Ch 5 Solutions: I’m Up and Down, and Round and Round Class 9 Math NCERT

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Exercise Set 5.1

Q1. Draw ΔABC with AB = 5 cm, ∠A = 70°, and ∠B = 60°. Draw the
circumcircle of ΔABC. Is the centre inside or outside the triangle?

Solution:

Q2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the
circumcircle of ΔABC. Is the centre inside or outside the triangle?

Solution:

Q3. Draw ΔABC, with AB = 6 cm, BC = 7 cm, and CA = 7 cm. Draw
the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.

Solution:

Q4. What is the least possible radius of a circle through two points A and B?

Solution:

Exercise Set 5.2

Q1. Show that the triangle formed by a chord and the centre of the
circle is isosceles.

Solution:

Q2. Show that if two such isosceles triangles (occurring in the previous question)
have equal base length, they are congruent to each other.

Solution:

Exercise Set 5.3

Q1. Can you explain why the converse to Theorem 4 is true, i.e., why
does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?
(Hint: Use Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)

Solution:

Q2. An isosceles triangle ABC is inscribed in a circle, with AB = AC.
Show that the altitude from A to BC passes through the centre of
the circle.

Solution:

Q3. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides
of the centre of a circle. If the radius of the circle is 5 cm, find the
distance between the midpoints of the chords.

Solution:

Exercise Set 5.4

Q1. Use the Baudhāyana–Pythagoras theorem to show why the Theorem
6 must be true.

Solution:

Q2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH,
and CE= CH, show that AB = GF.

Solution:

Q3. Solve the previous question using the Baudhāyana–Pythagoras theorem.

Solution:

Exercise Set 5.5

Exercise Set 5.6

Q2. In a circle with centre O, the central angle AOB is 60°. If the radius
of the circle is 12 cm, what is the length of the chord AB?

Q3. Find x in Fig. 5.26.

End-of-Chapter Exercises

Q1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?

Given

• Radius of the circle $r = 13 \text{ cm}$
• Perpendicular distance of chord from centre $d = 5 \text{ cm}$

We know:

$\text{Chord Length}=2\sqrt{r^2-d^2}$

Substitute the values:$$\text{Chord Length} =2\sqrt{13^2-5^2}$$
$$=2\sqrt{169-25}$$
$$=2\sqrt{144}$$
$$=2\times 12$$
$$=24 \text{ cm}$$

24 cm Ans

Q2. An arc of a circle subtends an angle of 70° at the centre. What is
The measure of the angle subtended by the arc at a point on the circle?

Solution:

Given

The angle subtended by the arc at the centre is:$$70^\circ$$

We know the theorem:

$\angle\text{ at centre }=2\times\angle\text{ at circumference}$

Let the angle subtended at a point on the circle be $x$

Then,$$70^\circ = 2x$$

Divide both sides by 2:$$x=\frac{70^\circ}{2}$$
$$x=35^\circ$$

35. Ans

Q3. The diameter of a circle is 26 cm. A chord of length 24 cm is
drawn in the circle. Find the distance from the centre of the circle
to the chord.

Given

• Diameter of the circle $= 26 \text{ cm}$

So, radius:$$r=\frac{26}{2}=13 \text{ cm}$$

• Length of the chord $=24 \text{ cm}$

Half of the chord:$$\frac{24}{2}=12 \text{ cm}$$

Let:

• $O$ be the centre of the circle
• $AB$ be the chord
• $M$ be the midpoint of the chord $AB$

The perpendicular from the centre to a chord bisects the chord.

Therefore,$$AM=12 \text{ cm}$$

In right triangle $OMA$
$$OA=13 \text{ cm}$$
$$AM=12 \text{ cm}$$

Using the Baudhāyana–Pythagoras theorem:

$OA^2=OM^2+AM^2$

Substituting the values:$$13^2=OM^2+12^2$$
$$169=OM^2+144$$
$$OM^2=169-144$$
$$OM^2=25$$
$$OM=5 \text{ cm}$$

5 cm. Ans