How to find the length of a chord - SAT Math

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Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{CX}$ is twice as long as $\overline{AX}$ , , and .

Give the length of $\overline{AX}$ .

Answer

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{CX}$ is twice this, or . The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot 20 = 2t \cdot 10$

This statement is identically true. Therefore, without further information, we cannot determine the value of - the length of $\overline{AX}$ .

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Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{BX}$ is 12 units longer than $\overline{AX}$ , , and .

Give the length of $\overline{AX}$ (nearest tenth, if applicable)

Answer

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{BX}$ is . The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot (t+12) = 8 \cdot 10$

$t \cdot t+t \cdot 12 = 8 0$

$t ^{2}+12 t = 8 0$

This quadratic equation can be solved by completing the square; since the coefficient of is 12, the square can be completed by adding

$\left (\frac{12}{2} \right ) ^{2} = 6 ^{2} = 36$

to both sides:

$t ^{2}+12 t + 36 = 8 0+ 36$

Restate the trinomial as the square of a binomial:

$(t+6 )^{2 } = 116$

Take the square root of both sides:

$t+6 = \pm \sqrt{116}$

$t+6 \approx - 10.8$ or $t+6 \approx 10.8$

Either

$t+6 \approx - 10.8$ ,

in which case

$t+6 - 6 \approx - 10.8 - 6$

$t \approx -16.8$ ,

or

$t+6 \approx 10.8$

in which case

$t+6 - 6 \approx 10.8 - 6$

$t \approx 4.8$ ,

Since is a length, we throw out the negative value; it follows that $t \approx 4.8$ , the correct length of $\overline{AX}$ .

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Question

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at point . and . Give the length of $\overline{CD}$ (nearest tenth, if applicable).

Answer

A diameter of a circle perpendicular to a chord bisects the chord. Therefore, the point of intersection is the midpoint of $\overline{CD}$ , and

$CX = DX = \frac{1}{2} CD$ .

Let stand for the common length of $\overline{CX}$ and $\overline{DX}$ ,

The figure referenced is below.

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$CX \cdot DX = AX \cdot BX$

Set and , and ; substitute and solve for :

$t \cdot t = 12 \cdot 20$

$t^{2 } = 240$

$t = \sqrt{240} \approx 15.5$

This is the length of $\overline{CX}$ ; the length of $\overline{CD}$ is twice this, so

$CD = 2 \cdot t \approx 2 \cdot 15.5 = 31.0$

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Question

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at a point .

What is the diameter of the circle?

Answer

In a circle, a diameter perpendicular to a chord bisects the chord. This makes the midpoint of $\overline{CD}$ ; consequently, $CX =DX = \frac{1}{2} CD = \frac{1}{2} \cdot 20 = 10$ .

The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting , and solving for :

$8 \cdot BX = 10 \cdot 10$

$8 \cdot BX = 100$

$8 \cdot BX \div 8 = 100 \div 8$

,

the correct length.

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Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point .

Give the length of $\overline{AX}$ .

Answer

Let , in which case ; the figure referenced is below (not drawn to scale).

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting , and solving for :

$AX \cdot BX = CX \cdot DX$

$t \cdot 12 = (t+6) \cdot 10$

$\frac{2t}{2} = \frac{60}{2}$

,

which is the length of $\overline{AX}$ .

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Question

Figure is not drawn to scale

In the provided diagram, the ratio of the length of $\overarc{BD}$ to that of $\overarc{AC}$ is 7 to 2. Evaluate the measure of $\angle BND$ .

