Ncert Solutions For Class 7 Maths Ex 6.5

Ncert Solutions For Class 7 Maths Chapter 6 Ex 6.5

Question 1:

 ABC is a right angled triangle with B=90. If AB = 12mm and BC = 5mm, find AC.

Answer:

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Here AB = 12mm and BC = 5mm (given)

Assuming AC =  y mm.

AC2=AB2+BC2In ABC by applying Pythagoras theorem, we get

y2=122+52  thereforey2=144+25 y2=169 y=169=13mm

Thus, AC = 13mm

 

Question 2:

XYZ is a right angled triangle with Y=90. If XZ = 5cm and YZ = 3cm, find XY.

Answer:

Assuming XY =  acm.

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Here XZ = 5cm and YZ = 3cm (given)

In XYZ by applying

Pythagoras theorem, we get

XZ2=XY2+YZ2  thereforea2=5232  thereforea2=259 a2=16 a=16=4cm

Thus, XY = 4cm

 

Question 3:

A 25cm long stick is rested on a wall at a point 24cm high on the wall. If the stick is at a distance x cm from wall on the ground, find the distance x.

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Answer:

 18

 

Let PR be the stick and P be the point where the stick touches the wall.

So, it becomes right angled triangle with Q=90

In PQR by applying Pythagoras theorem, we get

x2=252242PR2=PQ2+QR2

 thereforex2=625576 x2=49 x=49=7cm

Thus, QR = 7cm

 

Question 4:

Which of the given below could be the lengths of the side of a right angled triangle?

  1. 5mm, 1.2mm, .9mm
  2. 3mm, 2mm, 1mm
  3. 1mm, 4mm, 9mm

 

Answer:

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Assuming that the hypotenuse is the side with largest length so as per Pythagoras Theorem,

(Hypotenuse)2=(Perpendicular)2+(Base)2

5mm, 1.2mm, 0.9mm

Let PQR be the right angles triangle with Q=90.

So, by Pythagoras theorem
1.22+0.92=2.25PR2=PQ2+QR2

1.52=2.25

R.H.S = L.H.S

Therefore, 1.5mm, 1.2mm, 0.9mmare

The lengths of the side of a right angled triangle.

3mm, 2mm, 1mm

Let PQR be the right angles triangle with Q=90.

So, by Pythagoras theorem

PR2=PQ2+QR2  therefore22+12=5 32=9 R.H.SL.H.S

Therefore, 3mm, 2mm, 1mm are not

the lengths of the side of a right angled triangle.

 

1mm, 4mm, 0.9mm

Let PQR be the right angles triangle with Q=90.

So, by Pythagoras theorem

PR2=PQ2+QR2  therefore42+0.92=16.81 4.12=16.81

R.H.S = L.H.S

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Therefore, 4.1mm, 4mm, 0.9mm are

the lengths of the side of a right angled triangle.

 

Question 5:

A pole is broken at the height of 500cm from the footpath and its peak point is in contact with the footpath at a distance of 1200cm from the base of the pole. Find the actual length when it was in normal condition.

Answer: Let PQRbe the right angles triangle with Q=90 representing the pole broken at point P and touches the footpath at point R and Q is the base of the pole.

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Here, PQ = 500cm, QR = 1200cm ,

Let PR = x cm

In PQR by applying Pythagoras theorem, we get

PR2=PQ2+QR2  thereforex2=5002+12002 x2=250000+1440000 x2=1690000 x=49=1300cm

Thus, PR = 1300cm

Now the actual length of the pole  = PQ + PR

= 500 + 1300

= 1800cm

Thus, the actual length of the pole is 1800cm.

 

Question 6:

In XYZ

Find out the correct one from equations given below, Y=35andZ=55

  1. XZ2=XY2+YZ2
  2. XY2=YZ2+XZ2
  3. YZ2=XY2+XZ2

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Answer: In XYZ

XYZ+YZX+YXZ=180
  • 35+55+YXZ=180
  • YXZ=180(35+55)
  • YXZ=18090
  • YXZ=90

So, XYZ is a right angled triangle with  X=90.

Thus, (Hypotenuse)2=(Perpendicular)2+(Base)2

  • YZ2=XY2+XZ2

Thus, (c) is the correct option.

 

Question 7:

For a rectangle it is given that, width of a rectangle is 15mm and diagonal is 17 mm, then find the perimeter of the rectangle.

Answer:  Let WXYZ be a rectangle.

Let the length of a rectangle is y mm.

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In XYZ by applying

Pythagoras theorem, we get

XZ2=YZ2+XY2 172=152+y2 y2=289225 y2=64 y=64=8mm

Thus, XY = 8mm

Thus, the length of a rectangle is 8mm.

Now, perimeter of rectangle = 2 (XY + YZ) = 2(8 + 15)

= 46mm

Thus, the perimeter of the rectangle is 46mm.

 

Question 8:

For a rhombus the length of diagonals are given 80mm and 18mm.Then determine the perimeter of the rhombus.

Answer: Let PQRS be the rhombus.

Let the length of side of rhombus is x mm.

Also PR = 18mm,  SQ = 80mm

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As the diagonals bisect each other at right angle in case of rhombus,

So, OP = PR/2 = 18/2 = 9mm

And  SO = SQ/2 = 80/2 = 40mm

In POS by applying

Pythagoras theorem, we get

PS2=OP2+SO2 x2=92+402 x2=81+1600 x2=1681 x=1681=41mm

Thus, PS = 41mm

Now, the perimeter of the rhombus is = 4×Length of side of rhombus = 4×PS

= 4×41 = 164mm

Thus, the perimeter of the rhombus is 164mm.