\displaystyle \textbf{Question 1: } \text{A chord of length }6\text{ cm is drawn in a circle of radius }5\text{ cm}.
\displaystyle \text{Calculate its distance from the center of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the distance from the center be }x\text{ cm.}
\displaystyle \text{Half the chord }=\frac{6}{2}=3\text{ cm}
\displaystyle x=\sqrt{5^2-3^2}=\sqrt{25-9}=4\text{ cm}
\\

\displaystyle \textbf{Question 2: } \text{A chord of length }8\text{ cm is drawn at a distance of }3\text{ cm}
\displaystyle \text{from the center of a circle. Calculate the radius of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius be }r\text{ cm.}
\displaystyle \text{Half the chord }=\frac{8}{2}=4\text{ cm}
\displaystyle r=\sqrt{4^2+3^2}=\sqrt{16+9}=5\text{ cm}
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\displaystyle \textbf{Question 3: } \text{The radius of a circle is }17\text{ cm and the perpendicular from the center}
\displaystyle \text{to a chord is }8\text{ cm. Calculate the length of the chord.}
\displaystyle \text{Answer:}
\displaystyle \text{Let half the chord be }x\text{ cm.}
\displaystyle x=\sqrt{17^2-8^2}=\sqrt{289-64}=15\text{ cm}
\displaystyle \therefore \text{Length of the chord}=2x=30\text{ cm}
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\displaystyle \textbf{Question 4: } \text{A chord of length }24\text{ cm is at a distance of }5\text{ cm from the center}
\displaystyle \text{of a circle. Find the length of the chord of the same circle which is at a distance of }12\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius be }r\text{ cm.}
\displaystyle \text{Half the first chord}=\frac{24}{2}=12\text{ cm}
\displaystyle r=\sqrt{12^2+5^2}=\sqrt{144+25}=13\text{ cm}
\displaystyle \text{Let half the required chord be }x\text{ cm.}
\displaystyle x=\sqrt{13^2-12^2}=\sqrt{169-144}=5\text{ cm}
\displaystyle \therefore \text{Length of the required chord}=2x=10\text{ cm}
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\displaystyle \textbf{Question 5: } \text{In the following figure, }AD\text{ is a straight line, }OP\perp AD
\displaystyle \text{and }O\text{ is the center of both circles. If }OA=34\text{ cm},\ OB=20\text{ cm}
\displaystyle \text{and }OP=16\text{ cm},\text{ find the length of }AB.
\displaystyle \text{Answer:} c1
\displaystyle \text{Since the perpendicular from the center to a chord bisects the chord,}
\displaystyle P\text{ is the midpoint of both chords }AD\text{ and }BC.
\displaystyle \text{In right-angled }\triangle OPB,
\displaystyle PB=\sqrt{OB^2-OP^2}
\displaystyle =\sqrt{20^2-16^2}=\sqrt{400-256}=12\text{ cm}
\displaystyle \text{In right-angled }\triangle OPA,
\displaystyle PA=\sqrt{OA^2-OP^2}
\displaystyle =\sqrt{34^2-16^2}=\sqrt{1156-256}=30\text{ cm}
\displaystyle AB=AP-BP=30-12=18\text{ cm}
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\displaystyle \textbf{Question 6: } O\text{ is the center of a circle of radius }10\text{ cm}.
\displaystyle P\text{ is a point inside the circle such that }OP=6\text{ cm}.
\displaystyle A\text{ moves along the circumference and }AP=x\text{ cm}.
\displaystyle \text{Find the least and greatest values of }x\text{ and the positions of }O,P,A.
\displaystyle \hfill\text{[ICSE 1992]}
\displaystyle \text{Answer:}
\displaystyle OA=10\text{ cm and }OP=6\text{ cm}
\displaystyle \text{The least value of }AP\text{ occurs when }O,P,A\text{ are collinear,}
\displaystyle \text{with }P\text{ lying between }O\text{ and }A.
\displaystyle AP=OA-OP=10-6=4\text{ cm}
\displaystyle \therefore \text{The least value of }x\text{ is }4\text{ cm}.
\displaystyle \text{The greatest value of }AP\text{ occurs when }O,P,A\text{ are collinear,}
\displaystyle \text{with }O\text{ lying between }P\text{ and }A.
\displaystyle AP=AO+OP=10+6=16\text{ cm}
\displaystyle \therefore \text{The greatest value of }x\text{ is }16\text{ cm}.
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\displaystyle \textbf{Question 7: } \text{In a circle of radius }17\text{ cm, two parallel chords of lengths}
\displaystyle 30\text{ cm and }16\text{ cm are drawn. Find the distance between the chords if they are:}
\displaystyle \text{(i) on opposite sides of the center}
\displaystyle \text{(ii) on the same side of the center.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance of the }30\text{ cm chord from the center}
\displaystyle =\sqrt{17^2-\left(\frac{30}{2}\right)^2}
\displaystyle =\sqrt{289-225}=\sqrt{64}=8\text{ cm}
\displaystyle \text{Distance of the }16\text{ cm chord from the center}
\displaystyle =\sqrt{17^2-\left(\frac{16}{2}\right)^2}
\displaystyle =\sqrt{289-64}=\sqrt{225}=15\text{ cm}
\displaystyle \text{(i) When the chords are on opposite sides of the center,}
\displaystyle \text{distance between them}=8+15=23\text{ cm}
\displaystyle \text{(ii) When the chords are on the same side of the center,}
\displaystyle \text{distance between them}=15-8=7\text{ cm}
\\

