The DAV Class 6 Maths Book Solutions and **DAV Class 6 Maths Chapter 15 Brain Teasers **Solutions of Perimeter and Area offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 15 Brain Teasers Solutions

Question 1.

A. Tick (✓) the correct answer.

(a) The length of the side of a square whose perimeter is 32 cm is:

(i) 16 cm

(ii) 8 cm

(iii) 48 cm

(iv) 128 cm

Answer:

(ii) 8 cm.

Perimeter of the square = 32 cm.

Side of the square = \(\frac{\text { Perimeter of square }}{4}\)

= \(\frac{32}{4}\)

= 8 cm

(b) Perimeter of the following rectangle is 800 cm, then the value of x is:

(i) 150 cm

(ii) 250 cm

(iii) 25 cm

(iv) 5 cm

Answer:

(i) 150 cm.

Perimeter of rectangle = 2 (l + b)

⇒ 800 cm = 2(2.5 m + x)

⇒ 800 cm = 2(2.5 x 100 cm + x)

⇒ 800 cm = 2(250 cm + x)

⇒ 250 cm + x = \(\frac{800}{2}\) = 400 cm

⇒ x = 400 – 250 = 150 cm.

(c) Sum of the areas of regions A, B and C is 140 cm^{2}, then x is:

(i) 30 cm

(ii) 6 cm

(iii) 8 cm

(iv) 40 cm

Answer:

(ii) 6 cm.

Regions A, B and C form a rectangle.

Thus, area of rectangular region

= length × breadth

Here, length – 10 cm + x + 12 cm

= 22 cm + x

And breath = 5 cm.

Area of rectangular regions = (22 cm + x) × 5 cm

140cm = (22 cm + x) 5 cm

⇒ 22 cm + x = \(\frac{140 \mathrm{~cm}^2}{5 \mathrm{~cm}}\) = 28 cm

⇒ x = 28 cm – 22 cm = 6 cm.

(d) If the length of a rectangle is doubled keeping the breath the same, then its area is—

(i) doubled

(ii) unchanged

(iii) halved

(iv) squared

Answer:

(i) doubled.

Area of rectangle = l × b

on doubling the length keeping the breadth the same.

Area of new rectangle = 2l × b

Thus,

\(\frac{\text { Area of rectangle }}{\text { Area of new rectangle }}=\frac{l \times b}{2 l \times b}=\frac{1}{2}\)

⇒ Area of new rectangle

= 2 × Area of rectangle

(e) The perimeter of the given figure is:

(i) 26 units

(ii) 28 units

(iii) 30 units

(iv) 32 units

Answer:

(iii) 30 units

Perimeter of the given figure = sum of all lengths

= 5 + 2 + 2 + 3 + 1 + 4 + 2 + 1 + 1 + 6 + 1 + 2

= 30 units.

B. Answer the following questions.

(a) Into how many square of side 2 cm each can you divide a rectangle of length 8 cm and breadth 4 cm?

Answer:

Length of rectangle = 8 cm

Breadth of rectangle = 4 cm

Area of the rectangle = l × b

= 8 cm × 4 cm

= 32 cm^{2}

Side of the square = 2 cm

Area of the square = (side)^{2} = (2 cm)^{2}

= 4 cm^{2}

No. ot square = \(\frac{\text { Area of rectangle }}{\text { Area of square }}\)

= \(\frac{32 \mathrm{~cm}^2}{4 \mathrm{~cm}^2}\)

= 8

(b) What is the side of the square whose area is 36 cm^{2}?

Answer:

Area of the square = 36 cm^{2}

Side of the square = \(\sqrt{\text { Area of the square }}\)

= \(\sqrt{36}=\sqrt{6 \times 6}=\sqrt{6^2}\)

= 6 cm.

(c) A piece of string is 120 cm long. If the string is bent to form a square, what will be the length of each side?

Answer:

Perimeter of square = 120 cm

Length of each side of square

= \(\frac{\text { Perimeter of the square }}{4}\)

= \(\frac{120 \mathrm{~cm}}{4}\) = 30 cm

(d) Find the area of the shaded region.

