1. Ab is a chord of a circle with center o and radius 52 cm . point m divides the chord ab such that am = 63 cm and mb=33 cm find om
2. A circle is inscribed in a triangle whose sides are 10, 10 and 12 units . a second smaller circle is inscribed tangent to the first circle and to the equal sides of the triangle. Find the radius of the second triangle.
Help me guys thnk you its very much appreciated
1. Make a reasonable sketch.
draw radii to A and B
Triangle ABO is isosceles with sides 52 , 52 , and 96
By the cosine law we can find cosA
52^2 = 52^2 + 96^2 - 2(96)(52)cosA
cosA = 96^2/(2*52*96) = 12/13
In triangle AMO
MO^2 = 33^2 + 52^2 - 2(33)(52)cosA
= 1089 + 2704 - 3432(12/13) = 625
OM = √625 = 25
Give me a bit of time for #2
#2
Notice the centre of the inscribed circle must lie on the bisectors of the angles.
We have an isosceles triangle, with sides 10,10 and 12
If we place the triangle on the x-y grid, we can have the base from (-6,0) to (6,0) and the third vertex on the y-axis.
The y-axis becomes the altitude. Notice that in the right-angled triangles, we have a base of 6 and a hypotenuse of 10. You should recognize double the 3-4-5 triangle, so the height is 8 and the third vertex is (0,8)
label (-6,0) as B and A is (0,8)
slope of AB = (8-0)/(0-(-6)) = 4/3
An important property in trig is that the slope of a line is equal to the tangent of the angle that line makes with the x-axis
so we can find the angle at B
tan B = 4/3
B = 53.1301....
Since we need the angle of the bisector of angle B
that angle would be 53.1301../2
and the slope of that bisector is
tan (53.1301.../2) = 1/2 exactly
(are you surprised ?? We could show that it is exactly 1/2 by using Tan (2Ø) = 2tanØ/(1-tan^2 Ø) )
So the equation of the bisector from B is
y = (1/2)x + b, with (-6,0) on it
0 = (1/2)(-6) + b
b = 3
ahhh, so we have the centre of the large circle at (0,3) , the y-intercept
So the big circle cuts the y-axis at (0,6) , its diameter being 6
Label that point P , draw a horizontal line at P cut cut AB at Q
Because of parallel lines, angle AQP = angle B
and the slope of the angle bisector of angle AQP is also 1/2
How about the equation of the angle bisector of angle Q ??
We know its equation must be y = (1/2)x + k , ( I don't want to use b again)
So we need Q
equation of AB: y = (4/3)x + 8
at Q, y = 6
6 = (4/3)x + 8
-2 = (4/3)x
x = -2(3/4) = -3/2
So we have Q as (-3/2,6)
Q must lie on the angle bisector of angle Q, which is y = (1/2)x + k
ALL WE NEED IS K, AND WE ARE BASICALLY DONE
y = (1/2)x + k
6 = (1/2)(-3/2) + k
k = 6 + 3/4
so the centre of the little circle is at (0 , 6 3/4 )
but the large circle covered 6 of that distance
So the radius of the little circle is 3/4
(you might want to print out that solution for easier reading.)
The solution looks very long and complicated, but the essential work is really quite short, most of the solution consists of explanations.
Perhaps one of the other tutors might have a quicker way, sometimes it is easy to miss the obvious once your mind locks in on a method.
1. Well, if we have a circle with a radius of 52 cm and a chord AB, we can use the intersecting chord theorem to find the length of OM. But since I'm a Clown Bot, I prefer to take a funny approach.
Let's imagine point M running a race with point A and point B. They start at the center of the circle, point O. Point M says, "Watch me divide the chord and beat you both to the finish line!"
Now, point M decides to jog first and covers a distance of 63 cm, while point A just watches in awe. Then point M gets tired and decides to walk the rest of the way, covering a distance of 33 cm. Meanwhile, point B has been fast asleep, not moving at all.
Finally, point M reaches the finish line, which is 52 cm away from the starting point. And guess what? Point M wins the race! Hooray!
So, the distance from point O to point M (OM) is the radius of the circle (52 cm) minus the distance point M covered in the race (63 cm + 33 cm = 96 cm). In other words, OM = 52 cm - 96 cm = -44 cm.
But wait a minute, negative distances don't make sense in real life, right? So, we made a mistake somewhere or the race was just too confusing. Let's try a different approach to finding OM!
2. Now, let's move on to the second question. We have a triangle with sides measuring 10, 10, and 12 units. And there are these two circles, one inside the other, let's call them Big Circle and Small Circle.
Imagine the Big Circle saying to the Small Circle, "Hey, I can fit you perfectly in this triangle!" The Small Circle replies, "Oh really? Let me just squeeze in and prove you wrong!"
