NCERT Class 10 Chapter 6: TrianglesExercise 6.11. Fill in the blanks using the correct word given in brackets :
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2. Give two different examples of pair of
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3. State whether the following quadrilaterals are similar or not: Solution In the quadrilateral ABCD, all the angles are 90°. However, in the quadrilateral PQRS, the angles are not equal to 90°. Hence, the given quadrilaterals are not similar. Exercise 6.21. In Fig. 6.17, (i) and (ii), DE  BC. Find EC in (i) and AD in (ii). Solution (i) It is given that in ∆ABC, DE ‖ BC. Therefore, by using the basic proportionality theorem, we have: (ii) It is given that in ∆ABC, DE ‖ BC. Therefore, by using the basic proportionality theorem, we have: 2. E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF  QR :
Solution I. EF ‖ QR, if the basic proportionality theorem is satisfied, i.e. Since, LHS ≠ RHS, the theorem isn't satisfied. Hence, EF is not parallel to QR. II. EF ‖ QR, if the basic proportionality theorem is satisfied, i.e. Since, LHS = RHS, the theorem is satisfied. Hence, EF is parallel to QR. III.EF ‖ QR, if the basic proportionality theorem is satisfied, i.e. Since, LHS = RHS, the theorem is satisfied. Hence, EF is parallel to QR. 3. In Fig. 6.18, if LM  CB and LN  CD, prove that AM/AB = AN/AD. Solution It is given that LM is parallel to CB in ∆ABC, therefore: 4. In Fig. 6.19, DE  AC and DF  AE. Prove that BF/FE = BE/EC. Solution 5. In Fig. 6.20, DE  OQ and DF  OR. Show that EF  QR. Solution 6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB  PQ and AC  PR. Show that BC  QR. Solution 7. Using Theorem 6.1, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX). For reference: Solution Let the triangle be ABC with DE as the line drawn through D as midpoint of side AB and E as a point on line AC. To Prove : AE = EC Proof : 8. Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX). For reference: Solution Let the triangle be ABC, where DE is the line formed by joining the D which is the midpoint of AB and E which is the midpoint of AC. 9. ABCD is a trapezium in which AB  DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO. Solution 10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium. Solution Therefore, by using the converse basic proportionality theorem we can conclude that EO ‖ AB. It is also known that EO ‖ CD, which implies that AB ‖ CD. Since two of the sides of the given quadrilateral are parallel, so it satisfies the condition for being a trapezium. Hence, ABCD is a trapezium. Exercise 6.31. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : Solution (i) In ∆ABC and ∆PQR: Hence, by AAA criterion ∆ABC ~ ∆PQR. (ii) In ∆ABC and ∆PQR: Hence, by SSS criterion ∆ABC ~ ∆PQR. (iii) In ∆MPL and ∆DEF: The ratios obtained are not same. Hence, ∆MPL is not similar to ∆DEF. (iv) In ∆LMN and ∆PQR: Hence, by SAS criterion ∆LMN ~ ∆PQR. (v) In ∆ABC and ∆DEF: Hence, ∆ABC is not similar to ∆DEF. (vi) By Angle Sum property of a triangle, Hence, by AAA criterion ∆DEF ~ ∆PQR. 2. In Fig. 6.35, ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB. Solution ∠DOC and ∠BOC form a linear pair. Therefore Hence ∠DOC = 55°, ∠DCO = 55° and ∠OAB = 55°. 3. Diagonals AC and BD of a trapezium ABCD with AB  DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD. Solution In ∆AOB and ∆COD, ∠BAO = ∠OCD (Alternate interior angles) ∠ABO = ∠ODC (Alternate interior angles) ∠AOB = ∠COD (Vertically opposite angles) Hence, by the AAA similarity criterion, ∆AOB ~ ∆COD. Since, ∆AOB ~ ∆COD Therefore, the corresponding sides will be proportional Hence, proved that OB/OD = OA/OC. 4. In Fig. 6.36, QR/QS = QT/PR and ∠ 1 = ∠ 2. Show that ∆ PQS ~ ∆ TQR. Solution In ∆PQR, ∠PQR = ∠PRQ Therefore, it is an isosceles triangle which implies that PQ = PR. Using QR/QS = QT/PQ and ∠Q = ∠Q in ∆PQS and ∆TQR, we can conclude that the triangles are similar (by SAS similarity criterion). Hence, proved that ∆PQS ~ ∆TQR. 5. S and T are points on sides PR and QR of ∆ PQR such that ∠ P = ∠ RTS. Show that ∆ RPQ ~ ∆ RTS. Solution In ∆ RPQ and ∆ RTS, ∠RTS = ∠QPS (Given) ∠R = ∠R (Common Angle) Hence, by the AA similarity criterion ∆ RPQ ~ ∆ RTS. 6. In Fig. 6.37, if ∆ ABE ≅ ∆ ACD, show that ∆ ADE ~ ∆ ABC. Solution In ∆ ABE and ∆ ACD AB = AC (Using c.p.c.t) ... Equation (I) AD = AE (Using c.p.c.t) ... Equation (II) 7. In Fig. 6.38, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that:
