Question

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A yacht race follows a triangle course that has been mapped on to a Cartesian plane. The scale on the axes indicates distances in kilometres. The race begins and ends at the origin?
a. Calculate the length of each leg of the race?
b. Calculate the total racing distance?
c. If an observer’s boat is located at the midpoint of the second leg of the race, calculate the distance between this boat and the finishing point?

Answer 1

Let the points be A,B,C

• A(0,0)
• B(24,16)
• C(8,20)

#A

$\\ \bull\sf\leadsto AB=\sqrt{(24-0)^2+(16-0)^2}=\sqrt{24^2-16^2}=\sqrt{576-256}=\sqrt{320}=17.4units$(leg3)

$\\ \bull\sf\leadsto AC=\sqrt{(8-0)^2+(20-0)^2}=\sqrt{8^2+20^2}=\sqrt{64+400}=\sqrt{464}=21.4units$(leg1)

$\\ \bull\sf\leadsto BC=\sqrt{(24-8)^2+(26-20)^2}=\sqrt{16^2+(-4)^2}=\sqrt{256+16}=\sqrt{272}=16.4units$(leg2)

#B

We have to find perimeter

$\\ \bull\sf\leadsto Perimeter=AB+AC+BC$

$\\ \bull\sf\leadsto Periemter=17.4+16.4+21.4$

$\\ \bull\sf\leadsto Perimeter=55.2units$

#C

The length of leg3=17.4units

Answer 2

Answer:

Solution given;

points be A,B,C

P(0,0)

Q(24,16)

R(8,20)

a. Calculate the length of each leg of the race?

each leg's length

By using distance formula we can find length

I.e. d=$\sqrt{(x_2- x_1)²+(y_2-y1)²}$

leg 1

PR

=$\sqrt{(8-0)²+(20-0)²}=\sqrt{464}=4\sqrt{29}$

=21.54 km

leg 2

QR

$\sqrt{(8-24)²+(20-16)²}=\sqrt{272}=4\sqrt{17}$

=16.49 km

leg 3

PQ

=$\sqrt{(24-0)²+(16-0)²}=\sqrt{832}=8\sqrt{13}$

=28.84 km

b. Calculate the total racing distance?

The total racing distance=Sum of all side

=PR+QR+PR

=21.54+16.49+28.84

=66.87 km

c. If an observer’s boat is located at the midpoint of the second leg of the race, calculate the distance between this boat and the finishing point?

Solution given:

distance between this boat and the finishing point=$\frac{leg\:2}{2}=\frac{16.49}{2}=8.25units$

=8.25 km