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Summarizing the discussion regarding circle

Description: P-346-1.tif
Property: In a circle (or congruent circles) equal chords are made by equal arcs. {OP = OQ} = {O’R = O’S) PQ = RS and PQ = RS


Description: P-346-2.tif

Property: Equal arcs (or chords) subtend equal angles at the centre PQ = AB (or PQ = AB) POQ =AOB
Description: P-346-3.tif
Property: The perpendicular from the centre of a circle to a chord bisects the chord i.e., if OD ⊥ AB (OD is perpendicular to AB).

Description: P-346-4.tif

Property: The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord. AD = DBOD

Description: P-346-5.tif
Property: Perpendicular bisector of a chord passes through the centre i.e., OD  AB and AD = DB ∴ O is the centre of the circle

Description: P-346-6.tif
Property: Equal chords of a circle (or of congruent circles) are equidistant from the centre
∴ AB = PQ
 OD = OR

Description: P-346-7.tif
Property: Chords of a circle (or of congruent circles) are equidistant from the centre
 OD = OR
 AB = PQ

Description: P-347-1.tif
Property: The angle subtended by an arc (the degree measure of the arc) at the centre of a circle is twice the angle subtended by the arc at any point on the remaining part of the circle. mAOB = 2m ACB.

Description: P-347-2.tif
Property: Angle in a semicircle is a right angle.

Description: P-347-3.tif
Property: Angles in the same segment of a circle are equal i.e., ACB = ADB

Description: P-347-4.tif
Property: If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, then the four points lie on the same circle. ACB = ADB
∴ Points A, C, D, B are concyclic i.e., lie on the circle

Description: P-347-5.tif
Property: The sum of pair of opposite angles of a cyclic quadrilateral is 180°.
DAB + BCD = 180°
and ABC + CDA = 180° (Inverse of this theorem is also true)

Description: P-347-6.tif
Property: Equal chords (or equal arcs) of a circle (or congruent circles) subtended equal angles at the centre.
AB = CD (or AB = CD) AOB = COD
(Inverse of this theorem is also true)

Description: P-347-7.tif
Property: If a side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle. m

Description: P-348-1.tif
Property: A tangent at any point of a circle is perpendicular to the radius through the point of contact.
(Inverse of this theorem is also true)

Description: P-348-2.tif
Property: The lengths of two tangents drawn from an external point to a circle are equal i.e., AP = BP

Description: P-348-3.tif
Property: If two chords AB and CD of a circle, intersect inside a circle (outside the circle when produced at a point E), then AE × BE = CE × DE

Description: P-348-4.tif
Property: If PB be a secant which intersects the circle at A and B and PT be a tangent at T then PA × PB = (PT)2

Description: P-348-5.tif
Property: From an external point from which the tangents are drawn to the circle with centre O, then (a) they subtend equal angles at the centre (b) they are equally inclined to the line segment joining the centre of that point AOP = BOP and APO = BPO

Description: P-348-6.tif
Property: If P is an external point from which the tangents to the circle with centre O touch it at A and B then OP is the perpendicular bisector of AB.
OP  AB and AC = BC

Description: P-348-7.tif
Property: If from the point of contact of a tangent, a chord is drawn then the angles which the chord makes with the tangent line are equal respectively to the angles formed in the corresponding alternate segments. In the adjoining diagram.
BAT = ∠BCA and BAP = BDA

Description: P-349-1.tif
Property: The point of contact of two tangents lies on the straight line joining the two centres.
(a) When two circles touch externally then the distance between their centres is equal to sum of their radii, i.e., AB = AC + BC
(b) When two circles touch internally the distance between their centres is equal to the difference between their radii. i.e., AB = AC – BC

Description: P-349-2.tif
Property: For the two circles with centre X and Y and radii r1 and r2. AB and CD are two Direct Common Tangents (DCT), then the length of DCT Description: 5305.png

Description: P-349-3.tif
Property: For the two circles with centre X and Y and radii r1 and r2 PQ and RS are two transverse common tangent, then length of TCT Description: 5289.png

Some important points

  1. 1. If three circles, each of radius r, are so kept that each circle touches the other two, then the area of the shaded region is r2 [√3 – (π/2)] which is approximately equal to (4/25) r2.
    Description: P-349-4.tif
  2. If four circles, each of the radius r, are so kept that each circle touches the other two, then the area of the shaded region is r² (4 – π) which is equal to (6/7)r2.
    Description: P-349-5.tif
  3. In the diagram given below, if the radius (OA) of the outer circle is R and the radius (OB) of the inner circle is r, then the width (w) of the ring is (R − r) and the area of the shaded region is π(R2 − r2) or π(R − r) (R + r) or π(w) (R + r)
    Description: P-349-6.tif
Two identical circles intersect so that their centers, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles, is
  1. Description: 5242.png
  2. Description: 5234.png
  3. Description: 5227.png
  4. Description: 5219.png
Description: P-350-1.tif
Shaded area = 2 × (area of sector ADC – area of ∆ADC)
Description: 5204.png
Description: 5197.png
Hence, option (b)
Four points A, B, C and D lie on a straight line in the X – Y plane, such that AB = BC = CD, and the length of AB is 1 cm. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance the ant must traverse to reach the sugar particle (in m) is
  1. Description: 5190.png
  2. 1 + 
  3. Description: 5183.png
  4. 5
In the drawn figure, ant will go along APQD since it cannot be within a distance of 1 cm from the repellents kept at B and C.
Description: P-350-2.tif
Description: 5168.png 
Also AP = QD Description: 5161.png
So the minimum distance = AP + PQ + QD
Description: 5153.png 

Two circles C1 and C2, having the same radius of 2 cm and centers at P and Q respectively, inter­sect each other such that the line of centers PQ intersects C1 and C2 at F and E respectively. EF = 1 cm. The whole assembly is enclosed in a rectangle of the minimum area. The perimeter of the rectangle is


Description: P-350-3.tif


Hence, breadth of the rectangle = 4 cm
And length = 7 cm
Perimeter = 2 × (7 + 4) = 22 cm

Semi-circle C1 is drawn with a line segment PQ as its diameter with centre at R. Semicircles C2 and C3 are drawn with PR and QR as diameters respectively, both C2 and C3 lying inside C1. A full circle C4 is drawn in such a way that it is tangent to all the three semicircles C1, C2 and C3. C4 lies inside C1 and outside both C2 and C3. The radius of C4 is
  1. Description: 5137.png
  2. Description: 5129.png
  3. Description: 5122.png
  4. Description: 5115.png
Description: P-350-4.tif
Assume that the radius of C4 = r and PQ = k.
Now, PR = k/2 = RQ = RO
⇒ RS = (k/2) – r
RT = k/4
ST = (k/4) + r
Applying Pythagoras theorem in triangle STR.
Description: 5100.png 
⇒ r = k/6 = PQ/6
On a semicircle with diameter AD, chord BC is parallel to AD. Further each of the chords AB and CD has length 2 units, while AD = 8 units. What is the length of BC?
  1. 7.5
  2. 7
  3. 7.75
  4. 8
Description: P-351-1.tif
Finding area of DABD Description: 5085.png AB × BD Description: 5078.png AD × BE
So, BE = (√15)/2
Hence AE Description: 5071.png
Now, BC = EF = 8 – Description: 5064.png = 7
Alternatively, this question can be solved by using Ptolemy’s theorem also.
The adjacent sides AB, BC of a square of side ‘a’ units are tangent to a circle. The vertex D of the square lies on the circumference of the circle. The radius of the circle could be:
  1. a(2 – √2)
  2. a(√2 – 1)
  3. a(√2 + 1.5)
  4. a(√2 + 1)
Description: P-351-2.tif
OD = r, OB = r√2
Hence, r + r√2 = √2 a
Hence, r = a(2 −√2)
M, N, O and P are centres of four inter­secting circles each having a radius of 15 cm. If AB = 7 cm, CD = 5 cm, EF = 6 cm, GH = 8 cm, what is the perimeter of the quadrilateral MNOP?
Description: P-351-3.tif
  1. 84 cm
  2. 45 cm
  3. 94 cm
  4. 124 cm
MN = 30 – 6 = 24 cm
NO = 30 – 5 = 25 cm
OP = 30 – 7 = 23 cm
PM = 30 – 8 = 22 cm
So, the perimeter = 24 + 25 + 23 + 22 = 94 cm
Description: P-351-4.tif
AB = BC = AC = CD. Find ∠a.
  1. 30°
  2. 60°
  3. 15°
  4. None of these
ACB = 60° ACD = 180 – 60 = 120° CAD = CDA = 30° α = 30°
In the figure shown here QS = SR, QU = SU, PW = WS and ST||RV. What is the value of the Description: 5032.png
Description: P-351-5.tif
  1. Description: 5017.png
  2. Description: 5010.png
  3. Description: 5003.png
  4. Description: 4996.png
Description: P-351-6.tif
Draw a line form U such that it is parallel to ST (and hence RV, also) joining AS, to get ∆QSA.
Now, let us first find the area of ∆SUX.
PW = WS (W is the mid point of PS)
Consider the ∆PXS, SX||YW and PX and PS are transversals to those parallel lines, we must have
Description: 4980.png 
⇒ PY = YX
Similarly we also get
PV = TV (in ∆PTS; VW||ST) and QA = AT (in ∆QTS:
AU||ST) and QT = VT (in ∆VQR; ST||VR)
Combining all these QA = AT Description: 4973.png (VT)+ Description: 4966.png (VP)
Now, in ∆RUY
Description: 4959.png ⇒ UX Description: 4952.png (XY)
so, in ∆PUS, UX Description: 4945.png XY Description: 4937.png PYDescription: 4930.png
⇒ Area (∆SUXDescription: 4923.png Area (∆PUS)
Description: 4916.png  Area (∆SUXDescription: 4909.png (SUX), then
Area (∆PSX) = Area (∆PSU) – area (∆SUX) Description: 4900.png (∆ PQR)
Through T, the mid-point of the side QR of a ∆PR, a straight line is drawn to meet PQ produced to S and PR at U, so that PU = PS. If length of UR = 2 units then the length of QS is:

Description: P-352-1.tif 
  1. 2 Description: Image4278.JPG Units
  2. Description: Image4285.JPG Units
  3. 2 Units
  4. Cannot be determined
We have QT = TR and PU = PS and UR = 2 units
Description: P-352-2.tif
Draw RV||PS that meets SU extended at V.
Now, in ∆QST and ∆TVR
∠QST = ∠TVR (alternate angles as PS||VR) and
∴ DQST and DTVR are congruent.
∴ QS = VR … (1)
Now ∠QST = ∠PUS = ∠VUR = ∠UVR
In ∆UVR, ∠VUR = ∠RVU
or, RV = UR = 2 …(2)
From (1) and (2)
QS = VR = UR = 2 units
ABCD is a parallelogram and P is any point within it. If the area of the Parallelogram ABCD is 20 units, then what is the sum of the areas of the DPAB and DPCD?
  1. 5 units
  2. 10 units
  3. 12 units
  4. Cannot be determined
Description: P-352-3.tif
Let AB = CD = a and xy be the lengths of the per­pendiculars from P on AB and CD respectively, then
Area of (∆PAB + ∆PCD) Description: 4852.png ax + Description: 4845.png ay
Description: 4838.png a(x + yDescription: 4831.png ah

