chapter 2 a) - s.n.m.h.s.s · since two angles are equal these triangles are similar. so sides are...
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Chapter 2
A) Draw a right triangle with hypotenuse 5 cm.Answer :We can draw many trianglesusing set square.All these triangle form a circle.
B) Prove that “ Angle in a semicircle is a rightangle”
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Proof :Join P to centre O of the circle.
Consider triangle AOP and triangle BOPLet < APO = x0.Triangle AOP is an isosceles triangle.Base angles are equal. Ie <A = x0.Let <BPO =y0.Then <B = y0.
Now consider triangle APBSum of angles = 1800.ie <A + <P + <B = 1800.
ie x + (x+y) + y = 1800.
ie 2x +2y = 1800.
ie 2( x+y) = 1800.
x+y = 900.<P = 900. ie Angle in a semicircle is right.
C) What is the angle inside the circle in the picture?
Answer :To find <APB extend the line AP to AQ, which
meet at Q on the circle.<AQB = <APB = <Q + <B (By Exterior angle
theorem )= 90 + <B , which is more than 90
ie <APB is obtuse angle.
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D) Find <APB when the point P is outside the circle.Answer :<AQB is an exterior angle of triangle BQP.
There fore,<AQB =<P+<B
<P and <B both less than 90ie <APB < 90 < Pis acute angle.
Answer:
<ACB = > 90 so C is inside thecircle
<ADB = so Point D lies on the circle.
<AEB < so Point E is outside thecircle.
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Answer :Suppose drawing a circle withdiameter BD.<A = > 90 . so A is inside thecircle.
<C = >90 , so C is inside thecircle.
When we draw a circle with diameterAC.
<B = < 90 So B lies outside thecircle.
<D = 360 – (105+110+55)
= 360 -270= 90
ie D is the point on the circle.
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Answer : 5, 12, 13 are sides ofright angled triangle.
<A = so , A is on the circlewith diameter BC.
<C < so, C is out side thecircle with diameter AB
<B < so, B is out side thecircle with diameter AC
Answer :Draw BC and OD
ACB and ADO are right trianglesSince two angles are equal these triangles are similar.
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So sides are proportional.
=
AO is the radius of the circle.
= =
ie 2 AD = AC
D is the midpoint of AC.
Answer :
ABC is an isosceles triangle.
The circles with diameter AC and BCpasses through the point D.
Now, CD is a common side
ADC = (Angle in a semi circle )BDC =
given AC = BC so ,
ADC = BDCie AD =BD, D is the midpoint.
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Answer :
diameter = = =6.708Perimeter =2 r =3.14 x 6.708 = 21.06Area = = 3.14 x 3.354 x3.354 = 35.32 sq.cm
Answer :7 (i)
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PBA = Angle in a semicircle,QBA = So P, B, Q are in the same position.
7(ii)Answer :Centres are mid points ie this line is Parallel to baseand length is half of PQ
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Answer :
The diagonals of a rhombusare perpendicular bisectors.That is intersecting point ofdiagonalslies on each circle withdiameter sides of rhombus.
Similarly in second pictureadjacent sides are equal.These sides makes anisosceles triangle.There fore the diagonals are perpendicular to each other. All circles withdiameter as sides passesthrough the meeting point of diagonals.
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Answer :
Area of blue crescent = Area of blue semicircle - Area of small segmentArea of red crescent = Area of red semicircle - Area of big segment.Area of small and big segments = Area of yellow semicircle- Area of triangle.
Area of blue crescent+ Area of red crescent = Area of blue semicircle+Area of red semi circle -( Area of small segment +Area of big segment)
=Area of blue semicircle+Area of redcircle - (Area of yellow semicircle- Area of triangle)
=Area of blue semicircle+Area of redsemi circle - Area of yellow semicircle+ Area of triangle)= Area of triangle.
(because by Pythagoras theorem sum of areasof semicircles on perpendicular segments ofrighttriangle = Area of Semicircle on Hypotenuse.)
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Answer : For figure 1
AOB is an isosceles triangle. BAO = , base angles are
equal.Similarly OAC =
A = 20+30 = 500 So BOC =1000 , Central angleof arc A
BOC is also an isoscelestriangle.So OBC = OCB =
= 400
B = 20+40 =600
C = 30+40 =700
in second picture try your self
Answers :
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A = 500 , B = 600, , C = 700 O = 1000 OCA =400
in the third picture.
Central angle BOC =700.
ie CAB = =350.
Central angle AOC =400
ie ABC = = 200
C =
=
=
= 180-55=1250
OBC is isosceles triangle.
So OCB = OBC =
= 550
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Answer :Total angle of the clock = 3600
Angle between adjacent numbers = = 300
Central angle of arc 1 to 4 = 3 x 30 =900
So angle at number 8 = =45
Central angle of arc from 4 to 8 = 4 x 30 = 1200
So angle at 1 = =600
Third angle at 4 is equal to 750
Now to count equilateral triangles. For drawing equilateral triangle four gapes needed for each line.So 4 distinct triangles can be drawn. Points are 12, 4,8
1,5,92,6,103,7,11
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iii) Let angle on one part =2x, angle on other part =xx +2x = 1803x =180x =600
then central angle must be 1200
iv)Cyclic quadrilateral sum of opposite angles supplimentary.