Solution I. In ΔAEP and ΔCDP, ∠AEP = ∠CDP (Right Angles) ∠APE = ∠CPD (Vertically Opposite Angles) Therefore, by AA similarity criterion ΔAEP ~ ΔCDP. II. In ΔABD and ΔCBE, ∠ADB = ∠CEB (Right Angles) ∠ABD = ∠CBE& (Common Angle) Therefore, by AA similarity criterion ΔABD ~ ΔCBE. III. In ΔAEP and ΔABD, ∠AEP = ∠ADB (Right Angles) ∠PAE = ∠DAB (Common Angle) Therefore, by AA similarity criterion ΔAEP ~ ΔABD. IV. In ΔPDC and ΔBEC, ∠PDC = ∠BEC (Right Angles) ∠PCD = ∠BCE (Common Angle) Therefore, by AA similarity criterion ΔPDC ~ ΔBEC. 8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ ABE ~ ∆ CFB. Solution In ∆ABE and ∆CFB, ∠BAE = ∠BCF (Opposite angles of a parallelogram) ∠ABE = ∠CFB (Alternate interior angles) Therefore, by AA similarity criterion ΔABE ~ ΔCFB. 9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
Solution I. In ΔABC and ΔAMP, ∠ABC = ∠AMP (Right Angles) ∠BAC = ∠MAP (Common Angle) Therefore, by AA similarity criterion ΔABC ~ ΔAMP. II. Since, ∆ABC is similar to ∆AMP, their corresponding sides will have an equal ratio. Hence, 10. CD and GH are respectively the bisectors of ∠ACB and ∠ EGF such that D and H lie on sides AB and FE of ∆ ABC and ∆ EFG respectively. If ∆ABC ~ ∆ FEG, show that:
Solution It is given that ∆ABC ~ ∆ FEG, therefore ∠A = ∠F ∠B = ∠E ∠ACB = ∠FGE Divide both sides by 2 ∠ACB/2 = ∠FGE/2 ∠ACD = ∠FGH (Bisected Angles) I. Now, in ∆ACD and ∆FGH: ∠A = ∠F (As shown above) ∠ACD = ∠FGH (As shown above) Therefore, by the AA similarity criterion ∆ACD ~ ∆FGH. Which implies that II. Similarly, ∠DCB = ∠HGE In ∆DCB and ∆HGE: ∠B = ∠E (As Shown above) ∠DCB = ∠HGE (As shown above) Therefore, by the AA similarity criterion ∆DCB ~ ∆HGE. III. In ∆DCA and ∆HGF ∠A = ∠F (As shown above) ∠DCA = ∠HGF (As shown above) Therefore, by the AA similarity criterion ∆DCA ~ ∆HGF. 11. In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ ABD ~ ∆ ECF. Solution In an isosceles triangles, angles opposite to equal sides are also equal. Therefore, ∠ABD = ∠ACD = ∠ECF In ∆ ABD and ∆ ECF: ∠ADB = ∠EFC (Right Angles) ∠ABD = ∠ECF (As Shown above) Therefore, by the AA similarity criterion ∆ ABD ~ ∆ ECF. 12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see Fig. 6.41). Show that ∆ ABC ~ ∆ PQR. Solution It is given that Therefore, by the SSS similarity criterion, ∆ABD ~ ∆PQM. This implies that ∠ABD = ∠PQM (Corresponding angles of similar triangles) ∠ABC = ∠PQR In ∆ABC and ∆PQR AB/PQ = BC/QR (Given) ∠ABC = ∠PQR (As shown above) Hence, by the SAS similarity criterion ∆ABC ~ ∆PQR. 13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. Show that CA^{2} = CB.CD. Solution In ∆ADC and ∆BAC: ∠ADC = ∠BAC (Given) ∠ACD = ∠BCA (Common Angles) Therefore, by the AA similarity criterion ∆ADC ~ ∆BAC. This implies that CA/CB = CD/CA (Corresponding sides of similar triangles) CA^{2} = CB.CD 14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆ PQR. Solution First, we will produce AD and PM lines to exterior points E and N respectively such that AD = DE and PM = MN, then join E and N with C and R respectively. In ∆ABD and ∆CDE: . ∠ADB = ∠ CDE (Vertically opposite angles) AD = DE (By construction) BD = DC (D is the midpoint of BC) Therefore, by the SAS congruency criterion ∆ABD ≅ ∆CDE. This implies that AB = CE (By c.p.c.t.) In ∆PQM and ∆MNR: . ∠PMQ = ∠NMR (Vertically opposite angles) PM = MN (By construction) QR = MR (M is the midpoint of QR) Therefore, by the SAS congruency criterion ∆PMQ ≅ ∆NMR. This implies that PQ = RN (By c.p.c.t.) It is given that Therefore, by the SSS similarity criterion ∆ACE ~ ∆PRN. This implies that ∠CAE = ∠RPN (Corresponding angles of similar triangles) ∠CAD = ∠RPM Similarly, ∠BAD = ∠QPM. Add both the above equations ∠CAD + ∠BAD = ∠RPM + ∠QPM ∠BAC = ∠QPR In ∆ABC and ∆PQR: ∠BAC = ∠QPR (As shown above) AB/PQ = AC/PR (Given) Hence, by the SAS similarity criterion ∆ABC ~ ∆PQR. 15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. Solution Let AB be the pole whose shadow is BC and let the