IF YOU GET THIS RIGHT I WILL MARK BRAIN. LY DO NOT ANSWER IF YOU DO NOT KNOW OR YOU WILL GET REPORTED
This is geometry through and through. Plus a little trig thrown in for fun. If you inscribe an equilateral triangle inside a circle and the triangle has side lengths of 12, you have part of what you need to use Pythagorean's Theorem to find the hypotenuse of the triangle which is also the radius of the circle. First, use the formula 360/3 to find that the central angle measure of each angle INSIDE a triangle is 120. So you have 3 triangles within the large one, each with a top angle of 120 and a base of 12. If you extract that one triangle and then split it in half, you have a right triangle with a base of 6. This is a 30-60-90 triangle and this is important so you can check your work. Now use the apothem formula for a right triangle as it relates to a side in an equilateral triangle, which is a = sqrt3/6 * s. Our values are a = sqrt3/6(12) which simplifies to 2sqrt3. That's our apothem. If you're familiar with a 30-60-0 triangle, you could check this to see it's correct. Now you have the base leg of 6 and the height of 2sqrt3, now use Pythagorean's Theorem to find the hypotenuse, which is also the radius of the circle. This was really a difficult one to explain.

Side of equilateral Triangle =12 inch

Let r be the radius of Smallest circle which fits inside the triangle.

Theorem Which will be Used

→The Point of intersection of end point of radius and tangent through that end point makes an angle of 90°.

→Angle Subtended by an arc at the center is twice the angle subtended at any point on the circle.

→Perpendicular from the center of the circle to the chord bisects the chord.

→ Area of Equilateral Triangle

→Area of Triangle

⇒ Radius of the smallest circle

→ Area of Equilateral Triangle having side 12 inches

                                (1)

→Area of 3 Triangle having base =12 inches and height =r inches

                                                   (2)

Equating (1) and (2), that is both the Areas are equal.

⇒Radius of the largest circle

       Each Interior angle of equilateral Triangle=60°

∠AOC=2×60°=120°Angle Subtended by an arc at the center is twice the angle subtended at any point on the circle.

r=Radius of circle

→∠AOC+∠CAO+∠ACO=180°---Angle sum property of triangle.

→ 120°+2∠CAO=180°[OA=OC, →∠CAO=∠OCAAngle Opposite to equal side of a triangle are equal.]

→2∠CAO=180°-120°

→2∠CAO=60°

→∠CAO=30°Dividing both sides by, 2

→∠MAO+∠AMO+∠AOM=180°---Angle sum property of triangle.

→∠MAO+90°+60°=180°

→∠MAO=180°-150°

→∠MAO=30°

In ΔOMA

AC=12 inches

AM=MC=6 inches--Perpendicular from the center of the circle to the chord bisects the chord.

:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph