Triangles EX 6.5 Solutions
Question - 11 : - An aeroplane leaves an airport and flies duenorth at a speed of 1,000 km per hour. At the same time, another aeroplaneleaves the same airport and flies due west at a speed of 1,200 km per hour. Howfar apart will be the two planes after
hours?
Answer - 11 : -
Given,
Speed of first aeroplane = 1000 km/hr
Distance covered by firstaeroplane flying due north inhours (OA) = 100 × 3/2 km = 1500 kmSpeed of second aeroplane = 1200 km/hr
Distance covered by secondaeroplane flying due west inhours (OB) = 1200 × 3/2 km = 1800 km
In right angle ΔAOB, by Pythagoras Theorem,
AB2 = AO2 +OB2
⇒ AB2 = (1500)2 +(1800)2
⇒ AB = √(2250000 + 3240000)
=√5490000
⇒ AB = 300√61 km
Hence, the distance between two aeroplaneswill be 300√61 km.
Question - 12 : - Two poles of heights6 m and 11 m stand on a plane ground. If the distance between the feet of thepoles is 12 m, find the distance between their tops.
Answer - 12 : -
Given, Two poles of heights 6 m and 11 m standon a plane ground.
And distance between the feet of the poles is12 m.
Let AB and CD be the poles of height 6m and11m.
Therefore, CP = 11 – 6 = 5m
From the figure, it can be observed that AP = 12m
By Pythagoras theorem for ΔAPC, we get,
AP2 = PC2 +AC2
(12m)2 + (5m)2 = (AC)2
AC2 = (144+25) m2 = 169 m2
AC = 13m
Therefore, the distance between their tops is 13 m.
Question - 13 : - D and E are pointson the sides CA and CB respectively of a triangle ABC right angled at C. Provethat AE2 + BD2 = AB2 + DE2.
Answer - 13 : -
Given, D and E are points on the sides CA andCB respectively of a triangle ABC right angled at C.
By Pythagoras theorem in ΔACE, we get
AC2 + CE2 =AE2 ………………………………………….(i)
In ΔBCD, by Pythagoras theorem, we get
BC2 + CD2 =BD2 ………………………………..(ii)
From equations (i) and (ii),we get,
AC2 + CE2 +BC2 + CD2 = AE2 +BD2 …………..(iii)
In ΔCDE, by Pythagoras theorem, we get
DE2 = CD2 +CE2
In ΔABC, by Pythagoras theorem, we get
AB2 = AC2 +CB2
Putting the above two values in equation (iii), we get
DE2 + AB2 = AE2 + BD2.
Question - 14 : - The perpendicularfrom A on side BC of a Δ ABC intersects BC at D such that DB =3CD (see Figure). Prove that 2AB2 = 2AC2 +BC2.
Answer - 14 : - 
Given, the perpendicular from A on side BC ofa Δ ABC intersects BC at D such that;
DB = 3CD.
In Δ ABC,
AD ⊥BC and BD = 3CD
In right angle triangle, ADB and ADC, byPythagoras theorem,
AB2 = AD2 +BD2 ……………………….(i)
AC2 = AD2 +DC2 ……………………………..(ii)
Subtracting equation (ii) from equation (i), we get
AB2 – AC2 = BD2 –DC2
= 9CD2 – CD2 [Since,BD = 3CD]
= 8CD2
= 8(BC/4)2 [Since, BC =DB + CD = 3CD + CD = 4CD]
Therefore, AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.
In an equilateraltriangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 =7AB2.
Answer - 15 : -
Given, ABC is an equilateral triangle.
And D is a point on side BC such that BD =1/3BC
Let the side of the equilateral trianglebe a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given, BD = 1/3BC
∴ BD = a/3
DE = BE – BD = a/2 – a/3 = a/6
In ΔADE, by Pythagorastheorem,
AD2 = AE2 + DE2
⇒ 9 AD2 = 7AB2
Question - 16 : - In an equilateral triangle, prove that threetimes the square of one side is equal to four times the square of one of itsaltitudes.
Answer - 16 : -
Given, an equilateral triangle say ABC,
Let the sides of the equilateral triangle beof length a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
In ΔABE, by Pythagoras Theorem, we get
AB2 = AE2 + BE2
4AE2 = 3a2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)
Hence, proved.
Question - 17 : - Tick the correctanswer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
Answer - 17 : -
(A) 120° (B) 60° (C) 90° D) 45°
Solution
Given, in ΔABC, AB = 6√3 cm, AC = 12 cmand BC = 6 cm.
We can observe that,
AB2 = 108
AC2 = 144
And, BC2 = 36
AB2 + BC2 = AC2
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct answer is (C).