NCERT Class 10 Maths Solutions Chapter - 6 Triangles, Ex - 6.5

Ex - 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. Given that sides are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides we get 49, 576, and 625.

Clearly, 49 + 576 = 625 or 7² + 24² = 25² .

Therefore, given triangle is satisfying Pythagoras theorem. So, it is a right triangle. The longest side in a right angled triangle is the hypotenuse.

Therefore length of hypotenuse of this triangle = 25 cm.

ii .Given that sides are 3 cm, 8 cm, and 6 cm. Squaring the lengths of these sides we may get 9, 64, and 36. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, 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 may get 2500, 6400, and 10000. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, 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 may get 169, 144, and 25. Clearly, 144 +25 = 169 Or, 12² + 5² = 13².

Therefore given triangle is satisfying Pythagoras theorem. So, it is a right triangle.

The longest side in a right angled triangle is the hypotenuse.

Therefore length of hypotenuse of this triangle = 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 x MR.

Solution

Question 3.  In figure 6.53, ABD is a triangle right angled at A and AC  BD. Show that

(i)    AB² = BC x BD

(ii)    AC² = BC x DC

(iii)    AD² = BD x CD

Solution

iii.     In DCA & DAB 

DCA = DAB = 90º
         CDA = ADB            (common angle)
         DAC = DBA            (remaining angle)

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

Solution

Given that ABC is an isosceles triangle.
Therefore AC = CB
Applying Pythagoras theorem in ABC (i.e. right angled at point C)

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

Solution

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

Solution

Let AD be the altitude in given equilateral triangle ABC.    
We know that altitude bisects the opposite side. 
So, BD = DC = a
 
Since in an equilateral triangle, all the altitudes are equal in length.
So, length of each altitude will be 

Question 7.  Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Solution

In AOB, BOC, COD, AOD
Applying Pythagoras theorem

Question 8.  In figure 6.54, 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

^{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

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

Let OB be the pole and AB be the wire.
Therefore by Pythagoras theorem,

Question 11.  An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after  hours?

Solution

Let these distances are represented by OA and OB respectively.
Now applying Pythagoras theorem

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

Let CD and AB be the poles of height 11 and 6 m.
Therefore CP = 11 -  6 = 5 m
From the figure we may observe that AP = 12m
In   APC, by applying Pythagoras theorem 
 
Therefore distance between their tops = 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

  ^In ACE, 

^{Question 14.}  The perpendicular from A on side BC of a ABC intersect BC at D such that DB = 3 CD.  Prove that 2 AB² = 2 AC² + BC^2.

Solution

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

^Solution

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 side of equilateral triangle be a. And AE be the altitude of ABC

Now, in ABE by applying Pythagoras theorem 
AB² = AE² + BE²

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

Solution

Given that AB =  cm, AC = 12 cm and BC = 6 cm
We may observe that
AB² = 108
AC² = 144
And BC² = 36
AB² +BC² = AC² 

Thus the given triangle ABC is satisfying Pythagoras theorem 

Therefore triangle is a right angled triangle right angled at B
Therefore B = 90°.

Hence, (c).

Ex-6.6