\displaystyle \textbf{Question 8: } \text{Two parallel chords are drawn in a circle of diameter }30\text{ cm}.
\displaystyle \text{The length of one chord is }24\text{ cm and the distance between the chords is }21\text{ cm}.
\displaystyle \text{Find the length of the other chord.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the circle}=\frac{30}{2}=15\text{ cm}
\displaystyle \text{Half the }24\text{ cm chord}=\frac{24}{2}=12\text{ cm}
\displaystyle \text{Let its distance from the center be }x\text{ cm}.
\displaystyle x=\sqrt{15^2-12^2}=\sqrt{225-144}=9\text{ cm}
\displaystyle \text{Since the chords are on opposite sides of the center,}
\displaystyle \text{distance of the other chord from the center}=21-9=12\text{ cm}
\displaystyle \text{Let half the other chord be }y\text{ cm}.
\displaystyle y=\sqrt{15^2-12^2}=\sqrt{225-144}=9\text{ cm}
\displaystyle \therefore \text{Length of the other chord}=2y=18\text{ cm}
\\

\displaystyle \textbf{Question 9: } \text{A chord }CD\text{ of a circle with centre }O\text{ is bisected at }P
\displaystyle \text{by a diameter }AB.\text{ Given }OA=OB=15\text{ cm and }OP=9\text{ cm},
\displaystyle \text{calculate the lengths of (i) }CD\text{ (ii) }AD\text{ and (iii) }CB.
\displaystyle \text{Answer:} c2
\displaystyle \text{Since the diameter bisects the chord, }AB\perp CD.
\displaystyle CP=PD
\displaystyle \text{In right-angled }\triangle OPC,
\displaystyle CP=\sqrt{OC^2-OP^2}
\displaystyle =\sqrt{15^2-9^2}=\sqrt{225-81}=12\text{ cm}
\displaystyle \text{(i) }CD=2CP=2\times12=24\text{ cm}
\displaystyle \text{Also, }AP=AO+OP=15+9=24\text{ cm}
\displaystyle \text{(ii) In right-angled }\triangle APD,
\displaystyle AD=\sqrt{AP^2+PD^2}
\displaystyle =\sqrt{24^2+12^2}=\sqrt{720}=12\sqrt5\text{ cm}
\displaystyle \text{Therefore, }AD\approx26.83\text{ cm}
\displaystyle \text{Also, }PB=OB-OP=15-9=6\text{ cm}
\displaystyle \text{(iii) In right-angled }\triangle CPB,
\displaystyle CB=\sqrt{CP^2+PB^2}
\displaystyle =\sqrt{12^2+6^2}=\sqrt{180}=6\sqrt5\text{ cm}
\displaystyle \text{Therefore, }CB\approx13.42\text{ cm}
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\displaystyle \textbf{Question 10: } \text{The figure shows a circle with centre }O\text{ in which diameter }AB
\displaystyle \text{bisects chord }CD\text{ at }E.\text{ If }CE=ED=8\text{ cm and }EB=4\text{ cm},
\displaystyle \text{find the radius of the circle.}
\displaystyle \text{Answer:} c3
\displaystyle \text{Since the diameter bisects the chord, }AB\perp CD.
\displaystyle \therefore \angle OED=90^\circ
\displaystyle \text{Let }OE=x\text{ cm}.
\displaystyle \text{Then }OB=OE+EB=(x+4)\text{ cm}.
\displaystyle \text{Also, }OD=OB=(x+4)\text{ cm}\qquad\text{(radii of the same circle)}
\displaystyle \text{In right-angled }\triangle OED,
\displaystyle OD^2=OE^2+ED^2
\displaystyle (x+4)^2=x^2+8^2
\displaystyle x^2+8x+16=x^2+64
\displaystyle 8x=48
\displaystyle x=6
\displaystyle \therefore \text{Radius}=OB=x+4=6+4=10\text{ cm}
\\

