NCERT10th MathematicsExercise 6.5Topic: 6 Triangles
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EXERCISE 6.5
Q1. The sides of the triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm(ii) 3 cm, 8 cm, 6 cm(iii) 50 cm, 80 cm, 100 cm(iv) 13 cm, 12 cm, 5 cm
Solution:
1) In a right-angled triangle ABC,
a) ∠ ABC = 900
b) (AC) is hypotenuse.
2) By the theorem of Pythagoras, we have,
(AB)2 + (BC)2 = (AC)2 ------- equation 1.
Note: Here AC is the largest side, so always take the greatest side as the value of (AC)
(i) 7 cm, 24 cm, 25 cm
1) By the theorem of Pythagoras, we have,
(AB)2 + (BC)2 = (AC)2 ------- equation 1.
2) Let (AB) = 7, (BC) = 24, and (AC) = 25.
3) Let us check these values in Equation 1.
(AB)2 + (BC)2 = (AC)2
LHS = (AB)2 + (BC)2
LHS = (7)2 + (24)2
LHS = 49 + 576
LHS = 625 ------- equation 2
RHS = (AC)2RHS = (25)2
RHS = 625 ------- equation 3
4) From equations 2 and 3, we have,
LHS = RHS
5) So, here, 7 cm, 24 cm, and 25 cm are the sides of the right-angled triangle.
6) Here the length of the hypotenuse is 25 cm.
(ii) 3 cm, 8 cm, 6 cm
1) By the theorem of Pythagoras, we have,
(AB)2 + (BC)2 = (AC)2 ------- equation 1.
2) Let (AB) = 3, (BC) = 6, and (AC) = 8.
3) Let us check these values in Equation 1.
(AB)2 + (BC)2 = (AC)2
LHS = (AB)2 + (BC)2
LHS = (3)2 + (6)2
LHS = 9 + 36
LHS = 45 ------- equation 2
RHS = (AC)2RHS = (8)2
RHS = 64 ------- equation 3
4) From equations 2 and 3, we have,
LHS ≠ RHS
5) So, 3 cm, 8 cm, and 6 cm are not the sides of the right-angled triangle.
(iii) 50 cm, 80 cm, 100 cm
1) By the theorem of Pythagoras, we have,
(AB)2 + (BC)2 = (AC)2 ------- equation 1.
2) Let (AB) = 50, (BC) = 80, and (AC) = 100.
3) Let us check these values in Equation 1.
(AB)2 + (BC)2 = (AC)2
LHS = (AB)2 + (BC)2
LHS = (50)2 + (80)2
LHS = 2500 + 6400
LHS = 8900 ------- equation 2
RHS = (AC)2RHS = (100)2
RHS = 10000 ------- equation 3
4) From equations 2 and 3, we have,
LHS ≠ RHS
5) So, 50 cm, 80 cm, and 100 cm are not the sides of the right-angled triangle.
(iv) 13 cm, 12 cm, 5 cm
1) By the theorem of Pythagoras, we have,
(AB)2 + (BC)2 = (AC)2 ------- equation 1.
2) Let (AB) = 5, (BC) = 12, and (AC) = 13.
3) Let us check these values in equation 1.
(AB)2 + (BC)2 = (AC)2
LHS = (AB)2 + (BC)2
LHS = (5)2 + (12)2
LHS = 25 + 144
LHS = 169 ------- equation 2
RHS = (AC)2RHS = (13)2
RHS = 169 ------- equation 3
4) From equations 2 and 3, we have,
LHS = RHS
5) So, here, 5 cm, 12 cm, and 13 cm are the sides of the right-angled triangle.
6) Here the length of the hypotenuse is 13 cm.
Q2. PQR is a triangle right angled at P and M is a point on QR such that
PM ⟂ QR. Show that PM2 = QM x MR.