=180
5x = 360
x= = 720
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Answer :
Angle in the Blue picture = =360
So central angle in green circle = 720
Part of the circle inside the angle = =
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Answer :OAC =x0
then OCA =x0
and AOC = 180-2x
ie B = = 90-x
OAC + =x+ 90-x = 900
Answer :
For seeing the video Click Here
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Answer :
Let BOD =2xThen BCD =x ,Angle subtended is half the central angle.
CBA =90-xthen COA =180-2xNow AOC + BOD = 180-2x+2x =180ie arc APC and arc BQD joined togetherwould make half the circle.
Answer : Let AOB =2x then ACB =x = ADB (Angle subtended is half of centre angle )
PDQ = 180-x andPCQ =180-x (linear pairs are supplementary )
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P + CQD+ PCQ+ PDQ = P + CQD +180-x+180-x ie 360 = P + Q + 360-2x P + Q =360-(360 2x) P + Q = 2x
= AOB
Circle and quadrilateral
A) prove that B+ D is less than 180.
answer : B+ AEC =180, -----(1)
Opposite anglessupplementary.
D+ EAD = AEC Exterior angle theorem------------(2)
ie D < AEC -------------(3)from (1) and (3)
B + D < 180
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B) Prove that B+ ADC >180
Answer :
B+ E =180 -------(1)EAD + E = ADC
ie ADC > E ---------(2)
from (1) and (2)B+ ADC > 180
hence the result.
We get another not from this point :-
Definition A cyclic quadrilateral is a quadrilateral with opposite angles supplementary.
C) Prove that isosceles trapezium is a cyclicquadrilateral.
Answer :In isosceles trapezium ABCD
A = B A + D = 180
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ie B + D =180 similarly A + C =180opposite angles are supplementary. ABCD is cyclic.
Answer :
CBD = 300
ie CAD =300, angles subtendedin same arc
A = 30 +45 =750
A and C are supplementary.
A+ C =18075+50 + ACD =180
ACD =55ie C =50+55 =105
then ABD =550
B =30+55 =85then D =180-85 =95
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Answer :ABCD is a cyclicquadrilateral.Let CBE =xthen ABC =180-xABCD is cyclic.
D + ABC =180D = 180- ABCD =180-(180-x)D =x Which is equal to
exterior angle.
Answer :ABCD is a parallelogram.If ABCD is cyclic.
A + C =180But in a parallelogramopposite angles are equal.
A = Cie A =90
ie ABCD is a rectangle.
Answer:ABCD is
non-isosceles trapezium.AD ≠BC
A + D =180, co-interior angle of trapezium.A ≠ B
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ie B+ D ≠ 180ABCD is not cyclic.
Answer :in PDC
P =180 - ( DP and CP are bisectors of and
respectively.in ARB
R =180 -
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P + R =180 - + 180 -
P+ R= 360 -[ + ]
= 360 -
= 360 -
= 360-180= 180
Opposite angles are supplementary. PQRS is cyclic.
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APQ and CPQ are linear pairs.Ie CPQ =x0PQDC is cyclic.
D =180-xB and D are supplementary.
Ie AB CDABCD is isosceles trapezium.Isosceles trapezium are cyclic.
Answer :6(ii)
Let A =x, ABPQ is cyclic.
BQP =180-xBQP and SQP are linear pairs.SQP =x
PQRS is cyclicPRS =180-xPRS and CRS are linear pairsSRC =xD=180-xA and D are supplementary
ie ABCD is cyclic.
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Answer :
Let <A =XAQOR is cyclic
QOR = 180-Xsimilarly POR and POQ 180-Y and 180-Z respectively.
QOR+ POR+ POQ =180- X+180-Y+180-Z = 540-(X+Y+Z)
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=540-180=360ie angle around point O is 360. ie circles meet at same point.
Two chords
A) In the figure prove that
=
Answer :
APC = BPD opposite anglesC = B angle in same arc
So APC and BPD aresimiler.Sides are proportional.
=
ie PA x PB = PC x PD
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B) Draw a rectangle with sides a and b. Draw another rectangle with same area.
See video.
When chords areperpendicularPA x PB = PC2.
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Answer: 1( i)
ABCD is cyclic.If A =x then PDB =xIf B=y then PBD =yand P is commonSo angles in APC and PBD are same.
Answer :1 (ii)Angles in APC and PBD are equal, ie these triangles are similar. Sides are proportional.
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ie PA X PB = PC X PD
Answer 1(iii)if PD =PB then PA =PC and PBD is isosceles triangle and base angles are equal.Ie x =yAC BDAB = PA-PB =PC-PD =CDie ABCD is an isosceles trapezium.
Answer :see video
Answer: see video
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