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Circles Tangents Problems (view source)
Revision as of 11:42, 29 October 2019
, 11:42, 29 October 2019added Category:Circles using HotCat
#The students have to prove thne angle PAQ=twise the angle OPQ.
#The students have to prove thne angle PAQ=twise the angle OPQ.
===Geogebra file===
===Geogebra file===
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<span> </span>
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==Concepts used==
==Concepts used==
∴∠QPB=x˚ (∵PQ=BQ)<br>
∴∠QPB=x˚ (∵PQ=BQ)<br>
Now Let ∠PAQ=x˚<br>
Now Let ∠PAQ=y˚<br>
∠QPA=y˚ (∵ PQ=AQ)<br>
∴In △PAB<br>
∠PAB+∠PBA+∠APB=180˚<br>
y˚+x˚+(x˚+y˚)=180˚<br>
2x˚+2y˚=180˚<br>
2(x˚+y˚)=180˚<br>
x˚+y˚=90˚<br>
∴ ∠APB=90˚
=problem 3 [Ex-15.2 B.7]=
=problem 3 [Ex-15.2 B.7]=
[[File:fig3.png|200px]]
[[File:fig3.png|200px]]
==Interpretation of problem==
==Interpretation of problem==
=Problem 5=
# In the quadrilateral ABCD sides BC , DC & QB are given .
#AD⊥DC.
# Asked to find the radius OS or OP
==Concepts used==
#Tangents drawn from an external point to a circle are equal.
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent
==Algorithm==
In the fig BC=38 cm and BQ=27 cm <br>
BQ=BR=27 cm (∵ by concept 1) <br>
∴CR=BC-BR=38-27=11 cm <br>
CR=SC=11 cm (∵ by concept 1) <br>
DC=25 cm <br>
∴ DS=DC-SC=25-11=14 cm <br>
DS=DP=14 cm (∵ by concept 1) <br>
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br>
∠D=∠S=∠P=90˚ <br>
⇒ ∠O=90˚ <br>
∴ DSOP is a Square <br>
SO=OP=14 cm <br> hence Radius of given circle is 14 cm
=Problem 5 [Ex-15.2-B8]=
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].
[[File:123.png|300px]]
[[File:123.png|300px]]
==Algorithm==
In the figure AQ , AR and BC are tangents to the circle with center O.<br>
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br>
Perimeter of △ABC=AB+BC+CA<br>
=AB+(BP+PC)+CA<br>
=AB+BQ+CR+CA ------ (From eq-1)<br>
=(AB+BQ)+(CR+CA)<br>
=AQ+AR ----- (From fig)<br>
=AQ+AQ -- --- (∵AQ=AR)<br>
=2AQ<br>
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]
=Problem-6 [Ex-15.4-B3]=
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]
==Algorithm==
In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br>
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC
=Problem-7 [Ex-15.4-A3]=
In the figure AB=10cm,AC=6cm and the radius of the smaller circle is xcm. find x.
[[File:circle.png|400px]]
==Interpretation of the problem==
#In the given figure two circles touch internally.
#OB and OF are the radii of the semicircle with center "O".
#PC and PF are the radii of the circle with center "P".
==Ex 4.4.2==
#Suppose two chords of a circle are equidistant from the centre of the circle, prove that the chords have equal length.
'''DATA''' :- Let AB & CD are the two chords which are equidistant from the centre 'O' of the circle. [ Here OP is the perpendicular distance from the centre O to the chord AB and OQ is the perpendicular distance from the centre O to the chord CD] OP = OQ.
'''TO PROVE :-''' AB = CD,
'''CONSTRUCTION :-''' Join OA & OD.
'''PROOF :-'''
{[Consider In ∆AOP & ∆DOQ OA = OD OP = OQ Angle APO = Angle DQO ∆AOP ≡ ∆DOQ AP = DQ Let AB = AP + BP = AP + AP = 2AP AB = 2DQ ---------- 1. and CD = CQ + DQ = DQ + DQ CD = 2DQ --------- 2. From equtn 1 & equtn 2 AB = CD
Radii of the circle Equi distances from circle
SAS Axiom
Acording to properties of SAS axiom.
Perpendicular drawn from centre to chord which bisect the chord, i.e. AP = BP.
Perpendicular drawn from centre to chord which bisect the chord, i.e. CQ = DQ Acording to AXIOM-1]}
angle
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|'''Explanation'''
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|[[Image:solution.png|300px|link=http://karnatakaeducation.org.in/KOER/en/index.php/File:Solution.png]]
|Explanation for thestep
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|Write the step
|Explanation for thestep
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[[Category:Circles]]