tower be PQ with QR as the shadow. If we connect the top of pole and tower to the end of their respective shadows, we can obtain triangles ABC and PQR. Since, the shadows BC and QR are being cast at the same time, the sun's elevation will be the same. Therefore, ∠ACB = ∠PRQ In ∆ABC and ∆PQR: ∠ACB = ∠PRQ (As shown above) ∠ABC = ∠PQR (Right Angles) Therefore, by the AA similarity criterion ∆ABC ~ ∆PQR. This implies that AB/PQ = BC/QR (Corresponding sides of similar triangles) AB = 6 m, BC = 4 m QR = 28 m 6/PQ = 4/28 6/PQ = 1/7 PQ = 42 m Hence, the height of the tower is 42 m. 16. If AD and PM are medians of triangles ABC and PQR, respectively where ∆ ABC ~ ∆ PQR, prove that AB/PQ = AD/PM. Solution It is given that ∆ ABC ~ ∆ PQR, which implies that ∠ABC = ∠PQR (Corresponding Angles of similar triangles) ∠ABD = ∠PQM In ∆ABD and ∆PQM: AB/PQ = BD/QM (As shown above) ∠ABD = ∠PQM (As shown above) Therefore, by the SAS similarity criterion ∆ABD ~ ∆PQM. This implies that AB/PQ = AD/PM (Corresponding sides of similar triangles) Hence, proved that AB/PQ = AD/PM. Exercise 6.41. Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64 cm^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC. Solution Since ∆ ABC ~ ∆ DEF, therefore Hence, BC = 11.2 cm^{2}. 2. Diagonals of a trapezium ABCD with AB  DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD. SOLUTION In ∆AOB and ∆COD: ∠AOB = ∠COD (Vertically opposite angles) ∠ABO = ∠CDO (Alternate interior angles) Therefore, by the AA similarity criterion ∆AOB ~ ∆COD. This implies that Hence, the ratio of the areas of triangles AOB and COD is 4 : 1. 3. In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar (ABC)/ar (DBC) = AO/DO. Solution We need to construct two lines AP and DM where P and M are points on BC such that AP and DM are perpendicular to BC. Area of triangle = ½ × Base × Altitude Area of ∆ABC = ½ × BC × AP Area of ∆DBC = ½ × BC × DM Ratio of Areas of ∆ABC and ∆DBC = (½ × BC × AP)/ (½ × BC × DM) ar (∆ABC)/ ar (∆DBC) = AP/DM ... Equation (I) In ∆AOP and ∆DOM: ∠AOP = ∠DOM (Vertically opposite angles) ∠APO = ∠DMO& (Right Angles) Therefore, by the AA similarity criterion ∆AOP ~ ∆DOM. This implies that AO/DO = PA/MD (Corresponding sides of similar triangles) AO/DO = AP/DM ... Equation (II) From Equation (I) and Equation (II), ar (∆ABC)/ ar (∆DBC) = AO/DO 4. If the areas of two similar triangles are equal, prove that they are congruent. Solution Let there be two similar triangles ABC and PQR where Area of ABC = Area of PQR. AB/PQ = 1 AB = PQ Similarly, BC = QR AC = PR Therefore, by the SSS congruency criteria ∆ABC ≅ ∆PQR. Hence, proved that two similar triangles with equal areas are congruent. 5. D, E and F are respectively the midpoints of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC. Solution Since D and F are midpoints of sides AB and AC, therefore, by the midpoint theorem DF ‖ BC. In ∆ABC and ∆ADF: ∠BAC = ∠DAF (Common Angles) ∠ABC = ∠ADF (Corresponding Angles) Therefore, by the AA similarity criterion ∆ABC ~ ∆ADF. This implies that From the above equations, we can conclude that Ar (∆ADF) = ¼(Ar (∆ABC)) Ar (∆BED) = ¼(Ar (∆ABC)) Ar (∆CEF) = ¼(Ar (∆ABC)) Now, Hence, the ratio of the areas of triangles DEF and ABC is 1 : 4. 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. Solution Let there be two similar triangles ABC and PQR, where AD and PM are the respective medians of the triangles. To Prove : ar (∆ABC)/ ar (∆PQR) = AD^{2}/ PM^{2} Proof : It is given that ∆ABC ~ ∆PQR, which implies that ∠ABC = ∠PQR (Corresponding angles of similar triangles) ∠ABD = ∠PQM Now, in ∆ABD and ∆PQM: ∠ABD = ∠PQM (As shown above) AB/PQ = BD/QM (As shown above) Therefore, by the SAS similarity criterion ∆ABD ~ ∆PQM. This implies that Hence, proved that ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. 