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1 Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? In ABC and ABD, AC = AD (Given) CAB = DAB (AB bisects A) AB = AB (Common) ABC ABD (By SAS congruence rule) BC = BD (By CPCT) Therefore, BC and BD are of equal lengths. Question 2: ABCD is a quadrilateral in which AD = BC and DAB = CBA (See the given figure). Prove that (i) ABD BAC (ii) BD = AC (iii) ABD = BAC.
2 In ABD and BAC, AD = BC (Given) DAB = CBA (Given) AB = BA (Common) ABD BAC (By SAS congruence rule) BD = AC (By CPCT) And, ABD = BAC (By CPCT) Question 3: AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB. In BOC and AOD, BOC = AOD (Vertically opposite angles) CBO = DAO (Each 90º) BC = AD (Given) BOC AOD (AAS congruence rule)
3 BO = AO (By CPCT) CD bisects AB. Question 4: l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that ABC CDA. In ABC and CDA, BAC = DCA (Alternate interior angles, as p q) AC = CA (Common) BCA = DAC (Alternate interior angles, as l m) ABC CDA (By ASA congruence rule) Question 5: Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A (see the given figure). Show that: (i) APB AQB (ii) BP = BQ or B is equidistant from the arms of A.
4 In APB and AQB, APB = AQB (Each 90º) PAB = QAB (l is the angle bisector of A) AB = AB (Common) APB AQB (By AAS congruence rule) BP = BQ (By CPCT) Or, it can be said that B is equidistant from the arms of A. Question 6: In the given figure, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. It is given that BAD = EAC BAD + DAC = EAC + DAC BAC = DAE In BAC and DAE, AB = AD (Given) BAC = DAE (Proved above) AC = AE (Given) BAC DAE (By SAS congruence rule) BC = DE (By CPCT) Question 7: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB (See the given figure). Show that (i) DAP EBP
5 (ii) AD = BE It is given that EPA = DPB EPA + DPE = DPB + DPE DPA = EPB In DAP and EBP, DAP = EBP (Given) AP = BP (P is mid-point of AB) DPA = EPB (From above) DAP EBP (ASA congruence rule) AD = BE (By CPCT) Question 8: In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: (i) AMC BMD (ii) DBC is a right angle. (iii) DBC ACB (iv) CM = AB
6 (i) In AMC and BMD, AM = BM (M is the mid-point of AB) AMC = BMD (Vertically opposite angles) CM = DM (Given) AMC BMD (By SAS congruence rule) AC = BD (By CPCT) And, ACM = BDM (By CPCT) (ii) ACM = BDM However, ACM and BDM are alternate interior angles. Since alternate angles are equal, It can be said that DB AC DBC + ACB = 180º (Co-interior angles) DBC + 90º = 180º DBC = 90º (iii) In DBC and ACB, DB = AC (Already proved) DBC = ACB (Each 90 ) BC = CB (Common) DBC ACB (SAS congruence rule) (iv) DBC ACB AB = DC (By CPCT) AB = 2 CM CM = AB
7 Exercise 7.2 Question 1: In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersect each other at O. Join A to O. Show that: (i) OB = OC (ii) AO bisects A (i) It is given that in triangle ABC, AB = AC ACB = ABC (Angles opposite to equal sides of a triangle are equal) ACB = ABC OCB = OBC OB = OC (Sides opposite to equal angles of a triangle are also equal) (ii) In OAB and OAC, AO =AO (Common) AB = AC (Given) OB = OC (Proved above) Therefore, OAB OAC (By SSS congruence rule) BAO = CAO (CPCT) AO bisects A. Question 2: In ABC, AD is the perpendicular bisector of BC (see the given figure). Show that ABC is an isosceles triangle in which AB = AC.
8 In ADC and ADB, AD = AD (Common) ADC = ADB (Each 90º) CD = BD (AD is the perpendicular bisector of BC) ADC ADB (By SAS congruence rule) AB = AC (By CPCT) Therefore, ABC is an isosceles triangle in which AB = AC. Question 3: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal. In AEB and AFC, AEB and AFC (Each 90º) A = A (Common angle) AB = AC (Given) AEB AFC (By AAS congruence rule)
9 BE = CF (By CPCT) Question 4: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that (i) ABE ACF (ii) AB = AC, i.e., ABC is an isosceles triangle. (i) In ABE and ACF, ABE and ACF (Each 90º) A = A (Common angle) BE = CF (Given) ABE ACF (By AAS congruence rule) (ii) It has already been proved that ABE ACF AB = AC (By CPCT) Question 5: ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ABD = ACD.
