Here we provide NCERT Solutions for Class 9 MathsChapter 10 – Circles for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Class 9 Maths solution Chapter 10 pdf. Now you will get step by step solution to each question.
OP2 = OR2 + PR2
⇒ 52 = (4-x)2 + PR2
⇒ 25 = 16 + x2 – 8x + PR2
⇒ PR2 = 9 – x2 + 8x — (i)
PS2 = PR2 + RS2
⇒ 32 = PR2 + x2
⇒ PR2 = 9 – x2 — (ii)
Equating (i) and (ii),
9 – x2 + 8x = 9 – x2
⇒ 8x = 0
⇒ x = 0
Putting the value of x in (i) we get,
PR2 = 9 – 02
⇒ PR = 3cm
Length of the cord PQ = 2PR = 2×3 = 6cm
2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
AB and CD are chords intersecting at E.
AB = CD
AE = DE and CE = BE
OM ⊥ AB and ON ⊥ CD. OE is joined.
OM bisects AB (OM ⊥ AB)
ON bisects CD (ON ⊥ CD)
As AB = CD thus,
AM = ND — (i)
and MB = CN — (ii)
In ΔOME and ΔONE,
∠OME = ∠ONE (Perpendiculars)
OE = OE (Common)
OM = ON (AB = CD and thus equidistant from the centre)
ΔOME ≅ ΔONE by RHS congruence condition.
ME = EN by CPCT — (iii)
From (i) and (ii) we get,
AM + ME = ND + EN
⇒ AE = ED
From (ii) and (iii) we get,
MB – ME = CN – EN
⇒ EB = CE
3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
AB and CD are chords intersecting at E.
AB = CD, PQ is the diameter.
∠BEQ = ∠CEQ
OM ⊥ AB and ON ⊥ CD. OE is joined.
In ΔOEM and ΔOEN,
OM = ON (Equal chords are equidistant from the centre)
OE = OE (Common)
∠OME = ∠ONE (Perpendicular)
ΔOEM ≅ ΔOEN by RHS congruence condition.
∠MEO = ∠NEO by CPCT
⇒ ∠BEQ = ∠CEQ
4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).
OM ⊥ AD is drawn from O.
OM bisects AD as OM ⊥ AD.
⇒ AM = MD — (i)
also, OM bisects BC as OM ⊥ BC.
⇒ BM = MC — (ii)
From (i) and (ii),
AM – BM = MD – MC
⇒ AB = CD
5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m 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 6m each, what is the distance between Reshma and Mandip?
Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.
AB = 6cm and BC = 6cm.
Radius OA = 5cm
BM ⊥ AC is drawn.
ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
Applying Pythagoras theorem in ΔOAM,
OA2 = OM2 + AM2
⇒ 52 = x2 + y2 — (i)
Applying Pythagoras theorem in ΔAMB,
AB2 = BM2 + AM2
⇒ 62 = (5-x)2 + y2 — (ii)
Subtracting (i) from (ii), we get
36 – 25 = (5-x)2 – x2 –
⇒ 11 = 25 – 10x
⇒ 10x = 14 ⇒ x= 7/5
Substituting the value of x in (i), we get
y2 + 49/25 = 25
⇒ y2 = 25 – 49/25
⇒ y2 = (625 – 49)/25
⇒ y2 = 576/25
⇒ y = 24/5
AC = 2×AM = 2×y = 2×(24/5) m = 48/5 m = 9.6 m
Distance between Reshma and Mandip is 9.6 m.
6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle a metres then BD = a/2 m.
Applying Pythagoras theorem in ΔABD,
AB2 = BD2 + AD2
⇒ AD2 = AB2 – BD2
⇒ AD2 = a2 – (a/2)2
⇒ AD2 = 3a2/4
⇒ AD = √3a/2
OA = 2/3 AD ⇒ 20 m = 2/3 × √3a/2
⇒ a = 20√3 m
Length of the string is 20√3 m.
Page No: 184
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.
∠AOC = ∠AOB + ∠BOC
⇒ ∠AOC = 60° + 30°
⇒ ∠AOC = 90°
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.
∠ADC = 1/2∠AOC = 1/2 × 90° = 45°
Page No: 185
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.
AB is equal to the radius of the circle.
OA = OB = AB = radius of the circle.
Thus, ΔOAB is an equilateral triangle.