\displaystyle \textbf{Question 11: } \text{The figure shows two concentric circles with centre }O.
\displaystyle AD\text{ is a chord of the larger circle and intersects the smaller circle at }B\text{ and }C.
\displaystyle \text{Prove that }AB=CD. c4
\displaystyle \text{Answer:}
\displaystyle \text{Draw }OE\perp AD,\text{ meeting }AD\text{ at }E.
\displaystyle \text{Since the perpendicular from the centre of a circle bisects its chord,}
\displaystyle AE=DE\qquad\text{and}\qquad BE=CE
\displaystyle AB=AE-BE
\displaystyle CD=DE-CE
\displaystyle \text{Since }AE=DE\text{ and }BE=CE,
\displaystyle AE-BE=DE-CE
\displaystyle \therefore AB=CD
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\displaystyle \textbf{Question 12: } \text{A straight line cuts two equal circles and passes through the midpoint }M
\displaystyle \text{of the line joining their centres }O\text{ and }O'.\text{ Prove that the intercepted chords}
\displaystyle AB\text{ and }CD\text{ are equal.} c5
\displaystyle \text{Answer:}
\displaystyle \text{Draw }OP\perp AB\text{ and }O'P'\perp CD,c12.jpg
\displaystyle \text{where }P\text{ and }P'\text{ lie on the given straight line.}
\displaystyle \text{Consider }\triangle OPM\text{ and }\triangle O'P'M.
\displaystyle \angle OPM=\angle O'P'M=90^\circ
\displaystyle \angle OMP=\angle O'MP'\qquad\text{(vertically opposite angles)}
\displaystyle OM=O'M\qquad\text{(since }M\text{ is the midpoint of }OO'\text{)}
\displaystyle \therefore \triangle OPM\cong\triangle O'P'M\qquad\text{(AAS congruence)}
\displaystyle \therefore OP=O'P'\qquad\text{(CPCTC)}
\displaystyle \text{Thus, the chords }AB\text{ and }CD\text{ of the equal circles are equidistant}
\displaystyle \text{from their respective centres.}
\displaystyle \therefore AB=CD
\\

\displaystyle \textbf{Question 13: } M\text{ and }N\text{ are the midpoints of two equal chords }AB\text{ and }CD
\displaystyle \text{respectively, of a circle with centre }O.\text{ Prove that:}
\displaystyle \text{(i) }\angle BMN=\angle DNM
\displaystyle \text{(ii) }\angle AMN=\angle CNM c6
\displaystyle \text{Answer:}
\displaystyle \text{Join }OM\text{ and }ON.
\displaystyle \text{Since }M\text{ and }N\text{ are the midpoints of chords }AB\text{ and }CD,
\displaystyle OM\perp AB\quad\text{and}\quad ON\perp CDc13
\displaystyle \text{Also, }AB=CD\qquad\text{(given)}
\displaystyle \therefore OM=ON
\displaystyle \text{Since equal chords are equidistant from the centre.}
\displaystyle \therefore \triangle OMN\text{ is isosceles.}
\displaystyle \therefore \angle OMN=\angle ONM\qquad\ldots\text{(1)}
\displaystyle \text{(i) Since }OM\perp AB,
\displaystyle \angle BMN=90^\circ-\angle OMN
\displaystyle \text{Since }ON\perp CD,
\displaystyle \angle DNM=90^\circ-\angle ONM
\displaystyle \text{Using (1),}
\displaystyle \therefore \angle BMN=\angle DNM
\displaystyle \text{(ii) Since }OM\perp AB,
\displaystyle \angle AMN=90^\circ+\angle OMN
\displaystyle \text{Since }ON\perp CD,
\displaystyle \angle CNM=90^\circ+\angle ONM
\displaystyle \text{Using (1),}
\displaystyle \therefore \angle AMN=\angle CNM
\\

