Under which of the following conditions are you most likely to fall sick? (a) When you are taking examinations. (b) When you have travelled by bus and train for two days. (c) When your friend is suffering from measles. Why? Answer I will be most likely to fall sick when my friend is suffering from measles. This is because in this condition, I will visit my friend and will be likely to get infected with measles. Measeles is an infectious as well as an air-borne disease. When my friend will cough or sneeze, small droplets from his mouth containing microbes will mix in the air.
These microbes may be present in the air I will breathe. So, the chance of getting infection will increase. Concept insight: Recall the types of diseases-communicable and non communicable diseases. Under which of the following conditions is a person most likely to fall sick? (a) When she is recovering from malaria. (b) When she has recovered from malaria and is taking care of someone suffering from chicken-pox. (c) When she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chicken-pox.
Why? A person is most likely to fall sick when she is on a four-day fast after recovering from alaria and is taking care of someone suffering from chiken-pox. This is because immediately after suffering from malaria, the body is still weak, as during malaria a person has loss of appetite, vomiting and fever. Moreover, the various organ systems are also most likely not properly recovered and the immune system is also comparatively weak. So, she is likely to get infected with chicken-pox herself which can lead to serious health problems.
Concept insight: Recall the types of diseases-communicable and non communicable diseases. In the given figure, A, B and C are three points on a circle with centre O such hat BOC = 300 and AOB = 600. If D is a point on the circle other than the arc ABC, find ADC. We may observe that AOC = AOB + BOC = 600 + 300 We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle. A chord of a circle is equal to the radius of the circle.
Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. In OAB AB = OA = 0B = radius OAB is an equilateral triangle. So, each interior angle of this triangle will be of 600 AOB = 600 Now, In cyclic quadrilateral ACBD ACB + ADB = 1800 (Opposite angle in cyclic quadrilateral) ADB = 1800 – 300 = 1 500 So, angle subtended by this chord at a point on major arc and minor arc are 300 and 1 500respectively. In the given figure, PQR = 1000, where P, Q and R are points on a circle with centre O.
Find OPR. Consider PR as a chord of circle. Take any point S on major arc of circle. Now PQRS is a cyclic quadrilateral. PQR + PSR = 1800 1000 = 800 (Opposite angles of cyclic quadrilateral) PSR = 1800 – it any point on the remaining part of the circle. POR = 2PSR = 1600 In POR OP=OR (radii of same circle) OPR = ORP Angles opposite equal sides ofa triangle) OPR + ORP + POR = 1800 (Angle sum property of a triangle) 2 OPR + 1600= 1800 OPR = 100 In the given figure, ABC = 690, ACB = 310, find BDC.
In ABC (Angle sum property of a triangle) BAC +690 + 310 BAC + ABC + ACB = 1800 = 1800- 100?„o BAC = 800 (Angles in same segment of circle are equal) Q: 5 BDC = BAC = 800 BAD = BAC + CAD = 300 + 700 = 1000 = 1800 In the given fgure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that BEC = 1300 and ECD = 200. Find BAC. In CDE CDE + DCE = CEB CDE + 200 = 1300 CDE = 1100 But BAC = CDE (Exterior angle) (Angles in same segment of circle) BAC = 1100 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E.
If DBC = 700, BAC 300, findBCD. Further, if AB = SC, find ECD. CBD = CAD BCD + BAD = 1800 BCD = 800 of a triangle) BCA For chord CD (Angles in same segment) CAD = 700 (Opposite angles of a cyclic quadrilateral) BCD + 1000 = 1800 (given) BCA = CAB = 300 (Angles opposite to equal sides we have BCD = 800 BCA + ACD = 800 300 + ACD = 800 ACD = 500 ECD = 500 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices f the quadrilateral, prove that it is a rectangle.
Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O. (Consider BD as a chord) (cyclic quadrilateral) BCD= 1800 – 900 = 900 (Considering AC as a chord) ADC + ABC = 1800 ABC = 900 (Cyclic quadrilateral) 900 + ABC = 1800 Here, each interior angle of cyclic quadrilateral is of 900. Hence it is a rectangle. Q: 8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic. consider a trapezium ABCD with AB I ICD and SC = AD Draw AM CD and BN CD In AMD and BNC
AD = BC (Given) AMD = BNC (By construction each is 900) AM = BM (Perpendicular distance between two parallel lines is same) AMD BNC ADC = BCD (RHS congruence rule) (CPCT) (1) BAD and ADC are on same side of transversal AD BAD + ADC = 1800 BAD + BCD = 1800 [I-Jstng equation This equation shows that the opposite angles are supplementary. So, ABCD is a cyclic quadrilateral. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure).
Prove that ACP = QCD Join chords AP and DQ For chord AP PBA = ACP DBQ = QCD (Angles in same segment) segments intersecting at B. PBA = DBQ (Vertically opposite angles) and (3), we have ACP = QCD For chord DQ ABD and PBQ are line From equations (1), (2) If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side. Consider a ABC Two circles are drawn while taking AB and AC as diameter. Let they intersect each other at D and let D does not lie on BC.
Join AD ADS = 900 (Angle subtend by semicircle) ADC = 900 Angle subtend by serntctrcle) BDC = ADS + ADC = 900+900 = 1800 Hence BDC is straight line and our assumption was wrong. Thus, Point D lies on third side BC of ABC Q: 11 ABC and ADC are two right triangle with common hypotenuse AC. Prove that CAD = CBD. ABC + BCA + CAB BCA + CAB = 900 In ADC CDA + ACD + DAC ACD + DAC = 900 = 1800 (Angle sum property ofa triangle) 900+ BCA + CAB = 1800 (Angle sum property ofa triangle) 900 + ACD+ DAC Adding equations (1) and (2), we have BCA + CAB + ACD + DAC = 1800 (BCA + ACD) + (CAB + DAC) = 1800 BCD + DAB = 1800 B + D = 900+900 = 1800 .. 3) But it is given that From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 1800. So, it is a cyclic quadrilateral. Consider chord CD. NOW, CAD = CBD Q: 12 Prove that a cyclic parallelogram is a rectangle. Let ABCD be a cyclic parallelogram. A + C = 1800 (Opposite angle of cyclic quadrilateral) opposite angles of a parallelogram are equal A = C and B = D From equation (1) A+C=1800 A+A=1800 We know that A = 900 Parallelogram ABCD is having its one of interior angles as 900, so, it is a rectangle.