Solution:
1) In ∆ PQR, and ∆ MQP,
Note: For ∆ PQR and ∆ MQP, points P ↔ M, Q ↔ Q, and R ↔ P. That is
i) point P is associated with poin M,
ii) point Q is associated with poin Q,
iii) point R is associated with point P,
a) ∠ QPR = ∠ QMP = 900 -------- (given)b) ∠ PQR = ∠ MQP (same angles)
2) By AA similarity test,
1) In ∆ PQR, and ∆ MQP,
Note: For ∆ PQR and ∆ MQP, points P ↔ M, Q ↔ Q, and R ↔ P. That is
i) point P is associated with poin M,
ii) point Q is associated with poin Q,iii) point R is associated with point P,
a) ∠ QPR = ∠ QMP = 900 -------- (given)b) ∠ PQR = ∠ MQP (same angles)
2) By AA similarity test,
a) ∆ PQR ~ ∆ MQP --------- equation 1
3) In ∆ PQR, and ∆ MPR,
Note: For ∆ PQR and ∆ MPR, points P ↔ M, Q ↔ P, and R ↔ R. That is
i) point P is associated with poin M,
ii) point Q is associated with poin P,
iii) point R is associated with point R,
a) ∠ QPR = ∠ PMR = 900 -------- (given)b) ∠ PRQ = ∠ MRP (same angles)
4) By AA similarity test,
3) In ∆ PQR, and ∆ MPR,
Note: For ∆ PQR and ∆ MPR, points P ↔ M, Q ↔ P, and R ↔ R. That is
i) point P is associated with poin M,
ii) point Q is associated with poin P,iii) point R is associated with point R,
a) ∠ QPR = ∠ PMR = 900 -------- (given)b) ∠ PRQ = ∠ MRP (same angles)
4) By AA similarity test,
a) ∆ PQR ~ ∆ MPR --------- equation 2
5) From equations 1 and 2 we have,
∆ MQP ~ ∆ MPR --------- equation 3
6) As corresponding sides of similar triangles are proportional, we have,
a) (PM)/(RM) = (QM)/(PM)
b) (PM) x (PM) = (QM) x (RM)
c) (PM)2 = (QM) x (RM) hence proved.
Q3. In the following fig., ABD is a triangle right-angled at A and AC ⟂ BD.
Show that
(i) AB2 = BC . BD(ii) AC2 = BC . DC(iii) AD2 = BD . CD
Solution:
(i) AB2 = BC . BD
1) In ∆ ADB and ∆ CAB,
a) ∠ DAB = ∠ ACB = 900 -------- (given)b) ∠ ABD = ∠ CBA (same angles)
2) By AA similarity test,
(i) AB2 = BC . BD
1) In ∆ ADB and ∆ CAB,
a) ∠ DAB = ∠ ACB = 900 -------- (given)b) ∠ ABD = ∠ CBA (same angles)
2) By AA similarity test,
a) ∆ ADB ~ ∆ CAB --------- equation 1
3) As corresponding sides of similar triangles are proportional, we have,
a) (AB)/(CB) = (DB)/(AB)
b) (AB) x (AB) = (CB) x (DB)
c) (AB)2 = (CB) x (DB) hence proved.
(ii) AC2 = BC . DC
1) As per the following theorem,
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other. So we have,
a) ∆ ABC ~ ∆ DAC --------- equation 1
2) As corresponding sides of similar triangles are proportional, we have,
a) (AC)/(DC) = (BC)/(AC)
b) (AC) x (AC) = (BC) x (DC)
c) (AC)2 = (BC) x (DC) hence proved.
(iii) AD2 = BD . CD
1) In ∆ ACD and ∆ BAD,
a) ∠ ACD = ∠ BAD = 900 -------- (given)b) ∠ ADC = ∠ BDA (same angles)
2) By AA similarity test,
a) ∠ ACD = ∠ BAD = 900 -------- (given)
b) ∠ ADC = ∠ BDA (same angles)
a) ∆ ACD ~ ∆ BAD --------- equation 1
3) As corresponding sides of similar triangles are proportional, we have,
a) (AD)/(BD) = (CD)/(AD)
b) (AD) x (AD) = (BD) x (CD)
c) (AD)2 = (BD) x (CD) hence proved.
