TOPICS
Exercise - 6.5

Question-1 :-  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.
(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 :-
(i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.
    Squaring the lengths of these sides, we will obtain 49, 576, and 625.
    However, 49 + 576 = 625 Or, 72 + 242 = 252
    The sides of the given triangle are satisfying Pythagoras theorem.
    Therefore, it is a right triangle.
    We know that the longest side of a right triangle is the hypotenuse.
    Therefore, the length of the hypotenuse of this triangle is 25 cm.
    
(ii) It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.
    Squaring the lengths of these sides, we will obtain 9, 64, and 36.
    However, 9 + 36 ≠ 64 Or, 32 + 62 ≠ 82
    Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
    Therefore, the given triangle is not satisfying Pythagoras theorem. 
    Hence, it is not a right triangle.
    
(iii)Given that sides are 50 cm, 80 cm, and 100 cm.
    Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000. 
    However, 2500 + 6400 ≠ 10000 Or, 502 + 802 ≠ 1002
    Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
    Therefore, the given triangle is not satisfying Pythagoras theorem.
    Hence, it is not a right triangle.
    
(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
    Squaring the lengths of these sides, we will obtain 169, 144, and 25.
    Clearly, 144 + 25 = 169 Or, 122 + 52 = 132
    The sides of the given triangle are satisfying Pythagoras theorem.
    Therefore, it is a right triangle.
    We know that the longest side of a right triangle is the hypotenuse.
    Therefore, the length of the hypotenuse of this triangle is 13 cm.
    

Question-2 :-  PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR.

Solution :-
    triangle
    Let ∠MPR = x
    In Δ MPR,
    ∠MRP = 180o - 90o - x
    ∠MRP = 90o - x

    Similarly, In Δ MPQ,
    ∠MPQ = 90o - ∠MPR
    ∠MPQ = 90o - x
    ∠MQP = 180o - 90o - (90o - x)
    ∠MQP = x

    In Δ QMP and Δ PMR
    ∠MPQ = ∠MRP
    ∠PMQ = ∠RMP
    ∠MQP = ∠MPR
    Δ QMP ∼ Δ PMR           (By using AAA similarity rule)
    Therefore, QM/PM = PM/MR
    PM2 = QM x MR
    

Question-3 :-  In Figure, 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 triangle

Solution :-
(i) In Δ ADB and Δ CAB,
    ∠DAB = ∠ACB             (Each 90o)
    ∠ABD = ∠CBA             (Common angle)
    Δ ADB ∼ Δ CAB           (By using AA similarity rule)
    AB/CB = BD/AB
    AB2 = CB x BD
    
(ii) Let ∠CAB = x
    In Δ CBA, 
    ∠CBA = 180o - 90o - x
    ∠CBA = 90o - x
    Similarly, In Δ CAD
    ∠CAD = 90o - ∠CAB
    ∠CAD = 90o - x
    ∠CDA = 180o - 90o - (90o - x)
    ∠CDA = x

    In Δ CBA and Δ CAD
    ∠CBA = ∠CAD
    ∠CAB = ∠CDA
    ∠ACB = ∠DCA             (Each 90o)
    Δ CBA ∼ Δ CAD           (By using AAA similarity rule)
    Therefore, AC/DC = BC/AC
    AC2 = DC x BC
    
(iii) In Δ DCA and Δ DAB
     ∠DCA = ∠DAB            (Each 90o)
     ∠CDA = ∠ADB            (Common angle)
     Δ DCA ∼ Δ ∠DAB         (By using AA similarity rule)
     Therefore, DC/DA = DA/DB
     AD2 = BD x CD
    

Question-4 :-  ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.

Solution :-
    triangle
Given that : ABC is an isosceles triangle right angled at C.
Prove that : AB2 = 2AC2
Proof : Δ ABC is an isosceles triangle.         (Given)
        Therefore, AC = CB
        Applying Pythagoras theorem in Δ ABC, we get
        AC2 + CB2 = AB2
        AC2 + AC2 = AB2        (AC = CB)
        2 AC2 = AB2
    

Question-5 :- ABC is an isosceles triangle with AC = BC. If AB2 = 2 AC2, prove that ABC is a right triangle.

Solution :-
    triangle
Given that : ABC is an isosceles triangle with AC = BC and AB2 = 2 AC2.
Prove that : ABC is a right triangle.
Proof : AB2 = 2 AC2       (Given)
        AB2 = AC2 + AC2
        AB2 = AC2 + BC2        (AC = BC)
        This triangle satisfying the pythagoras theorem.
        Therefore, the given triangle is right-angled triangle.
    

