I’m Up and Down and Round and Round Class 9 Important Questions Maths Chapter 5

During revision, students quickly go through Class 9 Maths Important Questions and Ganita Manjari Class 9 Maths Chapter 5 I’m Up and Down and Round and Round Important Questions with Solutions for clarity.

Class 9 I’m Up and Down and Round and Round Important Questions

Important Questions of I’m Up and Down and Round and Round Class 9

Very Short Answer Type Questions

Question 1.
In a circle, a chord is 10 cm away from the centre. If the radius of the circle is 26 cm, what is the length of the chord?
Solution:

In the above figure, CM = 10 cm, AC = 26 cm
Therefore, by using the Baudhayana-Pythagoras Theorem, we get
AM² = AC² – CM²
⇒ AM² = 26² – 10² = 676 – 100 = 576
⇒ AM = √576 = 24 cm
Now, AB = 2 × AM (perpendicular from the centre bisects the chord)
So, AB = 2 × 24 = 48 cm.
Therefore, the required length of the chord is 48 cm.

Question 2.
An arc of a circle subtends an angle of 120° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
Solution:
Since the angle subtended by an arc at the centre is twice the angle subtended by it at any point in the alternate segment.
Here, the angle subtended by the arc at the centre is 120°.
Therefore, the angle subtended by the arc at a point on the circle will be \(\frac{120^{\circ}}{2}\) = 60°

Question 3.
A circle has centre O and two chords AB and DE such that ∠AOB = 70° and ∠DOE = 70°. A student claims AB = DE. Is this claim correct?
Solution:

In ΔAOB and ΔDOE,
OA = OD = r [radii of the same circle]
OB = OE = r [radii of the same circle]
∠AOB = ∠DOE = 70° [Given]
∴ ΔAOB ≅ ΔDOE [By SAS congruence]
∴ AB = DE [Corresponding parts of congruent triangles]
∴ Yes, the claim of the student is correct.

Question 4.
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Solution:
Given: In two congruent circles, AB = PQ
To prove: ∠AOB = ∠PRQ

Proof:
In ΔAOB and ΔPRQ,
AB = PQ [Given]
OA = PR, OB = QR [Radii of congruent circles]
∴ ΔAOB ≅ ΔPRQ [By SSS congruence rule]
∴ ∠AOB = ∠PRQ [CPCT]

Question 5.
Write true or false and justify your answer in the following:
Two chords AB and CD of a circle are each at distances 4 cm from the centre. Then AB = CD.
Solution:
True
Justification: Chords equidistant from the centre of a circle are equal in length.

Question 6.
Write true or false and justify your answer in the following.
If AOB is a diameter of a circle and C is a point on the circle, then AC² + BC² = AB².
Solution:
True

Justification:
AB is a diameter.
∴ ∠ACB = 90° [Angle in a semicircle]
In ΔABC, ∠C = 90°
∴ AB² = AC² + BC² [By Baudhayana-Pythagoras theorem]

Question 7.
In the given figure, A, B, C, and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Solution:
∠ABD = ∠ACD = 20° [Angles in the same segment]
Now, ∠BEC is the exterior angle of ΔBAE.
∴ ∠BEC = ∠ABE + ∠BAE
⇒ 130° = ∠ABD + ∠BAC
⇒ 130° = 20° + ∠BAC
⇒ ∠BAC = 130° – 20° = 110°

Class 9 Maths I’m Up and Down and Round and Round Important Questions

Short Answer Type Questions

Question 1.
Two adjacent angles of a cyclic quadrilateral are of measures 30° and 57°. Find the measure of the other two angles of the cyclic quadrilateral.
Solution:
Let ∠A and ∠B be the adjacent angles of the given cyclic quadrilateral such that ∠A = 30° and ∠B = 57°.
Also, suppose that ∠C is opposite ∠A and ∠D is the angle opposite ∠B.
Also, the sum of opposite angles of a cyclic quadrilateral is 180°.
So, ∠A + ∠C = 180°
⇒ ∠C = 180° – ∠A = 180° – 30° = 150°
and also, ∠B + ∠D = 180°
⇒ ∠D = 180° – ∠B = 180° – 57° = 123°.
So, the other two angles of the cyclic quadrilateral are 150° and 123°.