Q4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution:
1) By the theorem of Pythagoras, we have,(AB)2 = (AC)2 + (BC)2 ------- equation 1.
2) It is given that ∆ ABC is an isosceles triangle, so,
1) By the theorem of Pythagoras, we have,
(AB)2 = (AC)2 + (BC)2 ------- equation 1.
2) It is given that ∆ ABC is an isosceles triangle, so,
(BC) = (AC) ------- equation 2.
3) From equations 1 and 2, we have,
(AB)2 = (AC)2 + (BC)2
(AB)2 = (AC)2 + (AC)2
(AB)2 = 2(AC)2, hence proved.
Q5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2,
prove that ABC is a right triangle.
Solution:
1) It is given that ∆ ABC is an isosceles triangle, so,
1) It is given that ∆ ABC is an isosceles triangle, so,
(BC) = (AC) ------- equation 1.
2) It is given that,
(AB)2 = 2(AC)2
(AB)2 = (AC)2 + (AC)2
(AB)2 = (AC)2 + (BC)2 ------- equation 2.
3) From equation 2, by Pythagoras theorem, we can say that AB is the hypotenuse,
and AC and BC are other two sides of a right-angled triangle, hence proved.
Q6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
1) In ∆ ABC an equilateral triangle of side 2a.2) AD ⟂ BC and AB = 2a, BD = a.3) In ∆ ABD, by Pythagoras theorem, we have,
(AB)2 = (AD)2 + (BD)2
(2a)2 = (AD)2 + (a)2
(AD)2 = (2a)2 - (a)2
(AD)2 = 4a2 - a2
(AD)2 = 3a2
AD = √3 a
4) So, the altitude of a triangle ABC is √3 a.
Q7. 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:
1) 囗 ABCD is a rhombus, so AC ⟂ BD.2) In ∆ AOB, by Pythagoras theorem, we have,
(AB)2 = (AO)2 + (BO)2------- equation 1.
3) In ∆ BOC, by Pythagoras' theorem, we have,
(BC)2 = (BO)2 + (CO)2------- equation 2.
4) In ∆ COD, by Pythagoras theorem, we have,
(CD)2 = (CO)2 + (DO)2------- equation 3.
5) In ∆ DOA, by Pythagoras' theorem, we have,
(DA)2 = (DO)2 + (AO)2------- equation 4.
6) Adding equations 1, 2, 3, and 4, we get,
(AB)2 + (BC)2 + (CD)2 + (DA)2 = 2[(AO)2 + (BO)2 + (CO)2 + (DO)2]
(AB)2 + (BC)2 + (CD)2 + (DA)2 = 2[(AC/2)2 + (BD/2)2 + (AC/2)2 + (BD/2)2]
(AB)2 + (BC)2 + (CD)2 + (DA)2 = 2[(AC)2/4 + (BD)2/4 + (AC)2/4 + (BD)2/4]
(AB)2 + (BC)2 + (CD)2 + (DA)2 = 2/4[(AC)2 + (BD)2 + (AC)2 + (BD)2]
(AB)2 + (BC)2 + (CD)2 + (DA)2 = 2/4[2(AC)2 + 2(BD)2]
(AB)2 + (BC)2 + (CD)2 + (DA)2 = 4/4[(AC)2 + (BD)2]
(AB)2 + (BC)2 + (CD)2 + (DA)2 = [(AC)2 + (BD)2], hence proved.
Q8. In the following fig., O is a point in the interior of a triangle ABC,
OD ⟂ BC, OE ⟂ AC, and OF ⟂ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2,(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Solution:
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
1) In ∆ AOF, ∆ BOD, and ∆ COE by Pythagoras theorem, we have,
(OA)2 = (AF)2 + (OF)2------- equation 1.
