Advertisements
Advertisements
Question
In the given figure PT is a tangent to the circle. Chord BA produced meets the tangent PT at P.
Given PT = 20 cm and PA = 16 cm.
- Prove ΔPTB ~ ΔPAT
- Find the length of AB.
Solution
a. In ΔPTB and ΔPAT,
∠ PTA = ∠PBT ...(Alternate segment theorem)
∠TPA = ∠BPT ...(Common ∠)
∴ ΔPTB ~ ΔPAT ...(AA axiom)
b. PA × PB = PT2
`\implies` 16(16 + AB) = 400
`\implies` 16 + AB = 25
`\implies` AB = 9 cm
APPEARS IN
RELATED QUESTIONS
If the sides of a parallelogram touch a circle in following figure, prove that the parallelogram is a rhombus.
From the given figure, prove that : AP + BQ + CR = BP + CQ + AR.
Also show that : AP + BQ + CR = `1/2` × Perimeter of ΔABC.
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:
i)`∠`QPR .
In the following figure, PQ is the tangent to the circle at A, DB is the diameter and O is the centre of the circle. If ∠ADB = 30° and ∠CBD = 60°, calculate:
- ∠QAB,
- ∠PAD,
- ∠CDB.
If PQ is a tangent to the circle at R; calculate:
- ∠PRS,
- ∠ROT.
Given O is the centre of the circle and angle TRQ = 30°.
In the adjoining figure, O is the centre of the circle and AB is a tangent to it at point B. ∠BDC = 65°. Find ∠BAO.
In the given figure, PT touches a circle with centre O at R. Diameter SQ when produced to meet the tangent PT at P. If ∠SPR = x° and ∠QRP = y°; Show that x° + 2y° = 90°
In Fig. AP is a tangent to the circle at P, ABC is secant and PD is the bisector of ∠BPC. Prove that ∠BPD = `1/2` (∠ABP - ∠APB).
In the Figure, PT is a tangent to a circle. If m(∠BTA) = 45° and m(∠PTB) = 70°. Find m(∠ABT).
In the given figure, AC is a tangent to circle at point B. ∆EFD is an equilateral triangle and ∠CBD = 40°. Find:
- ∠BFD
- ∠FBD
- ∠ABF