\displaystyle \textbf{Question 14: } \text{In the following figure, }P\text{ and }Q\text{ are the points of intersection}
\displaystyle \text{of two circles with centres }O\text{ and }O'.\text{ The straight lines }APB\text{ and }CQD
\displaystyle \text{are parallel to }OO'.\text{ Prove that:}
\displaystyle \text{(i) }OO'=\frac{1}{2}AB\qquad\text{(ii) }AB=CD c7.jpg
\displaystyle \text{Answer:}
\displaystyle \text{Draw }OM\perp AB,\ O'N\perp AB,\ OM'\perp CD\text{ and }O'N'\perp CD.
\displaystyle \text{Since }AB\parallel OO'\text{ and }OM\perp AB,\text{ we have }OM\perp OO'.
\displaystyle \text{Similarly, }O'N\perp OO'.c14
\displaystyle \therefore OMNO'\text{ is a rectangle.}
\displaystyle \therefore MN=OO'\qquad\ldots\text{(1)}
\displaystyle \text{Since the perpendicular from the centre bisects a chord,}
\displaystyle MP=\frac{1}{2}AP\quad\text{and}\quad PN=\frac{1}{2}PB
\displaystyle MN=MP+PN
\displaystyle =\frac{1}{2}AP+\frac{1}{2}PB
\displaystyle =\frac{1}{2}(AP+PB)=\frac{1}{2}AB
\displaystyle \text{Using (1),}
\displaystyle \therefore OO'=\frac{1}{2}AB\qquad\ldots\text{(2)}
\displaystyle \text{Similarly, }OM'N'O'\text{ is a rectangle.}
\displaystyle \therefore M'N'=OO'
\displaystyle \text{Also, }M'Q=\frac{1}{2}CQ\quad\text{and}\quad QN'=\frac{1}{2}QD
\displaystyle M'N'=M'Q+QN'
\displaystyle =\frac{1}{2}CQ+\frac{1}{2}QD=\frac{1}{2}CD
\displaystyle \therefore OO'=\frac{1}{2}CD\qquad\ldots\text{(3)}
\displaystyle \text{From (2) and (3),}
\displaystyle \frac{1}{2}AB=\frac{1}{2}CD
\displaystyle \therefore AB=CD
\\

\displaystyle \textbf{Question 15: } \text{Two equal chords }AB\text{ and }CD\text{ of a circle with centre }O
\displaystyle \text{intersect each other at }P\text{ inside the circle. Prove that:}
\displaystyle \text{(i) }AP=CP\qquad\text{(ii) }BP=DP c8
\displaystyle \text{Answer:}
\displaystyle \text{Draw }OM\perp AB\text{ and }ON\perp CD.
\displaystyle \text{The perpendicular from the centre of a circle bisects the chord.}
\displaystyle \therefore AM=MB=\frac{1}{2}ABc15
\displaystyle \text{and }CN=ND=\frac{1}{2}CD
\displaystyle \text{Since }AB=CD,
\displaystyle MB=ND
\displaystyle \text{Equal chords of a circle are equidistant from its centre.}
\displaystyle \therefore OM=ON
\displaystyle \text{Consider }\triangle OPM\text{ and }\triangle OPN.
\displaystyle \angle OMP=\angle ONP=90^\circ
\displaystyle OP=OP\qquad\text{(common)}
\displaystyle OM=ON
\displaystyle \therefore \triangle OPM\cong\triangle OPN\qquad\text{(RHS congruence)}
\displaystyle \therefore PM=PN\qquad\text{(CPCTC)}
\displaystyle \text{(ii) }BP=BM+MP
\displaystyle DP=DN+NP
\displaystyle \text{Since }BM=DN\text{ and }MP=NP,
\displaystyle \therefore BP=DP
\displaystyle \text{(i) }AB=AP+PB\quad\text{and}\quad CD=CP+PD
\displaystyle AP=AB-PB
\displaystyle CP=CD-PD
\displaystyle \text{Since }AB=CD\text{ and }PB=PD,
\displaystyle \therefore AP=CP
\\