Answer

The measure of the angle formed by the two secants to the circle from a point outside the circle is equal to half the difference of the two arcs they intercept; that is,

$m \angle BND = \frac{1}{2} ( \overarc{BD} - \overarc{AC})$

The ratio of the degree measure of $\overarc{BD}$ to that of $\overarc{AC}$ is that of their lengths, which is 7 to 2. Therefore,

$\frac{m \overarc{BD} }{m \overarc{AC} } = \frac{7}{2}$

Letting $t = m \overarc{AC}$ :

$\frac{m \overarc{BD} }{t } = \frac{7}{2}$

$\frac{m \overarc{BD} }{t } \cdot t = \frac{7}{2} \cdot t$

$m \overarc{BD} = \frac{7}{2} t$

Therefore, in terms of :

$m \angle BND = \frac{1}{2} \left ( \frac{7}{2} t - t \right ) = \frac{1}{2} \left ( \frac{5}{2} t \right ) =\frac{5}{4} t$

Without further information, however, we cannot determine the value of or that of $m \angle BND$ . Therefore, the given information is insufficient.

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Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{CX}$ is twice as long as $\overline{AX}$ , , and .

Give the length of $\overline{AX}$ .

Answer

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{CX}$ is twice this, or . The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot 20 = 2t \cdot 10$

This statement is identically true. Therefore, without further information, we cannot determine the value of - the length of $\overline{AX}$ .

Compare your answer with the correct one above

Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{BX}$ is 12 units longer than $\overline{AX}$ , , and .

Give the length of $\overline{AX}$ (nearest tenth, if applicable)

Answer

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{BX}$ is . The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot (t+12) = 8 \cdot 10$

$t \cdot t+t \cdot 12 = 8 0$

$t ^{2}+12 t = 8 0$

This quadratic equation can be solved by completing the square; since the coefficient of is 12, the square can be completed by adding

$\left (\frac{12}{2} \right ) ^{2} = 6 ^{2} = 36$

to both sides:

$t ^{2}+12 t + 36 = 8 0+ 36$

Restate the trinomial as the square of a binomial:

$(t+6 )^{2 } = 116$

Take the square root of both sides:

$t+6 = \pm \sqrt{116}$

$t+6 \approx - 10.8$ or $t+6 \approx 10.8$

Either

$t+6 \approx - 10.8$ ,

in which case

$t+6 - 6 \approx - 10.8 - 6$

$t \approx -16.8$ ,

or

$t+6 \approx 10.8$

in which case

$t+6 - 6 \approx 10.8 - 6$

$t \approx 4.8$ ,

Since is a length, we throw out the negative value; it follows that $t \approx 4.8$ , the correct length of $\overline{AX}$ .

Compare your answer with the correct one above

Question

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at point . and . Give the length of $\overline{CD}$ (nearest tenth, if applicable).

Answer

A diameter of a circle perpendicular to a chord bisects the chord. Therefore, the point of intersection is the midpoint of $\overline{CD}$ , and

$CX = DX = \frac{1}{2} CD$ .

Let stand for the common length of $\overline{CX}$ and $\overline{DX}$ ,

The figure referenced is below.

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$CX \cdot DX = AX \cdot BX$

Set and , and ; substitute and solve for :

$t \cdot t = 12 \cdot 20$

$t^{2 } = 240$

$t = \sqrt{240} \approx 15.5$

This is the length of $\overline{CX}$ ; the length of $\overline{CD}$ is twice this, so

$CD = 2 \cdot t \approx 2 \cdot 15.5 = 31.0$

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Question

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at a point .

What is the diameter of the circle?

Answer

In a circle, a diameter perpendicular to a chord bisects the chord. This makes the midpoint of $\overline{CD}$ ; consequently, $CX =DX = \frac{1}{2} CD = \frac{1}{2} \cdot 20 = 10$ .

The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting , and solving for :

$8 \cdot BX = 10 \cdot 10$

$8 \cdot BX = 100$

$8 \cdot BX \div 8 = 100 \div 8$

,

the correct length.

Compare your answer with the correct one above

Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point .

Give the length of $\overline{AX}$ .

Answer

Let , in which case ; the figure referenced is below (not drawn to scale).

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting , and solving for :

$AX \cdot BX = CX \cdot DX$

$t \cdot 12 = (t+6) \cdot 10$

$\frac{2t}{2} = \frac{60}{2}$

,

which is the length of $\overline{AX}$ .