∠AOC = 60°
∠ACB = 1/2 ∠AOB = 1/2 × 60° = 30°
ACBD is a cyclic quadrilateral,
∠ACB + ∠ADB = 180° (Opposite angles of cyclic quadrilateral)
⇒ ∠ADB = 180° – 30° = 150°
Thus, angle subtend by the chord at a point on the minor arc and also at a point on the major arc are 150° and 30° respectively.
3. In Fig. 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Reflex ∠POR = 2 × ∠PQR = 2 × 100° = 200°
∴ ∠POR = 360° – 200° = 160°
OP = OR (radii of the circle)
∠OPR = ∠ORP
∠OPR + ∠ORP +∠POR = 180° (Sum of the angles in a triangle)
⇒ ∠OPR + ∠OPR + 160° = 180°
⇒ 2∠OPR = 180° – 160°
⇒ ∠OPR = 10°
4. In Fig. 10.38, ∠ABC = 69°, ∠ ACB = 31°, find ∠BDC.
∠BAC = ∠BDC (Angles in the segment of the circle)
∠BAC + ∠ABC + ∠ACB = 180° (Sum of the angles in a triangle)
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠BAC = 180° – 100°
⇒ ∠BAC = 80°
Thus, ∠BDC = 80°
5. In Fig. 10.39, 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.
∠BAC = ∠CDE (Angles in the segment of the circle)
∠CEB = ∠CDE + ∠DCE (Exterior angles of the triangle)
⇒ 130° = ∠CDE + 20°
⇒ ∠CDE = 110°
Thus, ∠BAC = 110°
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.
For chord CD,
∠CBD = ∠CAD (Angles in same segment)
∠CAD = 70°
∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠BCD + 100° = 180°
⇒ ∠BCD = 80°
AB = BC (given)
∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)
∠BCA = 30°
also, ∠BCD = 80°
∠BCA + ∠ACD = 80°
⇒ 30° + ∠ACD = 80°
∠ACD = 50°
∠ECD = 50°
7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Let ABCD be a cyclic quadrilateral and its diagonal AC and BD are the diameters of the circle through the vertices of the quadrilateral.
∠ABC = ∠BCD = ∠CDA = ∠DAB = 90° (Angles in the semi-circle)
Thus, ABCD is a rectangle as each internal angle is 90°.
8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
ABCD is a trapezium where non-parallel sides AD and BC are equal.
DM and CN are perpendicular drawn on AB from D and C respectively.
ABCD is cyclic trapezium.
In ΔDAM and ΔCBN,
AD = BC (Given)
∠AMD = ∠BNC (Right angles)
DM = CN (Distance between the parallel lines)
ΔDAM ≅ ΔCBN by RHS congruence condition.
∠A = ∠B by CPCT
also, ∠B + ∠C = 180° (sum of the co-interior angles)
⇒ ∠A + ∠C = 180°
Thus, ABCD is a cyclic quadrilateral as sum of the pair of opposite angles is 180°.
Page No: 186
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.
Chords AP and DQ are joined.
For chord AP,
∠PBA = ∠ACP (Angles in the same segment) — (i)
For chord DQ,
∠DBQ = ∠QCD (Angles in same segment) — (ii)
ABD and PBQ are line segments intersecting at B.
∠PBA = ∠DBQ (Vertically opposite angles) — (iii)
By the equations (i), (ii) and (iii),
∠ACP = ∠QCD
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.
Two circles are drawn on the sides AB and AC of the triangle ΔABC as diameters. The circles intersected at D.
AD is joined.
D lies on BC. We have to prove that BDC is a straight line.
∠ADB = ∠ADC = 90° (Angle in the semi circle)
∠ADB + ∠ADC = 180°
⇒ ∠BDC is straight line.
Thus, D lies on the BC.
11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
AC is the common hypotenuse. ∠B = ∠D = 90°.
∠CAD = ∠CBD
Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi circle and AC is the diameter of the circle.
⇒ Points A, B, C and D are concyclic.
Thus, CD is the chord.
⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)
12. Prove that a cyclic parallelogram is a rectangle.
ABCD is a cyclic parallelogram.
ABCD is rectangle.
∠1 + ∠2 = 180° (Opposite angles of a cyclic parallelogram)
also, Opposite angles of a cyclic parallelogram are equal.
∠1 = ∠2
⇒ ∠1 + ∠1 = 180°
⇒ ∠1 = 90°
One of the interior angle of the paralleogram is right angled. Thus, ABCD is a rectangle.
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