10 Let us join AD. In ABD and ACD, AB = AC (Given) BD = CD (Given) AD = AD (Common side) ABD ACD (By SSS congruence rule) ABD = ACD (By CPCT) Question 6: ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that BCD is a right angle.
11 In ABC, AB = AC (Given) ACB = ABC (Angles opposite to equal sides of a triangle are also equal) In ACD, AC = AD ADC = ACD (Angles opposite to equal sides of a triangle are also equal) In BCD, ABC + BCD + ADC = 180º (Angle sum property of a triangle) ACB + ACB + ACD + ACD = 180º 2( ACB + ACD) = 180º 2( BCD) = 180º BCD = 90º Question 7: ABC is a right angled triangle in which A = 90º and AB = AC. Find B and C. It is given that AB = AC C = B (Angles opposite to equal sides are also equal) In ABC,
12 A + B + C = 180º (Angle sum property of a triangle) 90º + B + C = 180º 90º + B + B = 180º 2 B = 90º B = 45º B = C = 45º Question 8: Show that the angles of an equilateral triangle are 60º each. Let us consider that ABC is an equilateral triangle. Therefore, AB = BC = AC AB = AC C = B (Angles opposite to equal sides of a triangle are equal) Also, AC = BC B = A (Angles opposite to equal sides of a triangle are equal) Therefore, we obtain A = B = C In ABC, A + B + C = 180 A + A + A = A = 180 A = 60 A = B = C = 60
13 Hence, in an equilateral triangle, all interior angles are of measure 60º. Exercise 7.3 Question 1: ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that (i) ABD ACD (ii) ABP ACP (iii) AP bisects A as well as D. (iv) AP is the perpendicular bisector of BC. (i) In ABD and ACD, AB = AC (Given) BD = CD (Given) AD = AD (Common) ABD ACD (By SSS congruence rule) BAD = CAD (By CPCT) BAP = CAP. (1) (ii) In ABP and ACP,
14 AB = AC (Given) BAP = CAP [From equation (1)] AP = AP (Common) ABP ACP (By SAS congruence rule) BP = CP (By CPCT) (2) (iii) From equation (1), BAP = CAP Hence, AP bisects A. In BDP and CDP, BD = CD (Given) DP = DP (Common) BP = CP [From equation (2)] BDP CDP (By S.S.S. Congruence rule) BDP = CDP (By CPCT) (3) Hence, AP bisects D. (iv) BDP CDP BPD = CPD (By CPCT). (4) BPD + CPD = 180 (Linear pair angles) BPD + BPD = BPD = 180 [From equation (4)] BPD = 90 (5) From equations (2) and (5), it can be said that AP is the perpendicular bisector of BC. Question 2: AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects A.
15 (i) In BAD and CAD, ADB = ADC (Each 90º as AD is an altitude) AB = AC (Given) AD = AD (Common) BAD CAD (By RHS Congruence rule) BD = CD (By CPCT) Hence, AD bisects BC. (ii) Also, by CPCT, BAD = CAD Hence, AD bisects A. Question 3: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR (see the given figure). Show that: (i) ABM PQN (ii) ABC PQR (i) In ABC, AM is the median to BC.
16 BM = BC In PQR, PN is the median to QR. QN = QR However, BC = QR BC = QR BM = QN (1) In ABM and PQN, AB = PQ (Given) BM = QN [From equation (1)] AM = PN (Given) ABM PQN (SSS congruence rule) ABM = PQN (By CPCT) ABC = PQR (2) (ii) In ABC and PQR, AB = PQ (Given) ABC = PQR [From equation (2)] BC = QR (Given) ABC PQR (By SAS congruence rule) Question 4: BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
17 In BEC and CFB, BEC = CFB (Each 90 ) BC = CB (Common) BE = CF (Given) BEC CFB (By RHS congruency) BCE = CBF (By CPCT) AB = AC (Sides opposite to equal angles of a triangle are equal) Hence, ABC is isosceles. Question 5: ABC is an isosceles triangle with AB = AC. Drawn AP BC to show that B = C. In APB and APC, APB = APC (Each 90º) AB =AC (Given) AP = AP (Common) APB APC (Using RHS congruence rule) B = C (By using CPCT) Exercise 7.4
18 Question 1: Show that in a right angled triangle, the hypotenuse is the longest side. Let us consider a right-angled triangle ABC, right-angled at B. In ABC, A + B + C = 180 (Angle sum property of a triangle) A + 90º + C = 180 A + C = 90 Hence, the other two angles have to be acute (i.e., less than 90º). B is the largest angle in ABC. B > A and B > C AC > BC and AC > AB [In any triangle, the side opposite to the larger (greater) angle is longer.] Therefore, AC is the largest side in ABC. However, AC is the hypotenuse of ABC. Therefore, hypotenuse is the longest side in a right-angled triangle. Question 2: In the given figure sides AB and AC of ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.