Compare your answer with the correct one above

Question

Figure is not drawn to scale

In the provided diagram, the ratio of the length of $\overarc{BD}$ to that of $\overarc{AC}$ is 7 to 2. Evaluate the measure of $\angle BND$ .

Answer

The measure of the angle formed by the two secants to the circle from a point outside the circle is equal to half the difference of the two arcs they intercept; that is,

$m \angle BND = \frac{1}{2} ( \overarc{BD} - \overarc{AC})$

The ratio of the degree measure of $\overarc{BD}$ to that of $\overarc{AC}$ is that of their lengths, which is 7 to 2. Therefore,

$\frac{m \overarc{BD} }{m \overarc{AC} } = \frac{7}{2}$

Letting $t = m \overarc{AC}$ :

$\frac{m \overarc{BD} }{t } = \frac{7}{2}$

$\frac{m \overarc{BD} }{t } \cdot t = \frac{7}{2} \cdot t$

$m \overarc{BD} = \frac{7}{2} t$

Therefore, in terms of :

$m \angle BND = \frac{1}{2} \left ( \frac{7}{2} t - t \right ) = \frac{1}{2} \left ( \frac{5}{2} t \right ) =\frac{5}{4} t$

Without further information, however, we cannot determine the value of or that of $m \angle BND$ . Therefore, the given information is insufficient.

Compare your answer with the correct one above

Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{CX}$ is twice as long as $\overline{AX}$ , , and .

Give the length of $\overline{AX}$ .

Answer

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{CX}$ is twice this, or . The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot 20 = 2t \cdot 10$

This statement is identically true. Therefore, without further information, we cannot determine the value of - the length of $\overline{AX}$ .

Compare your answer with the correct one above

Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{BX}$ is 12 units longer than $\overline{AX}$ , , and .

Give the length of $\overline{AX}$ (nearest tenth, if applicable)

Answer

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{BX}$ is . The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot (t+12) = 8 \cdot 10$

$t \cdot t+t \cdot 12 = 8 0$

$t ^{2}+12 t = 8 0$

This quadratic equation can be solved by completing the square; since the coefficient of is 12, the square can be completed by adding

$\left (\frac{12}{2} \right ) ^{2} = 6 ^{2} = 36$

to both sides:

$t ^{2}+12 t + 36 = 8 0+ 36$

Restate the trinomial as the square of a binomial:

$(t+6 )^{2 } = 116$

Take the square root of both sides:

$t+6 = \pm \sqrt{116}$

$t+6 \approx - 10.8$ or $t+6 \approx 10.8$

Either

$t+6 \approx - 10.8$ ,

in which case

$t+6 - 6 \approx - 10.8 - 6$

$t \approx -16.8$ ,

or

$t+6 \approx 10.8$

in which case

$t+6 - 6 \approx 10.8 - 6$

$t \approx 4.8$ ,

Since is a length, we throw out the negative value; it follows that $t \approx 4.8$ , the correct length of $\overline{AX}$ .

Compare your answer with the correct one above

Question

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at point . and . Give the length of $\overline{CD}$ (nearest tenth, if applicable).

Answer

A diameter of a circle perpendicular to a chord bisects the chord. Therefore, the point of intersection is the midpoint of $\overline{CD}$ , and

$CX = DX = \frac{1}{2} CD$ .

Let stand for the common length of $\overline{CX}$ and $\overline{DX}$ ,

The figure referenced is below.

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$CX \cdot DX = AX \cdot BX$

Set and , and ; substitute and solve for :

$t \cdot t = 12 \cdot 20$

$t^{2 } = 240$

$t = \sqrt{240} \approx 15.5$

This is the length of $\overline{CX}$ ; the length of $\overline{CD}$ is twice this, so

$CD = 2 \cdot t \approx 2 \cdot 15.5 = 31.0$

Compare your answer with the correct one above

Question

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at a point .