Question 2.
In the following figure, ∠ABC = 45°, prove that OA ⊥ OC.

Solution:
∠ABC = \(\frac {1}{2}\) ∠AOC
(since the angle subtended at any point in the alternate segment is half of that subtended at the centre.)
⇒ ∠AOC = 2∠ABC
= 2 × 45°
= 90°
⇒ OA ⊥ OC proved.

Question 3.
In ΔLMN, LM = 7.2 cm, ∠M = 105°, MN = 6.4 cm, then draw ΔLMN and construct its circumcircle.
Solution:

Steps of Construction:

1. Construct ΔLMN of the given measurements.
2. Draw the perpendicular bisectors of side MN and side ML of the triangle.
3. Name the point of intersection of the perpendicular bisectors as point C.
4. With C as centre and CM as radius, draw a circle which passes through the three vertices of the triangle.

Question 4.
Construct ΔDEF such that DE = EF = 6 cm, ∠F = 45° and construct its circumcircle.
Solution:

Steps of Construction:

1. Construct ΔDEF of the given measurements.
2. Draw the perpendicular bisectors of side DE and side EF of the triangle.
3. Name the point of intersection of the perpendicular bisectors as point C.
4. With C as centre and CE as radius, draw a circle which passes through the three vertices of the triangle.

Question 5.
AB and AC are two equal chords of a circle. Show that the bisector of the angle BAC passes through the centre of the circle.
Solution:
Given: In a circle with centre O, AB = AC
To show: Bisector of ∠BAC passes through the centre O.
Construction: Draw OD ⊥ AB and OE ⊥ AC.

Proof:
AB = AC [Given]
∴ \(\frac {1}{2}\)AB = \(\frac {1}{2}\)AC [Multiplying both sides by \(\frac {1}{2}\)]
∴ AD = AE …….(i)
[Perpendicular from the centre of a circle to a chord bisects the chord]
In ∆AOD and ∆AOE,
AD = AE [From(i)]
AO = AO [Common side]
∠ADO = ∠AEO = 90° [By construction]
∴ ∆AOD ≅ ∆AOE [By RHS congruence]
∴ ∠DAO = ∠EAO [Corresponding parts of congruent triangles]
i.e. ∠BAO = ∠CAO
∴ AO is the bisector of ∠BAC.
Hence, the bisector of ∠BAC passes through the centre O.

Question 6.
If a line segment joining the midpoints of two chords of a circle passes through the centre of the circle, show that the two chords are parallel.
Solution:
Given: M and N are the midpoints of AB and CD respectively.
MN passes through O.
To show: AB || CD

Proof:
M is the midpoint of AB [Given]
OM ⊥ AB [The line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord]
i.e. MN ⊥ AB
Similarly, we can show that,
MN ⊥ CD
∴ AB || CD [Two lines perpendicular to the same line are parallel to each other]

Question 7.
If two equal chords of a circle intersect within the circle, show that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
Given: AB and CD are two chords of a circle intersecting at E, and AB = CD.
To show: ∠AEO = ∠DEO
Construction: Draw OM ⊥ AB, and ON ⊥ CD.
Join OE.

Proof:
AB = CD [Given]
∴ OM = ON …….(i)
[Equal chords of a circle are equidistant from the centre]
In ∆OME and ∆ONE,
OM = ON [From (i)]
∠OME = ∠ONE = 90° [By construction]
OE = OE [Common side]
∴ ∆OME ≅ ∆ONE [By RHS congruence rule]
∴ ∠MEO = ∠NEO [ Corresponding parts of congruent triangles]
∴ ∠AEO = ∠DEO

Question 8.
In the given figure, AB = CD, and AB ⊥ CD. Points E and F are midpoints of AB and CD, respectively. Show that OEGF is a square.

