Maharashtra Board Class 10 Math Part – 2 Solution Chapter 3 Practice Set 3.4 – Circle
Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 3: Circle. Marathi or English Medium Students of Class 10 get here Circle full Exercise Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 2 Solution |
Chapter |
Circle |
Practice Set |
3.4 |
Practice Set – 3.4
(1) In figure 3.56, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following.
(1) ∠AOB
(2) ∠ACB
(3) arc AB
(4) arc ACB
Solution:
In the adjacent fig it is given that.
O is the center of the circle.
∴ OA = OB = radius
AB is a chord equal to the radius of the circle.
∴ OA = OB = AB
∴ ∆AOB is are equilateral triangle
(1) ∴ ∠AOB = ∠OAB = ∠OBA = 60°
(2) m(arc AB) = 60° [∵∠AOB = 60°]
∴ ∠ACB = 1/2 m (arc AB) [Inscribed angle theorem]
(3) m arc (AB) = 60°
(4) ∴ m (arc ACB) = 360° – m (arc AB) [definition of measure of arc]
Or, m (arc ACB) = 360° – 60° = 300°
(2) In figure 3.57, PQRS is cyclic. Side PQ ≅ side RQ.∠PSR = 110°, Find –
(1) Measure of ∠PQR
(2) m (arc PQR)
(3) m (arc QR)
(4) Measure of ∠PRQ
Solution:
In the adjacent fig. It is given that, PQRS is cyclic.
PQ ≅ RQ, ∠PSR = 110°
(1) m (arc PSR) = 2×∠PSR [inscribed angle theorem]
Or, m (arc PSR) = 2×110
= 220°
∴ m (arc PSR) = 360° – m (arc PQR) [definition of answer of arc]
Or, m (arc PSR) = 360° – 220° = 140°
∴ ∠PQR = 1/2 m (arc PSR) [inscribed angle theorem]
= 1/2 × 140 = 70°
(2) m (arc PQR) = 140°
(3) In ∆PQR,
PQ = QR [Given]
∴ ∠QPR = ∠QRP = [Opposite angle of equal side of triangle are equal]
∴ ∠QPR = 180° – ∠PQR/2
= 180° – 70°/2 = 110/2 = 55°
∴ m (arc QR) = 2 × ∠QPR [inscribed angle theorem]
= 2 × 55°
= 110°
(4) ∠PRQ = ∠QPR = 55°
(3) MRPN is cyclic, ∠R = (5x – 13)°, ∠N = (4x + 4)°. Find measures of ∠R and ∠N.
Solution:
In the adjacent fig. It is given that, MRPN is cyclic.
∠N = (4x + 4)°, ∠R = (5x – 13)°
∴ m (arc MRP) = 2×∠N [Inscribed circle theorem]
Or, m (arc MRP) = 2 (4x + 4)°
= (8x + 8)°
Now, on (arc MNP) = 2×∠R [Inscribed circle theorem]
= 2 × (5x – 13)°
= (10x – 26)°
Also, m (arc MRP) + m (arc MNP) = 360 [Definition of measure of arc]
Or, 8x + 8 + 10x – 26 = 360°
Or, 18x – 18 = 360
Or, 18 (x – 1) = 360°
Or, x – 1 = 360°/18
Or, x = 20+1
= 21
∴ ∠N = 4x + 4
= 4 × 21 + 4
= 84 + 4
= 88°
∠R = 5x – 13
= 5 × 21 – 13
= 105 – 13
= 92°
(4) In figure 3.58, seg RS is a diameter of the circle with centre O. Point T lies in the exterior of the circle. Prove that ∠RTS is an acute angle.
Solution:
In the adjacent fig. It is given that, RS is diameter,
Now, join RT & ST and let’s assume.
RT meets the circle at point P.
∴ We know, diameter of a circle divides the circle into two equal arc.
∴ m (arc RS) = 360°/2 [definition of measure of arc]
= 360°/2 = 180°
∴ ∠RPS = 1/2 m (arc RS) [inscribed angle theorem]
= ½ × 180° = 90°
∴ In ∆PST,
∠RPS = ∠PTS + ∠TSP [Exterior angle theorem, since ∠RPS is an exterior angle to ∆PTS].
∴ ∠PTS < ∠RPS = 90°
∴ ∠PTS <90°
∴ ∠PTS is an acute angle.
(5) Prove that, any rectangle is a cyclic quadrilateral.
Solution:
Let, consider the adjacent rectangle ABCD.
We know,
∠A = ∠B = ∠C = ∠D = 90° [Angle of rectangle]
∴ ∠A + ∠C = 90+90 = 180°
Now, If a pair of opposite angle is supplementary than the quadrilateral is a cyclic quadrilateral.
∴ ABCD is a cyclic quadrilateral [converse of cyclic quadrilateral theorem]
(6) In figure 3.59, altitudes YZ and XT of ∆WXY intersect at P. Prove that,
(1) WZPT is cyclic
(2) Points X, Z, T, Y are concyclic.
Solution:
In the adjacent fig.
XT & YZ are altitudes of ∆NXY,
∴ ∠PTW = 180 – ∠PTW [Collinear angle]
= 180° – 90
= 90°
(1) ∠PZW = 180° – ∠XZY [collinear angle]
= 180° – 90
= 90°
∴ InWZPT,
∠PZW = ∠PTW = 90°
∴ ∠PZW + ∠PTW = 90° + 90° = 180°
WZPT is a cyclic quadrilateral [Converse of cyclic quadrilateral theorem]
∵ ∠PZW & ∠PTW are opposite angles of quad WZPT they are supplementary
(2) ∠XZY = ∠XTY = 90° [∵ ZY & XT are altitudes of ∆WXY]
XY is a line which subtends right angles on the same side of line XY at points Z & T.
∴ Point x, Y, Z, T are concyclic points.
(7) In figure 3.60, m (arc NS) = 125°, m (arc EF) = 37°, find the measure ∠NMS.
Solution:
In the adjacent fig. It is given that,
m (arc NS) = 125°
m (arc EF) = 37°
Construction, joint point ES
∠NES = ½ × m (arc NS) [inscribed angle theorem]
= ½ × 125 = 125°/2
∠ESF = ½ × m (arc EF)
= 1/2 × 37
= 37°/2
∴ In ∆ESM
∠NES is an exterior angle.
∠ESM = ∠ESF = 37°/2 [∵ M – F]
∴ ∠ENS + ∠ESM = 180° – ∠NES [Exterior angle theorem]
Or, ∠ENS = 180° – 125/2 – 37/2
= 360° – 125 – 37/2 = 99°
∴ ∠NMS = ∠EMS = 99° [∵ N – E – M]
(8) In figure 3.61, chords AC and DE intersect at B. If ∠ABE = 108°, m (arc AE) = 95°, find m (arc DC).
Solution:
In the adjacent fig. It is given then,
∠ABE = 108°
m (arc AE) = 95°
∠ADE = 1/2 m (arc AE) [intersect angle theorem]
= 1/2 × 95
= 95°/2
In ∆ADB,
∠ABE = 108° is exterior angle.
∠ADB = ∠ADE = 95°/2 [D – B – E]
∴ ∠ABE = ∠ADB + ∠DAB [exterior angle theorem]
Or, 108° = 95°/2 + ∠DAB
Or, ∠DAB = 108° – 95/2
= 216-95/2 = 121°/2
Now, ∠DAB = ∠DAC = 121°/2 [∵ A – B – C]
∴ m (arc DC) = 2 × ∠DAB [Inscribed angle theorem]
Or, m (arc DC) = 2 × 121/2
= 121°
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 3 Circle Practice Set 3.4
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