
The distance between two parallel chords of lengths \[8\] cm and \[6\] cm in a circle of diameter \[10\] cm if the chords are on the same side of the center is
A.\[1\] cm
B.\[2\] cm
C.\[3\] cm
D.\[4\] cm
Answer
598.2k+ views
\[\]
Hint: We will drop a perpendicular line to both the chords from the center of the circle. We will then indirectly use the Pythagoras Theorem to find the distance between the two parallel chords by drawing two radii towards the end of each chord.
Complete step by step answer:
We are given the lengths of the two parallel chords, i.e., \[8\] cm and \[6\] cm respectively, and the diameter of the circle is given to be \[10\] cm.
If the diameter\[ = 10\] cm, then the radius of the circle\[ = \dfrac{{10}}{2} = 5\] cm.
We start with drawing the required diagram. First, we will draw a circle with the radius\[5\] cm. We then draw the two parallel chords on the same side of the center. Let the first parallel chord be\[AB\]and the second parallel chord be\[CD\]. So,\[AB = 8\] cm and\[CD = 6\] cm. Let the center of the circle be\[O\].
Let us now draw a line \[OP\]so that\[OP \bot AB\]and then join\[BO\].
Now, \[PB = \dfrac{1}{2}AB = \dfrac{1}{2} \times 8 = 4\] cm, since the perpendicular drawn divides the chord\[AB\]in two equal parts.
In \[\Delta BPO,\angle P = {90^ \circ }\]
Applying the Pythagoras Theorem, we get
\[
O{B^2} = P{B^2} + O{P^2} \\
\Rightarrow O{P^2} = O{B^2} - P{B^2} \\
\Rightarrow O{P^2} = {(5)^2} - {(4)^2} \\
\Rightarrow O{P^2} = 25 - 16 \\
\Rightarrow O{P^2} = 9 \\
\Rightarrow OP = \sqrt 9 \\
\Rightarrow OP = \pm 3 \\
\]
We only take the positive value of \[OP\]as the length of a side cannot be negative. \[\therefore OP = 3\] cm
Let us now extend the line\[OP\]till \[OQ\] so that\[OQ \bot CD\]and then join\[DO\].
Now, \[QD = \dfrac{1}{2}CD = \dfrac{1}{2} \times 6 = 3\] cm, since the perpendicular drawn divides the chord\[CD\]in two equal parts.
In \[\Delta DQO,\angle Q = {90^ \circ }\]
Applying the Pythagoras Theorem, we get
\[
O{D^2} = Q{D^2} + O{Q^2} \\
\Rightarrow O{Q^2} = O{D^2} - Q{D^2} \\
\Rightarrow O{Q^2} = {(5)^2} - {(3)^2} \\
\Rightarrow O{Q^2} = 25 - 9 \\
\Rightarrow O{Q^2} = 16 \\
\Rightarrow OQ = \sqrt {16} \\
\Rightarrow OQ = \pm 4 \\
\]
We only take the positive value of \[OQ\]as the length of a side cannot be negative. \[\therefore OQ = 4\] cm
\[\therefore \]The distance between the chords is \[PQ = OQ - OP = 4 - 3 = 1\] cm.
Thus, the answer is option A.
Note: In these types of questions, we need to remember that we have to use the Pythagoras Theorem to solve the problem. Also, when perpendicular is dropped from the center of the circle to any chord, it divides the chord into two equal parts.
Hint: We will drop a perpendicular line to both the chords from the center of the circle. We will then indirectly use the Pythagoras Theorem to find the distance between the two parallel chords by drawing two radii towards the end of each chord.
Complete step by step answer:
We are given the lengths of the two parallel chords, i.e., \[8\] cm and \[6\] cm respectively, and the diameter of the circle is given to be \[10\] cm.
If the diameter\[ = 10\] cm, then the radius of the circle\[ = \dfrac{{10}}{2} = 5\] cm.
We start with drawing the required diagram. First, we will draw a circle with the radius\[5\] cm. We then draw the two parallel chords on the same side of the center. Let the first parallel chord be\[AB\]and the second parallel chord be\[CD\]. So,\[AB = 8\] cm and\[CD = 6\] cm. Let the center of the circle be\[O\].
Let us now draw a line \[OP\]so that\[OP \bot AB\]and then join\[BO\].
Now, \[PB = \dfrac{1}{2}AB = \dfrac{1}{2} \times 8 = 4\] cm, since the perpendicular drawn divides the chord\[AB\]in two equal parts.
In \[\Delta BPO,\angle P = {90^ \circ }\]
Applying the Pythagoras Theorem, we get
\[
O{B^2} = P{B^2} + O{P^2} \\
\Rightarrow O{P^2} = O{B^2} - P{B^2} \\
\Rightarrow O{P^2} = {(5)^2} - {(4)^2} \\
\Rightarrow O{P^2} = 25 - 16 \\
\Rightarrow O{P^2} = 9 \\
\Rightarrow OP = \sqrt 9 \\
\Rightarrow OP = \pm 3 \\
\]
We only take the positive value of \[OP\]as the length of a side cannot be negative. \[\therefore OP = 3\] cm
Let us now extend the line\[OP\]till \[OQ\] so that\[OQ \bot CD\]and then join\[DO\].
Now, \[QD = \dfrac{1}{2}CD = \dfrac{1}{2} \times 6 = 3\] cm, since the perpendicular drawn divides the chord\[CD\]in two equal parts.
In \[\Delta DQO,\angle Q = {90^ \circ }\]
Applying the Pythagoras Theorem, we get
\[
O{D^2} = Q{D^2} + O{Q^2} \\
\Rightarrow O{Q^2} = O{D^2} - Q{D^2} \\
\Rightarrow O{Q^2} = {(5)^2} - {(3)^2} \\
\Rightarrow O{Q^2} = 25 - 9 \\
\Rightarrow O{Q^2} = 16 \\
\Rightarrow OQ = \sqrt {16} \\
\Rightarrow OQ = \pm 4 \\
\]
We only take the positive value of \[OQ\]as the length of a side cannot be negative. \[\therefore OQ = 4\] cm
\[\therefore \]The distance between the chords is \[PQ = OQ - OP = 4 - 3 = 1\] cm.
Thus, the answer is option A.
Note: In these types of questions, we need to remember that we have to use the Pythagoras Theorem to solve the problem. Also, when perpendicular is dropped from the center of the circle to any chord, it divides the chord into two equal parts.
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