Given area of the parallelogram = 20 units
Area of (∆PAB + ∆PCD) Description: 4823.png units
Nayantara bought a triangular piece of land of area 150 m2. He took a piece of rope and mea­sured the two sides of the plot and found the largest side to be 50 m and another side to be 10 m. What is the exact length of the third side?
  1. Description: 4816.png
  2. Description: 4809.png
  3. Description: 4802.png
  4. 32 m
Description: P-352-4.tif
Let the triangle be ABC.
Now, area of ∆ABC = (1/2) (BC) (AD) (Where D is a point on BC such that AD⊥BC)
Now, AD has to be equal to Description: 4787.png
AD = Description: 4780.png m = 6 m
Now ∆BDA is a right-angled triangle
∴ BD2 = AB2 – AD2 = 102 – 62 = 82 m2
⇒ BD = 8 m and DC = BC – BD = 42 m
In a right-angled ∆ADC, AC2 = AD2 + DC2 = (6)2 + (42)2
⇒ AC = Description: 4772.png
Alternatively, going through the options
Description: 4765.png 
Description: 4758.png 
Description: 4751.png
∴ Using the principle that the sum of any two sides of a triangle is greater than the third side. The two given sides are 50 and 10.
From the choices the above condition is satisfied only for choice (b).
In Choice (a) 10 + 50 < 68
In Choice (c) 39 + 10 < 50
In Choice (d) 32 + 10 < 50
C1 and C2 are two concentric circles with radii 5 cm and 9 cm respectively. If A, B and C are points on C2 such that AB and AC are tangent to C1 at how many points does BC intersect C1?
  1. 0
  2. 1
  3. 2
  4. Cannot be determined
Description: P-353-1.tif
Consider two concentric circles with radii r and R (r < R). If r Description: 4735.png and A, B and C are points on the outer circle such that AB and AC are tangent to the inner circle, then BC is also a tangent.
(In this case, C1 and C2 become the incircle and circumcircle of ABC respectively, which will be an equilateral triangle)
If r Description: 4728.png BC does not intersect or touch the inner circle and if r Description: 4721.png then BC intersects the inner circle at two points.
A rectangle MNOQ is drawn and length ‘NO’ is extended to point R and a triangle QPR is drawn with QP Description: 4714.png QM. Angle QRP = 45° and side QR Description: 4707.png S and T are the midpoints of sides QR and PR respectively. If ST = 6 units, the area (in sq. cm) of the rectangle is
Description: P-353-2.tif
  1. 112
  2. 144
  3. 288
  4. 256
Since the line joining the mid-points of two sides of a triangle is parallel and equal to half the third side, we have PQ = 2(ST), ⇒ PQ = 12 cm
Since, PQ Description: 4692.png QM
Now, ∠ORP = 45°
Draw PV⊥ON
In ∆RVP, tan 45° = Description: 4685.png
(Since OV = PQ = 12 cm) also, breadth2 + OR2 = (QR)2
In ∆ROQ, ⇒ Breadth = x + 12 where OR = x
⇒ (x + 12)2 =Description: 22748.png − x2 (given that QR =Description: 22756.png
⇒ 2x2 + 24x – 128 = 0
⇒ x2 + 12x – 24 = 0
⇒ x = 4 or −16
Since x > 0, x = 4
Breadth = x + 12 i.e., 16 cm, and the area of MNOQ
= 16 × 18 = 288 sq. cm.
∆ABC has sides AB, AC measuring 2001 and 1002 units respectively. How many such triangles are possible with all integral sides?
  1. 2001
  2. 1002
  3. 2003
  4. 1004
Value of BC will lie in between 999 and 3003. Hence 999 < BC < 3003.
So, the total values possible for BC = 2003
Ellipse The path of a moving point which moves in such a way that its distance from a fixed point (focus) bears a constant ratio with its distance from a fixed line (directrix)
Description: P-354-1.tif
Given: OB = OD = a (Semi-major axis)
OA = Ol = b (Semi = minor axis)
  • Area = πab
  • Perimeter = π(a + b)

Maths behind the formula (Area = πab) One way to see why the formula is true is to realize that the ellipse is just a unit circle that has been stretched by a factor ‘
a’ in the x-direction and by factor ‘b’ in the y-direction. Hence the area of the ellipse is just a × b times the area of unit circle.

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