the circumference of the circle.

Since, diameter of circle = 15 m, its radius(r) = [tex]\frac{15}{2}[/tex] = 7.5 m.

The circumference of the circle = 2×π×r = 47.123 m.

Half of the circumference = [tex]\frac{47.123}{2}[/tex] = 23.5615 m.

Distance from A to C passing through circumference = 79.339 m + 23.5615 m = 102.9005 m.

As we have to calculate the shortest distance from A to C, one cannot pass  from A to C either through B or D,

since the distance ABC or ADC = 50+80 = 130 m.

Therefore, shortest distance from A to C = 102.9005 m.

Diameter of circle is two times of its radius.

The shortest distance from A to C across the park is 102.91 meters.

Given that,  length of the park is 80 m.

  AC = 80 m

and  breadth of the park is 50 m.

    CD = 50 m

In right triangle ADC apply Pythagoras theorem

[tex]AC=\sqrt{AD^{2}+CD^{2} } \\\\AC=\sqrt{(80)^{2}+(50)^{2} }=94.34m[/tex]

 We have to calculate the shortest distance from A to C across the park that pass only through the lines shown in diagram.

So, we have to subtract the diameter of 15 m from AC

 [tex]94.34 m - 15 m = 79.34 m.[/tex]

Shortest distance is = 79.34 + half of circumference of circle.

Since, diameter of circle = 15 m,  radius(r) [tex]r=\frac{15}{2} = 7.5 m.[/tex]

Value of , [tex]\pi=\frac{22}{7}[/tex]

The circumference of the circle = [tex]2*\pi*r = 47.14 m.[/tex]

Half of the circumference, [tex]=\frac{47.14}{2} = 23.57 m.[/tex]

Thus, Shortest Distance from A to C passing through circumference is,

             [tex]79.34 m + 23.57 m = 102.91 m.[/tex]

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