(OB)2 = (BD)2 + (OD)2------- equation 2.
(OC)2 = (CE)2 + (OE)2------- equation 3.
2) Adding equations 1, 2, and 3, we get,
(OA)2 + (OB)2 + (OC)2 = (AF)2 + (OF)2 + (BD)2 + (OD)2 + (CE)2 + (OE)2
(OA)2 + (OB)2 + (OC)2 - (OE)2 - (OD)2 - (OF)2 = (AF)2 + (BD)2 + (CE)2
Hence proved.
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
1) In ∆ AOF, ∆ BOD, and ∆ COE by Pythagoras theorem, we have,
(OA)2 = (AF)2 + (OF)2------- equation 1.
(OB)2 = (BD)2 + (OD)2------- equation 2.
(OC)2 = (CE)2 + (OE)2------- equation 3.
2) Adding equations 1, 2, and 3, we get,
(AF)2 + (OF)2 + (BD)2 + (OD)2 + (CE)2 + (OE)2 = (OA)2 + (OB)2 + (OC)2
(AF)2 + (BD)2 + (CE)2 = (OA)2 + (OB)2 + (OC)2 - (OF)2 - (OD)2 - (OE)2
(AF)2 + (BD)2 + (CE)2 = [(OA)2 - (OE)2] + [(OC)2 - (OD)2] + [(OB)2 - (OF)2]
(AF)2 + (BD)2 + (CE)2 = (AE)2 + (CD)2 + (BF)2
Hence proved.
Q 9. A ladder 10 m long reaches a window 8 m above the ground.
Find the distance of the foot of the ladder from the base of the wall.
Solution:
1) By the theorem of Pythagoras, we have,(AB)2 + (BC)2 = (AC)2
1) By the theorem of Pythagoras, we have,
(AB)2 + (BC)2 = (AC)2
(8)2 + (BC)2 = (10)2
(BC)2 = (10)2 - (8)2
(BC)2 = 100 - 64
(BC)2 = 36(BC) = 6
2) The distance between the foot of the ladder from the base of the wall is 6 m.
Q 10. A guy wire attached to a vertical pole of height 18 m is 24 m long and
(BC)2 = (10)2 - (8)2
(BC)2 = 100 - 64
(BC)2 = 36(BC) = 6
2) The distance between the foot of the ladder from the base of the wall is 6 m.
Q 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:
1) By the theorem of Pythagoras, we have,(AB)2 + (BC)2 = (AC)2
1) By the theorem of Pythagoras, we have,
(AB)2 + (BC)2 = (AC)2
(18)2 + (BC)2 = (24)2
(BC)2 = (24)2 - (18)2
(BC)2 = (24 - 18) x (24 + 18)
(BC)2 = (6) x (42)
(BC)2 = (24)2 - (18)2
(BC)2 = (24 - 18) x (24 + 18)
(BC)2 = (6) x (42)
(BC)2 = (6) x (6 x 7)
(BC) = 6√7
2) The distance between the pole and the stake is 6√7 m.
Q 11. An aeroplane leaves an airport and flies due north at a speed of 1000 km
2) The distance between the pole and the stake is 6√7 m.
Q 11. An aeroplane leaves an airport and flies due north at a speed of 1000 km
per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1½ hours?
Solution:
1) It is given that,
a) The speed of the aeroplane travelling to north is 1000 km per hour.
b) The speed of the aeroplane travelling to west is 1200 km per hour.
2) We know that the Speed = (Distance)/(Time), so Distance = Speed x Time.
