6. Circle

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Solution:

Problem Set 6 : Problem 1.1 : Textbook Page 86

In right \(\triangle\)AMO,

OA² = OM² + AM² ... (Pythgoras’ Theorem)

∴ 10² = 6² + AM²

∴ 100 = 36 + AM²

∴ 100 − 36 = AM²

∴ 64 = AM²

i.e. AM² = 64

∴ AM = \(\sqrt{64}\)

∴ AM = 8 cm ... (i)

Now, AB = 2AM

∴ AB = 2 × 8

∴ AB = 16 cm ... (ii)

∴ The correct option is: (C) 16 cm

The correct option is: (C) incentre

The correct option is: (A) circumcircle

Solution:

Problem Set 6 : Problem 1.4 : Textbook Page 86

AB = 24 cm ... (Given)

And OM \(\perp\) AB ... (Given)

∴ AM = \(\displaystyle \frac{1}{2}\) AB ... (Perpendicular drawn from the centre of a circle to a chord bisects the chord)

∴ AM = \(\displaystyle \frac{1}{2}\) AB

∴ AM = \(\displaystyle \frac{1}{2}\) × 24

∴ AM = 12 cm ... (i)

In right \(\triangle\)AMO,

OA² = OM² + AM² ... (Pythgoras’ Theorem)

∴ OA² = 5² + 12²

∴ OA² = 25 + 144

∴ OA² = 169

∴ OA = \(\sqrt{169}\)

∴ OA = 13 cm

∴ The correct option is: (B) 13 cm

Solution:

The longest chord of a circle is its diameter.

Diameter = 2 × Radius

Diameter = 2 × 2.9 cm

Diameter = 5.8 cm

∴ The correct option is: (D) 5.8 cm

Solution:

Radius of the circle = 4 cm

Distance of point P from the centre O = l(OP) = 4.2 cm

Since, 4.2 cm > 4 cm, point P lies outside the circle.

∴ The correct option is: (C) Outside the circle

Solution:

Problem Set 6 : Problem 1.7 : Textbook Page 86

AB = 6 cm ... (Given)

And OM \(\perp\) AB ... (Given)

∴ AM = \(\displaystyle \frac{1}{2}\) AB ... (Perpendicular drawn from the centre of a circle to a chord bisects the chord)

∴ AM = \(\displaystyle \frac{1}{2}\) AB

∴ AM = \(\displaystyle \frac{1}{2}\) × 6

∴ AM = 3 cm ... (i)

Also, CD = 8 cm ... (Given)

And ON \(\perp\) CD ... (Given)

∴ CN = \(\displaystyle \frac{1}{2}\) CD ... (Perpendicular drawn from the centre of a circle to a chord bisects the chord)

∴ CN = \(\displaystyle \frac{1}{2}\) × 8

∴ CN = 4 cm ... (ii)

Now, in right \(\triangle\)AMO,

OA² = OM² + AM² ... (Pythgoras’ Theorem)

∴ 5² = OM² + 3²

∴ 25 = OM² + 9

∴ 25 − 9 = OM²

∴ 16 = OM²

i.e. OM² = 16

∴ OM = \(\sqrt{16}\)

∴ OM = 4 cm ... (iii)

and, in right \(\triangle\)CNO,

OA² = ON² + CN² ... (Pythgoras’ Theorem)

∴ 5² = ON² + 4²

∴ 25 = ON² + 16

∴ 25 − 16 = ON²

∴ 9 = ON²

i.e. ON² = 9

∴ ON = \(\sqrt{9}\)

∴ ON = 3 cm ... (iv)

Now, M – O – N is a straight line.

∴ MN = OM + ON

∴ MN = 4 + 3 ... [From (iii) and (iv)]

∴ MN = 7 cm ... (v)

∴ The correct option is: (D) 7 cm

2. Construct incircle and circumcircle of an equilateral \(\triangle\)DSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.

ProblemSet6 : Problem 2 : Textbook Page 86

Solution:

Draw SP = 7.5 cm.
Centre S, radius 7.5 cm, draw an arc.
Centre P, radius 7.5 cm, draw an arc intersecting the previous arc at point D
Draw seg DS and seg DP.
Thus, \(\triangle\)DSP is drawn
Draw perpendicular bisector of side DS
Draw perpendicular bisector of side DP
Let the bisectors intersect at point O.
With O as centre and OS as radius, draw a circle.
This is the required circumcircle.
With O as centre and OM as radius, draw a circle.
This is the required incircle.
Measure the radius of circumcircle (OS) and incircle (OM).
OS = 4.4 cm and OM = 2.2 cm

Ratio of radius of circumcircle to the radius of incircle

\(\displaystyle = \frac{\text{OS}}{\text{OM}}\)

\(\displaystyle = \frac{4.4}{2.2}\)

\(\displaystyle = \frac{44}{22}\)

\(\displaystyle = \frac{2}{1}\)

∴ The ratio of radius of circumcircle to the radius of incircle is 2 ∶ 1.

