250 Allied The Leading Mathematics-10 Construction 251 Geometry 11.1 Construction of Triangles or Quadrilaterals with Equal Area At the end of this topic, the students will be able to: ¾ construct a special quadrilateral (same or different) equal to the given special quadrilateral in area. ¾ construct a triangle equal to the given triangle in area. Learning Objectives I Construction of parallelogram/Rectangle equal to Area of Given Special Quadrilateral We can easily construct a parallelogram or rectangle equal to the area of the given regular quadrilateral (parallelogram or rectangle or rhombus, or square) by using the following steps: Step-1: Construct a parallelogram from the given information. Step-2: Also, construct another parallelogram on the same base AB and between the same parallel lines AB and CD as given instruction. Hence, the area of ABCD is equal to the area of ABFE. Note : We can construct the following constructions: B E F C D A B E F C D A Area of rectangle = Area of parallelogram. Area of parallelogram = Area of rectangle. B E F C D A B E D F C A Area of rhombus = Area of parallelogram. Area of rhombus = Area of rectangle. B E F C D A C B E X H G D A Area of square = Area of parallelogram. Area of quadrilateral = Area of parallelogram.
252 Allied The Leading Mathematics-10 Construction 253 Geometry Example-1 Construct a rectangle ABCD with the sides AB = 6 cm and BC = 4 cm. (a) Construct a parallelogram ABEF having an angle 60o equal to the area of the rectangle ABCD. (b) Why is the area of the rectangle ABCD equal to the area of the parallelogram ABEF? Give reason. Solution: For rectangle ABCD, AB = 6 cm, BC = 4 cm and ∠BAD = 90o . (a) Construct a rectangle ABCD and then construct a parallelogram ABEF having ∠FAB = 60o on the same base AB and between the same parallel lines AB and DC, we get D F C E A B 4 cm 6 cm 4 cm 60° 6 cm ∴ Area of rectangle ABCD = Area of ABEF. (b) The area of rectangle ABCD is equal to the area of the parallelogram ABEF because they stand on the same base AB and between AB and DE. II Construction of Triangle equal to Area of given Triangle We can easily construct a triangle equal to the area of the triangle by using the following steps: Step-1: Construct a triangle ABC from the given dimensions. Step-2: Construct a line EF through the vertex C and parallel to AB by drawing equal angles with ∠ABC or ∠CAB by using compass. Step-3: Construct another triangle from the given instruction on the base AB of ∆ABC and between the parallel lines AB and EF. Hence, the area ∠ABC is equal to the area of ∆ADB. C D E F A B B E F C 6 cm D A 60° Rough Sketch 4 cm 4 cm
252 Allied The Leading Mathematics-10 Construction 253 Geometry Example-2 Construct a triangle PQR in which PQ = 4 cm, QR = 5 cm and RP = 3 cm. (a) construct another triangle having a side 6 cm equal to the area of ∆PQR. (b) Why is the area of the triangles PQR and PQS? Give reason. Solution: For ∆PQR, PQ = 4 cm, QR = 5 cm and RP = 3 cm. (a) Now, constructing ∆PQR and then constructing ∆PQS on the same base PQ and between the parallels PQ and RS, we get P 4 cm Q 3 cm 5 cm 6 cm E F R S ∴ Area of ∆PQR = Area of ∆PQS (b) The area of the triangles PQR and PQS are equal because they stand on the same base PQ and between PQ and RS. PRACTICE 11.1 1. (i) Construct a parallelogram ABCD with AB = 6 cm, AD = 4 cm and ∠ABC = 45o . (a) Construct another parallelogram with a side 4.5 cm equal to the area of the parallelogram ABCD. (b) Why is the area of the rectangle ABCD equal with the area of the later constructed parallelogram ? Give reason. (ii) (a) Construct a parallelogram with a side 5 cm and the angles made by the diagonals to the side as 30o and 60o . (b) Construct another parallelogram of a diagonal 7 cm equal to the area of the previous constructed parallelogram. (c) Why are both parallelograms equal in area? Give reason. 2. (i) (a) Construct a rectangle equal to the area of a parallelogram PQRS with PQ = 6 cm, PS = 4 cm and ∠PQR = 30o . (b) Why are the area of the rectangle and parallelogram constructed in (a) equal ? Give reason. P 4 cm Q 3 cm 5 cm 6 cm E F R S Rough Sketch
254 Allied The Leading Mathematics-10 Construction 255 Geometry (ii) (a) Construct a rhombus with a side 5 cm and a diagonal 6.4 cm. (b) Construct a parallelogram with a side 6 cm equal area with the above rhombus. (c) Why are both rhombus and parallelogram equal in area? Give reason. 3. (i) Construct a triangle PQR in which PQ = 4.5 cm, QR = 5.2 cm and ∠PQR = 45°. (a) Construct another triangle PQS having a side 5.5 cm equal area with ∆PQR. (b) Why is the area of the triangles PQR and PQS equal? Give reason. (ii) Construct a triangle KLM having KL = 4.5 cm, LM = 5.2 cm and KM = 4.8 cm. (a) Construct another triangle with a side 5.8 cm equal to the area of ∆KLM. (b) Write the reason of equal area of both triangles. 4. (i) Construct a triangle ABD with a side 6 cm equal to the area of the triangle with AB = 4.5 cm, ∠CAB = 60o and ∠ABC = 45o . (ii) Construct a triangle with a side 6 cm equal to the area of the equilateral triangle with a side 5.3 cm. (iii) Construct an isosceles triangle equal area to a triangle having any two-sides 4.5 cm and 5.2 cm and the angle between them 60°. Consult with your teacher. Answers
254 Allied The Leading Mathematics-10 Construction 255 Geometry 11.2 Construction of Triangles or Quadrilaterals with Equal Area At the end of this topic, the students will be able to: ¾ construct a a triangle equal area to the given quadrilateral and its types. ¾ construct a parallelogram or rectangle equal to the given triangle in area. Learning Objectives I Construction of Triangle equal to Quadrilateral in Area (i) Construction of a Triangle equal to the Area of Special Quadrilaterals We can easily construct a triangle equal to the area of the special quadrilateral (parallelogram, rhombus, rectangle and square) by using the following steps: Step - 1: Construct the special quadrilateral, say parallelogram from the given data. Step - 2: Produce the side AB up to E and take a point E such that AB = BE. Step - 3: Draw ∆ADE by joining D and E. Hence, the area of ABCD is equal to the area of ∆ADE. Why? Discuss it. Example-1 Construct a parallelogram PQRS with PQ = 4 cm, QR = 3 cm and ∠PQR = 120o. . (a) Construct a triangle PAB having a side 5 cm equal area to PQRS. (b) Prove that: Ar( BDGH) = Ar(∆PAB) Solution: At first, draw the rough sketch as alongside. Now, constructing the parallelogram PQRS by given conditions. (a) Constructing ∆PAB having PA = 5 cm by constructing SQ//RB. P Q 120° 3 cm 5 cm 3 cm 4 cm 4 cm B X R S A Hence, Area of PQRS = Area of ∆PAB. “Alternatively” At first, draw the rough sketch as given here. Now, construct the parallelogram in PQRS by given conditions. D A B E C D A B OR E X C P Q 120° 3 cm 5 cm 3 cm 4 cm 4 cm B X S A R Rough Sketch
256 Allied The Leading Mathematics-10 Construction 257 Geometry Again, construct ∆PAB having PA = 5 cm on the double base of PQ and between the parallel lines PB and SR. P Q 120° 3 cm 5 cm 3 cm 4 cm 4 cm B X S A R Rough Sketch P Q 120° 3 cm 5 cm 3 cm 4 cm 4 cm B X R S A Hence, the area of PQRS = the area of ∆PAB. (b) Given: In the given figure, PQRS is a parallelogram. To prove: Ar. ( PQRS) = Ar. (∆PAB) Construction: Join AQ. Proof: SN Statements SN Reasons 1. Ar. (∆PAQ) = 1 2 × Ar. ( PQRS) 1. Standing on the same base PQ and between the same parallels PQ and SR. 2. Ar. (∆PAQ) = 1 2 × Ar. (∆PAB) 2. Being median AQ of ∆ABC 3. Ar. ( PQRS) = Ar. (∆PAB) 3. From statements (i) and (ii). Proved. (ii) Construction of Triangle equal to Irregular Quadrilateral in Area We can easily construct a triangle equal to the quadrilateral in area by using the following steps: Step – 1: At first, draw a rough sketch proportional to the given measures. Step – 2: Construct quadrilateral ABCD from the given information. Step – 3: Draw diagonal BD and produce AB up to X. Step – 4: Measure ∠DBC by a compass and construct ∠BCE such that ∠BCE = ∠DBC. So, DB//CE because of equal alternate angles. Step – 5: Join D and E. The required ∆ADE is formed equal to the quadrilateral ABCD in area. Why are they equal? Discuss. ∴ Area of quad. ABCD = Area of ∆ADE. D C A B E X Note: If we can construct ∆BCE in left, we can draw the diagonal AC and follow the steps 3 and 4.
