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given n =11
middle term = (a1 + a11)/2 = 30
a1 + a11 = 60 .......... i
Sn = n/2(a1 + a11) ......... ii
= 11/2 (60) ... from i
= 11 x 30
= 330
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Let 1st term of AP be 'a',
Let the common difference be 'd',
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or, 7 [a+6d] = 11 [a+10d]
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or, a + 17d = 0 .......(1)
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Pratik Sabre asked in Math
PQ is a chord of length 8cm to a circle of radius 5cm. The tangents at P and Q intersect at a point T. Find the length TP.
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Rayya Halim answered this
<big id="4134851_vote_span">1732 helpful votes in Math, Class XII-Science</big>
Given, PQ is the chord of the circle and PT and QT are the tangents drawn at the end points of the chord PQ. PQ = 8cm and OP = 5cm.
OT ⊥ PQ,
∴ PR = RQ = 
(Perpendicular from the centre of the circle to a chord bisect the chord)
In right ΔOPR,
OP2 = PR2 + OR2
⇒ OR2 = OP2 – PR2
⇒ OR2 = (5cm)2 – (4cm2) = (25 – 16)cm2
⇒ OR2 = 9cm2
⇒ OR = 3cm
In right ΔPTR,
PT2 = TR2 + PR2 ...(1)
We know that, the tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠OPT = 90º
In right ΔOPT,
OT2 = PT2 + OP2 ...(2)
From (1) and (2), we get
OT2 = (TR2 + PR2) + OP2
⇒ (TR + OR)2 = (TR2 + PR2) + OP2
⇒ TR2 + OR2 + 2 × TR × OR = TR2 + PR2 + OP2
⇒ 9 + 6 TR = 16 + 25
⇒ 6TR = 25 + 16 – 9 = 32
⇒ TR = 
∴ PT2 = TR2 + PR2 (Using (1))
⇒ PT2 = 
+ (4cm)2
⇒ PT2 = 
⇒ PT 
Thus, the length of tangent PT is
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The area of triangle ABC=0
1/2(x(0-b)+a(b-y) +0(y-0))=0
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