Class 9 NCERT Solutions- Chapter 10 Circles - Exercise 10.5

In Class 9 mathematics, the study of circles is a fundamental topic that forms a cornerstone of the geometric principles. Chapter 10, "Circles" in the NCERT textbook delves into the various properties and theorems related to circles. Exercise 10.5 specifically focuses on the practical applications and problem-solving related to these properties. Understanding this exercise is crucial as it reinforces key concepts of circle geometry and prepares students for the more advanced topics in mathematics.

Circles

A circle is a fundamental shape in geometry defined as the set of all points in a plane that are equidistant from the fixed point known as the center. Chapter 10 of the Class 9 NCERT textbook explores the various properties and theorems related to the circles including the tangents, chords, and angles subtended by arcs. Exercise 10.5 challenges students to apply these concepts through practical problems enhancing their problem-solving skills and understanding of the geometric principles.

Question 1.In fig. 10.36, A, B, and C are three points on a circle with Centre O such that ∠BOC=30° and ∠AOB=60°. If D isa point on the circle other than the arc ABC, find ∠ADC.

Solution:

Given: ∠BOC=30° and ∠AOB=60°
To find: ∠ADC
Solution: ∠AOC=2∠ADC ---------[The angle subtended by an arc at the centre is double the angle the angle subtended by it any point on the remaining part of the circle.]
∠AOB+∠BOC=2∠ADC
60°+30°=2∠ADC
90+30=2∠ADC
90/2=∠ADC
45=∠ADC

Question 2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord ata point on the minor arc and also at a point on the major arc.

Solution:

Given: PQ=OP
To find: Angle on major arc is ∠A=?
Angle on the minor arc is ∠B=?
Since, =PO=OQ 
∴∠POQ=60° 
∠POQ=2∠PAQ [The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining point of the circle]
Reflex ∠POQ=360°-60°
Reflex ∠POQ=300°
Reflex ∠POQ=2∠POQ
300°=2∠PBQ
300°/2=∠PBQ
150°=∠PBQ

Question 3. In fig. 10.37, ∠PQR=100°,where P, Q and R are the points on a circle with centre O. Find ∠OPR.

Solution:

Given: ∠PQR=100°
To find: ∠OPR=?
Reflex ∠POR=2∠PQR --------[ The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining point of the circle]
Reflex ∠PQR=2*100
=200°
∠POR=360°-200°
Now in ∆POR,OP=QR [ Radii of same circle]
∠P=∠R and let each =x.
∴∠P+∠O+∠R=180° [angle sum property of ∆]
x+160°+x=180°-160°
2x+160°=180°
x=20°/2=10°
∴∠OPR=10°

Question 4. In fig. 10.38, ∠ADC=69°,∠ACB=31°,find ∠BDC.

Solution:

Given: ∠ABC=69°,∠ACB=31°
To find: ∠BDC=?
Solution: In ∆ABC
∠A+∠B+∠C=180° ---------[Angle sum property of ∆]
∠A+69°+31°=180°
∠A=180°-100°
∠A=80°
∠A and ∠D lie on the same segment therefore,
∠D=∠A
∠D=80°
∠BDC=80°

Question 5. In fig., 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:

Given: ∠BEC=130°,∠ECD=20°
To find: ∠BAC?
Solution: In ∆EDC
∠E=180°-130° ---------[linear pair]
∠E=50°
∠E+∠C+∠D=180° ------[angle sum property of triangle]
50°+20°+∠D=180°
70°+∠D=180°
∠D=180/70=110°
Since, ∠A and ∠D line in the same segment 
∴∠A=∠D 
∠A=110°
∠BAC=110°

Question 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC =70°, ∠BAC is 30°, find ∠BCD. Further, if AB=BC, find ∠ECD.

Solution:

Given: ABCD is a cyclic quadrilateral diagonal intersect at E ∠DBC=70°, ∠BAC is 30°. If AB=BC.
To find: ∠BCD and ∠ECD
∠BDC=∠BAC=30° -------[angle in the same segment]
In ∆BCD,
∠B+∠C+∠D=180° --------[angle sum property of triangle]
∠C+100°=180°
∠C=180°-100°=80°
∴∠BCD=80°
If AB=BC,
Then, ∠BAC=∠BCA
30°=∠BCA
Now, ∠BCA+∠ECD=∠BCD
30°+∠ECD=80°
∠ECD=80°-30°
∴∠ECD=50°

Question 7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:

Given: ABCD is a cyclic quadrilateral. Diagonals of ABCD are also diameters of circle.
To prove: ABCD is a rectangle
AC=BD ----------[diameters of same circle]
OA=OA ---------[radii of the same circle]
OA=OC=1/2AC ---------2
OB=OD=1/2BD ----------2
From I and 2 diagonals are equal and bisect each other
∴ABCD is a rectangle

Question 8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:

Draw DL perpendicular AB and EF perpendicular AB
In ∆DEA and ∆CEB
∠E=∠F --------[each 90°]
AD=BC --------[given]
DE=CF --------[distance between || lines is same every line]
∴∆DEA≅∆CFB --------[R.H.S]
∠A=∠B ---------[by c.p.c.t.] 1
∠1=∠2 (from 1)
Adding 90° on each sides
∠1+90°=∠2+90°
∠1+∠EDC=∠2+FCD
∠ADC=∠BCD
∠D=∠C 2
Now,
∠A+∠A+∠C+∠C=360°
2∠A+2∠C=360°
2(∠A+∠C)=360°
∠A+∠C=360°/2=190°
Because sum of opposite angles is 180°.
ABCD is parallelogram.

Question 9. Two circles intersect at two points B and C. Through B, two-line segments ABD and PBQ are drawn to intersect the circles at A, D, and P, Q respectively (see fig. 10.40). Prove that ∠ACP=∠QCD.

Solution:

To prove: ∠ACP=∠QCD or ∠1=∠2
∠1=∠2 ------ [angles in the same segment are equal] 1
∠ 3=∠ 4 ------- [angles in the same segment are equal] 2
∠2=∠4 ------- [vertically opposite angles] 3
From 1 2 and 3
∠1=∠3
∴∠ACP=∠QCB

Question 10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution:

Given: ABC is ∆ and AB and AC are diameters of two circles
To prove: Point of intersection is D, lies on the BC.
Construction: Join AD
∠ADB=90° -------[angles in semicircle] 1
∠ADC=90 ° ------[angles in semicircle] 2
Adding 1 and 2
∠ADB+∠ADC=90°+90°
∠BDC=180°
BDC is a straight line therefore D lies on BC.

Question 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD=∠CBD.

Solution:

Given: ABC and ADC are two right angle triangles with common hypotenuse AC.
To prove: ∠ADB=∠CBD
Solution: ∠ABC=∠ADC=90°
Circle drawn by taking AC as diameter passes through B and D.
For chord CD 
∠CAD=∠CBD -------[angle in the same segment]

Question 12. Prove that a cyclic parallelogram is rectangle.

Solution:

Given: ABC is a cyclic ||gm 
To prove: ABCD is a rectangle.
Because ABCD is a cyclic ||gm
∴∠A+∠C=180°
∠A=∠C [opposite angle of ||gm]
∴∠A=∠C=(180°)/2=90° 
∠A=90°
∠C=90°
Similarly,
∠B+∠D=180°
∴∠B=∠D =(180°)/2=90° ----------[opposite of a ||gm]
Each angle of ABCD is 90° 
∠B=90°
∠D=90°
Thus, ABCD is a rectangle.

Read More:

• Important points about Circle
• Circles in Maths