7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. Solution Let there be a square ABCD with equilateral triangles ABD and ACE described on its one side and one diagonal. To Prove : ar (∆ABD) = ½ (ar (∆ACE)) Proof : By using the Pythagoras theorem in ABC, we get: AC^{2} = AB^{2} + BC^{2} AB = BC as ABCD is a square AC^{2} = 2AB^{2} ... Equation (I) It is known that ABD and ACE are both equilateral triangles. Therefore, by the AAA similarity criterion ∆ABD ~ ∆ACE. This implies that Hence, proved that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. Tick the correct answer and justify:8. ABC and BDE are two equilateral triangles such that D is the midpoint of BC. Ratio of the areas of triangles ABC and BDE is (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 Solution D is the midpoint of BC, therefore BD = DC = ½(BC) ... Equation (I) AB = BC (Sides of an equilateral triangle) ... Equation (II) Since, ∆ABC and ∆BDE are both equilateral triangles, therefore by the AAA similarity criterion ∆ABC ~ ∆BDE. This implies that Hence, the correct answer is (C) 4 : 1. 9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81 Solution Let there be two similar triangles ABC and DEF. Then, according to question Since, it is given that the triangles ABC and DEF are similar. Therefore, Hence, the correct answer is (D) 16 : 81. Exercise 6.51. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
Solution I. Squares of the smaller two sides = 7^{2} + 24^{2} = 49 + 576 = 625 cm^{2} Square of the longest side = 25^{2} = 625 cm^{2} Therefore, the triangle satisfies the Pythagoras theorem and is a right angled triangle. The length of hypotenuse of this triangle is 25 cm. II. Squares of the smaller two sides = 3^{2} + 6^{2} = 9 + 36 = 45 cm^{2} Square of the longest side = 8^{2} = 64 cm^{2} Therefore, the triangle does not satisfy the Pythagoras theorem and is not a right angled triangle. III. Squares of the smaller two sides = 50^{2} + 80^{2} = 2500 + 6400 = 8900 cm^{2} Square of the longest side = 100^{2} = 10000 cm^{2} Therefore, the triangle does not satisfy the Pythagoras theorem and is not a right angled triangle. IV. Squares of the smaller two sides = 5^{2} + 12^{2} = 25 + 144 = 169 cm^{2} Square of the longest side = 13^{2} = 169 cm^{2} Therefore, the triangle satisfies the Pythagoras theorem and is a right angled triangle. The length of hypotenuse of this triangle is 13 cm. 2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM^{2} = QM . MR. Solution By using Pythagoras Theorem in ∆PQM, we get PQ^{2} = PM^{2} + QM^{2} PM^{2} = PQ^{2}  QM^{2 ... Equation (I)} By using Pythagoras Theorem in ∆PRM, we get PR^{2} = PM^{2} + MR^{2} PM^{2} = PR^{2}  MR^{2} ... Equation (II) Add Equation (I) and Equation (II) 2PM^{2} = PQ^{2}  QM^{2} + PR^{2}  MR^{2} 2PM^{2} = (PQ^{2} + PR^{2})  (QM^{2} + MR^{2}) 2PM^{2} = (QR^{2})  (QM^{2} + MR^{2}) [PQ^{2} + PR^{2} = QR^{2} by Pythagoras Theorem] 2PM^{2} = QR^{2}  QM^{2}  MR^{2} 2PM^{2} = (QM + MR)^{2}  QM^{2}  MR^{2} [QR = QM + MR] 2PM^{2} = QM^{2} + MR^{2} + 2QM.MR  QM^{2}  MR^{2} 2PM^{2} = 2QM.MR PM^{2} = QM.MR Hence, proved that PM^{2} 3. In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that
Solution I. In ∆ABD and ∆CBA: ∠BAD = ∠BCA (Right Angles) ∠ABD = ∠ABC (Common Angles) Therefore, by the AA similarity criterion ∆ABD ~ ∆CBA. This implies that AB/CB = BD/AB (Corresponding Sides of similar triangles) AB^{2} = BC.BD II. Since, we have proven that ∆ABD ~ ∆CBA, therefore ∠CAB = ∠BDA (Corresponding angles of similar triangles) ∠CAB = ∠CDA In ∆CBA and ∆CAD: ∠ACB = ∠ACD (Right Angles) ∠CAB = ∠CDA (As shown above) Therefore, by the AA similarity criterion ∆CBA ~ ∆CAD. This implies that AC/BC = DC/AC (Corresponding Sides of similar triangles) AC^{2} = BC.DC III. In ∆DCA and ∆DAB: ∠DCA = ∠DAB (Right Angles) ∠CDA = ∠ADB (Common Angles) Therefore, by the AA similarity criterion ∆DCA ~ ∆DAB. This implies that CD/AD = AD/BD (Corresponding Sides of similar triangles) AD^{2} = BD.CD 4. ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2} . Solution It is given that ∆ABC is an isosceles triangle. This implies that the sides opposite to the equal angles are equal. Since, ∠ACB is given to be the right angle. Therefore, ∠ABC and ∠BAC are the equal angles with sides AC = BC. By using Pythagoras Theorem in ABC, we get AB^{2} = AC^{2} + BC^{2} AB^{2} = AC^{2} + AC^{2} AB^{2} = 2AC^{2} Hence, proved that AB^{2} = 2AC^{2}. 5. ABC is an isosceles triangle with AC = BC. If AB^{2} = 2 AC^{2} , prove that ABC is a right triangle. Solution It is given that AB^{2} = 2AC^{2} AB^{2} = AC^{2} + AC^{2} AB^{2} = AC^{2} + BC^{2} (Since AC = BC is given) The above equation satisfies the Pythagoras Theorem. Hence, ∆ABC is a right angles triangle. 