Question
HELP ME PLEASE.
1.An isosceles triangle with each leg measuring 13 cm is inscribed in a circle . if the altitude to the base is 12 cm find the radius of the circle
2. Circles a and b are tangent at point c. p is on circle a and q is on circle b such that pq is tangent to both circles. Given ac= 3 cm and bc= 8 cm, find pq
3.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
4. 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.
PLEASE ATLEAST ONE PLEASE THANKS
Answers
1. Make a sketch by placing the triangle on the x-y gird so that the altitude falls along the y-axis with (0,12) the top of the triangle and (0,) the middle of the base.
Look at the righ-angled triangle in the first quadrant.
You know the hypotenuse is 13 and the height is 12, so by Pythagoras the base must be 5
So our circle must pass through A(0,12) and B(5,0) and the centre must lie on the y-axis or on x = 0
You also know that the centre must lie on the right bisector of AB
Slope of AB = -12/5
so slope of right bisector is 5/12
midpoint of AB is (5/2 , 6)
equation of right-bisector:
y = (5/12)x + b , with (5/2 , 6) on it, so
6 = (5/12)(5/2) + b
b = 6 - 25/24 = 119/24
so the centre is at P(0 , 119/24)
and the radius is 12 - 119/24 = 169/24
or appr 7.04
2. Please use capital letters for vertices
I assume the centres of your circles are A and B
Since the circles are tangents to each other, AB is straight line and AB = 11
Also you will have right angles at P and Q
From A draw a line parallel to PQ to hit BQ at D
APQD will be a rectangle.
You will have a right-angled triangle and you can find AD, thus PQ
Give the others a try, I will not be available for the rest of the afternoon
Look at the righ-angled triangle in the first quadrant.
You know the hypotenuse is 13 and the height is 12, so by Pythagoras the base must be 5
So our circle must pass through A(0,12) and B(5,0) and the centre must lie on the y-axis or on x = 0
You also know that the centre must lie on the right bisector of AB
Slope of AB = -12/5
so slope of right bisector is 5/12
midpoint of AB is (5/2 , 6)
equation of right-bisector:
y = (5/12)x + b , with (5/2 , 6) on it, so
6 = (5/12)(5/2) + b
b = 6 - 25/24 = 119/24
so the centre is at P(0 , 119/24)
and the radius is 12 - 119/24 = 169/24
or appr 7.04
2. Please use capital letters for vertices
I assume the centres of your circles are A and B
Since the circles are tangents to each other, AB is straight line and AB = 11
Also you will have right angles at P and Q
From A draw a line parallel to PQ to hit BQ at D
APQD will be a rectangle.
You will have a right-angled triangle and you can find AD, thus PQ
Give the others a try, I will not be available for the rest of the afternoon
equation 2x-3y=10 touches the circle with centre m (-2,4)