Solution:
Given: AB = CD, and AB ⊥ CD
E and F are midpoints of AB and CD
To show: OEGF is a square
Proof:
E and F are the midpoints of chords AB and CD, respectively.
∴ OE ⊥ AB, OF ⊥ CD ……(i)
[The line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord]
Also, AB ⊥ CD [Given]
∴ EG ⊥ GF ……..(ii)
Also, AB = CD [Given]
∴ OE = OF ……(iii)
[Equal chords of a circle are equidistant from the centre]
∴ OEGF is a square. [From (i), (ii), and (iii)]

Question 9.
Aatish is a comic book fan. Inspired by the superhero team Avengers, he creates an emblem for himself. It is as shown in the figure.

If ∠ABO = 20° and ∠ACO = 30°, then find ∠BOC.
Solution:

In ∆OAB, we have,
OA = OB [radii of the same circle]
∴ ∠OAB = ∠OBA = 20° [angles opposite to equal sides are equal]
In ∆OAC,
OA = OC [radii of the same circle]
∴ ∠OAC = ∠OCA = 30° [angles opposite to equal sides are equal]
Now, ∠BAC = ∠OAB + ∠OAC
= 20° + 30°
= 50°
∠BOC = 2∠BAC [The angle subtended by an arc at the centre is double the angle subtended by it at any other point on the remaining part of the circle]
= 2 × 50°
= 100°

Question 10.
Two chords AB and AC of a circle subtend angles equal to 90° and 150°, respectively at the centre. Find ∠BAC, if AB and AC lie on the opposite sides of the centre.
Solution:

∠BOC = 360° – (∠AOB + ∠AOC)
= 360° – ∠AOB – ∠AOC
= 360° – 90° – 150°
= 120°
Now, ∠BAC = \(\frac {1}{2}\)∠BOC [The angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle]
= \(\frac {1}{2}\) × 120°
= 60°

I’m Up and Down and Round and Round Important Questions Class 9

Long Answer Type Questions

Question 1.
In the following figure, O is the centre of the circle, ∠BCO = 30°. Find x and y.

Solution:
We know that the angle subtended by an arc at the centre is twice the angle subtended by it at any point in the alternate segment.
Therefore, ∠COD = 2∠CBD = 2 × y
Also, ∠OMC = 90°, ∠C = 30°
So, in ∆OMC, ∠MOC = 180° – ∠OMC – ∠C
= 180° – 90° – 30°
= 60°
Also, ∠MOD = 90° (given)
⇒ ∠COD + ∠MOC = 90°
⇒ 2y + 60° = 90°
⇒ 2y = 90° – 60° = 30°
⇒ y = 15°
∠ABD = \(\frac {1}{2}\)∠AOD (these are the angles subtended by the arc AD at a point on the circle and at the centre, respectively)
Therefore, ∠ABD = \(\frac {1}{2}\) × 90° = 45°
So, ∠ABM = ∠ABD + ∠y
= 45° + 15°
= 60°
Now, in ∆ABM, ∠AMB = 90°, ∠ABM = 60°
So, ∠MAB = 180° – ∠AMB – ∠ABM
= 180° – 90° – 60°
= 30°
Therefore, x = 30° and y = 15°.

Question 2.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm. Draw its circumcircle and measure its radius.
Solution:


Steps of Construction:

1. Draw base BC = 8 cm.
2. Draw the perpendicular bisector of BC, which intersects BC at P.
3. Take point P, draw an arc of 4 cm on the perpendicular bisector.
4. Join BA and CA. Thus, ABC is the required triangle.
5. Draw the perpendicular bisector of AB. It should intersect at P.
6. Taking P as centre and PA, PB or PC as radius, draw a circle which is the required circumcircle.

Question 3.
In the given figure, ∠OAB = 30° and ∠OCB = 57°. Find ∠BOC and ∠AOC.

Solution:
In ∆BOC, OB = OC [Radii of the circle]
∴ ∠OCB = ∠OBC = 57° [Angles opposite to equal sides]
Now, ∠BOC + ∠OCB + ∠OBC = 180° [Angle sum property of a triangle]
⇒ ∠BOC + 57° + 57° = 180°
⇒ ∠BOC + 114° = 180°
⇒ ∠BOC = 66°
Similarly, we can prove that
∠OAB = ∠OBA = 30°
In ∆AOB,
∠AOB + ∠OAB + ∠OBA = 180° [Angle sum property of a triangle]
⇒ ∠AOB + 30° + 30° = 180°
⇒ ∠AOB + 60° = 180°
⇒ ∠AOB = 120°
But, ∠AOB = ∠AOC + ∠BOC
⇒ 120° = ∠AOC + 66°
⇒ ∠AOC = 120° – 66°
⇒ ∠AOC = 54°