6. ABC is an equilateral triangle of side 2a. Find each of its altitudes. Solution Let us draw the altitude AD in the triangle ABC. In ∆ADB and ∆ADC: ∠ADB = ∠ADC (Right Angles) AD = AD (Common Side) AB = AC (Sides of equilateral triangle ABC) Therefore, by the RHS congruency criteria ∆ADB ≅ ∆ADC. This implies that BD = CD (By c.p.c.t.) Therefore, BC = 2BD BD = BC/2 BD = a By applying Pythagoras theorem in ∆ADB, we get AB^{2} = BD^{2} + AD^{2} AD^{2} = AB^{2}  BD^{2} AD^{2} = 4a^{2}  a^{2} AD^{2} = 3a^{2} AD = ?3a Hence, the length of each of the altitude of the triangle ABC is ?3a. 7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Solution Let there be a rhombus ABCD whose diagonals bisect each other at O. To Prove : AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2} Proof : We know that the diagonals of a rhombus bisect each other at right angles. Therefore, AOB, AOD, BOC, COD are all right angles triangles. This also implies that AO = CO and BO = DO as O is the point of bisection. By applying Pythagoras Theorem to ∆AOB, we get AB^{2} = AO^{2} + BO^{2} Similarly, BC^{2} = BO^{2} + CO^{2} CD^{2} = CO^{2} + DO^{2} AD^{2} = AO^{2} + DO^{2} Adding all of the above obtained equations, we get AB^{2} + BC^{2} + CD^{2} + AD^{2} = AO^{2} + BO^{2} + BO^{2} + CO^{2} + CO^{2} + DO^{2} + AO^{2} + DO^{2} AB^{2} + BC^{2} + CD^{2} + AD^{2} = 2(AO^{2} + BO^{2} + CO^{2} + DO^{2}) AB^{2} + BC^{2} + CD^{2} + AD^{2} = 2(AO^{2} + CO^{2} + BO^{2} + DO^{2}) Since, AO = CO and BO = DO. Therefore, AB^{2} + BC^{2} + CD^{2} + AD^{2} = 2(2AO^{2} + 2BO^{2}) AB^{2} + BC^{2} + CD^{2} + AD^{2} = 4AO^{2} + 4BO^{2} AB^{2} + BC^{2} + CD^{2} + AD^{2} = (2AO)^{2} + (2BO)^{2} O is the midpoint of AC and BD, so 2AO = AC and 2BO = BD. Therefore, >AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2} Hence, proved that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. 8. In Fig. 6.54, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥AC and OF ⊥AB. Show that
Solution Let us join OB, OA and OC. I. By applying Pythagoras Theorem in ∆AOF, we get OA^{2} = AF^{2} + OF^{2} By applying Pythagoras Theorem in ∆BOD, we get OB^{2} = OD^{2} + BD^{2} By applying Pythagoras Theorem in ∆COE, we get OC^{2} = OE^{2} + EC^{2} Adding the above obtained equations, we get OA^{2} + OB^{2} + OC^{2} = AF^{2} + OF^{2} + OD^{2} + BD^{2} + OE^{2} + EC^{2} OA^{2} + OB^{2} + OC^{2}  OD^{2}  OE^{2}  OF^{2 }= AF^{2} + BD^{2 }+ EC^{2} Hence, proved. II. OA^{2} + OB^{2} + OC^{2}  OD^{2}  OE^{2}  OF^{2 }= AF^{2} + BD^{2 }+ EC^{2} (OA^{2}  OE^{2}) + (OC^{2}  OD^{2}) + (OB^{2}  OF^{2}) = AF^{2} + BD^{2 }+ EC^{2} By applying Pythagoras Theorem in ∆AOE, we get OA^{2} = OE^{2} + AE^{2} OA^{2}  OE^{2} = AE^{2} Similarly, OC^{2}  OD^{2} = CD^{2} OB^{2}  OF^{2} = BF^{2} Hence, (OA^{2}  OE^{2}) + (OC^{2}  OD^{2}) + (OB^{2}  OF^{2}) = AF^{2} + BD^{2 }+ EC^{2} AE^{2 }+ CD^{2 }+ BF^{2 }= AF^{2} + BD^{2 }+ EC^{2} 9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall. Solution Let AB be the ladder which reaches the window at A with the base of the wall at C. The wall will be perpendicular to the ground, therefore ABC will form a right triangle which is right angled at C. AB = 10 m , AC = 8m By applying Pythagoras Theorem in ABC, we get AB^{2} = AC^{2} + BC^{2} 10^{2} = 8^{2} + BC^{2} 100 = 64 + BC^{2} 36 = BC^{2} BC = 6 m Hence, the distance between foot of the ladder and the base of the wall is 6 m. 10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut? Solution Let the wire be AB and the pole be AC. BC is the distance between the stake and base of the pole. ∆ACB is a right triangle which is right angled at C. By Applying Pythagoras Theorem in ABC, we get AB^{2} = AC^{2} + BC^{2} 24^{2} = 18^{2} + BC^{2} 576  324 = BC^{2} 252 = BC^{2} BC = 6?7 m Hence, the stake needs to be driven 6?7 m away from the base of the pole to make the wire taut. 11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after 1*1/2 hours? Solution Let A be the current position of the plane flying to north and B be the current position of the plane flying to the west. Let O be the starting point for the two planes. AOB