Maharashtra Board Class 10 Math Part – 2 Solution Chapter 3 Practice Set 3.5 – 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 |
3.5 |
Practice set 3.5
(1) In figure 3.77, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.
Solution:
In the adjacent fig. It is given that,
PQ = 12, PR = 8
∴ By tangent scant segment theorem.
PQ2 = PR×PS
Or, 122 = 8×PS
Or, PS = 144/8 = 18
∴ RS = PS – PR = 18-8 = 10
(2) In figure 3.78, chord MN and chord RS intersect at point D.
1) If RD = 15, DS = 4,
MD = 8 find DN
2) If RS = 18, MD = 9,
DN = 8 find DS
Solution:
In the adjacent fig. It is given that,
(1) RD = 15, OC = 4, MD = 8
∴ by theorem of intersecting chords, MD × DN = RD × DS
Or, 8 × DN = 15 × 4
Or, DN = 15×4/8 = 15/2 = 7.5
(2) It is also given that,
RS = 18, MD = 9, DN = 8
Let, DS = x, ∴ RS = DS + RD
Or, 18 = x + RD
Or, RD = 18 – x
∴ By theorem of intersecting chords,
RD × DS = MD × DN
Or, (18 – x) = 9×
Or, -x2 + 18x = 72
Or, x2 – 18x = -72
Or, x2 – 18x + 72 = 0
Or, x2 – 6x – 12x + 72 = 0
Or, x (x – 6) – 12 (x – 6) = 0
Or, (x – 6) (x – 12) = 0
Either, x – 6 = 0
Or, x = 6
Or, x – 12 = 0
Or, x = 12
∴ DS = 6, 12
(3) In figure 3.79, O is the centre of the circle and B is a point of contact.
segOE⊥seg AD, AB = 12, AC = 8, find
(1) AD
(2) DC
(3) DE
Solution:
In the adjacent fig it is given that, O is the center of the circle.
OE⊥AD, AB = 12, AC = 8
∴ By tangent scant segment theorem,
(1) AB2 = AC×AD
Or, 122 = 8×AD
Or, AD = 144/8 = 18
(2) DC = AD – AC = 18 – 8 = 10
(3) Join seg OD and OC now, In ∆ODC
OD = OC [∵ OC & OD are radius of the circle]
∴ ∆ODC is an isosceles triangle with DC as base.
Also, OE is the altitude of ∆ODC
∴ DE = EC [∵ altitude to base of isosceles ∆bisects the base].
∴ DE + EC = DC
Or, DE + DE = DC [∵ DE = EC]
Or, 2DE = 10
Or, DE 10/2 = 5.
(4) In figure 3.80, if PQ = 6, QR = 10, PS = 8 find TS.
Solution:
In the adjacent fig. It is given that, PQ = 6, QR = 10, PS = 8
By theorem of intersecting chords outside the circle,
PR×PQ = PT × PS
Or, PR × 6 = (PS + ST) × PS
Or, (PQ + QR) × 6 = (8 + ST) × S
Or, (6+10) × 6 = 64+8×ST
Or, 96 – 64 = 8×ST
Or, ST = 32/8 = 4
(5) In figure 3.81, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r. Prove that, DE×GE = 4r2
Solution:
In the figure adjacent fig. It is given that, EF is the diameter, DF is a tangent radius of circle = r
∴ EF = 2r [∵ EF diameter]
Now, we know a tangent produced from one of the ends of a diameter is always perpendicular to the diameter.
∴ In ∆EFD
∠EFD = 90°
∴ By Pythagoras theorem,
EF2 + DF2 = DE2
Or, (2r)2 + DF2 = DE2
Or, DF2 = DE2 – 4r2 —- (i)
Or, DE2 – 4r2 = DG × DE [From equation (i)]
Or, 4r2 = DE2 – DG×DE
Or, 4r2 = DE (DE – DG)
Or, 4r2 = DE (DG + GE – DG) [∵ DE = DG + GE]
Or, 4r2 = DE × GE
Or, DE × GE = 4r2 (Proved)
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 3 Circle Practice Set 3.5
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