NCERT Solution class 7 Mathematics EXERCISE- 6.5

NCERT Solution class 7 Mathematics EXERCISE- 6.5

 

                    CHAPTER-6

                              EXERCISE – 6.5

Q1. PQR is a triangle , right angled at P. If PQ = 10 cm and PR = 24 cm , find QR.

SOLUTION:-

Let us draw a rough sketch of right – angled triangle

 

By the rule of Pythagoras theorem,

(QR)2= (RP)2 + (PQ)2

(QR)2= (24)2 + (10)2

(QR)2= 576 + 100

(QR)2= 676

QR = (676)2

QR = 26 cm.

Q2. ABC is a triangle , right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

SOLUTION:-

 

 

By the rule of Pythagoras theorem,

AB2 = AC2 + BC2

252 = 72 + BC2

625 = 49 +

BC2

625 – 49 = BC2

576 = BC2

(576)2 = BC

24 = BC

BC = 24 cm.

Q3. A 15 m long ladder reached a window 12 m high from the ground on placing  it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

SOLUTION:-

By the rule of Pythagoras theorem,

(15)2 = (12)2 + a2

225= 144 + a2

225 -144 = a2

81 = a2

(81)2 = a

9 = a

a = 9 cm.

Q4. Which of the following can be the sides of a right triangle?

(i) 2.5 cm , 6.5 cm , 6 cm.

(ii) 2 cm , 2 cm , 5 cm.

(iii) 1.5 cm , 2 cm , 2.5 cm.

In case of right – angled triangles, identify the right angles.

SOLUTION:-

(i) This triangle is  right – angled triangle.

(ii) This triangle is not right – angled triangle.

(iii) This triangle is right – angled triangle.

Q5. A tree is broken at a height of 5m from the ground and its top touches the ground at distance of 12m from the base of the tree. Find the original height of the tree.

SOLUTION:-

Let ABC is the triangle and B is the mid – point where tree is broken at the height 5 m from the ground.

Tree top touches the ground at a distance of AC = 12 m from the base of the tree,

By observing the figure we came to conclude that right angle triangle is formed at A.

From the rule of Pythagoras theorem,

BC2 = AB2 + AC2

BC2 = 52 + 122

BC2 = 25 + 144

BC2 = 169

BC = (169)2

BC = 13

Then, the original height of the tree = AB + BC

= 5 + 13

= 18 m

Q6. Angles Q and R of a ΔPQR are 250and 650 Write which of the following is true:

(i) PQ2 + QR2 = RP2

(ii) PQ2 + RP2 = QR2

(iii) RP2 + QR2 = PQ2

SOLUTION:-

Given that Angle B = 350, Angle C = 550

Then, Angle A = ?

We know that sum of the three interior angles of triangle is equal to 1800.

= Angle PQR + Angle QRP +  Angle RPQ = 1800

= 250 + 650 + Angle RPQ = 1800

= 900 + Angle RPQ = 1800

= Angle RPQ = 1800 – 900

= Angle RPQ = 900

Also, we know that side opposite to the right angle is the hypotenuse.

QR2 = PQ2 + PR2

Hence, (ii) is true

Q7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

SOLUTION:-

Let ABCD be the rectangular plot.

Then, AB = 40 cm and AC = 41 cm.

BC= ?

According to Pythagoras theorem,

From right angle triangle ABC, we have:

= AC2 = AB2 + BC2

= 412 = 402 + BC2

= BC2 = 412 – 402

= BC2 = 1681 – 1600

= BC2 = 81

= BC = 812

= BC = 9 cm

Hence, the perimeter of the rectangle plot = 2(length + breadth)

Where , length = 40 cm,

Breadth = 9 cm

Then,

= 2(40+9)

= 2(49)

= 98 cm.

Q8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

SOLUTION:-

Let PQRS be a rhombus , all sides of rhombus has equal length and its diagonal PR and SQ are intersecting each other at point O. Diagonals in rhombus bisect each other at 900

So, PO = (PR/2)

= 16/2

= 8 cm

And, SO = (SQ/2)

= 30/2

= 15 cm

Then, consider the triangle POS and apply the Pythagoras theorem,

PS2 = PO2 + SO2

PS2 = 82 + 152

PS2 = 64 +225

PS2 = 289

PS = 2892

PS = 17  cm

Hence , the length of side of rhombus is 17 cm.

Now,

Perimeter of rhombus = 4 x side of the rhombus

= 4 x 17

= 68 cm

Perimeter of rhombus is 68 cm.

 

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