will form a right triangle, right angled at O. Speed of plane going north = 1000 km/h = 1200 × 3/2 = 1800 km By Applying the Pythagoras theorem in AOB, we get AB^{2} = AO^{2} + BO^{2} AB^{2} = 1500^{2} + 1800^{2} AB^{2} = 2250000 + 3240000 AB^{2} = 5490000 AB = 300?61 km 12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops. SOLUTION Let AB be the taller pole and CD be the smaller pole. The distance between their tops is AC and the distance between their feet is BD. We will draw an imaginary line from C to O which is a point on the pole AB such that OC ‖ BD. The poles stand perpendicular to the ground, so ∠BDC is a right angle. ∠AOC = ∠BDC (Corresponding Angles) Therefore, ∆AOC is a right triangle. OCBD forms a rectangle as OC ‖ BD and AB ‖ CD. This implies that OC = BD = 12 m OB = CD = 6 m Now, AB = AO + OB AO = AB  OB = 11  6 = 5 m By Applying Pythagoras Theorem in AOC, we get AC^{2} = AO^{2} + OC^{2} AC^{2} = 5^{2} + 12^{2} AC^{2} = 25 + 144 AC^{2} = 169 AC = 13 m Hence, the distance between the top of the two poles is 13 m. 13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE^{2} + BD^{2} = AB^{2} + DE^{2}. By Applying Pythagoras Theorem in ABC, we get AB^{2} = AC^{2} + BC^{2} ... Equation (I) By Applying Pythagoras Theorem in AEC, we get AE^{2} = AC^{2} + EC^{2} By Applying Pythagoras Theorem in BDC, we get BD^{2} = BC^{2} + DC^{2} By Applying Pythagoras Theorem in DEC, we get DE^{2} = DC^{2} + EC^{2} ... Equation (II) Therefore, AE^{2} + BD^{2} = AC^{2} + EC^{2} + BC^{2} + DC^{2} AE^{2} + BD^{2} = (AC^{2} + BC^{2}) + (EC^{2} + DC^{2}) From Equation (I) and Equation (II), we get AE^{2} + BD^{2} = AB^{2} + DE^{2} Hence, proved. 14. The perpendicular from A on side BC of a ∆ ABC intersects BC at D such that DB = 3 CD (see Fig. 6.55). Prove that 2 AB^{2} = 2 AC^{2} + BC^{2}. Solution BC = BD + CD BC = 3CD + CD BC/4 = CD By applying Pythagoras Theorem in ADB, we get AB^{2} = AD^{2} + BD^{2} ... Equation (I) By applying Pythagoras Theorem in ADC, we get AC^{2} = AD^{2} + CD^{2} ... Equation (II) Subtract Equation (II) from Equation (I): AB^{2}  AC^{2} = BD^{2}  CD^{2} AB^{2}  AC^{2} = (3CD)^{2}  CD^{2} AB^{2}  AC^{2} = 9CD^{2}  CD^{2} AB^{2}  AC^{2} = 8CD^{2} AB^{2}  AC^{2} = 8(BC/4)^{2} AB^{2}  AC^{2} = BC^{2}/2 2(AB^{2}  AC^{2}) = BC^{2} 2AB^{2}  2AC^{2} = BC^{2} 2AB^{2} = 2AC^{2} + BC^{2} Hence, proved. 15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9 AD^{2} = 7 AB^{2}. Solution Let us construct altitude AE in ∆ABC. Since AE is the altitude of an equilateral triangle, therefore it will also act as a median. BE = CE = BC/2 By applying Pythagoras Theorem in ABE, we get AB^{2} = BE^{2} + AE^{2} AE^{2} = AB^{2}  BE^{2} AE^{2} = AB^{2}  BC^{2}/4 By applying Pythagoras Theorem in ADE, we get AD^{2} = AE^{2} + DE^{2} AD^{2} = AE^{2} + (BE  BD)^{2} AD^{2} = AB^{2}  BC^{2}/4 + (BC/2  BC/3)^{2} AD^{2} = AB^{2}  BC^{2}/4 + (BC/6)^{2} AD^{2} = AB^{2}  BC^{2}/4 + BC^{2}/36 AD^{2} = AB^{2}  8BC^{2}/36 AD^{2} = AB^{2}  2BC^{2}/9 AD^{2} = (9AB^{2}  2BC^{2})/9 9AD^{2} = 9AB^{2}  2AB^{2} (BC = AB as ABC is an equilateral triangle) 9AD^{2} = 7AB^{2} Hence, proved. 16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. Solution Let there be an equilateral triangle ABC with altitude AD and length of each side a. To prove : 3a^{2} = 4AD^{2} Proof : We know that the altitude of an equilateral triangle also acts as a median. Therefore, BD = CD = BC/2 By applying Pythagoras Theorem in ABD, we get AB^{2} = AD^{2} + BD^{2} AB^{2} = AD^{2} + BC^{2}/4 a^{2} = AD^{2} + a^{2}/4 3a^{2}/4 = AD^{2} 3a^{2} = 4AD^{2} Hence, proved that three times the square of one side is equal to four times the square of one of its altitudes. 17. Tick the correct answer and justify : In ∆ABC, AB = 6?3 cm, AC = 12 cm and BC = 6 cm. The angle B is : (A) 120° (B) 60° (C) 90° (D) 45° Solution AB^{2} = (6?3)^{2} = 108 AC^{2} = 12^{2} = 144 BC^{2} = 6^{2} = 36 It can be observed that AB^{2} + BC^{2} = 144 = AC^{2}. This implies that the triangle satisfies the Pythagoras Theorem with AC as the hypotenuse. Therefore, angle B is 90°. Hence, correct options is (C) 90°. Exercise 6.6 (Optional)1. In Fig. 6.56, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/SR = PQ/PR. Solution Let us produce the line QP to an external point T such that TR is parallel to PS. ∠QPS = ∠SPR as PS is the angle bisector for ∠QPR. ∠QPS = ∠QTR (Corresponding Angles) ∠SPR = ∠PRT (Alternate Interior Angles) Therefore, ∠PRT = ∠QTR. We know that if two angles of a triangle are equal then the triangle is isosceles and sides opposite to the equal angles are also equal. Therefore, PT = PR By using the basic proportionality theorem in QRT, we get QS/SR = PQ/PT QS/SR = PQ/PR Hence, proved. 2. In Fig. 6.57, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove that :