What is the diameter of the circle?

Answer

In a circle, a diameter perpendicular to a chord bisects the chord. This makes the midpoint of $\overline{CD}$ ; consequently, $CX =DX = \frac{1}{2} CD = \frac{1}{2} \cdot 20 = 10$ .

The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting , and solving for :

$8 \cdot BX = 10 \cdot 10$

$8 \cdot BX = 100$

$8 \cdot BX \div 8 = 100 \div 8$

,

the correct length.

Compare your answer with the correct one above

Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point .

Give the length of $\overline{AX}$ .

Answer

Let , in which case ; the figure referenced is below (not drawn to scale).

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting , and solving for :

$AX \cdot BX = CX \cdot DX$

$t \cdot 12 = (t+6) \cdot 10$

$\frac{2t}{2} = \frac{60}{2}$

,

which is the length of $\overline{AX}$ .

Compare your answer with the correct one above

Question

Figure is not drawn to scale

In the provided diagram, the ratio of the length of $\overarc{BD}$ to that of $\overarc{AC}$ is 7 to 2. Evaluate the measure of $\angle BND$ .

Answer

The measure of the angle formed by the two secants to the circle from a point outside the circle is equal to half the difference of the two arcs they intercept; that is,

$m \angle BND = \frac{1}{2} ( \overarc{BD} - \overarc{AC})$

The ratio of the degree measure of $\overarc{BD}$ to that of $\overarc{AC}$ is that of their lengths, which is 7 to 2. Therefore,

$\frac{m \overarc{BD} }{m \overarc{AC} } = \frac{7}{2}$

Letting $t = m \overarc{AC}$ :

$\frac{m \overarc{BD} }{t } = \frac{7}{2}$

$\frac{m \overarc{BD} }{t } \cdot t = \frac{7}{2} \cdot t$

$m \overarc{BD} = \frac{7}{2} t$

Therefore, in terms of :

$m \angle BND = \frac{1}{2} \left ( \frac{7}{2} t - t \right ) = \frac{1}{2} \left ( \frac{5}{2} t \right ) =\frac{5}{4} t$

Without further information, however, we cannot determine the value of or that of $m \angle BND$ . Therefore, the given information is insufficient.

Compare your answer with the correct one above

Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{CX}$ is twice as long as $\overline{AX}$ , , and .

Give the length of $\overline{AX}$ .

Answer

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{CX}$ is twice this, or . The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot 20 = 2t \cdot 10$

This statement is identically true. Therefore, without further information, we cannot determine the value of - the length of $\overline{AX}$ .

Compare your answer with the correct one above

Question

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{BX}$ is 12 units longer than $\overline{AX}$ , , and .

Give the length of $\overline{AX}$ (nearest tenth, if applicable)

Answer

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{BX}$ is . The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot (t+12) = 8 \cdot 10$

$t \cdot t+t \cdot 12 = 8 0$

$t ^{2}+12 t = 8 0$

This quadratic equation can be solved by completing the square; since the coefficient of is 12, the square can be completed by adding

$\left (\frac{12}{2} \right ) ^{2} = 6 ^{2} = 36$

to both sides:

$t ^{2}+12 t + 36 = 8 0+ 36$

Restate the trinomial as the square of a binomial:

$(t+6 )^{2 } = 116$

Take the square root of both sides:

$t+6 = \pm \sqrt{116}$

$t+6 \approx - 10.8$ or $t+6 \approx 10.8$

Either

$t+6 \approx - 10.8$ ,

in which case

$t+6 - 6 \approx - 10.8 - 6$

$t \approx -16.8$ ,

or

$t+6 \approx 10.8$

in which case

$t+6 - 6 \approx 10.8 - 6$

$t \approx 4.8$ ,

Since is a length, we throw out the negative value; it follows that $t \approx 4.8$ , the correct length of $\overline{AX}$ .

Compare your answer with the correct one above

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