Answer:
a)
- First we need to draw a perpendicular bisector of the side AB.
- Next, draw a perpendicular bisector of the side BC. We use the same method above.
- Now, put 0 where the bisectors intersect.
- Finally, put one part of the compass on 0 an the other part on C and draw the circle.
b)
![x=2.9 cm](https://img.qammunity.org/2021/formulas/mathematics/high-school/izx5xarpwk4tphxik631fcn7yeqe4qipvs.png)
c) R = 1 cm.
Explanation:
Let's define:
AB: distance of the side with length 3 cm
AC: distance of the side with length 4 cm
BC: distance of the side with length 5 cm
a)
- First we need to draw a perpendicular bisector of the side AB. We just put he compass in A, with any distance and trace two semicircle between AB.
- Next, draw a perpendicular bisector of the side BC. We use the same method above.
- Now, put 0 where the bisectors intersect.
- Finally, put one part of the compass on 0 an the other part on C and draw the circle.
b)
In a right triangle the hypotenuse of the triangle is the diameter of its circumscribe circle, so the radius of the circle is equal to half of the hypotenuse of the right triangle.
So if we draw a line from the center to the middle of the AB we will have new triangle, with sides BC/2, AB/2 and x, when x is the distance from the center to the length 3 cm. Therefore x will be:
![x=\sqrt{(5/2)^(2)+(3/2)^(2)}=2.9 cm](https://img.qammunity.org/2021/formulas/mathematics/high-school/zzedfullesnoxneinzcopgnageeuudf338.png)
c)
We need to find the radius of the circle to find the area.
So we can use this formula.
![A=(P\cdot R)/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/qfluuh0zmtrjuo82fsiqf5066a2wp53k7b.png)
A is the triangle area, it is AC*AB/2 = 6 cm²
P is the triangle perimeter, it is AC+AB+BC = 12 cm
R is the radius of the inner circle.
Then, R = 2*6/12 = 1 cm.
Have a nice day!