Question-6 :-  ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Solution :-
    triangle
    Let AD be the altitude in the given equilateral triangle, Δ ABC.
    We know that altitude bisects the opposite side.
    Therefore, BD = DC = a
    In Δ ADB,
    ∠ADB = 90o
    Applying Pythagoras theorem, we get
    AD2 + DB2 = AB2
    AD2 + a2 = (2a)2
    AD2 + a2 = 4a2
    AD2 = 4a2 - a2
    AD2 = 3a2
    AD = √3a
    In an equilateral triangle, all the altitudes are equal in length.
    Therefore, the length of each altitude will be √3a.
    

Question-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 :-
    rhombus
    In Δ AOB, Δ BOC, Δ COD, Δ AOD,
    Applying Pythagoras theorem, we obtain
    AB2 = AO2 + OB2 ....... (i)
    BC2 = BO2 + OC2 ....... (ii)
    CD2 = CO2 + OD2 ....... (iii)
    AD2 = AO2 + OD2 ....... (iv)
    By adding all equations, we get
    AB2 + BC2 + CD2 + AD2 = AO2 + OB2 + BO2 + OC2 + CO2 + OD2 + AO2 + OD2
    AB2 + BC2 + CD2 + AD2 = 2(AO2 + OB2 + OC2 + OD2)
    AB2 + BC2 + CD2 + AD2 = 2[(AC/2)2 + (BD/2)2 + (AC/2)2 + (BD/2)2]        (Diagonals bisect each other)
    AB2 + BC2 + CD2 + AD2 = 2[2 x AC2/4 + 2 x BD2/4]
    AB2 + BC2 + CD2 + AD2 = 2[AC2/2 + BD2/2]
    AB2 + BC2 + CD2 + AD2 = 2(AC2 + BD2)/2
    AB2 + BC2 + CD2 + AD2 = AC2 + BD2
    

Question-8 :-  In Figure, 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 = AE22 + CD2 + BF2. triangle

Solution :-
Given that : O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
Prove that : (i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
            (ii) AF2 + BD2 + CE2 = AE22 + CD2 + BF2.
Construction : Join OA, OB, and OC.
    triangle
Proof : (i) In Δ AOF, 
        Applying Pythagoras theorem in Δ AOF, we get
        OA2 = OF2 + AF2 .......(i)
        Similarly, In Δ BOD
        Applying Pythagoras theorem in Δ BOD, we get
        OB2 = OD2 + BD2 .......(ii)
        Similarly, In ΔCOE,
        Applying Pythagoras theorem in Δ COE, we get
        OC2 = OE2 + EC2 .......(iii)
        By Adding all equations, we get
        OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2
        OA2 + OB2 + OC2 - OF2 - OD2 - OE2 = AF2 + BD2 + EC2
    
(ii) From the above result, we get
     (OA2 - OE2) + (OB2 - OF2) + (OC2 - OD2) = AF2 + BD2 + EC2
     AE2 + CD2 + BF2 = AF2 + BD2 + EC2
    

Question-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 :-
    triangle
    Let OA be the wall and AB be the ladder.
    Length of ladder (AB) = 10 m
    Height of window from the ground (OA) = 8 m
    In Δ ABO,
    Therefore, by Pythagoras theorem,
    AB2 = OA2 + OB2
    102 = 82 + OB2
    100 = 64 + OB2
    OB2 = 100 - 64
    OB2 = 36
    OB = 6
    Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.
    

Question-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 :-
    triangle
    Let OB be the pole and AB be the wire.
    Height of Pole (OB) = 18 m
    Length of wire (AB) = 24 m
    In Δ ABO,
    Therefore, by Pythagoras theorem,
    AB2 = OA2 + OB2
    242 = OA2 + 182
    576 = OA2 + 324
    OA2 = 576 - 324
    OA2 = 252
    OA = 6√7
    Therefore, the distance from the base is 6√7 m.
    

Question-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 3/2 hours?

Solution :-
    triangle
    Let these distances be represented by OA and OB respectively.
    Distance travelled by the plane flying towards north in 3/2 hours (OA) = 1000 x 3/2 = 500 x 3 = 1500 km
    Similarly, distance travelled by the plane flying towards west in 3/2 hours (OB) = 1200 x 3/2 = 600 x 3 = 1800 km 
    
    Applying Pythagoras theorem, Distance b/w these planes after 3/2 hours
    AB2 = OA2 + OB2
    AB2 = (1500)2 + (1800)2
    AB2 = 2250000 + 3240000
    AB2 = 5490000
    AB = 300√61
    Therefore, the distance between these planes will be 300√61 km after 3/2 hours.
    