256 Allied The Leading Mathematics-10 Construction 257 Geometry Example-2 Construct a quadrilateral ABCD in which AB = 4 cm, BC = 3 cm, CD = 4.5 cm, DA = 5 cm and diagonal BD = 4 cm. (a) construct a triangle ADE equal to the area of the quadrilateral ABCD. (b) Why are the area of the quadrilateral ABCD and ∆ADE equal? Solution: At first, we draw the rough sketch as follows: Step-1: Draw a line segment AB = 4 cm. From A, draw an arc with radius AD = 5 cm and from B, draw another arc with radius BD = 4 cm which intersect atapoint,say Dand join D with A and B. From B and D draw the arcs with radius BC = 3 cm and DC = 4.5 respectively which intersect at a point,sayCand join C with B and D. Step-2: Draw the diagonal BD: The diagonal BD is already drawn and produced AB up to X. Step-3: Construct the alternate angles and parallel line with BD: (i) Take an arc with any radius between the sides BD and BC from B. (ii) From C, take the same arc on BC to left. (iii) Measure ∠DBC by compass and cut down the arc of (ii) from BC where it intersects with the arc of (ii). (iv) Join C with the point of intersection of two arcs on (iii) and produce it which meets at E producing AB. So, ∠DBC = ∠BCE and DB//CE. Step-4: Construct a triangle ADE: (i) Join D and E. Then it forms the shaded ∆ADE which is equal to the quadrilateral ABCD in area. 4.5 cm 5 cm D E B C A 4 cm ∴ Area of quad. ABCD = Area of ∆ADE. (b) The area of the quadrilateral ABCD and ∆ADE are equal because of adding ∆ABD with ∆BCD and ∆BDE standing on the same base BD and between the parallel lines BD and ED. 4.5 cm 5 cm 1.4 cm D B E C A 4 cm 4 cm 3 cm Rough Sketch 4.5 cm
258 Allied The Leading Mathematics-10 Construction 259 Geometry II Construction of Special Quadrilateral Equal Area with Given Triangle We can easily construct a parallelogram or a rectangle equal to the area of the given triangle by using the following steps: Step-1: Construct a triangle ABC from the given conditions. Step-2: Construct a line EF through the vertex C parallel to AB by drawing equal angle with ∠ABC or∠CAB by using a compass. Step-3: Find the mid-point D of AB by using a compass or bisect of AB with a compass. Step-4: Construct a parallelogram on the base AD or BD between the parallel lines AB and EF from the given instruction. Note : If we construct a rectangle equal to the area of the given triangle, draw the perpendicular bisector of the base AB as shown in the adjoining figure. Example-3 Construct a triangle ABC having AC = 5 cm, AB = 4 cm and ∠BAC = 45o . (a) Construct a parallelogram with a side 7 cm equal to the area of ∆ABC. (b) Prove that: Ar( ADGH) = Ar(∆ABC) Solution: For ∆ABC, AC = 5 cm, AB = 4 cm and ∠BAC = 45o (a) At first, construct ∆ABC and then construct a parallelogram with a side 7 cm equal to the area of ∆ABC. E C H G F B 5 cm 7 cm D 4 cm 7 cm A 45° ∴ Area of ∆ABC = Area of ADGH (b) Given: In the given figure, PQRS is a parallelogram. To prove: Ar. ( ADGH) = Ar. (∆ABC) Construction: Join CD. Proof: SN Statements SN Reasons 1. Ar.(∆ACD) = 1 2 ×Ar.( ADGH) 1. Standing on the same base AD and between the same parallels AD and EF. 2. Ar. (∆ACD) = 1 2 × Ar. (∆ABC) 2. Being median AQ of ∆ABC. 3. Ar. ( ADGH) = Ar. (∆ABC) 3. From statements (i) and (ii). Proved. E C G H F B D A E C H G F B 5 cm A 45° Rough Sketch
258 Allied The Leading Mathematics-10 Construction 259 Geometry Example-4 (a) Construct a rectangle equal to the area of a triangle ABC having BC = 5 cm, ∠ABC = 60o and ∠BCA = 30o . (b) Why are the rectangle and triangle ABC equal ? Solution: Here, For ∆ABC, BC = 5 cm, ∠ABC = 60o and ∠BCA = 30o Now, at first, constructing ∆ABC and then constructing rectangle equal to the area of ∆ABC, we get G F B D C 60° 30° E A 5 cm ∴ Area of ∆ABC = Area of CDEF The rectangle CDEF and ∆ABC are equal because ∆ACD is the half of both CDEF and ∆ABC when join AD. PRACTICE 11.2 1. Based on the following data, first construct the a special quadrilateral and then construct a triangle equal to the area of the quadrilateral and also prove that their areas are equal. (i) Parallelogram ABCD having AB = 5.5 cm, BC = 4 cm, ∠A = 30o . (ii) Parallelogram PQRS having PS = 5.6 cm, PQ = 6.3 cm, PR = 8.2 cm. (iii) Rectangle ABCD having AB = 6 cm, BC = 4 cm. 2. (i) Construct a parallelogram ABCD having the measures AC = 6.2 cm, DB = 5.0 cm and ∠AOD = 45o . (a) Construct a triangle ∆ADE, ∆AFE with a side equal area to ABCD. (b) Prove that: Ar( ABCD) = Ar(∆ADE) (ii) Construct a parallelogram ABCD having the measures AB = 5.2 cm, DB = 7.2 cm and ∠ABD = 30o . (a) Construct a triangle ∆BCE equal area to ABCD. (b) Prove that: Ar( ABCD) = Ar(∆BCE) (iii) Construct a rectangle PQRS with the diagonal QS = 8 cm, PR = 6 cm and the angle between the diagonal is 30o . (a) Construct a triangle ∆QMT having QM = 5 cm equal to the area of the rectangle PQRS. (b) Prove that: Ar( PQRS) = Ar(∆PST) Rough Sketch F G H D 5 cm B C A E 60° 30°
260 Allied The Leading Mathematics-10 Construction PB Geometry 3. Using the following data, first construct a quadrilateral ABCD and then construct a triangle having area equal to the area of the quadrilateral. Why are their areas equal ? Give reason. (i) AB = 5 cm, BC = 6 cm, CD = 6.4 cm, DA = 6.8 cm and diagonal AC = 6 cm. (ii) AB = BC = 5.4 cm, CD = DA = 4.6 cm, ∠DAB = 75o . 4. (i) Construct a triangle ∆BCE equal to the area of the quadrilateral ABCD having the measures AB = 4.2 cm, BC = 4.6 cm, CD = 5.5 cm, DA = 6 cm and BD = 6 cm. (ii) Construct a triangle ∆PST equal to the area of the quadrilateral PQRS having the measures PQ = 5.2 cm, QR = 3.6 cm, RS = 4.5 cm, SP = 5 cm and ∠PQR = 60o . 5. (i) Construct a triangle with dimensions AB = 4.5 cm, BC = 3.5 cm, ∠B = 60o . (a) Construct a parallelogram having a side 5 cm equal to the area of the triangle. (b) Prove that: Ar(∆ABC) = Ar(Parallelogram) (ii) Construct a triangle with sides a = 4 cm, b = 5 cm, c = 4.6 cm and then construct a parallelogram having a side 5.4 cm equal to the area of the triangle. Also, prove that their areas are equal. (iii) Construct a parallelogram having an angle 45o equal to the area of the triangle with AB = 4.5 cm, ∠A = 75o and ∠B = 60o . (iv) Construct a parallelogram having an angle 60o equal to the area of the triangle with AB = 4.5 cm and BC = 5 cm and ∠A = 45o . 6. (i) Construct a triangle with dimensions ∠C = 90o , AB = 5 cm, BC = 3 cm. (a) Construct a rectangle equal to the area of the triangle. (b) Prove that: Ar(∆ABC) = Ar(Rectangle) (ii) Construct a rectangle equal area to an isosceles triangle ABC in which AB = BC = 7 cm and CA = 6.5 cm. Also, prove that their areas are equal. (iii) Construct a rectangle of equal area of ∆ABC having ∠B = 45o , ∠C = 60o and a = 3.5 cm. (iv) Construct a rectangle equal to the area of the equilateral triangle with a side 4.5 cm. Consult with your teacher. Answers Project Work Draw the figures on: (i) a triangle equal area to the given quadrilateral having four sides and one diagonal or angle. (ii) a triangle equal area to the given parallelogram or rectangle or square or rhombus or kite or trapezium having given dimensions. (iii) a parallelogram or rectangle equal area to the given scalene or isosceles or equilateral or right-angled triangle having given dimensions. (iv) a parallelogram or rectangle equal area to the given parallelogram or rectangle or square or rhombus or kite or trapezium having given dimensions. (v) a triangle equal area to the given scalene or isosceles or equilateral or right-angled triangle having given dimensions.
PB Allied The Leading Mathematics-10 Circle 261 Geometry 12.1 Theorems on Arc and Its Corresponding Angles of Circle At the end of this topic, the students will be able to: ¾ search the relationship between the central angle and inscribed angle of a circle and their corresponding arc. ¾ establish experimentally and logically the relation between the central and inscribed angles on the same arc of a circle. ¾ solve the problems related to the angles of a circle and their corresponding arcs. Learning Objectives I Introduction We are all familiar with objects such as ring, bicycle wheels, coins and so on. They all have the same type of boundary, i.e., their shapes are circular. Now, we study allsuch shapes undersubheadings: (i) Basic Concepts (a) Circle, Centre, Radius and Circumference A circle is the set of all points in a plane that lie at a constant distance from a fixed point on the plane. The fixed point is called the centre of the circle and the line segment joining the centre to any point on the circle is called its radius. The constant distance is called the length of the radius of the circle. In the figure, a circle with centre at O and radius r is denote by (O, ABC) or simply by ABC or C(O, r), where A, B, and C are any three points on the circle. Obviously, each of the line segments OA, OB and OC are the radii of the circle. The circumference of a circle is the perimeter or boundary of the circle. (b) Interior and exterior points A point P or Q or R in the same plane as a circle C(O, r) is said; (a) to lie inside the circle or be an interior point, if OP < r. (b) to lie outside the circle or be an exterior point, if OQ > r. and (c) to lie on the circle or be a point of the circle, if OR = r. Circular Dam of Pharping Hydropower Nepal’s First Hydropower Circumference A O C B (a) O A (b) R r Q O Interior point Exterior point P CHAPTER 12 CIRCLE
