### Similar Triangles Class 10th Mathematics AP Board Solution

##### Question 1.Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.Answer:Need to prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals ABCD is a rhombus in which diagonals AC and BD intersect at point O.We need to prove AB2 + BC2 + CD2 + DA2 = AC2 + DB2⇒ In Δ AOB; AB2 = AO2 + BO2⇒ In Δ BOC; BC2 = CO2 + BO2⇒ In Δ COD; CD2 = DO2 + CO2⇒ In Δ AOD; AD2 = DO2 + AO2⇒ Adding the above 4 equations we get⇒ AB2 + BC2 + CD2 + DA2 = AO2 + BO2 + CO2 + BO2 + DO2 + CO2 + DO2 + AO2⇒ = 2(AO2 + BO2 + CO2 + DO2)Since, AO2 = CO2 and BO2 = DO2= 2(2 AO2 + 2 BO2)= 4(AO2 + BO2) ……eq(1)Now, let us take the sum of squares of diagonals⇒ AC2 + DB2 = (AO + CO)2 + (DO+ BO)2= (2AO)2 + (2DO)2= 4 AO2 + 4 BO2 ……eq(2)From eq(1) and eq(2) we get⇒ AB2 + BC2 + CD2 + DA2 = AC2 + DB2Hence, provedQuestion 2.ABC is a right triangle right angled at B. Let D and E be any points on AB and BC respectively.Prove that AE2 + CD2 = AC2 + DE2 . Answer:Given, ABC as a right angled triangleNeed to prove that AE2 + CD2 = AC2 + DE2⇒ In right angled triangle ABC and DBC, we have⇒ AE2 = AB2 + BE2 ………..eq(1)⇒ DC2 = DB2 + BC2 ……….eq(2)⇒ Adding equation 1 and 2 we have⇒ AE2 + DC2 = AB2 + BE2 + DB2 + BC2= (AB2 + BC2) + (BE2 + DB2)⇒ Since AB2 + BC2 = AC2 in right angled triangle ABC∴ AC2 + DE2Hence provedQuestion 3.Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.Answer: Given, an equilateral triangle ABC, in which AD perpendicular BCNeed to prove that 3 AB2 = 4AD2⇒ Let AB = BC = CA = a⇒ In Δ ABD and Δ ACD⇒ AB = AC, AD = AD and ∠ ADB = ∠ ADC∴ Δ ABD ≅ Δ ACD∴ BD = CD = ⇒ Now, in Δ ABD, ∠ D = 90°∴ AB2 = BD2 + AD2⇒ AB2 = + AD2= + AD23AB2 = 4 AD2Question 4.PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM.MR.Answer: ⇒ Let ∠MPR = x⇒ In Δ MPR, ∠MRP = 180-90-x⇒ ∠MRP = 90-xSimilarly in Δ MPQ,∠MPQ = 90-∠MPR = 90-x⇒ ∠MQP = 180-90-(90-x)⇒ ∠MQP = xIn Δ QMP and Δ PMR⇒ ∠MPQ = ∠MRP⇒ ∠PMQ = ∠RMP⇒ ∠MQP = ∠MPR⇒ Δ QMP ∼ Δ PMR⇒ = ⇒ PM2 = MR × QMHence provedQuestion 5.ABD is a triangle right angled at A and AC ⊥ BDShow thati. AB2 = BC . BDii. AC2 = BC . DCiii. AD2 = BD . CD Answer:Given, ABCD is a right angled triangle and AC is perpendicular to BD(i) consider two triangles ACB and DAB⇒ We have ∠ ABC = ∠ DBC⇒ ∠ ACB = ∠ DAB⇒ ∠ CAB = ∠ ADB∴ they are similar and corresponding sides must be proportionali.e, ∠ ADC = ∠ ADB⇒ ∴ AB2 = BC × CD(ii) ∠ BDA = ∠ BDC = 90°⇒ ∠ 3 = ∠ 2 = 90° ∠ 1⇒ ∠ 2 + ∠ 4 = 90° ∠ 2⇒ From AAA criterion of similarity we have in two triangles if the angles are equal, then sides opposite to the equal angles are in the same ratio (or proportional) and hence the triangles are similarTheir corresponding sides must be proportional⇒ ⇒ ⇒ = BC × DC(iii) In two triangles ADB and ABC we have∠ ADC = ∠ ADB⇒ ∠ DCA = ∠ DAB⇒ ∠ DAC = ∠ DBA⇒ ∠ DCA = ∠ DAB⇒ Triangle ADB and ABC are similar and so their corresponding sides must be proportion.⇒ = = ⇒ = ⇒ AD2 = DB × DCHence provedQuestion 6.ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.Answer:Since the triangle is right angled at C∴ the side AB is hypotenuse.⇒ Let the base of the triangle be AC and the altitude be BC.