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2 years ago

Given circle X with radius 5 units and chord AB with length 8 units, what is the

Mathematics

1 answer:

kvv77 [185]2 years ago

3 0

Answer: The correct option is (B) 3.

Step-by-step explanation: We are given a circle X with radius 5 units and chord AB with length 8 units.

We are to find the length of segment XC that bisects chord.

We know that the line segment drawn from the center of a circle to the midpoint of a chord is perpendicular to the chord.

So, in the given circle X, the segment XC is perpendicular to chord AB. Then, triangle XCB will be a right angled triangle with hypotenuse XB.

Since XC bisects AB, so the length of BC will be

$BC=\dfrac{AB}{2}=\dfrac{8}{2}=4~\textup{units}.$

And, radius, XB = 5 units.

Using Pythagoras theorem in triangle XCB, we have

$XB^2=XC^2+BC^2\\\\\Rightarrow XC^2=XB^2-BC^2\\\\\Rightarrow XC^@=5^2-4^2\\\\\Rightarrow XC^2=9\\\\\Rightarrow XC^2=3^2\\\\\Rightarrow XC=3.$

Thus, the length of the segment XC is 3 units.

Option (B) is CORRECT.

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Read 2 more answers

Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1

astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

$x^{2} - 24 = 1$

Given that resulting expression is a second order polynomial of the form $x^{2} - a^{2}$ , there are two real and distinct solutions. Roots of the expression are:

$x_{1} = -5$ and $x_{2} = 5$ .

Now, it is also required to determine which part of the interval $(x_{1}, x_{2})$ is equal to a number greater than zero (positive). That is:

$x^{2} - 24 > 0$

$x^{2} > 24$

$x < -4.899$ and $x > 4.899$ .

Therefore, exists two sub-intervals: $[-5, -4.899]$ and $\left[4.899,5\right]$ . Besides, $x^{2} - 24 > y = 1$ in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

$A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx$

$A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx$

$A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}$

$A = 2.525 -2.474+9.798 + 2.525 - 2.474$

$A = 9.9\,units^{2}$

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

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2 years ago