262 Allied The Leading Mathematics-10 Circle 263 Geometry (ii) Chords on Circle (a) Chord, diameter and semi-circle of a circle A line segment joining any two points on a circle is called the chord of a circle. In the following figure, the line segment AB joining two given points A and B is the chord AB of the circle. If a chord passes through the centre of a circle, it is called a diameter. In the above figure, MN passes through the centre O and is therefore a diameter of the circle. A diameter divides a circle into two equal parts, each of which is known as a semi-circle. Obviously, a circle can have many diameters. It can be shown that the diameter is the largest chord of a circle; and all diameters of the same circle are equal. Furthermore, a diameter (d) of a circle is twice as long as its radius (r). i.e., d = 2r. (b) Segment of a circle A chord of a circle divides the region enclosed by a circle into two portions. Each such part is a segment of the circular disc. One of them is called the alternate segment of the other. The segment APB containing the diameter MN is called the major segment and the other AQB not containing the diameter MN is called the minor segment as shown in the adjoining figure. (iii) Arcs on Circles (a) Major arc and minor arc An arc of a circle is the set of all points of the circle between any two given points. Since two points divide the circle into two sets of points, there will be two arcs corresponding to any two given points. One may be larger than the other. The larger one is called the major arc and the smaller one the minor arc. The two arcs (i.e., major and minor arcs) will be equal if they stand exactly on a straight line through the centre. In the figure, ADB and ACB are two arcs determined by the points A and B. Here, ACB is the major arc that is greater than semi-circle and ADB is the minor arc that is less than semi-circle. C Chord Chord Chord Semi-circle D B B A A O O N M Diameter P P Major segment Minor segment N M B A Q C Semi circle Major arc O A B D Minor arc C Semi circle Major arc O A B D Minor arc
262 Allied The Leading Mathematics-10 Circle 263 Geometry (b) Sector of a circle The sector of a circle is a portion inside a circle enclosed by an arc and the radii joining the ends of the arc to the centre of the circle. In the given figure, A and B are the two ends of the arc ACB of a circle with centre at O. Here the portion OACB is a sector of the circle. In the adjoining figure, ∠APB = y° also stands on the same arc AB as ∠AOB = x° at the centre; but APB is an angle at the circumference of the circle and on the sector APB. We further note that OA and OB are two radii of the circle with centre at O. Suppose the radii OA and OB are produced to X and Y. Then the arc ACB lies inside ∠AOB. (c) Central Angle and Inscribed Angle Any angle whose vertex is at the centre of a circle is called a central angle. Thus, the angle AOB standing on the arc ACB and the reflex angle AOB standing on the arc APB are the central angles of the circle with centre at O as shown in the above figure. An angle subtended by an intercepted arc or chord of a circle at its circumference is known as an inscribed angle. Thus, angle APB is an inscribed angle subtended by the arc ACB. The uses of central angle and inscribed angle are in making Pie chart, sun-dial, star, clock, watch, slide of circular pizza, bread, cake, etc. (d) Angles in a segment If an arc subtends a number of angles at different points of the circumference of a circle, the angles on the same side of the chord joining the two ends of the arc are said to be in the same segment of the circle. In the given figure, ∠BAC, ∠ BDC and ∠ BEC are all in the same segment of a circle standing on the same arc BC. (e) Concentric, congruent and intersecting circles Circles in a plane are said to be concentric (i.e., ‘co’ means ‘same’ centre) if they have the same centre. In case they have the same radii also, they are identical or congruent. In addition to that, circles having the same radii are said to be congruent. Congruent circle can be fitted exactly onto another, i.e., one can be superposed on the other. More accurately, “Two circles are said to be congruent, if and only if, they have equal radii.” In the given figure, the three circles with different radii OA, OB and OC are concentric with O as their common centre ( or co-centre). P Y O B C A X x x P O y B C A Pie Chart Sun Dial Star A D E B C C Concentric circles B A O
264 Allied The Leading Mathematics-10 Circle 265 Geometry Given two circles in a plane, they may intersect at two distinct points, two coincident points or may not intersect at all as shown in the figure below: P Q P Q A B P R R C R r r r PQ < R + r PQ = R + r PQ > R + r Q When two circles intersect each other at two different points then they are called intersecting circles. The straight line joining two intersecting points is known their common chord. In the figure, AB is the common chord. In intersecting circle, the distance between two centres P and Q is always less than the sum of their radii. i.e., PQ < R + r. If PQ = R + r, two circles touch at a point. If PQ > R + r, two circles are disjoint and they never intersect. (f) Concyclic Points and Circumscribed or Cyclic Quadrilateral A given set of points are said to be concyclic if they lie on the circumference of a given circle. In the given figure, A, B, C, D are concyclic and the quadrilateral ABCD lies wholly inside the circle. Thus ABCD is a figure inscribed in a circle, and is often known as a cyclic quadrilateral. We also say that the quadrilateral is circumscribed by the circle through A, B, C and D. II Basic Concept of Properties on Arc of Circle Equal arcs of a circle subtend equal angles at its centre. Property 1 By Observation* Step-1: Cut down a circle with suitable radius from the chart paper as shown in the adjoining figure. Fold it half along diagonal and mark arcs with overlapping the points on both semi-circles then open it and draw the central angles on the equal arcs. Step-2: Also, cut down these two central angles and overlap them, which are congruent. Hence, it is clear that the both central angles AOB and COD are equal. A D C B A D O B C D O C A B O A=D B=C O Chart paper
264 Allied The Leading Mathematics-10 Circle 265 Geometry In a circle, arcs that subtend equal angles at the centre are equal. Property 2 (CONVERSE OF Property 1) By Observation* Step-1: Cut down a circle with suitable radius from the chart paper as shown in the adjoining figure. Draw two equal central angles AOB and COD. Step-2: Also, cut down ∠AOB and ∠COD and overlap them, which are congruent. Hence, it is clear that the arcs of the same circle that subtend equal angles at its centre are equal. Note: If central angle AOB subtended by AB is measured in degree, then its corresponding arc AB also can be measured in degree. i.e., if ∠AOB = x° then AB ≡ x°. Hence, AB ≡ ∠AOB. The symbol ≡ or is generally read as equal in degree measure. Similarly, if ACB is the inscribed angle, then ∠AOB = 2∠ACB. ∴ AB ≡ 2∠ACB i.e., ∠ACB ≡ 1 2 AB It is to be remembered that AB is not equal to ∠AOB, but their measurements in degree are equal. The use of the above notation is illustrated below in establishing a theorem related to inscribed angle and corresponding arc. Twice the degree measure of an inscribed angle is equal to the degree measure of its corresponding arc. Theoretical Proof* Given: BAC is an inscribed angle subtended by the arc BC of a circle with centre O. To prove: 2∠BAC ≡ BC or ∠BAC ≡ 1 2 BC Construction: Join AO, BO and CO. SN Statements SN Reasons 1. OA = OB 1. Radii of the same circle. 2. ∠ABO = ∠BAO 2. Base angles of isosceles ∆AOB 3. ∠ABO + ∠BAO + ∠AOB = 180o 3. Sum of angles of a ∆AOB. 4. 2∠BAO = 180o – ∠AOB 4. From statements (2) and (3). 5. Similarly, 2∠OAC = 180o – ∠AOC 5. Same as above. 6. 2∠BAO + 2∠OAC = 360o – (∠AOB + ∠AOC) 6. Adding statements (4) and (5). 7. 2∠BAC = 360o – reflex ∠BOC 7. Addition axiom. 8. 2∠BAC ≡ Circumference – BAC 8. Reflex ∠BOC ≡ BAC and 360o ≡ circumference. B A D O C D O C A B O A = D B = C O A B C O
266 Allied The Leading Mathematics-10 Circle 267 Geometry 9. 2∠BAC ≡ BC or ∠BAC ≡ 1 2 BC 9. From statement (8). Hence, the degree measure of an inscribed angle is half of the degree measure of its corresponding arc. In a circle, equal chords cut off equal arcs. Property 3 By Observation* Step-1: Cut down a circle with suitable radius from the chart paper as shown in the adjoining figure. Draw two equal chords AB and CD in it. Step-2: Also, cut down the segments APB and CQD and then overlap them, which are congruent. Hence, in the circle, equal chords AB and CD cut off equal arcs AB and CD. If two arcs of a circle are equal, the corresponding chords are equal. Property 4 (CONVERSE OF Property 3) By Observation* Step-1: Cut down a circle with suitable radius from the chart paper as shown in the adjoining figure. Mark two equal arcs AB and CD by drawing equal central angles. Join AB and CD. Step-2: Also, cut down the segments APB and CQD and then overlap them, which are congruent. Hence, in the circle, the chords AB and CD are equal, intersected by equal arcs AB and CD. In a circle, parallel chords cut off equal arcs. Property 5 Hints : We can esily prove the theorem by taking the alternate angles and their corresponding arcs. If two arcs of a circle are equal, the chords between them are parallel. Property 6 (CONVERSE OF Property 5 Hints : We can easily prove the converse of theorem 7 by equating the alternate angles. O A P B D C Q Chart paper O A B D C Chart paper A B C D O