3) So after 1½ hours i.e. after 3/2 hours,
the distances traveled by the plans towards north
a) Distance of a plane = Speed x Timeb) Distance of a plane = 1000 x 3/2c) Distance of a plane = 500 x 3d) Distance of a plane = 1500 km. -------- equation 1
4) So after 1½ hours i.e. after 3/2 hours,
the distances traveled by the plans towards west
a) Distance of a plane = Speed x Timeb) Distance of a plane = 1200 x 3/2c) Distance of a plane = 600 x 3d) Distance of a plane = 1800 km. -------- equation 2
5) Here, (OP) = 1500 km, and (OQ) = 1800 km. and ∠ POQ = 900.
6) By the theorem of Pythagoras, we have,
(PQ)2 = (OP)2 + (OQ)2
(PQ)2 = (1500)2 + (1800)2
(PQ)2 = (3 x 5 x 100)2 + (3 x 6 x 100)2
(PQ)2 = [(3 x 100)2] x [(5)2 + (6)2]
(PQ)2 = [(3 x 100)2] x [25 + 36]
(PQ)2 = [(3 x 100)2] x [61]
(PQ) = (3 x 100) x √61
(PQ) = 300 x √61
7) The distance between two planes after 1½ hours will be 300√61 km.
Q 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:
1) The distance between the poles RS and PQ is 12 m.2) So, according to the figure, in ∆ SVQ,
(VQ) = 11 - 6 = 5 m and (SV) = 12 m.
3) By the theorem of Pythagoras, we have,
(QS)2 = (SV)2 + (VQ)2
(QS)2 = (12)2 + (5)2
(QS)2 = 144 + 25
(QS)2 = 169
(QS) = 13
4) So, the distance between the tips of the poles is 13 m.
Q 13. D and E are points on the sides CA and CB respectively of a triangle
ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Solution:
1) In ∆ ACE, by the theorem of Pythagoras, we have,(AE)2 = (EC)2 + (AC)2 ------- equation 1.
2) In ∆ BCD, by the theorem of Pythagoras, we have,(BD)2 = (BC)2 + (DC)2 ------- equation 2.
3) Adding equation 1 and equation 2, we get,(AE)2 + (BD)2 = [(EC)2 + (BC)2] + [(AC)2 + (DC)2] ------- equation 3.
4) Re-arranging the terms of equation 3, we get,
1) In ∆ ACE, by the theorem of Pythagoras, we have,
(AE)2 = (EC)2 + (AC)2 ------- equation 1.
2) In ∆ BCD, by the theorem of Pythagoras, we have,
(BD)2 = (BC)2 + (DC)2 ------- equation 2.
3) Adding equation 1 and equation 2, we get,
(AE)2 + (BD)2 = [(EC)2 + (BC)2] + [(AC)2 + (DC)2] ------- equation 3.
4) Re-arranging the terms of equation 3, we get,
(AE)2 + (BD)2 = [(EC)2 + (DC)2] + [(AC)2 + (BC)2] ------- equation 4.
5) In ∆ ABC, by the theorem of Pythagoras, we have,
(AB)2 = (BC)2 + (AC)2 ------- equation 5.
6) In ∆ DCE, by the theorem of Pythagoras, we have,
(DE)2 = (EC)2 + (DC)2 ------- equation 6.
7) From equations 4, 5, and 6, we have,
(AE)2 + (BD)2 = [(EC)2 + (DC)2] + [(AC)2 + (BC)2]
(AE)2 + (BD)2 = [(DE)2] + [(AB)2]
8) So, we have, AE2 + BD2 = DE2 + AB2], hence proved.
Q 14. The perpendicular from A on side BC of an ∆ ABC intersects BC at D such that DB = 3 CD (see the following fig). Prove that 2 AB2 = 2 AC2 + BC2.
Solution:
1) In ∆ ADB, by the theorem of Pythagoras, we have,
(AD)2 = (AB)2 - (DB)2 ------- equation 1.
2) In ∆ ADC, by the theorem of Pythagoras, we have,
(AD)2 = (AC)2 - (CD)2 ------- equation 2.