Solution It can be concluded from the figure that DBMN is a rectangle as its opposite sides are parallel and the angles are 90° each. This implies that DN = BM and DM = BN. (i). ∠CDM = ∠CDB  ∠BDM ∠CDM = 90°  ∠BDM ∠CDM + ∠BDM = 90° Also, in ∆CDM: ∠CDM + ∠CMD + ∠MCD = 180° ∠CDM + 90° + ∠MCD = 180° ∠CDM + ∠MCD = 90° Therefore, we can say that ∠CDM + ∠MCD = ∠CDM + ∠BDM (Both sides are equal to 90°) ∠MCD = ∠BDM In ∆CDM and ∆MDB: ∠MCD = ∠BDM (As shown above) ∠CMD = ∠BMD (Right Angles) Therefore, by the AA similarity criterion ∆CDM ~ ∆MDB. This implies that DM/BM = MC/DM DM^{2} = MC.BM We have already shown that BM = DN. DM^{2} = DN.MC (ii). ∠ADB = ∠ADN + ∠BDN ∠ADN + ∠BDN = 90° Also, in ∆ADN: ∠AND + ∠ADN + ∠DAN = 180° 90° + ∠ADN + ∠DAN = 180° ∠ADN + ∠DAN = 90° Therefore, we can say that ∠ADN + ∠DAN = ∠ADN + ∠BDN (Both sides equal to 90°) ∠DAN = ∠BDN In ∆ADN and ∆NDB: ∠DAN = ∠BDN (As shown above) ∠CMD = ∠BMD (Right Angles) Therefore, by the AA similarity criterion ∆ADN ~ ∆NDB. This implies that DN/BN = NA/DN DN^{2} = AN.BN We have already shown that BN = DM. DN^{2} = DM.AN Hence, proved. 3. In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC^{2} = AB^{2} + BC^{2} + 2 BC . BD. Solution By applying Pythagoras Theorem in ABD, we get AB^{2} = AD^{2} + BD^{2} By applying Pythagoras Theorem in ACD, we get AC^{2} = AD^{2} + CD^{2} AC^{2} = AD^{2} + (BD + BC)^{2} AC^{2} = AD^{2} + BD^{2} + BC^{2} + 2 BC.BD We already know that AD^{2} + BD^{2} = AB^{2}, therefore AC^{2} = AB^{2} + BC^{2} + 2 BC.BD Hence, proved. 4. In Fig. 6.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC^{2} = AB^{2} + BC^{2}  2 BC . BD. Solution By applying Pythagoras Theorem in ABD, we get AB^{2} = AD^{2} + BD^{2} By applying Pythagoras Theorem in ACD, we get AC^{2} = AD^{2} + CD^{2} AC^{2} = AD^{2} + (BC  BD)^{2} AC^{2} = AD^{2} + BD^{2} + BC^{2}  2 BC.BD We already know that AD^{2} + BD^{2} = AB^{2}, therefore AC^{2} = AB^{2} + BC^{2}  2 BC.BD Hence, proved. 5. In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
Solution It is given that AD is the median of triangle ABC, therefore BD = CD = BC/2 ... Equation (I) By applying Pythagoras Theorem in AMD, we get AD^{2} = AM^{2} + DM^{2} (i). By applying Pythagoras Theorem in ACM, we get AC^{2} = AM^{2} + CM^{2} AC^{2} = AM^{2} + (CD + DM)^{2} AC^{2} = AM^{2} + DM^{2} + CD^{2} + 2 CD.DM By Using Equation (I) AC^{2} = AM^{2} +DM^{2} + (BC/2)^{2} + 2(BC/2)DM We already know that AM^{2} + DM^{2} = AD^{2}, therefore AC^{2} = AD^{2} + (BC/2)^{2} + 2(BC/2)DM AC^{2} = AD^{2} + BC .DM + (BC/2)^{2} (ii). By applying Pythagoras Theorem in ABM, we get AB^{2} = AM^{2} + BM^{2} We know that AD^{2} = AM^{2} + DM^{2} AM^{2} = AD^{2}  DM^{2} Now, AB^{2} = AD^{2}  DM^{2} + BM^{2} AB^{2} = AD^{2}  DM^{2} + (BD  DM)^{2} AB^{2} = AD^{2}  DM^{2} + BD^{2} + DM^{2}  2BD.DM By Using Equation (I) AB^{2} = AD^{2} + (BC/2)^{2}  2 (BC/2).DM AB^{2} = AD^{2} + (BC/2)^{2}  BC .DM (iii). By applying Pythagoras Theorem in ABM, we get AB^{2} = AM^{2} + BM^{2} By applying Pythagoras Theorem in AMC, we get AC^{2} = AM^{2} + MC^{2} Adding the above two equations, we get AB^{2} + AC^{2} = 2AM^{2} + BM^{2} + MC^{2} AB^{2} + AC^{2} = 2AM^{2} + (BD  DM)^{2} + (DM + CD)^{2} AB^{2} + AC^{2} = 2AM^{2} + BD^{2} + DM^{2}  2 BD.DM + DM^{2} + CD^{2} + 2 DM.CD AB^{2} + AC^{2} = 2AM^{2} + 2DM^{2} + BD^{2} + CD^{2} + 2DM(CD  BD) AB^{2} + AC^{2} = 2(AM^{2} + DM^{2}) + BD^{2} + CD^{2} + 2DM(CD  BD) We know that AM^{2} + DM^{2} = AD^{2}, therefore AB^{2} + AC^{2} = 2AD^{2} + BD^{2} + CD^{2} + 2DM(CD  BD) By Using Equation (I) AB^{2} + AC^{2} = 2AD^{2} + (BC/2)^{2} + (BC/2)^{2} + 2DM(BC/2  BC/2) AB^{2} + AC^{2} = 2AD^{2} + BC^{2}/4 + BC^{2}/4 AB^{2} + AC^{2} = 2AD^{2} + BC^{2}/2 Hence, proved. 