Question 4.
AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre respectively, show that 4q² = p² + 3r².
Solution:
Given: AB = 2AC, OM = p, ON = q, AO = r
To show: 4q² = p² + 3r²

Proof:
AB = 2AC [Given]
∴ \(\frac {1}{2}\)AB = 2 × \(\frac {1}{2}\)AC [Multiplying both sides by \(\frac {1}{2}\)]
∴ AM = 2AN [Perpendicular from the centre of a circle to a chord bisects the chord]
∴ AM² = 4AN² …..(i) [Squaring both sides]
In right-angled ∆ANO,
AO² = AN² + ON² [By Baudhayana-Pythagoras theorem]
⇒ r² = AN² + q²
⇒ AN² = r² – q²
⇒ 4AN² = 4(r² – q²) [Multiplying both sides by 4]
⇒ 4AN² = 4r² – 4q² ……..(ii)
Similarly, in right-angled ∆AMO,
∴ AO² = AM² + OM² [By Baudhayana-Pythagoras theorem]
⇒ r² = AM² + p²
⇒ AM² = r² – p² ……(iii)
⇒ r² – p² = 4r² – 4q² [From (i), (ii), and (iii)]
⇒ 4q² = p² + 3r²

Question 5.
Three girls, Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Solution:
Let the positions of Reshma, Salma, and Mandip be represented by points R, S, and M respectively, and let RK be x.

Construction:
Draw OL ⊥ RS.
OL ⊥ RS [By construction]
∴ RL = \(\frac {1}{2}\)RS [Perpendicular from the centre of a circle to a chord bisects the chord]
= \(\frac {1}{2}\) × 6 m
= 3 m
In right-angled ∆OLR,
∴ OR² = OL² + RL² [By Baudhayana-Pythagoras theorem]
⇒ 5² = OL² + 3²
⇒ OL² = 25 – 9
⇒ OL² = 16 m²
⇒ OL = 4 m
Now, RS = SM = 6 m [Given]
OR = OM = 5 m [Radii of a circle]
∴ ROMS is a kite.
∴ RM ⊥ OS [Diagonals of a kite are perpendicular to each other]
i.e., RK ⊥ OS
Area of ∆ORS = Area of ∆ORK + Area of ∆RKS
⇒ \(\frac {1}{2}\) × RS × OL = \(\frac {1}{2}\) × OK × RK + \(\frac {1}{2}\) × KS × RK
⇒ 6 × 4 = x(OK + KS)
⇒ 24 = x × OS
⇒ 5x = 24
⇒ x = 4.8 m
∴ RM = 2RK [Perpendicular from the centre of a circle to a chord bisects the chord]
= 2x
= 2 × 4.8 m
= 9.6 m
∴ The distance between Reshma and Mandip is 9.6 m.

Important Questions from I’m Up and Down and Round and Round Class 9

Case-Study Based Questions

Question 1.
In a mid-sized town, Bhagalpur, there is a circular park. There are two gates, A and B, as marked in the figure. The circular park has its centre at O. The distance between A and X is equal to the radius of the park.

Based on the information given above, answer the following questions:
(i) What kind of arc is the arc AXB?
(ii) What kind of arc is the arc AYB?
(iii) What is the measure of ∠AOX? Justify your answer.
(iv) Compare the angles ∠AOB and reflex ∠AOB. What is the sum of these angles?
Solution:
(i) AXB is the minor arc as it is smaller than a semicircle.
(ii) AYB is the minor arc as it is larger than a semicircle.
(iii) ∠AOX = 60°
In ∆AOX, AX = radius of the circle
So, AX = OX = OA
⇒ ∆AOX is an equilateral triangle.
Hence, each angle is 60°.
So, ∠AOX = 60°.
(iv) ∠AOB < reflex(∠AOB) (major arc subtends a larger angle at the centre than the minor arc)
Also, ∠AOB + reflex(∠AOB) = 360°.

