In an isosceles triangle the length of the base is 10 cm
1 answer:
The computation shows the radius of the circle that is inscribed in the isosceles triangle will be 3.33cm.
<h3>How to calculate the radius?</h3>From the information given, the isosceles triangle the length of a base is 10 cm and the length of a leg is 13 cm.
Let A = area of the triangle
Let S = semi perimeter of the triangle.
The radius will be: = A/S
where,
The radius will be:
= 3.33cm
In conclusion, the radius is 3.33cm.
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The braking distance is the distance the car travels before coming to a stop after the brakes are applied
a. The braking distances are as follows;
- The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
- The braking distance at 55 mph, is approximately <u>298.35 ft.</u>
- The braking distance at 85 mph, is approximately <u>708.92 ft.</u>
b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s
Reason:
The given function for the braking distance is D = 2.6 + v²/22
a. The braking distance if the car is going 25 mph is therefore;
25 mph = 36.66339 ft./s
At 25 mph, the braking distance is approximately <u>63.7 ft.</u>
At 55 mph, the braking distance is given as follows;
55 mph = 80.65945 ft.s
At 55 mph, the braking distance is approximately <u>298.35 ft.</u>
At 85 mph, the braking distance is given as follows;
85 mph = 124.6555 ft.s
At 85 mph, the braking distance is approximately <u>708.92 ft.</u>
b. The speed of the car when the braking distance is 450 feet is given as follows;
v² = (450 - 2.6) × 22 = 9842.8
v = √(9842.2) ≈ 98.211 ft./s
The car was moving at v ≈ <u>98.211 ft./s</u>
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