TOPICS

Exercise - 6.5

Triangles

**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

(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, 7^{2}+ 24^{2}= 25^{2}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, 3^{2}+ 6^{2}≠ 8^{2}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, 50^{2}+ 80^{2}≠ 100^{2}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, 12^{2}+ 5^{2}= 13^{2}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 PM^{2} = QM . MR.

Let ∠MPR = x In Δ MPR, ∠MRP = 180^{o}- 90^{o}- x ∠MRP = 90^{o}- x Similarly, In Δ MPQ, ∠MPQ = 90^{o}- ∠MPR ∠MPQ = 90^{o}- x ∠MQP = 180^{o}- 90^{o}- (90^{o}- 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 PM^{2}= QM x MR

**Question-3 :-** In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB^{2} = BC . BD

(ii) AC^{2} = BC . DC

(iii) AD^{2} = BD . CD

(i) In Δ ADB and Δ CAB, ∠DAB = ∠ACB (Each 90^{o}) ∠ABD = ∠CBA (Common angle) Δ ADB ∼ Δ CAB (By using AA similarity rule) AB/CB = BD/AB AB^{2}= CB x BD

(ii) Let ∠CAB = x In Δ CBA, ∠CBA = 180^{o}- 90^{o}- x ∠CBA = 90^{o}- x Similarly, In Δ CAD ∠CAD = 90^{o}- ∠CAB ∠CAD = 90^{o}- x ∠CDA = 180^{o}- 90^{o}- (90^{o}- x) ∠CDA = x In Δ CBA and Δ CAD ∠CBA = ∠CAD ∠CAB = ∠CDA ∠ACB = ∠DCA (Each 90^{o}) Δ CBA ∼ Δ CAD (By using AAA similarity rule) Therefore, AC/DC = BC/AC AC^{2}= DC x BC

(iii) In Δ DCA and Δ DAB ∠DCA = ∠DAB (Each 90^{o}) ∠CDA = ∠ADB (Common angle) Δ DCA ∼ Δ ∠DAB (By using AA similarity rule) Therefore, DC/DA = DA/DB AD^{2}= BD x CD

**Question-4 :-** ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2}.

Given that : ABC is an isosceles triangle right angled at C. Prove that : AB^{2}= 2AC^{2}Proof : Δ ABC is an isosceles triangle. (Given) Therefore, AC = CB Applying Pythagoras theorem in Δ ABC, we get AC^{2}+ CB^{2}= AB^{2}AC^{2}+ AC^{2}= AB^{2}(AC = CB) 2 AC^{2}= AB^{2}

**Question-5 :-** ABC is an isosceles triangle with AC = BC. If AB^{2} = 2 AC^{2}, prove that ABC is a right triangle.

Given that : ABC is an isosceles triangle with AC = BC and AB^{2}= 2 AC^{2}. Prove that : ABC is a right triangle. Proof : AB^{2}= 2 AC^{2}(Given) AB^{2}= AC^{2}+ AC^{2}AB^{2}= AC^{2}+ BC^{2}(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.

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 = 90^{o}Applying Pythagoras theorem, we get AD^{2}+ DB^{2}= AB^{2}AD^{2}+ a^{2}= (2a)^{2}AD^{2}+ a^{2}= 4a^{2}AD^{2}= 4a^{2}- a^{2}AD^{2}= 3a^{2}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.

In Δ AOB, Δ BOC, Δ COD, Δ AOD, Applying Pythagoras theorem, we obtain AB^{2}= AO^{2}+ OB^{2}....... (i) BC^{2}= BO^{2}+ OC^{2}....... (ii) CD^{2}= CO^{2}+ OD^{2}....... (iii) AD^{2}= AO^{2}+ OD^{2}....... (iv) By adding all equations, we get AB^{2}+ BC^{2}+ CD^{2}+ AD^{2}= AO^{2}+ OB^{2}+ BO^{2}+ OC^{2}+ CO^{2}+ OD^{2}+ AO^{2}+ OD^{2}AB^{2}+ BC^{2}+ CD^{2}+ AD^{2}= 2(AO^{2}+ OB^{2}+ OC^{2}+ OD^{2}) AB^{2}+ BC^{2}+ CD^{2}+ AD^{2}= 2[(AC/2)^{2}+ (BD/2)^{2}+ (AC/2)^{2}+ (BD/2)^{2}] (Diagonals bisect each other) AB^{2}+ BC^{2}+ CD^{2}+ AD^{2}= 2[2 x AC^{2}/4 + 2 x BD^{2}/4] AB^{2}+ BC^{2}+ CD^{2}+ AD^{2}= 2[AC^{2}/2 + BD^{2}/2] AB^{2}+ BC^{2}+ CD^{2}+ AD^{2}= 2(AC^{2}+ BD^{2})/2 AB^{2}+ BC^{2}+ CD^{2}+ AD^{2}= AC^{2}+ BD^{2}

**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) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2},

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2}2 + CD^{2} + BF^{2}.