6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. Solution Let ABCD be a parallelogram with AB ‖ CD and altitude AE. We will produce BA to a point F such that DF ? BF. To Prove : AC^{2} + BD^{2} = AB^{2} + BC^{2} + CD^{2} + AD^{2} Proof : By applying Pythagoras Theorem in ADF, we get AD^{2} = AF^{2} + DF^{2} By applying Pythagoras Theorem in BDF, we get BD^{2} = BF^{2} + DF^{2} BD^{2} = (AB + AF)^{2} + DF^{2} BD^{2} = AB^{2} + AF^{2} + 2 AB.AF + DF^{2} We know that AF^{2} + DF^{2} = AD^{2}, therefore BD^{2} = AB^{2} + AD^{2} + 2 AB.AF; ... Equation (I) By applying Pythagoras Theorem in ADE, we get AD^{2} = AE^{2} + DE^{2} By applying Pythagoras Theorem in ACE, we get AC^{2} = AE^{2} + CE^{2} AC^{2} = AE^{2} + (CD  DE)^{2} AC^{2} = AE^{2} + CD^{2}  2 CD.DE + DE^{2} We know that AE^{2} + DE^{2} = AD^{2}, therefore AC^{2} = AD^{2} + CD^{2}  2 CD.DE ... Equation (II) Add Equation (I) and Equation (II) AC^{2} + BD^{2} = AD^{2} + CD^{2}  2 CD.DE + AB^{2} + AD^{2} + 2 AB.AF Since ABCD is a parallelogram, therefore AB = CD and BC = AD (Opposite sides of a parallelogram) Also, it can be observed that AEDF forms a rectangle, which implies that AF = DE; (Opposite sides of a rectangle) Therefore, AC^{2} + BD^{2} = AD^{2} + CD^{2} + AB^{2} + BC^{2 }+ 2 AB.AF  2 AB.AF AC^{2} + BD^{2} = AD^{2} + CD^{2} + AB^{2} + BC^{2} Hence, proved that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. 7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that :
Solution AP = BP = CP = DP as they are radii of the given circle. Therefore, AP/DP = 1 = CP/BP. (i). In ∆APC and ∆DPB: AP/DP = CP/BP (As shown above) ∠APC = ∠DPB (Vertically opposite angles) Therefore, by the SAS similarity criterion ∆APC ~ ∆DPB. (ii). We have already shown that AP/DP = CP/BP. This implies that AP.BP = CP.DP AP.PB = PC.DP Hence, proved. 8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
Solution (i). Since ABCD is a quadrilateral which is contained in a circle, therefore, ABCD forms a cyclic quadrilateral. This implies that, ∠PAC = ∠BDC (Exterior angle of a cyclic quadrilateral is equal to the opposite interior angle) In ∆PAC and ∆PDB: ∠PAC = ∠BDC (As shown above) ∠APC = ∠BPD (Common Angle) Therefore, by the AA similarity criterion ∆PAC ~ ∆PDB. (ii). We have already shown that ∆PAC ~ ∆PDB. This implies that AP/DP = PC/PB (Corresponding sides of similar triangles) AP.PB = PC.DP Hence, proved. 9. In Fig. 6.63, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC. Solution Let us produce BA to an external point P such that AP = AC. BC/ CD = AB/ AC (Given) BC/CD = AB/AP (AP = AC) Therefore, by using the converse of basic proportionality theorem, we can say that AD ‖ CP. ∠BAD = ∠APC (Corresponding Angles) ∠CAD = ∠ACP (Alternate Interior Angles) ∆ACP is an isosceles triangle as AP = AC. This implies that ∠APC = ∠ACP (Angles opposite to equal sides in an isosceles triangle) Therefore, ∠BAD = ∠CAD AD divides the ∠BAC in two equal parts. Hence, AD is the bisector of ∠BAC. 10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds? Solution The situation can be represented in form of a triangle ABC, where A is the tip of fishing rod, C is the end of the string and BC is the surface of water. By applying Pythagoras theorem in ABC, we get AC^{2} = AB^{2} + BC^{2} AC^{2} = (1.8)^{2} + (2.4)^{2} AC^{2} = 3.24 + 5.76 AC^{2} = 9 AC = 3 Therefore, Nazima has 3 m of her fishing rod string out. It is given that she is pulling the string by 5 cm per second. Therefore, Length of string pulled by Nazima in 1 second = 5 cm Length of string pulled by Nazima in 12 seconds = 12 × 5 = 60 cm = 0.6 m Let the new position of the fly be D which is a point on the water surface BC. Length of string remaining out after 12 seconds = AD = 3  0.6 m = 2.4 m By applying Pythagoras Theorem in ABD, we get AD^{2} = AB^{2} + BD^{2} (2.4)^{2} = (1.8)^{2} + BD^{2} BD^{2} = (2.4)^{2}  (1.8)^{2} BD^{2} = 5.76  3.24 BD^{2} = 2.52 BD = 1.59 m (Approx.) Horizontal distance of the fly = 1.2 + BD = 1.2 + 1.59 = 2.79 m Hence, the fly is at a horizontal distance of 2.79 m (approx. value) from Nazima.
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