Question-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 :-
    triangle
    Let CD and AB be the poles of height 11 m and 6 m.
    Therefore, CP = 11 − 6 = 5 m
    The distance between the feet of the poles (AP) = 12 m
    Applying Pythagoras theorem for ΔAPC, we obtain
    AP2 + PC2 = AC2
    122 + 52 = AC2
    144 + 25 = AC2
    AC2 = 169
    AC = 13
    Therefore, the distance between their tops is 13 m.
    

Question-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 :-
    triangle
    Applying Pythagoras theorem in Δ ACE, we obtain
    AC2 + CE2 = AE2 ........ (i)
    Applying Pythagoras theorem in Δ BCD, we obtain
    BC2 + CD2 = BD2 ........ (ii)
    By adding equations (i) & (ii), we get
    AC2 + CE2 + BC2 + CD2 = AE2 + BD2 .......(iii)

    Applying Pythagoras theorem in Δ CDE, we obtain
    CD2 + CE2 = DE2 ........ (iv)
    Applying Pythagoras theorem in Δ ABC, we obtain
    BC2 + AC2 = AB2 ........ (v)

    Putting the values of equations (v) & (iv) into equation (iii), we get
    AE2 + BD2 = AB2 + DE2
    

Question-14 :-  The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3 CD (see Figure). Prove that 2AB2 = 2AC2 + BC2. triangle

Solution :-
Given that : The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3 CD .
Prove that : 2AB2 = 2AC2 + BC2
Proof : Applying Pythagoras theorem for Δ ACD, we obtain
        AC2 = AD2 + DC2
        AD2 = AC2 - DC2 ........(i)
        Applying Pythagoras theorem in Δ ABD, we obtain
        AB2 = AD2 + DB2
        AD2 = AB2 - DB2 ........(ii)

        From equations (i) & (ii), we get
        AC2 - DC2 = AB2 - DB2 ........(iii)
        It is given that 3 DC = DB
        Therefore, DC = BC/4 and DB = 3BC/4
        Putting these values in equation (iii), we get
        AC2 - (BC/4)2 = AB2 - (3BC/4)2
        AC2 - BC2/16 = AB2 - 9 BC2/16
        16 AC2 - BC2 = 16 AB2 - 9 BC2
        16 AB2 - 16 AC2 = 8 BC2
        2 AB2 = 2 AC2 + BC2
    

Question-15 :-  In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD2 = 7 AB2.

Solution :-
    triangle
Given that : In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.
Prove that : 9 AD2 = 7 AB2
Proof : Let the side(BC) of the equilateral triangle be a, and AE be the altitude of Δ ABC.
        Therefore, BE = EC = BC/2 = a/2 and AE = a√3/2
        Given that BD = BC/3 = a/3
        DE = BE - BD = a/2 - a/3 = a/6
        Applying Pythagoras theorem in Δ ADE, we get
        AD2 = AE2 + DE2
        AD2 = (a√3/2)2 + (a/6)2
        AD2 = 3a2/4 + a2/36
        AD2 = 28a2/36
        AD2 = 7/9 x AB2
        9 AD2 = 7 AB2
    

Question-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 :-
    triangle
    Let the side(BC) of the equilateral triangle be a, and AE be the altitude of Δ ABC.
    Therefore, BE = EC = BC/2 = a/2
    Applying Pythagoras theorem in Δ ABE, we obtain 
    AB2 = AE2 + BE2
    a2 = AE2 + (a/2)2
    AE2 = a2 - a2/4
    AE2 = 3a2/4
    4AE2 = 3a2
    4 × (Square of altitude) = 3 × (Square of one side)
    

Question-17 :-  Tick the correct answer and justify : In Δ ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is :
(A) 120o
(B) 60o
(C) 90o
(D) 45o

Solution :-
    triangle
    Given that, AB = 6√3 cm, AC = 12 cm, and BC = 6 cm
    It can be observed that AB2 = 108, AC2 = 144 and BC2 = 36
    AB2 + BC2 = AC2
    108 + 36 = 144
    144 = 144
    LHS = RHS
    The given triangle, Δ ABC, is satisfying Pythagoras theorem.
    Therefore, the triangle is a right triangle, right-angled at B.
    Hence, ∠B = 90o
    Hence, the correct answer is (C).
    
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