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Class 9 NCERT Solutions- Chapter 10 Circles – Exercise 10.5
• Last Updated : 13 Jan, 2021

### 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°
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.] ### 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 remining 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 remining 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 ttriangle]
∠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°]
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)
∠1+90°=∠2+90°
∠1+∠EDC=∠2+FCD
∠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.
∠ADC=90 ° ——[angles in semicircle] 2
∠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.
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. My Personal Notes arrow_drop_up