Triangles NCERT Exercise 6.5 two Class Ten Mathematics

Triangle

Exercise 6.5 Part 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 the base of the wall.

Solution: In this case, the ladder makes the hypotenuse, the height of top of the ladder is perpendicular and the distance of foot of the ladder from the base of the wall is the base.

So, h = 10 m, p = 8 m and b = ?

From Pythagoras theorem;

`b^2 = h^2 – p^2`

Or, `b^2 = 10^2 – 8^2`

`= 100 – 64 = 36`

Or, `b = 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: Here; h = 24 m, p = 18 m and b = ?

From Pythagoras theorem;

`b^2 = h^2 – p^2`

Or, `b^2 = 24^2 – 18^2`

`= 576 – 324 = 252`

Or, `b = sqrt(252) = 6sqrt7  m`

Question 11: An aeroplane leaves and 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 1.5 hours?

Solution: Distance covered by the first plane in 1.5 hours = 1500 km

Distance covered by the second plane in 1.5 hours = 1800 km

The position of the two planes after 1.5 hour journey can be shown by a right triangle and we need to find the hypotenuse to know the aerial distance between them.

Here; h = ? p = 1800 km and b = 1500 km

From Pythagoras theorem;

`h^2 = p^2 + b^2`

Or, `h^2 = 1800^2 + 1500^2`

`= 3240000 + 2250000 = 5490000`

Or, `h = 300sqrt(61)  km`


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.

similar triangles exercise solution

Solution: In this figure; AB = 6 m, CD = 11 m and BC = AE = 12 m

In right triangle DEA;

`DE = CD – AB = 11 – 6 = 5` m, `AE = 12` m and `AD = ?`

Distance between the tops of the two poles can be calculated by using Pythagoras theorem;

`AD^2 = DE^2 + AE^2`

`= 5^2 + 12^2`

`= 25 + 144 = 169`

Or, `AD = 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.

similar triangles exercise solution

Solution: In ∆ ACE; `AE^2 = AC^2 + CE^2` ……… (1)

In ∆ DCB; `BD^2 = DC^2 + CB^2` ……… (2)

In ∆ ACB; `AB^2 = AC^2 + CB^2` ……… (3)

In ∆ DCE; `DE^2 = DC^2 + CE^2` ……… (4)

Adding equations (1) and (2), we get;

`AE^2 + BD^2 = AC^2 + CE^2 + DC^2 + CB^2` …….. (5)

Adding equations (3) and (4), we get;

`AB^2 + DE^2 = AC^2 + CB^2 + DC^2 + CE^2` ………. (6)

On comparing the RHS of equations (5) and(6), we get;

`AE^2 + BD^2 = AB^2 + DE^2` proved


Question 14: The perpendiculars from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2.

similar triangles exercise solution

Solution: `AB^2 = AD^2 + DB^2`

Or, `AB^2 = AD^2 + 9CD^2` (because DB = 3CD) …….. (1)

In ∆ ADC; `AD^2 = AC^2 – CD^2`

Substituting the value of AD in equation (1), we get;

`AB^2 = AC^2 – CD^2 + 9CD^2 = AC^2 + 8CD^2`

Or, `2AB^2 = 2AC^2 + 16CD^2` ……… (2)

Since `BC = CD + 3CD = 4CD`

Hence, `BC^2 = 16CD^2`

Substituting the value from above equation in equation (2), we get;

`2AB^2 = 2AC^2 + BC^2` proved

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

similar triangles exercise solution

Solution: Let us assume that each side of triangle ABC is ‘a’

Then, `BD = a/3`, `MC = a/2` and

`DM=(2a)/(3)-a/2=(4a-3a)/(6)=a/6`

Also, `AM=(a\sqrt3)/(2)`

According to Pythagoras theorem, in triangle ADM;

`AD^2=AM^2+DM^2`

Or, `AD^2=((a\sqrt3)/(2))^2+(a/6)^2`

`=(3a^2)/(4)+9a^2)/(36)=(27a^2+a^2)/(36)`

Or, `AD^2=(28a^2)/(36)=(7a^2)/(9)`

Or, `7AD^2=9a^2`

Or, `7AD^2=9AB^2` proved

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: Let us assume that each side of the triangle is ‘a’, then its altitude AM is as follows:

`AM=(a\sqrt3)/(2)`

Four times of square of altitude can be calculates as follows:

`4xx((a\sqrt3)/(2))^2=4xx(3a^2)/(4)=3a^2`

Hence; three times of square of a side = four times of square of altitude proved

Question 17: Tick the correct answer and justify: In Δ ABC, `AB = 6sqrt3` cm, AC = 12 cm and BC = 6 cm. The angle B is:

  1. 120°
  2. 60°
  3. 90°
  4. 45°

Solution: Here; the longest side is 12 cm. Let us check if the sides of the given triangle fulfill the criterion of Pythagoras triplet.

`AC^2=AB^2+BC^2`

Or, `12^2=(6sqrt3)^2+6^2`

Or, `144=108+36=144`

Here; LHS = RHS and hence the given triangle is a right triangle.

Hence; angle B = 90°, i.e. option (c) is the correct answer.



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