266 Allied The Leading Mathematics-10 Circle 267 Geometry III Theorems on Angles on the Same Arc of Circle The angle at the centre of a circle is double the angle at its circumference standing on the same arc. THEOREM 1 (A)* Observation Method Step-1: Cut down a circle having center O with suitable radius from the chart paper as shown in the adjoining figure. Mark an arc AB on it and take a point C on its circumference. Join A and B with O and C. Step-2: Cut down the sector AOB and fold it half. Then overlap it into inscribed angle ACB, which is exactly the same. Hence, it is clear that the central angle AOB is double of the inscribed angle ACB on the same arc of the circle. Or the inscribed angle ACB is half of the central angle AOB on the same arc of the circle. (B) Experimental Verification Step-1: Two circles of different radii more than 3 cm, each with centre labeled as O, are drawn. Step-2: ∠BOC at the centre and ∠BAC at circumference standing on the same arc BC are drawn in each circle. Step-3: The central ∠BOC and the inscribed ∠BAC are measured in each case and the results are tabulated below: Fig. ∠ BOC ∠BAC Remarks (i) 88o 44o ∠BOC = 2∠BAC (ii) 74° 37° ∠BOC = 2∠BAC Conclusion: From the above table it is clear that 'Angle at the centre is double the angle at the circumference standing on the same arc.' (C) Theoretical proof Given: In the circle with centre O, the central ∠BOC and the inscribed ∠BAC are standing on the same arc BC. To prove: ∠BOC = 2∠BAC. Construction: Join AO and produce it to any point D. C O A B O A=B O A B O A B C Fig. (ii) A B C O Fig. (i) O A B C Fig. (iii) B O B D C
268 Allied The Leading Mathematics-10 Circle 269 Geometry Proof: SN Statements SN Reasons 1. ∠ABO = ∠BAO 1. OA = OB in ∆AOB and base angles. 2. ∠BOD = ∠ABO + ∠BAO 2. Exterior angle property of a ∆AOB. 3. ∠BOD = 2∠BAO 3. From statements (1) and (2). 4. ∠DOC = 2∠OAC 4. Similar to (1) – (3) above. 5. ∠BOD + ∠DOC = 2∠BAO + 2∠OAC 5. By adding statement (3) and (5). 6. ∴ ∠BOC = 2∠BAC 6. Addition axiom. Hence, the angle at the centre of a circle is double the angle at its circumference standing on the same arc. Proved. “Alternatively” Proof: SN Statements SN Reasons 1. ∠BOC ≡ BC° 1. The relation between the central angle and its corresponding arc. 2. 2∠BAC ≡ BC° 2. The relation between the inscribed angle and its corresponding arc. 3. ∠BOC = 2∠BAC 3. From the statements (1) and (2). Hence, the angle at the centre of a circle is double the angle at its circumference standing on the same arc. Proved. Converse 8.1 If an angle is double of another angle standing on the same line segment, the angular point of smaller angle and the ends of the line segment lay a circle with center as angular point of the greater angle. Angles in the same segment of a circle are equal. ‘OR’ Inscribed (circumference) angles on the same arc are equal. THEOREM 2 (A)* Observation Method Step-1: Cut down a circle having center O with suitable radius from the chart paper as shown in the adjoining figure. Draw the inscribed angles ACB and ADB on the same arc AB. Step-2: Also, cut down the sector ACB and ADB arrange their corners which is exactly overlapped. Hence, it is clear that the inscribed angles ACB and ADB are equal in measure. C D O A B C A B D A B C=D A=B B=A A O B C
268 Allied The Leading Mathematics-10 Circle 269 Geometry (B) Experiment verification: Step-1: Two circles of different radii, each with centre at O, are drawn. Step-2: Inscribed angles ACB and ADB are drawn in the same segment ACDB of each circle. Step-3: In each figure, ∠ ACB and ∠ ADB are measured and tabulated below. Fig. ∠ ACB ∠ ADB Remark (i) 55o 55o ∠ACB = ∠ADB (ii) 34o 34o ∠ACB = ∠ADB Conclusion: From the above table it is clear that ‘Angles in the same segment of a circle are equal. (C) Theoretical proof Given: ∠BAC and ∠BDC are on the same arc BC of a circle with centre at O. To prove: ∠BAC = ∠BDC. Construction: Join OB and OC. Proof: SN Statements SN Reasons 1. ∠BOC = 2∠BAC 1. Relation between central angle and inscribed angle on the same arc AB. 2. ∠BOC = 2∠BDC 2. Same as above (1). 3. 2∠BAC = 2∠BDC 3. From statements (1) and (2). 4. ∠BAC = ∠BDC 4. By dividing both sides of statement (3) by 2. Hence, the angles in the same segment of a circle are equal. Proved. “Alternatively” Proof: SN Statements SN Reasons 1. ∠BAC ≡ 1 2 BC ° 1. The relation between the inscribed angle and its corresponding arc. 2. ∠BDC ≡ 1 2 BC ° 2. Same as above (1). 3. ∠BAC = ∠BDC 3. From the statements (1) and (2). Hence, inscribed angles on the same arc are equal. Proved. Corollary 9.1 Inscribed angles standing on the equal arcs are equal. O C D B A Fig. (i) O C D B A Fig. (ii) A D O B C
270 Allied The Leading Mathematics-10 Circle 271 Geometry Example-1 In the given circle with centre O, (a) Define central angle and inscribed angle. (b) Write the name of central angle and inscribed angle on the same arc. (c) Find the values of the unknown angles x and y. Solution: (a) An angle subtended at the cente of a circle by its arc is called a central angle and an angle subtended at the point of the circumference of a circle by its arc is called an inscribed angle. (b) The central angle and inscribed angle on the same arc AC are reflex angle AOC and ∠ABC respectively. (c) From the given figure, (i) x = 360° – 150° [ Complete angle at a point] ∴ x = 210o (ii) y = 1 2 of x = 1 2 × 210° = 105° [ Relation between central and inscribed angles on the same major arc AC.] Example-2 Find the values of the unknown angles x and y. (a) Why are ∠PRS and ∠PQS equal ? (b) Find the values of the unknown angles x and y. Solution: (a) ∠PRS and ∠PQS are equal because they stand on the same arc PS. (b) From the given figure, Now, we have (i) ∠PQS = ∠ QRS = 45° [ Alternate angles on PQ//SR.] (ii) ∠ PRS = ∠ PQS [ Inscribed angles on the same arc PS.] or, x = 45°. (iii) ∠QTR = ∠QSR + ∠PRS [ Exterior angle property of ∆RST.] = 45o + 45o = 90o (iv) ∠TQR + ∠QTR + ∠QRT = 180o [ Sum of interior angles of ∆RQT.] or, y + 50o + 90o = 180o or, y = 180o – 140o or, y = 40o A B C O y 150° x R Q P S 50° 45° y x
270 Allied The Leading Mathematics-10 Circle 271 Geometry Example-3 In the given circle with center O, two chords AB and CD intersect at the point E. (a) State the theorem on the relation between the inscribed angle and central angle standing on the same arc of a circle. (b) Prove that: ∠AOC + ∠BOD = 2∠BED (c) If ∠AOC = 28° and ∠BOD = 32°, find the value of ∠BED. Solution: (a) The inscribed angle and central angle standing on the same arc of a circle are equal. (b) Given: In the given circle with center O, if two chords AB and CD intersect at the point E. To Prove: ∠AOC + ∠BOD = 2∠BED Construction: Join BC. Proof: SN Statements SN Reasons 1. ∠AOC = 2∠ABC 1. Relation between central and inscribed angles on the same arc AC 2. ∠BOD = 2∠BCD 2. Same as above (1) 3. ∠AOC + ∠BOD = 2(∠ABC + ∠BCD) 3. By adding statements (1) and (2) 4. ∠BED = ∠ABC + ∠BCD 4. Exterior angle property of ∆BEC 5. ∠AOC + ∠BOD = 2∠BED 5. From statements (3) and (4) Proved. (c) Here, ∠AOC = 28° and ∠BOD = 32° Now, ∠BED = 1 2 (∠AOC + ∠BOD) = 1 2 (28° + 32°) = 1 2 × 60° = 30°. Example-4 Given figure is a circle in which AD//BC. (a) Which arcs are equal in the given circle ? Why ? (b) Prove that: ∠AYC = ∠BXD. Solution: (a) The arcs AB and CD are equal in the given circle because they are intersected by the parallel chords AD and BC. (b) Given: In the given circle, AD//BC. To Prove: ∠AYC = ∠BXD. Proof: SN Statements SN Reasons 1. AB = CD 1. Corresponding arcs of parallel chords 2. AB + BC = BC + CD 2. By adding on both sides of statement (1) 3. ABC = BCD 3. Whole parts axiom 4. ∠AYC = ∠BXD 4. Inscribed angles on equal arcs Proved. O D B E C A Y X A D B C
272 Allied The Leading Mathematics-10 Circle 273 Geometry Example-5 In the given circle, if AB = AC, show that BE = CD. Solution: Given: In the given circle, AB = AC. To Prove: BE = CD. Proof: SN Statements SN Reasons 1. In ∆ABE and ∆ACD, (i) ∠BAE = ∠CAD (A) (ii) AB = AC (S) (iii) ∠ABE = ∠ACD (A) 1. (i) Common angles. (ii) By given. (iii) Inscribed angles on the same arc DE. 2. ∆ABE ≅ ∆ACD 2. By ASA congruent condition. 3. BE = CD 3. Corresponding sides of congruent triangles. Proved. Example-6 In the given figure, two circles intersect at A and B. Through A two straight lines CAD and EAF are drawn terminated by the circumference. Prove that ∠CBE = ∠DBF. Solution: Given: In the given figure, A and B are the points of intersection of two circles. Two straight lines CAD and EAF pass through the point A. To Prove: ∠CBE = ∠DBF. Proof: SN Statements SN Reasons 1. ∠CBE = ∠CAE 1. Inscribed angles on the same arc CE. 2. ∠DAF = ∠DBF 2. Inscribed angles on the same arc DF. 3. ∠CAE = ∠DAF 3. Vertically opposite angles. 4. ∠CBE = ∠DBF 4. From statements (1), (2) and (3). Proved B D A E C B E D C F A
272 Allied The Leading Mathematics-10 Circle 273 Geometry PRACTICE 12.1 Read Think Understand Do Keeping Skill Sharp 1. (a) What is the relation between the central angle and inscribed angle standing on the same arc of a circle? (b) Write the relation between the angles at the circumference of a circle on the same arc. (c) What is the relation between an arc and its corresponding central angle of a circle ? (d) What is the degree measure of an arc of a circle if its corresponding inscribed angle is 32° ? 2. Circle ( ) the correct answer. (a) The central angle and inscribed angle are subtended by the same arc of a circle. If the measure of the central angle is 45°, which is the measure of inscribed angle ? (i) 90° (ii) 20° (iii) 22.5° (iv) 45° (b) What is the value of x in the given circle ? (i) 48° (ii) 96° (iii) 24° (iv) 38° (c) What is the value of a in the given circle ? (i) 30° (ii) 60° (iii) 120° (iv) 100° (d) What is the value of y in the given circle ? (i) 84° (ii) 44° (iii) 14° (iv) 42° (e) What is the value of p in the given circle ? (i) 136° (ii) 44° (iii) 112° (iv) 68° Check Your Performance 3. (i) (a) State the relation between the central angle POQ and the circumference angle PRQ standing on its same minor arc PQ of a circle with centre O, where R is on its major arc. (b) Prove that from the above information: ∠PRQ = 1 2 ∠POQ. (c) Take any one point S on its major arc. If ∠POQ = 82°, find the degree measures of ∠PRQ and ∠PSQ. Now, establish the relation among ∠PRQ and ∠PSQ. (ii) (a) State the theorem on the two inscribed angles on the same arc of a circle. (b) Prove the statement of the above theorem. (c) Justify your reasons by measuring the inscribed angles using a protractor. 60° O a 48° x Q. No. 2. (b) Q. No. 2. (c) y 14° 28° p 224° Q. No. 2. (d) Q. No. 2. (e)
