# Class 9 NCERT Solutions- Chapter 10 Circles – Exercise 10.5

**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 is** **a 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 at** **a 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.