# 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$\angle B = 90^{\circ}$. If AB = 12mm and BC = 5mm, find AC.

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

Assuming AC =  y mm.

AC2=AB2+BC2$AC^{2} = AB^{2} + BC^{2}$In ABC$\triangle ABC$ by applying Pythagoras theorem, we get

y2=122+52$y^{2} = 12^{2} + 5^{2}$  thereforey2=144+25$\ therefore y^{2} = 144 + 25$ y2=169$∴y^{2} = 169$ y=169=13mm$∴ y = \sqrt{169} = 13mm$

Thus, AC = 13mm

Question 2:

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

Assuming XY =  acm.

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

In XYZ$\triangle XYZ$ by applying

Pythagoras theorem, we get

XZ2=XY2+YZ2$XZ^{2} = XY^{2} + YZ^{2}$  thereforea2=5232$\ therefore a^{2} = 5^{2} – 3^{2}$  thereforea2=259$\ therefore a^{2} = 25 – 9$ a2=16$∴ a^{2} = 16$ a=16=4cm$∴ a = \sqrt{16} = 4cm$

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.

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$\angle Q = 90^{\circ}$

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

x2=252242$x^{2} =25^{2} – 24^{2}$PR2=PQ2+QR2$PR^{2} = PQ^{2} + QR^{2}$

thereforex2=625576$\ therefore x^{2} = 625 – 576$ x2=49$∴x^{2} = 49$ x=49=7cm$∴ x = \sqrt{49} = 7cm$

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

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

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

5mm, 1.2mm, 0.9mm

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

So, by Pythagoras theorem
1.22+0.92=2.25$1.2^{2} + 0.9^{2} = 2.25$PR2=PQ2+QR2$PR^{2} = PQ^{2} + QR^{2}$

1.52=2.25$∴ 1.5^{2} = 2.25$

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

So, by Pythagoras theorem

PR2=PQ2+QR2$PR^{2} = PQ^{2} + QR^{2}$  therefore22+12=5$\ therefore 2^{2} + 1^{2} = 5$ 32=9$∴ 3^{2} = 9$ R.H.SL.H.S$R.H.S \neq L.H.S$

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

the lengths of the side of a right angled triangle.

1mm, 4mm, 0.9mm

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

So, by Pythagoras theorem

PR2=PQ2+QR2$PR^{2} = PQ^{2} + QR^{2}$  therefore42+0.92=16.81$\ therefore 4^{2} + 0.9 ^{2} = 16.81$ 4.12=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 PQR$\triangle PQR$be the right angles triangle with Q=90$\angle Q = 90^{\circ}$ 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 PQR$\triangle PQR$ by applying Pythagoras theorem, we get

PR2=PQ2+QR2$PR^{2} = PQ^{2} + QR^{2}$  thereforex2=5002+12002$\ therefore x^{2} =500^{2} + 1200^{2}$ x2=250000+1440000$∴x^{2} = 250000 + 1440000$ x2=1690000$∴x^{2} = 1690000$ x=49=1300cm$∴ x = \sqrt{49} = 1300cm$

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 XYZ$\triangle XYZ$

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

1. XZ2=XY2+YZ2$XZ^{2} = XY^{2} + YZ^{2}$
2. XY2=YZ2+XZ2$XY^{2} = YZ^{2} + XZ^{2}$
3. YZ2=XY2+XZ2$YZ^{2} = XY^{2} + XZ^{2}$

Answer: In XYZ$\triangle XYZ$

XYZ+YZX+YXZ=180$\angle XYZ + \angle YZX + \angle YXZ = 180^{\circ}$
• 35+55+YXZ=180$35^{\circ} + 55^{\circ} +\angle YXZ = 180^{\circ}$
• YXZ=180(35+55)$\angle YXZ = 180^{\circ} – (35^{\circ} + 55^{\circ})$
• YXZ=18090$\angle YXZ = 180^{\circ} – 90^{\circ}$
• YXZ=90$\angle YXZ = 90^{\circ}$

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

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

• YZ2=XY2+XZ2$YZ^{2} = XY^{2} + XZ^{2}$

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 XYZ$\triangle XYZ$ by applying

Pythagoras theorem, we get

XZ2=YZ2+XY2$XZ^{2} = YZ^{2} + XY^{2}$ 172=152+y2$∴ 17^{2} =15^{2}  + y^{2}$ y2=289225$∴y^{2} = 289 – 225 $ y2=64$∴y^{2} = 64$ y=64=8mm$∴ y = \sqrt{64} = 8mm$

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 POS$\triangle POS$ by applying

Pythagoras theorem, we get

PS2=OP2+SO2$PS^{2} = OP^{2} + SO^{2}$ x2=92+402$∴x^{2} =9^{2}  + 40^{2}$ x2=81+1600$∴x^{2} = 81 + 1600$ x2=1681$∴x^{2} = 1681$ x=1681=41mm$∴ x = \sqrt{1681} = 41mm$

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.