Ncert Solutions For Class 7 Maths Ex 6.5

Question 1:

 ABC is a right angled triangle with $\angle B = 90^{\circ}$. If AB = 12mm and BC = 5mm, find AC.

Answer:

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

Assuming AC =  y mm.

$AC^{2} = AB^{2} + BC^{2}$In $\triangle ABC$ by applying Pythagoras theorem, we get

$y^{2} = 12^{2} + 5^{2}$ $\ therefore y^{2} = 144 + 25$ $∴y^{2} = 169$ $∴ y = \sqrt{169} = 13mm$

Thus, AC = 13mm

 

Question 2:

XYZ is a right angled triangle with $\angle Y = 90^{\circ}$. If XZ = 5cm and YZ = 3cm, find XY.

Answer:

Assuming XY =  acm.

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

In $\triangle XYZ$ by applying

Pythagoras theorem, we get

$XZ^{2} = XY^{2} + YZ^{2}$ $\ therefore a^{2} = 5^{2} – 3^{2}$ $\ therefore a^{2} = 25 – 9$ $∴ a^{2} = 16$ $∴ a = \sqrt{16} = 4cm$

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.

Answer:

 

 

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

So, it becomes right angled triangle with $\angle Q = 90^{\circ}$

In $\triangle PQR$ by applying Pythagoras theorem, we get

$x^{2} =25^{2} – 24^{2}$$PR^{2} = PQ^{2} + QR^{2}$

$\ therefore x^{2} = 625 – 576$ $∴x^{2} = 49$ $∴ x = \sqrt{49} = 7cm$

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:

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 $\triangle PQR$ be the right angles triangle with $\angle Q = 90^{\circ}$.

So, by Pythagoras theorem
$1.2^{2} + 0.9^{2} = 2.25$$PR^{2} = PQ^{2} + QR^{2}$

$∴ 1.5^{2} = 2.25$

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 $\triangle PQR$ be the right angles triangle with $\angle Q = 90^{\circ}$.

So, by Pythagoras theorem

$PR^{2} = PQ^{2} + QR^{2}$ $\ therefore 2^{2} + 1^{2} = 5$ $∴ 3^{2} = 9$ $R.H.S \neq L.H.S$

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

the lengths of the side of a right angled triangle.

 

1mm, 4mm, 0.9mm

Let $\triangle PQR$ be the right angles triangle with $\angle Q = 90^{\circ}$.

So, by Pythagoras theorem

$PR^{2} = PQ^{2} + QR^{2}$ $\ therefore 4^{2} + 0.9 ^{2} = 16.81$ $∴ 4.1^{2} = 16.81$

R.H.S = L.H.S

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 $\triangle PQR$be the right angles triangle with $\angle Q = 90^{\circ}$ representing the pole broken at point P and touches the footpath at point R and Q is the base of the pole.

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

Let PR = x cm

In $\triangle PQR$ by applying Pythagoras theorem, we get

$PR^{2} = PQ^{2} + QR^{2}$ $\ therefore x^{2} =500^{2} + 1200^{2}$ $∴x^{2} = 250000 + 1440000$ $∴x^{2} = 1690000$ $∴ x = \sqrt{49} = 1300cm$

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 $\triangle XYZ$

Find out the correct one from equations given below, $\angle Y\; = 35^{\circ} and\;\angle Z=55^{\circ}$

1. $XZ^{2} = XY^{2} + YZ^{2}$
2. $XY^{2} = YZ^{2} + XZ^{2}$
3. $YZ^{2} = XY^{2} + XZ^{2}$

 

Answer: In $\triangle XYZ$

$\angle XYZ + \angle YZX + \angle YXZ = 180^{\circ}$

• $35^{\circ} + 55^{\circ} +\angle YXZ = 180^{\circ}$
• $\angle YXZ = 180^{\circ} – (35^{\circ} + 55^{\circ})$
• $\angle YXZ = 180^{\circ} – 90^{\circ}$
• $\angle YXZ = 90^{\circ}$

So, $\triangle XYZ$ is a right angled triangle with  $\angle X = 90^{\circ}$.

Thus, $(Hypotenuse)^{2} = (Perpendicular)^{2} + (Base)^{2}$

• $YZ^{2} = XY^{2} + XZ^{2}$

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.

In $\triangle XYZ$ by applying

Pythagoras theorem, we get

$XZ^{2} = YZ^{2} + XY^{2}$ $∴ 17^{2} =15^{2}  + y^{2}$ $∴y^{2} = 289 – 225 $ $∴y^{2} = 64$ $∴ y = \sqrt{64} = 8mm$

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

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 $\triangle POS$ by applying

Pythagoras theorem, we get

$PS^{2} = OP^{2} + SO^{2}$ $∴x^{2} =9^{2}  + 40^{2}$ $∴x^{2} = 81 + 1600$ $∴x^{2} = 1681$ $∴ x = \sqrt{1681} = 41mm$

Thus, PS = 41mm

Now, the perimeter of the rhombus is = 4$\times$Length of side of rhombus = 4$\times$PS

= 4$\times$41 = 164mm

Thus, the perimeter of the rhombus is 164mm.

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