274 Allied The Leading Mathematics-10 Circle 275 Geometry 4. (i) (a) Write the relation between the measure of inscribed angles standing on the same arc. (b) Write the name of inscribed angles on the same arc. (c) Find the values of the unknown angles a and b. (ii) (a) Write the relation between the measure of central angle and inscribed angle standing on the same arc. (b) Write the pair of central angle and inscribed angle on the same arc. (c) Find the values of the unknown angles a and b. 5. (a) In the figure, XYZ is an equilateral triangle inscribed in a circle whose centre is O. Calculate the angle XOY. (b) XYZ is an isosceles triangle inscribed with XY = XZ in a circle whose centre is O. If ∠YOZ = 124°. Calculate the angle XYZ. 6. In the given figures (i) and (ii), A, B, C and D are concyclic points on a circle with centre O. (a) Write the corresponding central angles of the inscribed angles x and y in each circle. (b) Find the values of x and y in each figure. (c) Add the values of x and y in each figure. What do you find ? Write your conclusion 7. Find the values of the unknown angles x and y in the figures: (i) (ii) (iii) (iv) 8. (i) In the given figure, O is the center of the circle. If two chords QR and ST intersect at the point X, prove that ∠QXT = 1 2 (∠QOT + ∠ROS). (ii) In the given figure, O is the center of the circle. Prove that ∠ROQ = ∠RPQ + ∠RSQ. b Q O R P 60° S a O q B C A D 43° p X O 124° Y Z X O ? Y Z x O Q R S P y 3a 2a x 224° O A (i) D (ii) B C y C A B B 65° O 50° y x C A B 15° O x y 25° C 110° A B D O x y C A B D O 50° x y O S R Q T X O P R Q. No. 8. (i) Q. No. 8. (ii) Q S
274 Allied The Leading Mathematics-10 Circle 275 Geometry 9. (i) Given figure is a circle in which WX//ZY. Prove that: ∠WAY = ∠XBZ. (ii) Given figure is a circle in which ∠AYC = ∠BXD. Prove that: AD//BC 10. (i) In the figure, ∠AEC = ∠BFD. Prove that: (a) ∠ABC = ∠BCD (b) AB//CD (ii) In the given figure, AB//CD. Prove that: ∠AEC = ∠BFD 3. (c) 41° 4. (i) (c) 30°, 30° (ii) (c) 86°, 43° 5. (a) 120° (b) 59° 6. (i) (b) 68°, 112° (ii) (b) 72°, 108° 7. (i) 80°, 40° (ii) 40°, 90° (iii) 65°, 50° (iv) 140°, 20° Answers Project Work Show the following theorems by paper cutting and draw their figures. (i) The angle at the centre of a circle is twice the angle of its circumference standing on the same arc. (ii) The angles in the same segment of a circle are equal. Y X B W Z A Y X B W Q. No. 9. (i) Q. No. 9. (ii) Z A E F A B C D E F A B C D Q. No. 10. (i) Q. No. 10. (ii)
276 Allied The Leading Mathematics-10 Circle 277 Geometry 12.2 Theorem on Inscribed Angle on Semi-circle At the end of this topic, the students will be able to: ¾ show experimentally and logically the inscribed angle on the semi-circle as a right. ¾ solve the related problems. Learning Objectives Semi-circles There are many semi-circular structures in our surroundings. Some of them are as follows: When a diagonal is drawn in a circle, it becomes separating two halves. Each is called a semicircle. We can find many inscribed angles on the semi-circle. What is the measure of each inscribed angle on semi-circle? Can we find it easily? Thales's theorem: Angle in a semi-circle is a right angle. THEOREM 3 (A)* Observation Method Step-1: Cut down a circle having center O with suitable radius from the chart paper as shown in the adjoining figure. Make an inscribed angle ACB on the diameter at one side. Step-2: Also, cut down the circle along diameter AB and fold half one semi-circle which have not made inscribed angle, whose angular measure is 90o . Then overlap it into inscribed angle ACB, which is exactly the same. Hence, it is clear that the inscribed angle ACB is a right angle. A B O C A O B A=B O A C O B
276 Allied The Leading Mathematics-10 Circle 277 Geometry (B) Experimental Verification Step-1: Two circles of different radii, each with centre labeled as O, are drawn. Step-2: The diameter AB and ∠ACB in the semicircle are drawn in each circle. Step-3: The ∠ACB in a semi-circle is measured in each case and the results are tabulated below: Fig. ∠ ACB Remark (i) 90o ∠ACB = 90o (ii) 90o ∠ACB = 90o Conclusion: Hence, ‘Angle in a semi-circle is a right angle.’ Proved. (C) Theoretical proof Given: ∠ ACB is an angle in the semi-circle of the circle ABC with centre at O, i.e., angle standing on the diameter AB. To prove: ACB is a right angle. or, ∠ACB = 90o . Proof: SN Statements SN Reasons 1. ∠AOB = 2∠ACB 1. Relation between the central angle and the inscribed angle on the same arc. 2. ∠AOB = 180° 2. AOB as straight angle. 3. 2∠ACB = 180° 3. From statements (1) and (2). 4. ∴ ∠ACB = 90° 4. By dividing both sides by 2. Hence, the angle in a semi-circle is a right angle. Proved. Converse3.1: The arc of a circle subtending a right angle at any point of the circle in its alternate segment is a semi-circle. Corollary 3.1: Any angle subtended by a minor arc in the circumference of a circle is acute and any angle subtended by a major arc in the circumference of a circle is obtuse. In the figure, ∠AFC is the acute angle which is inscribed in minor arc AXC and ∠AHC is the obtuse angle which is inscribed in major arc AYC. Note : If the hypotenuse of a right triangle is the diameter of a circle, the circle passes through the opposite vertex of the hypotenuse. A O B C Fig. (ii) A O B C Fig. (i) A O B C X O Z Y A X H C F O B Y
278 Allied The Leading Mathematics-10 Circle 279 Geometry Example-1 Draw a circle with centre O. Also draw a diameter XOY and produce it up to Z. Take a point W on its circumference and join with X and Y. (a) What is the measure of ∠XWY ? Why ? Give reason. (b) Prove it theoretically. (c) If ∠WYZ = 132°, find the ∠WXY. Solution: Draw the figure from the given information. (a) The measure of ∠ XWY is 90o because it inscribes on semi-circle. (b) Please look the previous theoretical proof of theorem 3. (c) Here, ∠WYZ = 132° Now, ∠WXY + ∠XWY = ∠WYZ [ Exterior angle property of triangle.] or, ∠WXY + 90° =132° ∴ ∠WXY = 42° Example-2 In the adjoining figure, O is the centre of a circle. (a) State the theorem on the inscribed angle on semi-circle. (b) If ∠BDC = 63°, find the values of x and y. (c) Write the relation between of x and y. Solution: (a) The inscribed angle on semi-circle is a right angle or 90°. (b) From the given figure, (i) ∠ABC =∠BCD = 90o [ Inscribed angles on the semi-circles.] (ii) ∠BAC = ∠BDC = 63o [ Inscribed angles on the same arc BC.] (iii) ∠ACB + ∠ABC + ∠BAC = 180o [ Sum of an interior angle of ∆ABC.] or, x + 90° + 63o = 132° or, x = 180° – 153° ∴ x = 27° (iv) ∠ABO = ∠BAO [ Base angles of isosceles triangle AOB.] ∴ y = 63°. (c) The relation between x and y is x + y = 90°. Example-3 In the given circle with center O, AB is the diameter. (a) Why is the measure of ACD 90° ? (b) If DO⊥AB, prove that ∠AEC = ∠ODA. (c) If ∠BAC = 55°, find the value of ∠AEC. Solution: (a) The measure of ACD is 90° because it subtends on a semi-circle (b) Given: In the given circle with center O, AB is the diameter and DO⊥AB. Z Y X O W A B C D O 63° x y D C A B O E
278 Allied The Leading Mathematics-10 Circle 279 Geometry To prove: ∠AEC = ∠ADO. Proof: SN Statements SN Reasons 1. ∠ACB = 90o 1. Inscribed angle on semi-circle 2. In ∆ABC and ∆AOD, (i) ∠ACB = ∠AOD (ii) ∠BAC = ∠OAD (iii) ∠ADO =∠ABC 2. (i) Both right angles (ii) Common angles (iii) Remaining angles 3. ∠AEC = ∠ABC 3. Inscribed angles on the same arc AC 4. ∠AEC = ∠ADO 4. From the statements 2(iii) and (3) Proved. (c) Here, ∠BAC = 55°. Then, ∠ADO = 180° – 90° – 55° = 35° So, ∠AEC = ∠ADO = 35°. Example-4 In the given diagram, PQR is an isosceles triangle with PQ = PR. A circle is drawn with PQ as a diameter, prove that QS = RS. Solution: Given: In the given diagram, PQR is an isosceles triangle with PQ = PR. A circle is drawn with PQ as a diameter. To prove: QS = RS. Construction: Join PS. Proof: SN Statements SN Reasons 1. PS⊥QR 1. Inscribed ∠PSQ on semi-circle 2. In ∆PSQ and ∆PSR, (i) ∠PSQ = ∠PSR (R) (ii) PQ = PR (H) (iii) PS = PS (S) 2. (i) Both right angles (ii) By given (iii) Common side 3. ∆PSQ ≅ ∆PSR 3. By RHS congruent condition 4. QS = RS 4. By corresponding sides of congruent triangles Proved. Example-5 In the given circle with center O, BC = DE. Prove that AD⊥BE. Solution: Given: In the given circle with center O, BC = DE. To prove: AD⊥BE. Construction: Join EA. Proof: O P Q R S D E F C B O A D E F C B O A
280 Allied The Leading Mathematics-10 Circle 281 Geometry SN Statements SN Reasons 1. In ∆ABC and ∆AEF, (i) ∠BAC = ∠FAE (ii) ∠ACB = ∠AEF (iii) ∠ABC = ∠AFE 1. (i) Being inscribed angles on equal arcs. (ii) Being inscribed angles on the same arc AB. (iii) Being remaining angles. 2. ∠ABC = 90o 2. Being inscribed angle on semi-circle. 3. ∠AFE = 90o 3. From statements 1(iii) and (2). 4. AD⊥BE 4. From statement (3). Proved. PRACTICE 12.2 Read Think Understand Do Keeping Skill Sharp 1. (a) What is the measure of the angle standing on the semi-circle? (b) In the given figure, ∠ABC = 90°. Write the name of the arc ABC. (c) In the given figure, O is the centre of the semi-circle. What is the value of x? 2. Circle ( ) the correct answer. (a) In the given figure, ∠SPR = 90°. Which is the name of SR ? (i) Diameter (ii) Radius (iii) Circumference (iv) Centre (b) In the given figure, O is the center of the circle. The value of ∠MLN is ......... . (i) 40° (ii) 60° (iii) 50° (iv) 90° (c) What is the value of x in the given circle with centre O? (i) 60° (ii) 30° (iii) 90° (iv) 45° (d) If XYZ is a triangle with right angle at Y, which one of the following is false ? (i) XZ is a diameter of a circle. (ii) X, Y and Z pass through a circle. (iii) Y is the centre of a circle. (iv) The sum of ∠X and ∠Z is 90°. (e) Which is the value of ∠ABD in the given circle with centre P ? (i) 90° (ii) 60° (iii) 30° (iv) 50° A C B O P Q R 3x R O S P L O N M 40° x 60° O X Z Y D A P B 30° C