Question 2.
The Olympic flag consists of five interlinked rings. The rings symbolise the union of five continents. Answer the following questions with the help of the given figures.

(i) AD is the diameter of the circle with centre O and AB is a chord. If AB = 30 cm, then find AE.

(ii) If AD = 34 m, then find OM.
(iii) (a) If circles are drawn taking two sides of a triangle as diameter, then prove that the point of intersection of these circles lies on the third side.
OR
(b) Two circles whose centres are O and O’ intersect at P. Through P, a line parallel to OO’ intersecting the circles at C and D is drawn as shown in the figure. Prove that CD = 2OO’.

Solution:
(i) Given that O is the centre and OE ⊥AB
∴ AE = \(\frac {1}{2}\)AB [The perpendicular from the centre of a circle to a chord bisects the chord]
= \(\frac {1}{2}\) × 30 cm
= 15 cm
(ii) Diameter = AD = 34 cm
∴ Radius = AO = \(\frac {34}{2}\) = 17 cm
In ∆AOE,
AO² = AE² + OE² [By Baudhayana-Pythagoras theorem]
⇒ 17² = 15² + OE²
⇒ OE² = 289 – 225 = 64
⇒ OE = √64 cm = 8 cm
(iii)

Given that AB and AC are the diameters of the circles.
∴ ∠ADB = 90°, ∠ADC = 90° [Angle inscribed in a semicircle]
Now, ∠ADB + ∠ADC = 90° + 90° = 180°
∴ BDC is a straight line [Linear pair axiom]
∴ D lies on side BC.
OR
(iii) Construction: Draw OA ⊥ CP and OB ⊥ PD
Join OO’.

Given that CD || OO’
OA ⊥ CP and OB ⊥ PD [by construction]
∴ OABO’ is a rectangle
∴ OO’ = AB
∴ OO’ = AP + PB ……(i)
But, AP = \(\frac {1}{2}\)CP
PB = \(\frac {1}{2}\)PD
[The perpendicular from the centre of a circle to a chord bisects the chord]
∴ OO’ = \(\frac {1}{2}\)CP + \(\frac {1}{2}\)PD [From (i)]
⇒ OO’ = \(\frac {1}{2}\)(CP + PD)
⇒ 2OO’ = CD

I’m Up and Down and Round and Round Class 9 Practice Questions

Very Short Answer Type Questions

Question 1.
In the following figure, find ∠ADC.

Answer:
45°

Question 2.
In the following figure, find ∠POR.

Answer:
160°

Short Answer Type Questions

Question 1.
In figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Answer:
80°

Question 2.
In the following figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

Answer:
80°

Long Answer Type Questions

Question 1.
Prove that in a circle the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Answer:
Consider all three cases.

Question 2.
If the sum of a pair of opposite angles of a quadrilateral is 180°, then prove that it is a cyclic quadrilateral.
Answer:
Use the theorem ‘angle subtended by an arc at the centre is twice the angle subtended by it at any point in the alternate segment’.

Question 3.
Prove that in a circle, angles in the same segment are equal.
Answer:
Use the theorem ‘angle subtended by an arc at the centre is twice the angle subtended by it at any point in the alternate segment’.

Case Study Based Questions

Question 1.
In a circular park, there is a path all around it. The radius of the park is 26 m. The radius of the inner circle is √584 m. A virtual chord AD is at 10 m from the centre O of the park. OM ⊥ AD.

Based on the information given above, answer the following questions:
(i) Is ∠OCB = ∠OBC? Give a reason for your answer.
(ii) Is ∠ODA = ∠OCB? Give reason for your answer.
(iii) What is the length of BC?
(iv) What is the width of the path around the park?
Answer:
(i) Yes, ∠OCB = ∠OBC, as ∆OBC is an isosceles triangle and base angles are equal.
(ii) No, ∠ODA ≠ ∠OCB
∠ODA < ∠OCB
(iii) BC = 44 m
(iv) 2 m

The post I’m Up and Down and Round and Round Class 9 Important Questions Maths Chapter 5 appeared first on Learn CBSE.

from Learn CBSE https://ift.tt/CVkBrxI
via IFTTT