After some hilarious attempts, the Small Circle manages to fit perfectly inside the triangle, with its center touching the sides of equal length. And just like that, the Small Circle wins the challenge!
Now, let's find the radius of the Small Circle. To do that, we can use the formula for the inradius of a triangle: inradius = Area of the triangle / Semiperimeter of the triangle.
Since we have an isosceles triangle, we know that it can be divided into two congruent right-angled triangles. And because I know you love math, I'll spare you the funny details this time and do the calculations.
The area of the triangle is given by Heron's formula: Area = sqrt(s * (s - a) * (s - b) * (s - c)), where s is the semiperimeter and a, b, c are the side lengths.
In this case, a = b = 10 and c = 12. So, the semiperimeter is s = (a + b + c) / 2 = (10 + 10 + 12) / 2 = 16.
Now plug in the values into the formula: Area = sqrt(16 * (16 - 10) * (16 - 10) * (16 - 12)) = sqrt(16 * 6 * 6 * 4) = 48.
The semiperimeter is already given as 16, so let's calculate the inradius: inradius = 48 / 16 = 3 units.
And there you have it! The radius of the second circle, also known as the Small Circle, is 3 units.
I hope I managed to make you smile while providing the answers! If you have any more questions, feel free to ask.
Sure! I can help you with both questions.
First, let's solve the first problem:
1. We know that the radius of the circle is 52 cm and that AM = 63 cm and MB = 33 cm.
2. Since M is the midpoint of AB, we can use the midpoint formula to find the coordinates of M. The coordinates of A and B can be represented as (-a, 0) and (a, 0) respectively.
3. The midpoint formula is (x1 + x2)/2, (y1 + y2)/2. Applying this formula, we get the following:
-a + a/2 = 63 (x-coordinate of M)
0 + 0 = 0 (y-coordinate of M)
4. Simplifying the equation, we get:
a/2 = 63
5. Solving for a, we find:
a = 63 * 2
a = 126
6. Now, we can find the distance from O to M using Pythagoras' theorem:
OM^2 = OA^2 - AM^2
= 52^2 - 126^2
= 2704 - 15876
= -13172
7. The distance OM cannot be negative, so there is an error in the given information. Please double-check the lengths provided.
Moving on to the second question:
1. We have a triangle with sides of length 10, 10, and 12 units.
2. Let the radius of the larger inscribed circle be r1, and the radius of the smaller circle be r2.
3. The inradius (r1) of the larger circle can be found using the formula:
r1 = (s - a) / 2,
where s is the semi-perimeter, and a is one side length of the triangle.
4. The semi-perimeter is given by:
s = (10 + 10 + 12) / 2
= 32 / 2
= 16
5. Now, we can substitute the values into the formula to find r1:
r1 = (16 - 12) / 2
= 4 / 2
= 2 units
6. The radius of the smaller circle can be found using the formula:
r2 = (2 * Area) / (Perimeter),
where Area is the area of the triangle, and Perimeter is the perimeter of the triangle.
7. The area of the triangle can be found using Heron's formula:
Area = sqrt(s * (s - a) * (s - b) * (s - c)),
where s is the semi-perimeter, and a, b, and c are the side lengths.
8. Substituting the given values into the formula, we get:
Area = sqrt(16 * (16 - 10) * (16 - 10) * (16 - 12))
= sqrt(16 * 6 * 6 * 4)
= sqrt(6^2 * 4^2)
= 24 square units
9. Lastly, we can substitute the values into the formula to find r2:
r2 = (2 * 24) / (10 + 10 + 12)
= 48 / 32
= 1.5 units
Therefore, the radius of the second circle (r2) is 1.5 units.
1. To find the length of OM, we can use the following steps:
Step 1: Draw a diagram of the given problem with circle O, chord AB, point M, and other relevant information.
Step 2: Since AM and MB divide the chord AB into two equal parts, we can say that AM = MB = (1/2)AB. From the given information, we know that AM = 63 cm and MB = 33 cm, so we can write the equation as (1/2)AB = 63 cm.
Step 3: Solve the equation from Step 2 for AB: AB = (2)(AM) = (2)(63 cm) = 126 cm.
Step 4: We know that OM is perpendicular to AB because it is the radius of circle O. Therefore, OM bisects AB at point M and forms two right triangles, MOA and MOB.
Step 5: Apply the Pythagorean theorem to find the length of OM in triangle MOA: OM^2 = OA^2 - AM^2. The radius OA is given as 52 cm, so we have OM^2 = 52^2 - (63/2)^2.
Step 6: Calculate OM^2 using the values obtained in Step 5.
Step 7: Take the square root of OM^2 to find OM.