Answer:

Area of the region = side × side

= 10 cm × 10 cm

= 100 cm^{2}

Area of squares of four corners = 4 (side × side)

= 4 × (1 × 1)

= 4 × 1 cm^{2}

= 4 cm^{2}

Thus, area of shaded regions = 100 cm^{2} – 4 cm^{2}

= 96 cm^{2}.

(e) Find the perimeter of a rectangle of length 1.5 m and breadth 50 cm.

Answer:

Length of rectangle = 1.5 m

= 1.5 × 100 cm

= 150 cm

Breadth of rectangle = 50 cm

Thus, perimeter of rectangle = 2 (l + b)

= 2(150 + 50)

= 2 × 200 cm

= 400 cm.

Question 2.

Which of the following has larger area and by how much? A rectangle of length 42 cm and breadth 28 cm.

or

A square of side 39 cm.

Answer:

Area of the rectangle = length × breadth

= 42 cm × 28 cm

= 1176 cm^{2}

Area of the square = (side)^{2}

= (39)^{2}

= 1521 cm^{2}

It is clear that area of the square is greater than the area of the rectangle by 1521 – 1176

= 345 cm^{2}.

Question 3.

What will happen to the area of a square when

(a) its side is doubled?

(b) its side is halved?

Answer:

Let the side of the square be a cm its area = (side)^{2} = a^{2} cm^{2}

(a) If its side is 2a cm

then its area = (2a)^{2} = 4a^{2} cm^{2}

∴ its area becomes 4 times.

(b) If its side is \(\frac{a}{2}\) cm

then its area \(\left(\frac{a}{2}\right)^2=\frac{a^2}{4}\)cm^{2}

∴ its area will become one fourth.

Question 4.

The area of a rectangle is 1560 cm^{2} and its length is 60 cm. Find its breadth.

Answer:

Area of the rectangle = 1560 cm^{2}

Length of the rectangle = 60 cm

∴ Breadth = \(\frac{\text { Area }}{\text { Length }}\) = \(\frac{1560}{60}\) = 26 cm

Hence the required breadth = 26 cm

Question 5.

The area of a square of side 8 cm is the same as that of a rectangle of length 32 cm. What is the breadth of the rectangle?

Answer:

Area of the square = Area of the rectangle

(8)^{2} = Area of the rectangle

64 = Area of the rectangle

∴ Breadth of the rectangle = \(\frac{64}{\text { Length }}\)

= \(\frac{64}{32}\) = 2 cm

Hence the required breadth = 2 cm

Question 6.

A square tile is having side of length 15 cm. How many such tiles would be required to cover the square floor of a bathroom of side 3 m?

Answer:

Length of side of square tile =15 cm

∴ its area = 15 × 15

= 225 cm^{2}

Area of the floor of bathroom = 3 × 3

= 9 m^{2}

= 9 × 100 × 100 cm^{2}

= 90000 cm^{2}

Number of tiles required

= \(\frac{90000}{225}\) = 400

Hence the required number of tiles = 400

Question 7.

Find the area of a square lawn whose perimeter is 48 m.

Answer:

Perimeter of the square lawn = 48 m

∴ Side of the lawn = \(\frac{48}{4}\)m = 12 m

∴ Area of the square lawn = (Side)^{2} = (12)^{2} = 144 m^{2}

Hence the required area = 144 m^{2}

Question 8.

Fire broke out at point A. Fire extinguisher is at point X. Which route (in km) should take to reach point A?

Answer:

Distance of XDCA = 1 + 1.5 + 2 + 1.5 + 1 + 1.5 + 2 + 1 + 0.5 + 0.5

= 12.5 km

Distance of XGFEA = 2 + 0.5 + 1.5 + 3.5 + 3 + 0.5 + 2

= 13 km

Distance of XGFEA > Distance of XDCA.

So, fire extinguisher should take route XDCA

### DAV Class 6 Maths Chapter 1 HOTS

Question 1.

A street lane is to be paved with tiles of length 12 cm and breadth 10 cm. If the length of the lane is 240 m and its breadth is 12 m, find the number of tiles needed.