\displaystyle \textbf{Question 16: } \text{In the following figure, }OABC\text{ is a square. A circle with centre }O
\displaystyle \text{meets }OC\text{ at }P\text{ and }OA\text{ at }Q.\text{ Prove that:}
\displaystyle \text{(i) }\triangle OPA\cong\triangle OQC
\displaystyle \text{(ii) }\triangle BPC\cong\triangle BQA c9
\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }\triangle OPA\text{ and }\triangle OQC.
\displaystyle OP=OQ\qquad\text{(radii of the same circle)}
\displaystyle OA=OC\qquad\text{(sides of the square)}
\displaystyle \angle POA=\angle QOC=90^\circ
\displaystyle \therefore \triangle OPA\cong\triangle OQC\qquad\text{(SAS congruence)}
\displaystyle \text{(ii) }OC=OA\qquad\text{(sides of the square)}c16
\displaystyle OP=OQ\qquad\text{(radii of the same circle)}
\displaystyle OC-OP=OA-OQ
\displaystyle \therefore CP=QA
\displaystyle \text{Consider }\triangle BPC\text{ and }\triangle BQA.
\displaystyle BC=BA\qquad\text{(sides of the square)}
\displaystyle CP=QA
\displaystyle \angle PCB=\angle QAB=90^\circ
\displaystyle \therefore \triangle BPC\cong\triangle BQA\qquad\text{(SAS congruence)}
\\

\displaystyle \textbf{Question 17: } \text{The length of the common chord of two intersecting circles is }30\text{ cm}.
\displaystyle \text{If the diameters of the circles are }50\text{ cm and }34\text{ cm},
\displaystyle \text{calculate the distance between their centres.}
\displaystyle \text{Answer:} c17
\displaystyle \text{Let }P\text{ and }Q\text{ be the centres, and let the common chord }AB
\displaystyle \text{meet }PQ\text{ at }M.
\displaystyle \text{The line joining the centres bisects the common chord at right angles.}
\displaystyle \therefore AM=MB=\frac{30}{2}=15\text{ cm}
\displaystyle \text{Radius of the larger circle}=\frac{50}{2}=25\text{ cm}
\displaystyle \text{Radius of the smaller circle}=\frac{34}{2}=17\text{ cm}
\displaystyle \text{In right-angled }\triangle PMA,
\displaystyle PM=\sqrt{PA^2-AM^2}
\displaystyle =\sqrt{17^2-15^2}=\sqrt{289-225}=8\text{ cm}
\displaystyle \text{In right-angled }\triangle QMA,
\displaystyle QM=\sqrt{QA^2-AM^2}
\displaystyle =\sqrt{25^2-15^2}=\sqrt{625-225}=20\text{ cm}
\displaystyle \text{Since the centres lie on opposite sides of the common chord,}
\displaystyle PQ=PM+QM=8+20=28\text{ cm}
\displaystyle \therefore \text{The distance between the centres is }28\text{ cm}.
\\