3. Construct \(\triangle\)NTS where NT = 5.7 cm, TS = 7.5 cm and \(\angle\)NTS = 110° and draw incircle and circumcircle of it.

Problem Set 6 : Problem 3 : Textbook Page 86

Solution:

Draw TS = 7.5 cm.
At point T, make an angle of 110°.
With T as centre and radius 5.7 cm, draw an arc on the ray drawn in the previous step. Name the point N.
Draw seg NS.
Thus, \(\triangle\)NTS is constructed.
Draw the perpendicular bisector of NT.
Draw the perpendicular bisector of TS.
Let the bisectors intersect at point O.
With O as centre and OS as radius, draw a circle.
This is the required circumcircle.
Draw bisector of \(\angle\)N.
Draw bisector of \(\angle\)S.
Let the bisectors intersect at point I.
Draw IJ \(\perp\) NS.
With I as centre and IJ as radius, draw a circle.
This is the required incircle.

4. In the figure 6.19, C is the centre of the circle. seg QT is a diameter. CT = 13, CP = 5, find the length of chord RS.

Problem Set 6 : Problem 4 : Textbook Page 86

Solution:

Draw seg CR.

Now, CR = CT ... (Radii of the same circle)

But, CT = 13 ... (Given)

∴ CR = 13 units ... (i)

Now, in right \(\triangle\)CPR,

CR² = CP²+ PR² ... (Pythgoras’ Theorem)

∴ 13² = 5² + PR²

∴ 169 = 25 + PR²

∴ 169 − 25 = PR²

∴ 144 = PR²

i.e. PR² = 144

∴ PR = \(\sqrt{144}\)

∴ PR = 12 units

Also, CP \(\perp\) RS ... (Given)

∴ RS = 2 × PR ... (Theorem)

∴ RS = 2 × 12

∴ RS = 24 units ... (iii)

∴ The length of chord RS is 24 units.

5. In the figure 6.20, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If \(\angle\)AEP \(\cong\) \(\angle\)DEP then, prove that AB = CD.

Problem Set 6 : Problem 5 : Textbook Page 86

Proof:

Draw PM \(\perp\) AB and PN \(\perp\) CD.

In \(\triangle\)PME and \(\triangle\)PNE,

seg PE = seg PE ... (Common side)

\(\angle\)PME = \(\angle\)PNE ... (90° each – Construction)

∴ \(\angle\)MEP \(\cong\) \(\angle\)NEP ... (Given)

∴ \(\triangle\)PME \(\cong\) \(\triangle\)PNE ... (SAA test)

∴ seg PM = seg PN ... (c s c t)

Now, the chords of a circle equidistant from the centre of a circle are congruent. ... (Theorem)

∴ AB = CD ... (Chords equidistant from the centre are equal)

6. In the figure 6.21, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that \(\triangle\)ABC is an isosceles triangle.

Problem Set 6 : Problem 6 : Textbook Page 86

Proof:

OE \(\perp\) AB ... (Given)

∴ AE = BE ... (Theorem) ... (i)

Now, in \(\triangle\)ACE and \(\triangle\)BCE,

seg CE = seg CE ... (Common side)

\(\angle\)CEA = \(\angle\)CEB ... [(90° each – (Given)])

seg AE = seg BE ... [From (i)]

∴ \(\triangle\)ACE \(\cong\) \(\triangle\)BCE ... (SAS test)

∴ seg CA = seg CB ... (c s c t)

∴ \(\triangle\)ABC is an isosceles triangle.

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Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence the length of the chord is _____. 16 cm 8 cm 12 cm 32 cm

Solution:

The point of concurrence of all angle bisectors of a triangle is called the _____ . centroid circumcentre incentre orthocentre

The circle which passes through all the vertices of a triangle is called _____ . circumcircle incircle congruent circle concentric circle

Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is _____. 12 cm 13 cm 14 cm 15 cm

Solution:

The length of the longest chord of the circle with radius 2.9 cm is ____. 3.5 cm 6 cm 10 cm 5.8 cm

Solution:

Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie? On the centre Inside the circle Outside the circle On the circle

Solution:

The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is _____. 2 cm 1 cm 8 cm 7 cm

Solution:

2. Construct incircle and circumcircle of an equilateral \(\triangle\)DSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.

Solution:

3. Construct \(\triangle\)NTS where NT = 5.7 cm, TS = 7.5 cm and \(\angle\)NTS = 110° and draw incircle and circumcircle of it.

Solution:

4. In the figure 6.19, C is the centre of the circle. seg QT is a diameter. CT = 13, CP = 5, find the length of chord RS.

Solution:

5. In the figure 6.20, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If \(\angle\)AEP \(\cong\) \(\angle\)DEP then, prove that AB = CD.

Proof:

6. In the figure 6.21, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that \(\triangle\)ABC is an isosceles triangle.

Proof:

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