Answer:

Area of tiles = l × b

= 12 cm × 10 cm

= 120 cm^{2}

Length (L) of the lane = 240 m

= 240 × 100 cm

= 24000 cm

Breadth (B) of the lane = 12 m

= 12 × 100 cm

= 1200 cm

Area of the lane = L × B

= 24000 cm × 1200 cm

= 28800000 cm^{2}

Thus, the number of tiles should be needed to pave the lane

= \(\frac{\text { Area of the lane }}{\text { Area of a tile }}=\frac{28800000 \mathrm{~cm}^2}{120 \mathrm{~cm}^2}\)

= 2,40,000 tiles

Question 2.

A wire of length 20 cm is to be bent to form a rectangle of

(i) maximum area

(ii) minimum area If the length and breadth are taken always in natural numbers, find the same in each case.

Sol. Perimeter of rectangle = 20 cm

2(l + b) = 20 ⇒ l + b = = 10 cm

On putting b = 1 cm, 2 cm, 3 cm, 4 cm, 5 cm, …. respectively we get l = 9 cm, 8 cm, 7 cm, 6 cm, 5 cm, …. respectively

(i) Maximum area when l and b are equals, i.e. 1 = 5 cm, b = 5 cm

Maximum area = 5 × 5 = 25 cm^{2}

(ii) Minimum area when 1 = 9 cm and b = 1 cm

∴ Minimum area = l × b = 9 × 1 = 9 cm^{2}

Additional Questions

Question 1.

The length and breadth of a rectangular sheet are 24 cm and 16 cm respectively.

(i) Find the area of the sheet.

Answer:

Length of the sheet = 24 cm

Breadth of the sheet = 16 cm

Area of the sheet = l × b

= 24 cm × 16 cm

= 384 cm^{2}

(ii) If the sheet is made a square with side 16 cm, what will be the area of the square sheet?

Answer:

Each side of the square sheet =16 cm its area = (side)^{2}

= (16)^{2}

= 256 cm^{2}

(iii) What will be the area of the separated sheet?

Answer:

Length of the separated sheet = 16 cm and the breadth = 24 cm – 16 cm = 8 cm

Area of this sheet = l × b

= 16 cm × 8 cm

= 128 cm^{2}

(iv) What will be the ratio of the areas of the two sheets?

Answer:

Area of the square sheet = 256 cm^{2}

Area of the rectangular sheet = 128 cm^{2}

Ratio between two area = 256 cm^{2} : 128 cm^{2} = 2:1

Question 2.

The perimeter of a square is 64 cm. Find its area.

Answer:

Perimeter of a square = 64 cm

its side = \(\frac{\text { Perimeter }}{4}=\frac{64}{4}\) = 16 cm

Area of this square = (Side)^{2} = (16)^{2}

= 256 cm^{2}

Hence the required area = 256 cm^{2}

Question 3.

The area of a rectangular field is 324 m^{2}. If its length is 27 m, find its (i) Breadth (ii) Perimeter.

Answer:

Area of the rectangular field = 324 m^{2}

Length of the field = 27 m

(i) ∴ Breadth of the field = \(\frac{\text { Area }}{\text { Length }}=\frac{324}{27}\)

= 12 m

(ii) Perimeter of the rectangular field = 2 [l + b]

= 2[27 + 12] m

= 2 × 39 m

= 78 m

Question 4.

The length and breadth of a rectangular field are 22 m and 14 m respectively. Find the cost of fencing it at a rate of ? 15 per metre.

Answer:

Length of the field = 22 m

Breadth of the field = 14 m

Perimeter of the field = 2[l + b]

= 2 [22 + 14]

= 2 × 36

= 72 m

Cost of fencing the field = ₹ (15 × 72) = ₹ 1080

Hence the required cost = ₹ 1080

Question 5.

The cost of fencing of a rectangular field at the rate of ₹ 20 per metre is ₹ 1080. Find the perimeter of the field.

Answer:

₹ 20 is the cost of 1 metre fencing.

₹ 1080 will be the cost of \(\frac{1080}{20}\) meter of fencing = 54 m

Hence the perimeter of the field = 54 m.

Question 6.

Find the area of a square field whose perimeter is 96 m.

Answer:

Perimeter of the square field = 96 m

ite Side = \(\frac{\text { Perimeter }}{4}=\frac{96}{4}\) m = 24 m

∴ Area of this square field = (side)^{2} = (24)^{2} = 576 m^{2}

Hence the required area = 576 m^{2}