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
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
equation 2x-3y=10 touches the circle with centre m (-2,4)
1. To find the radius of the circle inscribed in an isosceles triangle, we can use the following formula:
Radius = (Leg Length * Altitude to Base) / (2 * Area of Triangle)
Given that each leg of the isosceles triangle measures 13 cm, and the altitude to the base is 12 cm, we first need to find the area of the triangle.
Using the formula for the area of a triangle:
Area = (Base * Altitude) / 2
Since the triangle is isosceles, the base can be determined by using the Pythagorean theorem:
Base = sqrt((Leg Length^2) - (Altitude^2))
= sqrt((13^2) - (12^2))
= sqrt(169 - 144)
= sqrt(25)
= 5 cm
Plugging in the values into the formula for the area of the triangle:
Area = (5 * 12) / 2
= 60 / 2
= 30 cm^2
Now, we can calculate the radius of the circle:
Radius = (13 * 12) / (2 * 30)
= 156 / 60
= 13/5 cm
= 2.6 cm
Therefore, the radius of the circle is 2.6 cm.
2. To find the length of pq, we can use the following theorem:
The length of a common external tangent to two circles is equal to the difference of the radii.
Given that ac = 3 cm and bc = 8 cm, we can find the difference in radii:
Radii difference = bc - ac
= 8 cm - 3 cm
= 5 cm
Therefore, pq is also 5 cm.
3. To find the length of om, we can use the Pythagorean theorem in the right triangle oam:
Applying the Pythagorean theorem:
(om)^2 = (am)^2 - (ao)^2
= (63 cm)^2 - (52 cm)^2
= 3969 cm^2 - 2704 cm^2
= 1265 cm^2
Taking the square root of both sides:
om = sqrt(1265 cm^2)
= 35.6 cm (approx)
Therefore, om is approximately 35.6 cm.
4. To find the radius of the second circle, we can use the formula:
Radius of the second circle = (2 * Area of the triangle) / (Perimeter of the triangle)
Given that the sides of the triangle are 10, 10, and 12 units, we can calculate the area:
Semi-perimeter = (10 + 10 + 12) / 2
= 32 / 2
= 16 units
Area of the triangle = sqrt(Semi-perimeter * (Semi-perimeter - a) * (Semi-perimeter - b) * (Semi-perimeter - c))
= sqrt(16 * (16 - 10) * (16 - 10) * (16 - 12))
= sqrt(16 * 6 * 6 * 4)
= sqrt(2304)
= 48 units
Now, we can calculate the radius of the second circle:
Radius of the second circle = (2 * 48) / (10 + 10 + 12)
= 96 / 32
= 3 units
Therefore, the radius of the second circle is 3 units.
Sure, I can help you with one of the questions. Let's start with question 1:
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.
To solve this question, we can use the properties of an isosceles triangle inscribed in a circle. In such a triangle, the altitude from the vertex angle to the base will also bisect the base.
We can use the Pythagorean theorem to find the length of the base of the triangle. Let's call it 'b'.
Using the Pythagorean theorem, we have:
b^2 = 13^2 - 12^2
b^2 = 169 - 144
b^2 = 25
b = 5 cm
Now, the base of the triangle is divided into two equal parts by the altitude, so each part will have a length of 2.5 cm.
In an isosceles triangle inscribed in a circle, the radius of the circle is the perpendicular bisector of the base. So, in this case, the radius would be half of the base length, which is 2.5 cm.
Therefore, the radius of the circle is 2.5 cm.
I hope this helps! Let me know if you have any other questions.