3) From equations 1 and 2, we have,
(AB)2 - (DB)2 = (AC)2 - (CD)2
(AB)2 - (AC)2 = (DB)2 - (CD)2 ------- equation 3.
4) It is given that DB = 3 CD ------- equation 4.
5) From equations 3 and 4, we have,
(AB)2 - (AC)2 = (DB)2 - (CD)2
(AB)2 - (AC)2 = (3 CD)2 - (CD)2
(AB)2 - (AC)2 = 9 (CD)2 - (CD)2
(AB)2 - (AC)2 = 8 (CD)2
2 (AB)2 - 2 (AC)2 = 16 (CD)2
2 (AB)2 - 2 (AC)2 = (4 CD)2
2 (AB)2 - 2 (AC)2 = (CD + 3 CD)2
2 (AB)2 - 2 (AC)2 = (CD + DB)2
2 (AB)2 - 2 (AC)2 = (BC)2
2 (AB)2 = 2 (AC)2 + (BC)2, hence proved.
Q 15. In an equilateral triangle ABC, D is a point on side BC such that
BD = 1/3 BC. Prove that 9 AD2 = 7 AB2.
Solution:
1) In ∆ ABC,
a) AB = BC = AC
b) AE = (√3/2) AB --------- equation 1
c) BE = (1/2) AB --------- equation 2
d) BD = (1/3) AB (given) --------- equation 3
2) In ∆ AED, by the theorem of Pythagoras, we have,
(AD)2 = (AE)2 + (DE)2
(AD)2 = (AE)2 + (BE - BD)2 --------- equation 4
3) From equations 1, 2, 3, and 4, we have,
(AD)2 = (AE)2 + (BE - BD)2
(AD)2 = [(√3/2) AB]2 + [((1/2) AB) - ((1/3) AB)]2
(AD)2 = [(3/4) (AB)2] + [((1/6) AB)]2
(AD)2 = [(3/4) (AB)2] + [(1/36) (AB)]2
(AD)2 = [(3/4) + (1/36)] [AB]2
(AD)2 = [(27/36) + (1/36)] [AB]2
(AD)2 = [(27 + 1)/36)] [AB]2
(AD)2 = [(28)/36)] [AB]2
(AD)2 = [(7)/9)] [AB]2
9 (AD)2 = 7 (AB)2, hence proved.
Q 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:
1) In ∆ ABC,
a) AD = (√3/2) AB --------- equation 1
b) BD = (1/2) AB --------- equation 2
2) In ∆ ADB, by the theorem of Pythagoras, we have,
(AD)2 = (AB)2 - (BD)2 --------- equation 3
3) from equations 1, 2, and 3, we have,
(AD)2 = (AB)2 - [(1/2) (AB)]2
(AD)2 = (AB)2 - (1/4) (AB)2
(AD)2 = [1 - (1/4)] (AB)2
(AD)2 = [(4/4) - (1/4)] (AB)2
(AD)2 = [(4 - 1)/4)] (AB)2
(AD)2 = (3/4) (AB)2
4 (AD)2 = 3 (AB)2,
4) 4 × (Square of altitude) = 3 × (Square of one side), hence proved.
Q 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:
1) In ∆ ABC, it is given that,
a) AB = 6√3
b) AC = 12
c) BC = 6
2) Now we will find (AB)2
a) (AB)2 = (6√3)2
b) (AB)2 = (36 x 3)
c) (AB)2 = 108 ------- equation 1
3) Now we will find (AC)2
a) (AC)2 = (12)2
b) (AC)2 = 144 ------- equation 2
4) Now we will find (BC)2
a) (BC)2 = (6)2
b) (BC)2 = 36 ------- equation 3
5) From equations 1, 2, and 3, we have,
144 = 108 + 36
(AC)2 = (AB)2 + (BC)2 ------- equation 4
6) From equation 4, we have,
This is a right angled triangle at point B, so angle B is 90° hence answer is C.
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