19 In the given figure, ABC + PBC = 180 (Linear pair) ABC = 180 PBC... (1) Also, ACB + QCB = 180 ACB = 180 QCB (2) As PBC < QCB, 180º PBC > 180º QCB ABC > ACB [From equations (1) and (2)] AC > AB (Side opposite to the larger angle is larger.) Question 3: In the given figure, B < A and C < D. Show that AD < BC. In AOB, B < A
20 AO < BO (Side opposite to smaller angle is smaller)... (1) In COD, C < D OD < OC (Side opposite to smaller angle is smaller)... (2) On adding equations (1) and (2), we obtain AO + OD < BO + OC AD < BC Question 4: AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that A > C and B > D. Let us join AC.
21 In ABC, AB < BC (AB is the smallest side of quadrilateral ABCD) 2 < 1 (Angle opposite to the smaller side is smaller)... (1) In ADC, AD < CD (CD is the largest side of quadrilateral ABCD) 4 < 3 (Angle opposite to the smaller side is smaller)... (2) On adding equations (1) and (2), we obtain < C < A A > C Let us join BD. In ABD, AB < AD (AB is the smallest side of quadrilateral ABCD) 8 < 5 (Angle opposite to the smaller side is smaller)... (3) In BDC, BC < CD (CD is the largest side of quadrilateral ABCD) 7 < 6 (Angle opposite to the smaller side is smaller)... (4) On adding equations (3) and (4), we obtain < D < B B > D
22 Question 5: In the given figure, PR > PQ and PS bisects QPR. Prove that PSR > PSQ. As PR > PQ, PQR > PRQ (Angle opposite to larger side is larger)... (1) PS is the bisector of QPR. QPS = RPS... (2) PSR is the exterior angle of PQS. PSR = PQR + QPS... (3) PSQ is the exterior angle of PRS. PSQ = PRQ + RPS... (4) Adding equations (1) and (2), we obtain PQR + QPS > PRQ + RPS PSR > PSQ [Using the values of equations (3) and (4)] Question 6: Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
23 Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle. In PNM, N = 90º P + N + M = 180º (Angle sum property of a triangle) P + M = 90º Clearly, M is an acute angle. M < N PN < PM (Side opposite to the smaller angle is smaller) Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
24 Exercise 7.5 Question 1: ABC is a triangle. Locate a point in the interior of ABC which is equidistant from all the vertices of ABC. Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together. In ABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of ABC. Question 2: In a triangle locate a point in its interior which is equidistant from all the sides of the triangle. The point which is equidistant from all the sides of a triangle is called the incentre of the triangle. Incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle.
25 Here, in ABC, we can find the incentre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of ABC. Question 3: In a huge park people are concentrated at three points (see the given figure) A: where there are different slides and swings for children, B: near which a man-made lake is situated, C: which is near to a large parking and exit. Where should an ice-cream parlour be set up so that maximum number of persons can approach it? (Hint: The parlor should be equidistant from A, B and C) Maximum number of persons can approach the ice-cream parlour if it is equidistant from A, B and C. Now, A, B and C form a triangle. In a triangle, the circumcentre is the only point that is equidistant from its vertices. So, the ice-cream parlour should be set up at the circumcentre O of ABC.
26 In this situation, maximum number of persons can approach it. We can find circumcentre O of this triangle by drawing perpendicular bisectors of the sides of this triangle. Question 4: Complete the hexagonal and star shaped rangolies (see the given figures) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles? It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles in it.
27 Area of OAB Area of hexagonal-shaped rangoli Star-shaped rangoli has 12 equilateral triangles of side 5 cm in it. Area of star-shaped rangoli =
28 Therefore, star-shaped rangoli has more equilateral triangles in it.
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