Given that : O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Prove that : (i) OA^{2}+ OB^{2}+ OC^{2}– OD^{2}– OE^{2}– OF^{2}= AF^{2}+ BD^{2}+ CE^{2}(ii) AF^{2}+ BD^{2}+ CE^{2}= AE^{2}2 + CD^{2}+ BF^{2}. Construction : Join OA, OB, and OC. Proof : (i) In Δ AOF, Applying Pythagoras theorem in Δ AOF, we get OA^{2}= OF^{2}+ AF^{2}.......(i) Similarly, In Δ BOD Applying Pythagoras theorem in Δ BOD, we get OB^{2}= OD^{2}+ BD^{2}.......(ii) Similarly, In ΔCOE, Applying Pythagoras theorem in Δ COE, we get OC^{2}= OE^{2}+ EC^{2}.......(iii) By Adding all equations, we get OA^{2}+ OB^{2}+ OC^{2}= OF^{2}+ AF^{2}+ OD^{2}+ BD^{2}+ OE^{2}+ EC^{2}OA^{2}+ OB^{2}+ OC^{2}- OF^{2}- OD^{2}- OE^{2}= AF^{2}+ BD^{2}+ EC^{2}

(ii) From the above result, we get (OA^{2}- OE^{2}) + (OB^{2}- OF^{2}) + (OC^{2}- OD^{2}) = AF^{2}+ BD^{2}+ EC^{2}AE^{2}+ CD^{2}+ BF^{2}= AF^{2}+ BD^{2}+ EC^{2}

**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.

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, AB^{2}= OA^{2}+ OB^{2}10^{2}= 8^{2}+ OB^{2}100 = 64 + OB^{2}OB^{2}= 100 - 64 OB^{2}= 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?

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, AB^{2}= OA^{2}+ OB^{2}24^{2}= OA^{2}+ 18^{2}576 = OA^{2}+ 324 OA^{2}= 576 - 324 OA^{2}= 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?

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 AB^{2}= OA^{2}+ OB^{2}AB^{2}= (1500)^{2}+ (1800)^{2}AB^{2}= 2250000 + 3240000 AB^{2}= 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.

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 AP^{2}+ PC^{2}= AC^{2}12^{2}+ 5^{2}= AC^{2}144 + 25 = AC^{2}AC^{2}= 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 AE^{2} + BD^{2} = AB^{2} + DE^{2}.

Applying Pythagoras theorem in Δ ACE, we obtain AC^{2}+ CE^{2}= AE^{2}........ (i) Applying Pythagoras theorem in Δ BCD, we obtain BC^{2}+ CD^{2}= BD^{2}........ (ii) By adding equations (i) & (ii), we get AC^{2}+ CE^{2}+ BC^{2}+ CD^{2}= AE^{2}+ BD^{2}.......(iii) Applying Pythagoras theorem in Δ CDE, we obtain CD^{2}+ CE^{2}= DE^{2}........ (iv) Applying Pythagoras theorem in Δ ABC, we obtain BC^{2}+ AC^{2}= AB^{2}........ (v) Putting the values of equations (v) & (iv) into equation (iii), we get AE^{2}+ BD^{2}= AB^{2}+ DE^{2}

**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 2AB^{2} = 2AC^{2} + BC^{2}.

Given that : The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3 CD . Prove that : 2AB^{2}= 2AC^{2}+ BC^{2}Proof : Applying Pythagoras theorem for Δ ACD, we obtain AC^{2}= AD^{2}+ DC^{2}AD^{2}= AC^{2}- DC^{2}........(i) Applying Pythagoras theorem in Δ ABD, we obtain AB^{2}= AD^{2}+ DB^{2}AD^{2}= AB^{2}- DB^{2}........(ii) From equations (i) & (ii), we get AC^{2}- DC^{2}= AB^{2}- DB^{2}........(iii) It is given that 3 DC = DB Therefore, DC = BC/4 and DB = 3BC/4 Putting these values in equation (iii), we get AC^{2}- (BC/4)^{2}= AB^{2}- (3BC/4)^{2}AC^{2}- BC^{2}/16 = AB^{2}- 9 BC^{2}/16 16 AC^{2}- BC^{2}= 16 AB^{2}- 9 BC^{2}16 AB^{2}- 16 AC^{2}= 8 BC^{2}2 AB^{2}= 2 AC^{2}+ BC^{2}

**Question-15 :-** In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD^{2} = 7 AB^{2}.

Given that : 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}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 AD^{2}= AE^{2}+ DE^{2}AD^{2}= (a√3/2)^{2}+ (a/6)^{2}AD^{2}= 3a^{2}/4 + a^{2}/36 AD^{2}= 28a^{2}/36 AD^{2}= 7/9 x AB^{2}9 AD^{2}= 7 AB^{2}

**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.

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 AB^{2}= AE^{2}+ BE^{2}a^{2}= AE^{2}+ (a/2)^{2}AE^{2}= a^{2}- a^{2}/4 AE^{2}= 3a^{2}/4 4AE^{2}= 3a^{2}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) 120^{o}

(B) 60^{o}

(C) 90^{o}

(D) 45^{o}

Given that, AB = 6√3 cm, AC = 12 cm, and BC = 6 cm It can be observed that AB^{2}= 108, AC^{2}= 144 and BC^{2}= 36 AB^{2}+ BC^{2}= AC^{2}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 = 90^{o}Hence, the correct answer is (C).

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