\displaystyle \textbf{Question 18: } \text{The line joining the midpoints of two chords of a circle}
\displaystyle \text{passes through the centre. Prove that the chords are parallel.}
\displaystyle \text{Answer:} c18.jpg
\displaystyle \text{Let }L\text{ and }M\text{ be the midpoints of chords }AB\text{ and }CD\text{ respectively.}
\displaystyle \text{Given that }L,\ O\text{ and }M\text{ are collinear.}
\displaystyle AL=LB\qquad\text{and}\qquad CM=MD
\displaystyle \text{Since the line from the centre to the midpoint of a chord is perpendicular to the chord,}
\displaystyle OL\perp AB
\displaystyle \text{Similarly, }OM\perp CD
\displaystyle \text{Since }OL\text{ and }OM\text{ lie on the same straight line }LM,
\displaystyle AB\perp LM\qquad\text{and}\qquad CD\perp LM
\displaystyle \text{Two lines perpendicular to the same line are parallel.}
\displaystyle \therefore AB\parallel CD
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\displaystyle \textbf{Question 19: } \text{In the following figure, the line }ABCD\perp PQ,\text{ where }P\text{ and }Q
\displaystyle \text{are the centres of the circles, and }O\text{ is the point of intersection of }ABCD\text{ and }PQ.
\displaystyle \text{Show that: (i) }AB=CD\qquad\text{(ii) }AC=BD.
\displaystyle \text{Answer:} c10.jpg
\displaystyle \text{Since }PO\perp AD,\text{ the perpendicular from the centre bisects chord }AD.
\displaystyle \therefore AO=OD\qquad\ldots\text{(1)}
\displaystyle \text{Since }QO\perp BC,\text{ the perpendicular from the centre bisects chord }BC.
\displaystyle \therefore BO=OC\qquad\ldots\text{(2)}
\displaystyle \text{(i) }AB=AO-BO
\displaystyle CD=OD-OC
\displaystyle \text{Using (1) and (2), }AO-BO=OD-OC
\displaystyle \therefore AB=CD
\displaystyle \text{(ii) }AC=AO+OC
\displaystyle BD=BO+OD
\displaystyle \text{Using (1) and (2), }AO+OC=OD+BO
\displaystyle \therefore AC=BD
\\

\displaystyle \textbf{Question 20: } AB\text{ and }CD\text{ are two equal chords of a circle with centre }O
\displaystyle \text{which intersect at right angles at }P.\text{ If }OM\perp AB\text{ and }ON\perp CD,
\displaystyle \text{show that }OMPN\text{ is a square.}
\displaystyle \text{Answer:} c19
\displaystyle \text{Since }OM\perp AB\text{ and }P,M\text{ lie on }AB,
\displaystyle \angle OMP=90^\circ
\displaystyle \text{Since }ON\perp CD\text{ and }P,N\text{ lie on }CD,
\displaystyle \angle ONP=90^\circ
\displaystyle \text{Also, }AB\perp CD
\displaystyle \therefore \angle MPN=90^\circ
\displaystyle \text{Thus, }\angle MON=90^\circ
\displaystyle \therefore OMPN\text{ is a rectangle.}
\displaystyle \text{Since }AB=CD,\text{ the equal chords are equidistant from the centre.}
\displaystyle \therefore OM=ON
\displaystyle \text{Thus, }OMPN\text{ is a rectangle with two adjacent sides equal.}
\displaystyle \therefore OMPN\text{ is a square.}
\\

\displaystyle \textbf{Question 21: } \text{In the given figure, }O\text{ is the centre of the circle. }AB\text{ and }CD
\displaystyle \text{are two chords such that }OM\perp AB\text{ and }ON\perp CD.
\displaystyle \text{If }AB=24\text{ cm},\ OM=5\text{ cm and }ON=12\text{ cm, find:}
\displaystyle \text{(i) the radius of the circle}
\displaystyle \text{(ii) the length of chord }CD.\hfill\text{[ICSE 2014]} c11
\displaystyle \text{Answer:}
\displaystyle \text{Since the perpendicular from the centre bisects a chord,}
\displaystyle AM=MB=\frac{AB}{2}=\frac{24}{2}=12\text{ cm}
\displaystyle \text{(i) In right-angled }\triangle AOM,
\displaystyle AO^2=AM^2+OM^2c20.jpg
\displaystyle AO^2=12^2+5^2=144+25=169
\displaystyle AO=\sqrt{169}=13\text{ cm}
\displaystyle \therefore \text{The radius of the circle is }13\text{ cm}.
\displaystyle \text{(ii) Since }ON\perp CD,\ N\text{ is the midpoint of }CD.
\displaystyle \text{In right-angled }\triangle CNO,
\displaystyle CO^2=CN^2+ON^2
\displaystyle CN^2=CO^2-ON^2
\displaystyle CN^2=13^2-12^2=169-144=25
\displaystyle CN=\sqrt{25}=5\text{ cm}
\displaystyle CD=2CN=2\times5=10\text{ cm}
\displaystyle \therefore \text{The length of chord }CD\text{ is }10\text{ cm}.
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