MATHEMATICS HIGH SCHOOL

Which of the following statements is true regarding the relationship between circles and triangles? A. There are many circles that can be circumscribed about a triangle. B. There are many triangles that can be inscribed in a given circle. C. There is only one unique triangle that can be inscribed in a given circle. D. There are many triangles that can be circumscribed about a given circle.

Answers

Answer 1
Answer:

The true statement regarding the relationship between circles and triangles is option B. There are many triangles that can be inscribed in a given circle.

A circle can be characterized by its center's location and its radius's length.

How to make an inscribed circle in a triangle?

There can be many different ways, one can include arc way, one can include angle bisector and perpendicular, and we can even try to discover some.

But usually, (for the angle bisector and perpendiculars), we do the following:

Divide one of the angles in half. Divide another angle in half.

The incenter, or point where they cross, is the inscribed circle's centre.

Construct a perpendicular from the triangle's centre point to one of its sides.

Draw an inscribed circle by placing the compass on the centre point and adjusting the length to where the perpendicular crosses the triangle.

The true statement regarding the relationship between circles and triangles is option B. There are many triangles that can be inscribed in a given circle.

Learn more about inscribing circle here:

brainly.com/question/17043518

#SPJ2

Answer 2
Answer: Its worth trying drawing circles and triangles to find the correct one.

If you do you'll find that the answer is  B.

Related Questions

HIGH SCHOOL

A group of 9 friends decides to buy a piece of land together. The cost of the land is $6,975. Each friend pays an equal share of the cost. How much does each friend pay?

Answers

$6975 / 9 = $775

take the total amount divided by the number of friends, this is the cost for each friend
HIGH SCHOOL

What is 89/10 as a mixed number

Answers

89/10 = (80 + 9)/10 = 80/10 + 9/10 = 8 9/10
MIDDLE SCHOOL

This scale drawing of a rectangular wall has dimensions 4.75 inches by 2.5 inches. The length of the shorter side of the actual wall is 36 inches. Enter the area of the actual wall.
I will mark the first one who answers me brainleist


_[blank]_ square inches

Answers

The area of the wall is the product of its dimension.

The area of the actual wall is  2462.4 square inches

The scale is given as:

The shorter side of the actual wall is 36 inches.

So, we have:

Express as fractions:

Multiply both sides by 36

So, we have:

The area is then calculated as:

Hence, the area of the actual wall is  2462.4 square inches

Read more about areas at:

brainly.com/question/2602591

Answer:

2462.4 square inches

Step-by-step explanation:

To find the length of the longer wall use the following ratio and solve for x.

x/36= 4.75/ 2.5

x= 68.4

Area of a rectangle = l(w)

68.4(36)= 2462.4 square inches

MIDDLE SCHOOL

You roll a number cube twice. What is the probability of rolling a 2 first and then rolling an odd number? 1/36,1/12,1/9,1/4

Answers

Answer:

Choice b is correct answer.

1/12

Step-by-step explanation:

We have to find the probability of happening of two events.

From question statement,

total outcomes = 6

Let

A = rolling of a 2 (favourable outcomes = 1)

B = rolling of an odd number(1,3,5)    (favourable outcomes = 3)

Probability is ratio of favourable outcomes to total outcomes.

P(A) = 1/6

P(B) = 3/6 = 1/2

The probability of two events occurring smiltaneously,

P(A and B ) = P(A) P(B)

Putting the values of P(A) and P(B) . we have

P(A and B ) =  (1/6)(1/2)

P(A and B ) =  1/12 which is the answer.



A = event of rolling a 2 on the first roll

P(A) = 1/6 because there is only one "2" out of six labels total on the number cube.

B = event of rolling an odd number on the second roll

P(B) = 3/6 = 1/2 since there are three odd numbers {1,3,5} out of six total

The events A and B are independent, so we can multiply the probabilities

P(A and B) = probability of both events happening simultaenously

P(A and B) = P(A)*P(B)

P(A and B) = (1/6)*(1/2)

P(A and B) = (1*1)/(6*2)

P(A and B) = 1/12

Final Answer: 1/12

Note: The fraction 1/12 is approximately equal to 0.0833

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