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Ask QuestionPosted by Mayank ... 6 years, 10 months ago
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Posted by Hiten Jain 6 years, 10 months ago
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Pranav Bhutada 6 years, 10 months ago
Posted by Aniket Dhama 6 years, 10 months ago
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Posted by Harsh Rathote 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Steps of construction:
- Draw circle with centre O and radius OA = 5 cm, .
- Mark another point B on the circle such that ∠AOB = 120°, supplementary to the angle between the tangents. Since the angle between the tangents to be constructed is 60°.
∴ ∠AOB = 180° – 60° = 120°.
3.Construct angles of 90° at A and B and extend the lines so as to intersect at point P.
4. Thus, AP and BP are the required tangents to the circle.
Posted by Vedansh Chauhan 6 years, 10 months ago
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Posted by Sandhya Chauhan 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Pythagoras Theorem
Statement: In a right angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
Given: A right triangle ABC right angled at B.
To prove: AC2 = AB2 + BC2
Construction: Draw BD
AC
Proof :
We know that: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
ADB
ABC
So,
(Sides are proportional)
Or, AD.AC = AB2 ... (1)
Also,BDCABC
So,
Or, CD. AC = BC2 ... (2)
Adding (1) and (2),
AD. AC + CD. AC = AB2 + BC2
AC (AD + CD) = AB2 + BC2
AC.AC = AB2 + BC2
AC2 = AB2 + BC2
Hence Proved.
Posted by Vasu Rathore 6 years, 10 months ago
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Posted by Dev Raj 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
We know that, the lengths of tangents drawn from an external point to a circle are equal.
∴ TP = TQ
In ΔTPQ,
TP = TQ
⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)
∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)
∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))
⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)
We know that, a tangent to a circle is perpendicular to the radius through the point of contact.
OP ⊥ PT,
∴ ∠OPT = 90º
⇒ ∠OPQ + ∠TPQ = 90º
⇒ ∠OPQ = 90º – ∠TPQ
⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)
From (1) and (2), we get
∠PTQ = 2∠OPQ
Parneet Kaur 6 years, 10 months ago
Vikram Singh 6 years, 10 months ago
Posted by Maanvendra Singh Rathore 6 years, 10 months ago
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Posted by Krish Agrawal 6 years, 10 months ago
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Posted by Pawan Othwal 6 years, 10 months ago
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Posted by Payal Thakur 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Given: Let, ABCD is a rhombus and since diagonals of a rhombus bisect each other at {tex} 90 ^ { \circ }{/tex}
To Prove: {tex} \therefore A B ^ { 2 } + B C ^ { 2 } + C D ^ { 2 } + A D ^ { 2 } = A C ^ { 2 } + B D ^ { 2 }{/tex}
Proof :
{tex}\therefore {/tex}{tex}A O = O C {/tex}
{tex}\Rightarrow A O ^ { 2 } = O C ^ { 2 }{/tex}
{tex}B O = O D {/tex}
{tex}\Rightarrow B O ^ { 2 } = O D ^ { 2 }{/tex}
and {tex}\angle A O B = 90 ^ { \circ }{/tex}
{tex}\therefore {/tex} {tex} A B ^ { 2 } = O A ^ { 2 } + B O ^ { 2 }{/tex}
Similarly, {tex} A D ^ { 2 } = x ^ { 2 } + y ^ { 2 } {/tex}
{tex}BC ^ { 2 } = x ^ { 2 } + y ^ { 2 } {/tex}
{tex}C D ^ { 2 } = x ^ { 2 } + y ^ { 2 } {/tex}
{tex} \therefore A B ^ { 2 } + B C ^ { 2 } + C D ^ { 2 } + D A ^ { 2 } = 4 A O ^ { 2 } + 4 D O ^ { 2 }{/tex}
{tex} = ( 2 A O ) ^ { 2 } + ( 2 D O ) ^ { 2 }{/tex}
{tex} = ( 2 x ) ^ { 2 } + ( 2 y ) ^ { 2 }{/tex}
{tex} \therefore A B ^ { 2 } + B C ^ { 2 } + C D ^ { 2 } + A D ^ { 2 } = A C ^ { 2 } + B D ^ { 2 }{/tex}
Hence proved.
Posted by Arsh Virdi 6 years, 10 months ago
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Posted by Rajpal Nadha 6 years, 10 months ago
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Posted by Abhishek Singh 6 years, 10 months ago
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Posted by Dalvir Singh 6 years, 10 months ago
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Posted by Adarsh Kumar 6 years, 10 months ago
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Posted by Anjali Soni 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Given: l and m are the tangent to a circle such that l || m, intersecting at A and B respectively.
To prove: AB is a diameter of the circle.
Proof:
A tangent at any point of a circle is perpendicular to the radius through the point of contact.
{tex}\therefore{/tex} {tex}\angle X A O = 90 ^ { \circ }{/tex}
and {tex}\angle Y B O = 90 ^ { \circ }{/tex}
Since {tex}\angle X A O + \angle Y B O = 180 ^ { \circ }{/tex}
An angle on the same side of the transversal is 180°.
Hence the line AB passes through the centre and is the diameter of the circle.
Posted by Poojamit Joshi 6 years, 10 months ago
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Payal Thakur 6 years, 10 months ago
Payal Thakur 6 years, 10 months ago
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