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

• Last Updated : 13 Sep, 2021

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

Solution:

Given: âˆ BOC=30Â° and âˆ AOB=60Â°
Solution: âˆ AOC=2âˆ 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 âˆ A=?
Angle on the minor arc is âˆ B=?
Since, =PO=OQ
âˆ´âˆ POQ=60Â°
âˆ POQ=2âˆ 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 âˆ POQ=360Â°-60Â°
Reflex âˆ POQ=300Â°
Reflex âˆ POQ=2âˆ POQ
300Â°=2âˆ PBQ
300Â°/2=âˆ PBQ
150Â°=âˆ PBQ

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

Solution:

Given: âˆ PQR=100Â°
To find: âˆ OPR=?
Reflex âˆ POR=2âˆ 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 âˆ PQR=2*100
=200Â°
âˆ POR=360Â°-200Â°
Now in âˆ†POR,OP=QR [ Radii of same circle]
âˆ P=âˆ R and let each =x.
âˆ´âˆ P+âˆ O+âˆ R=180Â° [angle sum property of âˆ†]
x+160Â°+x=180Â°-160Â°
2x+160Â°=180Â°
x=20Â°/2=10Â°
âˆ´âˆ OPR=10Â°

### Question 4. In fig. 10.38, âˆ ADC=69Â°,âˆ ACB=31Â°,find âˆ BDC.

Solution:

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

### 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 âˆ BEC=130Â° and âˆ ECD=20Â°. Find âˆ BAC.

Solution:

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

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

Solution:

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

### 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
âˆ´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 âˆ†DEA and âˆ†CEB
âˆ E=âˆ F ——–[each 90Â°]
DE=CF ——–[distance between || lines is same every line]
âˆ´âˆ†DEAâ‰…âˆ†CFB ——–[R.H.S]
âˆ A=âˆ B ———[by c.p.c.t.] 1
âˆ 1=âˆ 2 (from 1)
âˆ 1+90Â°=âˆ 2+90Â°
âˆ 1+âˆ EDC=âˆ 2+FCD
âˆ D=âˆ C 2
Now,
âˆ A+âˆ A+âˆ C+âˆ C=360Â°
2âˆ A+2âˆ C=360Â°
2(âˆ A+âˆ C)=360Â°
âˆ A+âˆ C=360Â°/2=190Â°
Because sum of opposite angles is 180Â°.
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 âˆ ACP=âˆ QCD.

Solution:

To prove: âˆ ACP=âˆ QCD or âˆ 1=âˆ 2
âˆ 1=âˆ 2 —— [angles in the same segment are equal] 1
âˆ  3=âˆ  4 ——- [angles in the same segment are equal] 2
âˆ 2=âˆ 4 ——- [vertically opposite angles] 3
From 1 2 and 3
âˆ 1=âˆ 3
âˆ´âˆ ACP=âˆ 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 âˆ† and AB and AC are diameters of two circles
To prove: Point of intersection is D, lies on the BC.
âˆ ADB=90Â° ——-[angles in semicircle] 1
âˆ ADC=90 Â° ——[angles in semicircle] 2
âˆ BDC=180Â°
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 âˆ CAD=âˆ 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
âˆ CAD=âˆ 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
âˆ´âˆ A+âˆ C=180Â°
âˆ A=âˆ C [opposite angle of ||gm]
âˆ´âˆ A=âˆ C=(180Â°)/2=90Â°
âˆ A=90Â°
âˆ C=90Â°
Similarly,
âˆ B+âˆ D=180Â°
âˆ´âˆ B=âˆ D =(180Â°)/2=90Â° ———-[opposite of a ||gm]
Each angle of ABCD is 90Â°
âˆ B=90Â°
âˆ D=90Â°
Thus, ABCD is a rectangle.

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