all the questions of this exercise
Answers
Exercise 10.4
1.
Answer
OP = 5cm, PS = 3cm and OS = 4cm.
also, PQ = 2PR
Let RS be x.
In ΔPOR,
OP2 = OR2 + PR2
⇒ 52 = (4-x)2 + PR2
⇒ 25 = 16 + x2 - 8x + PR2
⇒ PR2 = 9 - x2 + 8x --- (i)
In ΔPRS,
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.
Answer
Given,
AB and CD are chords intersecting at E.
AB = CD
To prove,
AE = DE and CE = BE
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.
Proof,
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.
Answer
Given,
AB and CD are chords intersecting at E.
AB = CD, PQ is the diameter.
To prove,
∠BEQ = ∠CEQ
Construction,
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.
Thus,
∠MEO = ∠NEO by CPCT
⇒ ∠BEQ = ∠CEQ
4.
Answer
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.
Answer
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
Thus,
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.
Answer
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.
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