QR is the tangent to the circle whose centre is P. If QA ||RP and AB is the diameter, prove that RB is a tangent to the circle. pls..help... exam tomorrow...
Answers
Answered by
7
tangent to the circle = RQ
radius at point of contact Q = PQ
∴ PQ⊥RQ
∠PQR = 90° --------- (1)
QA || RP and PQ is the transverse.
∠2 = ∠3 --------- (2) (Alternate interior angles)
But, PQ = PA
∴ In ∆PQA, ∠3 = ∠4 ------- (3) (Angles opposite to equal sides are equal)
From (2) and (3) we get
∠2 = ∠3 = ∠4
The angles at the center P are ∠BPQ and ∠BAQ
∴ ∠BPQ = 2 ∠BAQ
∠1 + ∠2 = 2∠4
∠1 + ∠2 = ∠4 + ∠4
as ∠2 = ∠4
∠1 = ∠4
So, ∠1 = ∠2 = ∠3 = ∠4
⇒ ∠1 = ∠2
In ∆BPR and ∆RPQ,
BP = PQ (radii of the circle)
∠1 = ∠2
RP = RP (common)
∴ ∆BPR ≅ ∆QPR (SAS congruency)
⇒ ∠PBR = ∠PQR (corresponding angles)
But ∠PBR = 90° from equation (1)
i.e., PB ⊥ BR
Therefore, RB is a tangent to the circle at point B.
radius at point of contact Q = PQ
∴ PQ⊥RQ
∠PQR = 90° --------- (1)
QA || RP and PQ is the transverse.
∠2 = ∠3 --------- (2) (Alternate interior angles)
But, PQ = PA
∴ In ∆PQA, ∠3 = ∠4 ------- (3) (Angles opposite to equal sides are equal)
From (2) and (3) we get
∠2 = ∠3 = ∠4
The angles at the center P are ∠BPQ and ∠BAQ
∴ ∠BPQ = 2 ∠BAQ
∠1 + ∠2 = 2∠4
∠1 + ∠2 = ∠4 + ∠4
as ∠2 = ∠4
∠1 = ∠4
So, ∠1 = ∠2 = ∠3 = ∠4
⇒ ∠1 = ∠2
In ∆BPR and ∆RPQ,
BP = PQ (radii of the circle)
∠1 = ∠2
RP = RP (common)
∴ ∆BPR ≅ ∆QPR (SAS congruency)
⇒ ∠PBR = ∠PQR (corresponding angles)
But ∠PBR = 90° from equation (1)
i.e., PB ⊥ BR
Therefore, RB is a tangent to the circle at point B.
Answered by
2
In right ΔOSP
By Pythagoras theorem
OP² = OS² + PS²
r² = (r/2)² + PS²
PS² = r² – (r/2)²
= 3r²/4
so, PS = (√3r/2)
PR = 2PS = √3r
area of rhombus = 1/2 x d₁ x d₂
area of rhombus OPQR = 1/2 x OQ x PR
32√3 = 1/2 x r x √3r
32 = 1/2 r²
r² = 64
r = 8
Area of circle = πr²
= 22/7 x 8²
= 201. 14 sq cm
Similar questions