⇒ Applying the Pythagorean theorem⇒ HYP2 = Base2 + Alt2⇒ AB2 = AC2 + BC2Since the triangle is isosceles triangle two of the sides shall be equal∴ AC = BCThus AB2 = AC2 + BC2AB2 = 2AC2Hence, provedQuestion 7.‘O’ is any point in the interior of a triangle ABC.OD ⊥ BC, OE ⊥ AC and OF ⊥ AB, show thati. OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2ii. AF2 + BD2 + CE2 = AE2 + CD2 + BF2. Answer:Given, Δ ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB,Need to prove OA2 + OB2 + OC2-OD2-OE2-OF2 = AF2 + BD2 + CE2⇒ Join point O to A,B and C(i) ∠AFO = 90°AO2 = AF2 + OF2⇒ AF2 = AO2 - OF2 …….eq(1)Similarly BD2 = BO2-OD2 …..eq(2)⇒ CE2 = CO2-OE2 …..eq(3)Adding eq(1), (2) and (3) we get⇒ AF2 + BD2 + CE2 = OA2 + OB2 + OC2-OD2-OE2-OF2(ii) AF2 + BD2 + CE2 = (AO2-OE2) + ( BO2-OF2) + ( CO2-OD2)= AE2 + CD2 + BF2Hence, provedQuestion 8.A wire attached to vertically pole of height 18m is24m long and has a stake attached to other end. How far from the base of the pole should the stake be driven so that the wire will be taut?Answer:Given, height of a pole is 18 and wire attached is 24mNeed to find the distance from the base to keep wire taut⇒ Let AB be a wire and pole be BC⇒ to keep the wire taut let it be fixed at A⇒ AB2 = AC2 + BC2⇒ 242 = AC2 + 182⇒ AC2 = 242 + 182⇒ AC2 = 576-324⇒ = 252⇒ AC = √ 252= √(36 × 7)= 6√ 7Hence, the stake may be placed at a distance of 6√ 7m the base of poleQuestion 9.Two poles of heights 6m and 11m stand on a plane ground. If the distance between the feet of the pole is 12m find the distance between their tops.Answer: Given, BC = 6m, AD = 11m, BC = EDAnd AE = AD-ED = 11-6 = 5mBE = CD = 12mNeed to find AB⇒ Now, In Δ ABE, ∠E = 90°⇒ AB2 = AE2 + BE2⇒ AB2 = 52 + 122 = 169⇒ AB2 = 169⇒ AB = 13mThe distance between their tops is 13mQuestion 10.In an equilateral triangle ABC, D is on a side BC such that Prove that 9AD2 = 7AB2.Answer: Given, ABC is a equilateral triangle where AB = BC = AC and BD = BCDraw AE perpendicular BC⇒ Δ ABE ≅ Δ ACE∴ BE = EC = ⇒ Now in Δ ABE, AB2 = BE2 + AE2⇒ also AD2 = AE2 + DE2∴ AB2 –AD2 = BE2 – DE2= BE2 – (BE-BD)2= ( )2 – = ( )2 – AB2-AD2 = 2 Or 7 AB2 = 9 AD2Hence, provedQuestion 11.In the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it.Prove that 8AE2 = 3AC2 + 5AD2. Answer:Given, ABC triangleNeed to prove 8AE2 = 3AC2 + 5AD2⇒ In Δ ABD, ∠B = 90°∴ AC2 = AB2 + BC2 ….eq(1)⇒ Similarly, AE2 = AB2 + BE2 …….eq(2)⇒ And AD2 = AB2 + BD2 …….eq(3)⇒ Form eq(1)⇒ 3AC2 = 3AB2 + 3 BC2 …eq(4)⇒ From eq(2)⇒ 5AD2 = 5AB2 + 5BD2 ……eq(5)Adding equation (4) and (5)3AC2 + 5AD2 = 8AB2 + 3 BC2 + 5BD2= 8AB2 + 3 + 5 = 8(AB2 + BE2)= 8AE2Hence, provedQuestion 12.ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ΔABE and ΔACD. Answer:Given, ABC is an isosceles triangle in which ∠B = 90°Need to find the ratio between the areas of Δ ABE and Δ ACD⇒ AB = BC⇒ By Pythagoras theorem, we have AC2 = AB2 + BC2⇒ since AB = BC⇒ AC2 = AB2 + AB2⇒ AC2 = 2 AB2 …..eq(1)⇒ it is also given that Δ ABE ∼ Δ ACD(ratio of areas of similar triangles is equal to ratio of squares of their corresponding sides)⇒ = ⇒ = from 1⇒ = ∴ ar(Δ ABC):ar(Δ ACD) = 1:2Hence the ratio is 1:2

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