280 Allied The Leading Mathematics-10 Circle 281 Geometry Check Your Performance 3. (i) Draw a semi-circle with centre O. Also, draw a diameter AOB. Take a point B on its circumference and join with A and C. (a) What is the measure of the angle ∠ABC ? Why ? Give reason. (b) Prove it theoretically. (c) If ∠ACB = 31°, find the ∠BAC. (ii) (a) Write a statement of the theorem of an angle inscribed in a semi-circle. (b) Prove experimentally it by drawing two semi-circles with radii more than 3 cm. (c) What is the sum of two acute angles in each inscribed right triangle? Write your conclusion. 4. Observe the given figures (i) and (ii). (a) What is the measure of the angle ∠SPT ? Why ? Give reason. (b) Find the values of a and b in each figure. (c) Write the relation between a and b in both figures. 5. (a) Find the values of the unknown angles x and y in the given figures. (b) Write the relation between x and y in both figures. (i) (ii) (iii) 6. (i) In the given figure, BD is a diameter of the circle ABCD and ∠ACB = 30o . What is the value of ∠DBA? (ii) In the figure, O is the centre of the circle and AB is a diameter. If ∠BEC = 55o ; find ∠CAD. 7. (a) In the given circle with center O, find the values of x and y. (b) In the given circle with center O, find the measure of ∠ADB. a b O 60° P S T (i) a b O P T S (ii) N K L M O 2x 3x y P R R O y x 3x A B C D O y 45° x D A B 30° C B A O E D 55° C O A O 40° D C B Q O y 80° x P R
282 Allied The Leading Mathematics-10 Circle 283 Geometry 8. (a) In the adjoining diagram, O is the centre of the circle. DO ⊥ AB. Prove that ∠OCA = ∠ODC. (b) In the given circle with center O, AB is the diameter and ∠AEC = ODA. Prove that DO⊥AB. 9. (a) In the given figure, PS is the diameter of the circle. If PQ = PR, prove that ∠QPS = ∠RPS. (b) NO is the bisector of ∠MNP, where O at the centre of the circle. Prove that MN = NP. 10. (a) O is the centre and AB the diameter of the circle. If AB = AC; prove that AD bisects BC, where D is the point of intersection of the circle and BC. (b) In the figure, O is the centre of the circle. OS⊥RQ and PQ is the diameter. Prove that PR = 2SO. 3. (i) (c) 59° 4. (b) 30°, 30° (ii) (b) 120°, 60° 5. (i) (a) 45°, 45° (ii) (a) 30°, 60° (iii) (a) 18°, 36° 6. (i) 60° (ii) 35° 7. (i) 40°, 40° (ii) 50° Answers D C A B O A B E C D O P Q S R N O P M P O S Q R A O B D C
282 Allied The Leading Mathematics-10 Circle 283 Geometry 12.3 Theorem on Cyclic Quadrilateral At the end of this topic, the students will be able to: ¾ prove experimentally and logically the relation among corresponding angles in a cyclic quadrilateral. ¾ solve the related problems. Learning Objectives Cyclic Quadrilaterals A quadrilateral is called cyclic if all the four vertices lie on a circle. Sometimes a cyclic quadrilateral is also called inscribed quadrilateral. If atleast one point of them is interior or exterior, the quadrilateral in not cyclic. A B O D C A B O D C A B O D C We draw different types of cyclic quadrilaterals as shown on the adjoining figures. Now, we measure the angles of the cyclic quadrilateral and identify its properties. The opposite angles of a cyclic quadrilateral are supplementary or two right angles (180°). THEOREM 4 (A)* Observation Method Step-1: Cut down a circle having center O with suitable radius from the chart paper as shown in the adjoining figure. Draw a cyclic quadrilateral ABCD. Step-2: Take out the cyclic quadrilateral from the circle by cutting and also cut down the same quadrilateral into four parts as shown in the adjoining figure. Then arrange the parts having the corners A and C as well as B and D in a straight line, which are exactly made a straight line by their edges. Hence, it is clear that the opposite angles of a cyclic quadrilateral are supplementary or two right angles (180o ). Which of the following is cyclic quadrilateral? D C B A A + C B + D
284 Allied The Leading Mathematics-10 Circle 285 Geometry (B) Experimental Verification Step-1: Two circles with different radii are drawn. One quadrilateral labeled as ABCD is inscribed in each circle. Fig. (i) A δ α β γ B O D C Fig. (ii) A δ α β γ O B D C Step-2: Pairs of opposite angles ∠BAD = α, ∠ABC = β, ∠BCD = γ and ∠ADC = δ are measured in each figure and tabulated below: Fig. ∠α ∠β ∠γ ∠δ ∠α + ∠γ ∠β + ∠δ Remark (i) 70o 85o 110o 95o 180o 180o ∠BAD+∠BCD=180o (ii) 90o 85o 90o 95o 180o 180o ∠BAD+∠BCD=180o Conclusion: From the above table it is clear that the opposite angles of a cyclic quadri–lateral are supplementary or 2 right angles (180o ). Proved. (C) Theoretical proof Given: ABCD is a quadrilateral inscribed in a circle whose centre is at O. To prove: (i) ∠BAD + ∠BCD = 180o and (ii) ∠ABC+ ∠ADC=180o . Construction: Join OB and OD. Proof: SN Statements SN Reasons 1. Reflex ∠BOD + Obtuse ∠BOD = 360° 1. Angle around a point is a complete angle. 2. (i) ∠BCD = 1 2 Reflex ∠BOD (ii) ∠BAD = 1 2 Obtuse ∠BOD 2. Relation between the central angle and the inscribed angle standing on the same arc 3. ∠BCD + ∠BAD = 1 2 (Reflex ∠BOD + Obtuse ∠BOD) 3. By adding property 2 (i) and (ii) 4. ∠BCD + ∠BAD = 1 2 360o = 180o 4. From statements (1) and (3) 5. ∠ABC + ∠ABD = 180o 5. By substitution property Hence, the opposite angles of a cyclic quadrilateral are supplementary or two right angles (180o ) Q.E.D. A B C D O
284 Allied The Leading Mathematics-10 Circle 285 Geometry “Next Method” (Not needed construction) SN Statements SN Reasons 1. (i) ∠BCD ≡ 1 2 Arc BAD (ii) ∠BAD ≡ 1 2 Arc BCD 1. Relation between the inscribed angles and corresponding arcs 2. ∠BCD + ∠BAD ≡ 1 2 (Arc BAD + Arc BCD) 2. By adding statements (i) and (ii) 3. ∠BCD + ∠BAD ≡ ABCD = 1 2 × 360o = 180o 3. By whole parts axiom and measure of complete angle 4. ∠ABC + ∠ABD = 180o 4. By substitution property Proved. Converse 11.1 The opposite angles of a quadrilateral are supplementary or two right angles (180o ), it becomes cyclic. Corollary 11.1 A cyclic parallelogram is a rectangle. An exterior angle made by producing one side of a cyclic quadrilateral is equal to the opposite angle of its adjacent angle. SUB-THEOREM 5 Solution: Here, Given: In the given figure, ABCD is a cyclic quadrilateral. To prove: ∠DCE = ∠BAD. Proof: SN Statements SN Reasons 1. ∠DCE + ∠BCD = 180° 1. Angles in straight line at a point. 2. ∠BAD + ∠BCD = 180° 2. Opposite angles of cyclic quadrilateral. 3. ∠DCE + ∠BCD = ∠BAD + ∠BCD or, ∠DCE = ∠BAD. 3. From statements (1) and (2). Q.E.D. Converse of Corollary of Theorem 5: If an exterior angle made by producing one side of a quadrilateral is equal to the opposite angle of its adjacent angle, the quadrilateral becomes a cyclic. A D B C E
286 Allied The Leading Mathematics-10 Circle 287 Geometry Example-1 In the given figure, O is the centre of a circle. (a) The sum of opposite angles of a cyclic quadrilateral is ....... . (b) What is the measure of Reflex ∠POR ? (c) Find the values of x and y. (d) Which angle is greater ∠PSR or ∠PQR ? Why ? Solution: (a) The sum of opposite angles of a cyclic quadrilateral is 180o . (b) Reflex ∠POR = 2 × ∠POR = 2 × 95o = 190o . (c) From the given figure, (i) x + 95o = 180o [ Sum of opposite angles of cyclic quadrilateral.] or, x = 180o – 95o = 85o (ii) y = 2x [ Relation between central and inscribed angles on the same arc PSR.] = 2 × 85o = 170o . (d) ∠PSR is greater than ∠PQR because ∠PSR is subtended on the major arc PQR. Example-2 In the given figure alongside, O is the centre of a circle. (a) Is the quadrilateral ABCO cyclic ? Give reason. (b) Find the measure of ∠ABC. (c) Which angle is smaller or greater between ∠ABC and Obt. ∠AOC by how much ? Solution: (a) No, the quadrilateral ABCO is not a cyclic quadrilateral because the point O does not lie on the circumference of the circle. (b) In the given figure, construct a inscribed angle ADC on the arc ABC. Now, (i) ∠ADC = 1 2 ∠AOC = 1 2 × 140o = 70o [ Relation between central and inscribed angles on the same arc PSR.] (ii) ∠ABC + ∠ADC = 180o [Sum of opposite angles of cyclic quadrilateral.] or, ∠ABC + 70o = 180o or, ∠ABC = 180o – 70o = 110o . (c) ∠ABC is smaller than Obt. ∠AOC by 140o – 110o = 30o . Example-3 In the given figure alongside, O is the centre of a circle. (a) Write the name of the acute angle PST. (b) Find the measure of ∠PST. (c) Write the relation between ∠PST and ∠PQR. Solution: (a) The name of the acute angle PST is the exterior angle of the cyclic quadrilateral PQRS. S R Q P 95° y O x D O 140° B A C O Q R S T P 52°
286 Allied The Leading Mathematics-10 Circle 287 Geometry (b) From the given figure, (i) ∠QRP = ∠QPR = 52o [ Base angles of an isosceles triangle.] (ii) ∠QRP + ∠QPR + ∠PQR = 180o [Sum of angles of a triangle.] or, 52o + 52o + ∠PQR = 180o or, ∠PQR = 180o – 104o = 76o . (iii) ∠PST = ∠PQR [ Exterior angle property of cyclic quadrilateral.] or, ∠PST = 76o . (c) ∠PST = ∠PQR. Example-4 ∆TPQ is an isosceles triangle in which TP = TQ. A circle passing through P and Q cuts TP and TQ at R and S respectively. (a) Prove that: PQ//RS. (b) Write the relation between the arcs PR and QS. (c) Is ∆TRS also an isosceles triangle ? Give reason. Solution: (a) Given: TPQ is an isosceles triangle in which TP = TQ. A circle passing through P and Q cuts TP and TQ at R and S respectively. To prove: PQ//RS. Proof: SN Statements SN Reasons 1. ∠PQT = ∠QPT 1. Being base angles of an isosceles ∆PQT 2. ∠PQT = ∠SRT 2. By property of cyclic quadrilateral 3. ∠QPT = ∠SRT 3. From statements (1) and (2) 4. PQ//RS 4. Equal corresponding angles in statement (3) (b) The arcs PR and QS are equal because they are cut off by the parallel chords PQ and RS. (c) Yes, ∆TRS is also an isosceles triangle because the equal arcs PR and QS make equal chords PQ and RS in the isosceles triangle TPQ. Proved. Example-5 In the given figure, P, Q and R are the mid-points of the sides AB, BC and AC of ∆ABC respectively and AS⊥BC. (a) Prove that PQRS is a cyclic quadrilateral. (b) Which angle is equal with ∠PQR in the cyclic quadrilateral PQRS? P Q S R T B P D Q B C S R
