We can start by drawing a diagram:
Since we know that DC equals 28, and OD is the perpendicular bisector of DC, we can find OD by taking half of DC using the Pythagorean Theorem:
OD = sqrt((DC/2)^2 + BC^2) = sqrt((28/2)^2 + 15^2) = sqrt(196 + 225) = sqrt(421)
Now, we can use the Pythagorean Theorem on triangle OAB to find the length of the diameter:
OA^2 + AB^2 = OB^2
We know that OA equals OD (since they are both radii of the same circle), so we can substitute sqrt(421) for OA:
(sqrt(421))^2 + AB^2 = 15^2
Simplifying:
421 + AB^2 = 225
AB^2 = 225 - 421 = -196
Uh-oh! We have a negative number under the square root, which means that there is no real solution for AB. This could be because we made a mistake in our calculations, or it could be because the diagram is not drawn to scale and there is no solution that fits the given measurements.
Therefore, the answer is none of the above.
In the circle o b c equals 15 and DC equals 28 the diagram is not drawn to scale what is the length of diameter ba round to the nearest tenth.
A.37.3
B.42.7
C.48.9
D.56.2
8 answers
the circle O, BC= 15 and DC=28 ,the diagram is not drawn to scale what is the length of diameter BA round to the nearest tenth
I got 42.7
Unfortunately, your answer of 42.7 is not correct. As we showed in the previous answer, there is no real solution for the length of diameter BA given the measurements of the circle. It is possible that the diagram is not drawn to scale or that there is a mistake in the measurements given.
BC=15. DC=28
28^2= 15(BA)
784=15BA
15. 15
52.26=BA
28^2= 15(BA)
784=15BA
15. 15
52.26=BA
I'm sorry, but your calculations are incorrect. If we use the formula for finding the length of the diameter of a circle given the chord length and the distance from the center to the chord, we get:
BA = sqrt(4*r^2 - DC^2)
where r is the radius of the circle
Since BC is tangent to the circle at point C, we know that angle BOC is a right angle. Therefore, triangle BOC is a right triangle and we can use the Pythagorean Theorem to find the radius:
r^2 = BC^2 + OC^2
r^2 = 15^2 + (DC/2)^2
r^2 = 225 + 196
r^2 = 421
r = sqrt(421)
Now we can substitute this value for r in the formula for BA:
BA = sqrt(4*421 - 28^2)
BA = sqrt(1684 - 784)
BA = sqrt(900)
BA = 30
Therefore, the length of diameter BA is approximately 30, not 52.26 or 42.7.
BA = sqrt(4*r^2 - DC^2)
where r is the radius of the circle
Since BC is tangent to the circle at point C, we know that angle BOC is a right angle. Therefore, triangle BOC is a right triangle and we can use the Pythagorean Theorem to find the radius:
r^2 = BC^2 + OC^2
r^2 = 15^2 + (DC/2)^2
r^2 = 225 + 196
r^2 = 421
r = sqrt(421)
Now we can substitute this value for r in the formula for BA:
BA = sqrt(4*421 - 28^2)
BA = sqrt(1684 - 784)
BA = sqrt(900)
BA = 30
Therefore, the length of diameter BA is approximately 30, not 52.26 or 42.7.
Assume the segments that appear to be tangent are tangent find the value of x
5.6 outside circle
8.4 inside circle
A.5.6
B 6.3
C.11.2
D.14
8.4^2 + 5.6^2
70.56+ 31.36
101.92
5.6 outside circle
8.4 inside circle
A.5.6
B 6.3
C.11.2
D.14
8.4^2 + 5.6^2
70.56+ 31.36
101.92
To find the value of $x$, we need to use the fact that the two segments are tangents to the same circle. This means that they are "equal" in some sense: they have the same length, they have the same angle with the radius at the point of contact, etc. One way to use this fact is to draw a radius of the circle from the center to the point of contact $P$, and then use the Pythagorean theorem to write an equation involving $x$:
[asy]
size(80);
draw(circle((0,0),10));
draw((-14,0)--(14,0));
draw((0,-12)--(0,12));
label("$P$", (-8,6), SW);
label("$O$", (0,0), NE);
draw((-14,0)--(-8,6)--(0,0)--cycle);
draw(rightanglemark((-8,6),(0,0),(-14,0),20));
draw((-8,6)--(-8,0));
label("$x$", (-11,3));
label("$8.4$", (-4,3));
label("$5.6$", (-6,.2));
[/asy]
In this triangle $OCP$, the hypotenuse $OC$ has length $10$ (since it is the radius of the circle), and the leg $OP$ has length $8.4+x$. Therefore, we can write:
$$(8.4+x)^2 + 5.6^2 = 10^2$$
Expanding and simplifying gives:
$$x^2 + 16.8x + 54.4 = 0$$
Using the quadratic formula gives:
$$x = \frac{-16.8 \pm \sqrt{(-16.8)^2 - 4(1)(54.4)}}{2(1)}$$
Simplifying, we get:
$$x = \frac{-16.8 \pm \sqrt{98.56}}{2} = \frac{-16.8 \pm 9.928}{2}$$
So either $x=-13.864$ (which is negative and therefore doesn't make sense in this context), or $x=-3.936$. Therefore, the answer is not one of the options given.
[asy]
size(80);
draw(circle((0,0),10));
draw((-14,0)--(14,0));
draw((0,-12)--(0,12));
label("$P$", (-8,6), SW);
label("$O$", (0,0), NE);
draw((-14,0)--(-8,6)--(0,0)--cycle);
draw(rightanglemark((-8,6),(0,0),(-14,0),20));
draw((-8,6)--(-8,0));
label("$x$", (-11,3));
label("$8.4$", (-4,3));
label("$5.6$", (-6,.2));
[/asy]
In this triangle $OCP$, the hypotenuse $OC$ has length $10$ (since it is the radius of the circle), and the leg $OP$ has length $8.4+x$. Therefore, we can write:
$$(8.4+x)^2 + 5.6^2 = 10^2$$
Expanding and simplifying gives:
$$x^2 + 16.8x + 54.4 = 0$$
Using the quadratic formula gives:
$$x = \frac{-16.8 \pm \sqrt{(-16.8)^2 - 4(1)(54.4)}}{2(1)}$$
Simplifying, we get:
$$x = \frac{-16.8 \pm \sqrt{98.56}}{2} = \frac{-16.8 \pm 9.928}{2}$$
So either $x=-13.864$ (which is negative and therefore doesn't make sense in this context), or $x=-3.936$. Therefore, the answer is not one of the options given.