288 Allied The Leading Mathematics-10 Circle 289 Geometry Solution: (a) Given: In the given figure, ABCD is a cyclic quadrilateral. To prove: PQRS is a cyclic quadrilateral. Construction: Join PS. Proof: SN Statements SN Reasons 1. QR//AB and PQ//BC 1. Being lines joining the mid–points of two sides of ∆ABC. 2. AD = DS 2. Being AP = BP and PD//BS as the midpoint theorem. 3. PD⊥AS. 3. Being PQ//BC and AD⊥BC. 4. PS = AP 4. From stat. (2) and (3) in ∆PAS. 5. PB = PS 5. Being AP = PB and PS = AP. 6. ∠PBS = ∠PSB 6. Base angles of the isosceles ∆PBS. 7. PQRB is a parallelogram. 7. Being parallel opposite sides in stat. (1) 8. ∠PBR = ∠PQR 8. Bing opposite angels of //gm PQRB. 9. ∠PSB = ∠PQR 9. From stat. (6) and (8). 10. PQRS is a cyclic quadrilateral. 10. From stat. (9) by exterior angle property of the cyclic quadrilateral. Proved. (b) ∠PSB is equal with ∠PQR in the cyclic quadrilateral PQRS. Example-6 In the figure alongside, DM⊥EF and FN⊥ED. (a) Prove that ∠EMN = ∠EDF. (b) If ∠ENM = 57°, what is the value of ∠DFE ? Solution: Given: In the given figure, DM⊥EF and FN⊥ED. To prove: ∠EMN = ∠EDF. Proof: SN Statements SN Reasons 1. ∠DNF = ∠DMF = 90o 1. By given. 2. DFMN is a cyclic quadrilateral. 2. Equal angles standing on the same base DF. 3. ∠EMN = ∠EDF 3. By exterior angle property of cyclic quadrilateral. Proved. (b) If ∠ENM = 57°, the value of ∠DFE = 57° because of exterior angle property of cyclic quadrilateral. D M N E F
288 Allied The Leading Mathematics-10 Circle 289 Geometry Example-7 In the given figure, BD is the bisector of ∠ABC. AE and BD meet at M and BC and DE meet at N. Prove that MN//AC. Solution: Given: In the given figure, BD is the bisector of ∠ABC. AE and BD meet at M and BC and DE meet at N. To prove: MN//AC. Construction: Join BE. Proof: SN Statements SN Reasons 1. ∠ABD = ∠CBD 1. By given, bisector BD of ∠ABC 2. ∠ABD = ∠AED 2. Being inscribed angles on the same arc AD 3. ∠CBD = ∠AED or, ∠NBM = ∠MEN 3. From statements (1) and (2) 4. BMNE is a cyclic quadrilateral. 4. Being equal angles on the same line segment MN 5. ∠BMN = ∠BEM 5. Being inscribed angles on the same arc BM 6. ∠BCA = ∠BEA 6. Being inscribed angles on the same arc AB 7. ∠BMN = ∠BCA 7. From statements (5) and (6) 8. MN//AC 8. Being equal corresponding angles in stat. (7) Proved. PRACTICE 12.3 Read Think Understand Do Keeping Skill Sharp 1. (a) What are concyclic points? (b) Define cyclic quadrilateral. (c) What is the measure of the sum of the opposite angle of a cyclic quadrilateral? (d) In the given figure, ABCD is a cyclic quadrilateral. Which angle is equal to ∠ADE? 2. Circle ( ) the correct answer. (a) A cyclic quadrilateral is formed by ....................... . (i) three points on a circle. (ii) any four points inside a circle. (iii) four concyclic points. (iv) four sides in a circle B E N C D A M B A D C E
290 Allied The Leading Mathematics-10 Circle 291 Geometry (b) The sum of opposite angles of a cyclic quadrilateral KMLN is ........ . (i) 90° (ii) 180° (iii) 360° (iv) 0° (c) The exterior angle formed by producing a side of a cyclic quadrilateral is equal to ........ (i) its adjacent angle (ii) its opposite angle (iii) non-adjacent co-interior angle (iv) opposite angle to its adjacent angle (d) What is the correct value of a in the given figure ? (i) 70° (ii) 110° (iii) 80° (iv) 180° (e) What is the correct value of ∠LMK in the given figure ? (i) 30° (ii) 110° (iii) 140° (iv) 180° Check Your Performance 3. Observe the following figures. (a) What is the sum of opposite angles of a cyclic quadrilateral ? (b) Which angle is greater between ∠ABC and ∠ADC ? Why ? (c) Find the values of unknown angles x and y in each figure. (i) (ii) (iii) 4. Observe the given figures. (a) Is the quadrilateral ABCD cyclic ? Give reason. (b) Find the measures of indicated unknown angles. (c) Which angle is smaller or greater between ∠ABC and ∠ADC by how much ? (i) (ii) (iii) a 110° N O L M K 40° O A B D C 125° x y A B D x45° C O O D B A C x 50° y A B C D A D C B E A B C E x y60° D 85° a 110° c O r s q 61° p 48°
290 Allied The Leading Mathematics-10 Circle 291 Geometry 5. Observe the following circles. (a) Write the name of the acute angle DEF. (b) Find the value of ∠DEF. (i) (ii) (iii) 6. Observe the following figures and find the vale of x in each figure. (i) (ii) (iii) 7. (i) (a) State the theorem on the sum of angles of opposite angles of a cyclic quadrilateral. (b) Prove that the theorem mentioned in (i). (c) If the measure of one angle the cyclic quadrilateral is 45°, how many degrees is its corresponding angle ? (ii) (a) How many degrees is the sum of angles of opposite angles of a cyclic quadrilateral? (b) Verify experimentally the above statement mentioned in (i) by drawing at least two circles with radii more than 3 cm. (c) What type of angle forms that is opposite to the acute angle of a cyclic quadrilateral? (iii) (a) Draw a cyclic quadrilateral in a circle with centre O. Write the relation of an exterior angle with the opposite angle of its adjacent angle when producing one side of the cyclic quadrilateral. (b) Prove the above relation mentioned in (i). 8. (i) In the given figure, PQ//RS. A circle passing through P and Q cuts TP and TQ at R and S respectively. (a) Prove that: TP = TQ. (b) Write the relation between the chords PR and QS. (c) Is ∆TRS also an isosceles triangle ? Give reason. (ii) TRS is an isosceles triangle in which TR = TS. A circle passing through P and Q cuts TP and TQ at R and S respectively. (a) Prove that PQ//RS. (b) Write the relation between the arcs PR and QS. (c) Is ∆PTQ also an isosceles triangle ? Give reason. A H E D F x P S O D x 150° E F O D M E F x 140° O P Q R 144° x x O R P Q x O R P Q R S P Q T Q P S R T
292 Allied The Leading Mathematics-10 Circle 293 Geometry 9. (i) In the adjoining figure, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that ∆QRS is an isosceles triangle. (ii) In the given figure, ABCD is a cyclic quadrilateral. A side CD is produced to the point P. If AB = AC, prove that AD is the bisector of ∠BDP. 10. (i) In the figure, ABCD is an isosceles trapezium in which AB = DC. Prove that ABCD is a cyclic trapezium. (ii) In the given figure, MN//AC. AE and BD meet at M and BC and DE meet at N. Prove that BD is the bisector of ∠ABC. 3. (c) (i) 45° (ii) 35°, 125° (iii) 40°, 50° 4. (i) (a) Yes (b) 55°, 125° (c) 70° (ii) (a) No (b) 190°, 35° (c) 5° (iii) (a) Yes (b) 48°, 48°, 48°, 96° (c) 38° 5. (b) (i) 60° (ii) 75° (iii) 70° 6. (i) 108° (ii) 120° (iii) 60° Answers Project Work Show the following theorems by paper cutting and drawing their figures. (i) The inscribed angle in semicircle is right angle. (ii) The sum of opposite angles of a cyclic quadrilateral is supplementary. R Q P T S P D C B A A D B C D M N E A B C
292 Allied The Leading Mathematics-10 Circle 293 Geometry CONFIDENCE LEVEL TEST - V Unit V : Geometry Class: 10, The Leading Maths Time: 45 mins. FM: 23 Attempt all questions. 1. In the given figure, ∆BCE and parallelogram ABCD are on the same base BC and between the same parallels AE and BC. (a) State the relation between area of ∆BCE and parallelogram ABCD.[1] (b) If the area of ∆BCE = 14 cm2 , find the area of ABCD. [1] (c) Construct a parallelogram with a side 4.5 cm equal to the area of the parallelogram ABCD with AB = 6 cm, AD = 4 cm and ABC = 45°. [3] 2. In the adjoining figure, D is the mid-point of the side AB of ∆ABC and ADEC is a parallelogram. (a) State the relation between area of ∆ACD and parallelogram ADEC. [1] (b) If the area of ∆ABC = 16 cm2 , find the area of parallelogram ADEC. [1] (c) Prove that: Ar(∆ABC) = Ar( ADEC). [3] 3. In the given rough sketch, ΔPQR (having sides PQ = 4 cm, QR = 5 cm and PR = 3 cm) and ΔPQS (having side PS = 6 cm) are standing on the same base PQ and between the same parallel lines PQ and EF. (a) State the relation between area of ∆PQR and ∆PQS. [1] (b) Find the area of ∆PQS. [1] (c) Construct the ΔPQS equal to the area of ∆PQR with the help of a compass. [2] 4. In the given figure, O is the center of circle and ∠BAC, ∠BDC and ∠BOC are standing on the same arc BC. (a) Write the name of ∠BAC and ∠BOC. [1] (b) Write the relation of inscribed angles standing on the same arc. [1] (c) Prove that angle at the center of a circle is double the angle at its circumference standing on the same arc with the help of the given figure. [2] 5. In the given figure, O is the center of circle. (a) Write the relation of the opposite angles of a cyclic quadrilateral. [1] (b) Experimentally verify that the relation of opposite angles of a cyclic quadrilateral by drawing at least two circles with radii more than 3 cm. [3] (c) What should be the value of ∠BCD for being OBCD a parallelogram ? [1] Best of Luck A B C F D E A C B D E R S E F P 4 cm Q 3 cm 5 cm 6 cm A B C D O A C B D O
294 Allied The Leading Mathematics-10 Circle 295 Geometry Additional Practice – V Theorem related to Area of Parallelograms 1. In the figure, the area of the rectangle ABEF is 38 cm2 . What is the area of //gm ABCD? 2. (a) In the figure alongside, PQRS is a rhombus with diagonals 8 cm and 12 cm. Find the area of the parallelogram PQCD. (b) In the figure alongside, ABCD and ABEF are parallelograms. Find the area of the ABEF. 3. (a) Prove that the area of the parallelograms stand on the bases and between the same parallel straight lines are equal. (b) In the given figure, prove that area of AEFD = Area of GBCH. 4. (a) In the given figure, ABRP and ABQS are parallelograms. Prove that: Area of ∆ABP = 1 2 × Area of ABQS (b) In the given figure, AC is a diagonal of the square ABCD and CDPQ is a parallelogram. Prove that: 2 × Ar. (∆ABC) = Ar. ( CDPQ) Theorem on Area of Triangle and Parallelogram 5. (a) If the area of BDEF is 40 cm2 . What is the area of ∆ABC? (b) In the figure, the area of ∆ABC is 36 cm2 . What is the area of ∆DEF? (c) In the figure, the area of ∆AEB is 25 cm2 . What is the area of ∆BFC? 6. In the adjoining figure, the area of the trapezium ABCD is 144 cm2 and the area of ∆ADC is 60 cm2 . What is the area of ∆DEC? 7. (a) In the given figure, WM//XY, WX//ZY, MN⊥XY, WM =7 cm and MN = 5 cm, find the area of ∆WXZ. (b) In the given figure, PQRS and QSRT are the parallelograms. If PX⊥SQ, PX = 4 cm and SQ = 10 cm, find the area of ∆RQT. F D E A B C (a) S R D C P Q (b) B A C D E F P 12 cm 6 cm A B D F H C E G P Q A B D C (a) (b) D C E F A B F D C E B A (c) D C E F A B A B E C D Z M Y N W X S R P Q X T
294 Allied The Leading Mathematics-10 Circle 295 Geometry (c) In the given figure, BD//FC, BD = 8 cm, AP = 4 cm and CQ = 6 cm, find the area of ∆ADE. 8. (a) Prove that the area of a triangle is equal to half the product of the base and corresponding altitude. (b) Show that the area of a triangle is half of the area of square standing on the same base and between the same parallel. 9. (a) Verify experimentally that the area of parallelogram is twice of the area of triangle standing on the same base and between the same parallel lines. (b) Verify experimentally that the area of triangle is half of the area of parallelogram on the same base and between the same parallel lines. 10. (a) In the adjoining figure, D is the mid-point of the side AB of ∆ABC and ADEF is a rectangle. Prove that: Ar. (∆ABC) = Ar. (ADEF) [Hint: Use median theorem in ∆ABC and use the theorem 3.] (b) In the given figure, S is the mid-point of the side UP of ∆PUT and PQRS is a rhombus. Prove that: Area of ∆PUT = Area of rhombus PQRS (c) In ∆PQR, the medians RS and QT intersect at a point O. Prove that: (i) Ar. (∆QOS) = Ar. (∆ROT) (ii) Ar. (Quad.PSOT) = Ar. (∆QOR) (d) In the given figure, NE = BN. Prove that: Ar. (∆MEN) = 1 4 Ar. ( ABCD) 11. (a) ABCD is rectangle and O is any point in its interior. Prove that Ar(∆AOB) + Ar(∆COD) = Ar(∆BOC) + Ar(∆AOD). [Hints: Draw straight line parallel to AB or CD through O and use the theorem 3.] (b) E, F, G and H are the mid-points of the sides AB, BC, CD and DA of ABCD. Prove that the area of EFGH is half of the area of ABCD. [Hints: Join FH and use the theorem 3.] D C T Q P B E F C E A D B T R Q U S P P T Q R O S E A B D C M N
296 Allied The Leading Mathematics-10 Circle 297 Geometry Theorem on Area of Triangles 12. In the figure, ΔPQS is a right-angled triangle having PQ = 6 cm and QS = 8 cm. What is the area of ΔPQR? 13. In the figure, the area of the ∆CDE is 54 cm2 and ABCD is a parallelogram. Find the area of ∆ABD. 14. In the adjoining figure, the area of ∆BDE = 22 cm2 and the area of ∆ABD = 17 cm2 . Find the area of the quadrilateral ABCD. 15. (a) Prove that the area of triangle and right-angled triangle standing on the same base and between the same parallel lines are equal. (b) Prove that the area of triangles standing on the equal bases and between the parallel lines are equal. 16. (a) In the figure, PQRS is a parallelogram and RP//SO. Prove that: Ar. ( PQRS) = Ar. (∆QOT) [Hints: Join PT and use the median theorem.] (b) In the given figure, KLMN is a rectangle and LN//KJ. Prove that: Ar. (KLMN) = Ar. (∆IJM) [Hints: Join IN and use the median theorem.] 17. (a) In the figure, AB//GC//EF, AD//CB and AF//DE. Prove that the area of parallelograms ABCD and DEFH are equal. (b) In the figure, PS//QR, QP//TS and PT//SR. Prove that the area of ∆PQT and ∆RST are equal. Construction of Triangle equal to Quadrilateral in Area 18. From the following data, first construct a quadrilateral ABCD and then construct a triangle having area equal to the area of the quadrilateral: AB = 6.0 cm, BC = 5.0 cm, CD = 4 cm, ∠ABC = 120o and ∠BCD = 60o . 19. Given the following data, first construct the given special quadrilateral and then construct a triangle equal to the area of the quadrilateral: Rhombus PQRS having PR = 5 cm and QS = 7 cm. R S 6 cm Q8 cm A D P B C E C D A B E S T R O P Q I L K M N J A B C G H E F D P S Q R T M N
296 Allied The Leading Mathematics-10 Circle 297 Geometry 20. (a) Construct a triangle ∆PST equal to the area of the rhombus ABCD having the measures AB = 4.2 cm and AC = 5.3 cm. (b) Construct a trapezium ABCD in which AB = 6.2 cm, BC = 5 cm, AD = 4 cm and ∠ABC = 60o . Also, construct a triangle BCE equal to the area of the trapezium. Construction of Triangles of Quadrilaterals with Equal Area 21. Construct a right-angled triangle with the hypotenuse 5 cm and a leg 4 cm. Also, construct another triangle with an angle 60° equal to the area of the right-angled triangle. 22. Construct a parallelogram with a side 6 cm equal to the area of a square with a side 5 cm. 23. Construct a parallelogram ABCD with AB = 7 cm, AD = 5 cm and ∠ABC = 45o . Also, construct another parallelogram with a side 5.5 cm equal to the area of the parallelogram ABCD. Theorems on Arc of Circle 24. Establish the relation between the corresponding arcs of the equal chords of a circle. 25. In the circle with center O, ACB = 27 cm and ADB = 45 cm. Find the measure of ∠AOB and ∠ADB. 26. (a) In the given intersection of two equal circles with centers P and R. Prove that PQRS is a rhombus. (b) In the given intersecting two equal circles with centers A and B, CE = DE . Prove that AC//BD. 27. In the adjoining figure, O is the center of a circle. Show that: ∠XOZ = 2(∠XZY + ∠YXZ). Theorems on Angles on Same Arc of Circle 28. In the given circle having center O, if ∠PRQ = a, what is the value of ∠POQ? D O A B C Q S P R A C E B D Z O Y X O P R a Q (b) Why are the inscribed angles standing on the same arc of a circle are equal ? Justify theoretically. (a)
298 Allied The Leading Mathematics-10 Circle 299 Geometry 29. Calculate the measure of ∠A and ∠B? (i) (ii) 30. (a) In the figure, AC is a diameter of the circle ABCD, ABE and DCE are straight lines, ∠A = 26o and ∠E = 23o . Find ∠ABD. (b) In the given figure, O is the centre of the circle and ∠ABC = 105o . Find ∠COD. 31. Verify experimentally that the angle at the circumference of a circle is half of the angle at its centre standing on the same arc. 32. (a) In the given figure, the points A, B, C and D lie on a circle and AB//DC. Prove that (i) AD = BC and (ii) AC = BD. (b) In the given figure, the points A, B, C and D lie on the same circle. If DE = AE, prove that AC = BD. Theorem on Inscribed Angle on Semi-circle 33. (a) What is the value of 'a' in the adjoining figure ? (b) What is the length of diameter of the given semi-circle ? 34. In the given circle with center O, find the values of ∠BCO and ∠AOC. 35. In the adjoining figure, O is the center of the circle. Find the values of x and y. 32x-20° 22x+10° A C D B O 40x+62° 110x-194° A B O C O A 26° 23° D C B E O A D C B 105° A B D C B A E D C C O a A B P R Q O 4 cm 3 cm B A C O ? 65° ? S P Q R O y x 50°
298 Allied The Leading Mathematics-10 Circle 299 Geometry 36. In the given figure, O is the center of the circle. Find the value of ∠COD. (i) (ii) 37. In the given figure, O is the center of the circle. Prove that the measure of ∠ABC is right angle. 38. In the given figure, OP is diameter of small circle and radius of the largest circle. Prove that PQ = QR. 39. In the given circle with center O, AD⊥BE. Prove that BC = DE. 40. (a) In the given circle with center O, DE⊥AB and arc AC = arc BD. Prove that ∠ABC = ∠BDE. (b) Two circles intersect at Q and S. A line segment through Q meets the circles at P and R. Prove that PS = RS. Theorem on Cyclic Quadrilateral 41. (a) In the given cyclic quadrilateral PQRS, what is the value of x? (b) In the given cyclic quadrilateral KLMN, what is the value of a? A B O 60° E D C 66° A B P C D O A B C O R Q P O E F D C O B A D C A B O E S Q R P R S Q R O x 105° L P N M K 85° a
300 Allied The Leading Mathematics-10 Circle PB Geometry 42. From the given figures, find the value ‘x’ or ‘a’: (i) (ii) (iii) 43. (a) Show that the sum of opposite angles of a cyclic quadrilateral is 180°. (b) Verify experimental method that the sum of corresponding angles of cyclic quadrilateral is supplementary. (c) Verify that the exterior angle formed by producing a side of cyclic quadrilateral is equal to the opposite angle of its adjacent angle by experimental method. 44. (a) In the given figure, AB and EF are parallel to each other. Prove that CDEF is a cyclic quadrilateral. (b) In the figure, PQRS is a parallelogram. Prove that TUQR is a concyclic point. 45. (a) In the figure, two equal circles having centers P and Q intersect at C and A. Prove that ∆ABD is an equilateral triangle. (b) In the adjoining figure, QBR is a straight line. Show that MALP is a cyclic quadrilateral. 1. 38 cm2 2. (a) 48 cm2 (b) 72 cm2 5. (a) 40 cm2 (b) 9 cm2 (c) 25 cm2 6. 25 cm2 7. (a) 37.5 cm2 (b) 20 cm2 (c) 40 cm2 12. 24 cm2 13. 54 cm2 14. 39 cm2 25. 135°, 67.5° 28. 2a 29. 76° 30. (a) 49° (b) 30° 33. (a) 90° (b) 5 cm 34. 25°, 50° 35. 40°, 40° 36. (a) 60° (b) 57° 41. (a) 75° (b) 85° 42. (a) 100° (b) 47° (c) 70° Answers S O R P Q T 50° x B A E C D x 94° A P Q N C M B 70° x T S R P U Q A D E B C F A C D P Q B B P L A M Q R