SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14B

 Exercise 14 B

Question 1 

Find the area of the shaded portion in each case. 

(i) (Image to be added)

Sol: Area of shaded portion = Area of outer square -Area of Inner square 

$=8^{2}-4^{2}$

$=64-16$

$=48 m^{2}$

(ii)  (Image to be added) 

Sol:  : Area of shaded portion = Area of outer square -Area of Inner square

$=(24 \times 20)-(22 \times 18)$

$=480-396$

$=84 \mathrm{~m}^{2}$

Question 2

A photograph of sides 35 cm by 22 cm is mounted into a frame of external dimension 45 cm by 30 cm . find the area of the border surrounding the photograph . 

Sol: Area of border = External area of frame - Area of photograph 

$=(45 \times 30)-(35 \times 22)$

$=1350-770$

$=580 \mathrm{~cm}^{2}$

Question 3

A Verandah 1.25 m wide is constructed all along the outside of  a room 5.5m long and 4m wide. find the cost of cementing the floor of this verandah at the rate of Rs 15 per sqm.

Sol: (IMAGE TO BE ADDED)

To find the cost cementing the floor of verandah we have to find the area. 

Outer width = $\begin{aligned} & 5.5+1.25+1.21 \\=& 8 m \end{aligned}$

Outer length = $4+1.25+1.25=6.5 \mathrm{~m}$ 

Area of verandah = Area of outer - Area of inner 

$=(8 \times 6.5)-(5.5 \times 4)$

$=52-22$

$=30 \mathrm{~m}^{2}$

Cost of cementing floor = $15 \times 30$

= Rs 450 Answer 

Question 4

A sheet of paper measuring 30 cm by 20 cm. A strip 4cm wide is cut from it all around . Find the area of remaining sheet and also the area of the strip cut out . 

Sol: Width after cutting 4cm= 30-4+4

$=30-8=22 \mathrm{~cm}$

Length = 20 - 4+4= $20-8=12 \mathrm{~cm}$

Area of remaining sheet =  L$\times$ B

$=22 \times 12=264 \mathrm{~cm}^{2}$

Area of strip = $(30 \times 20)-(264)$

$600-264=336 \mathrm{~cm}^{2}$

Question 5

Find the area of the crossroads at right angles to each other through the center of the field . 

Sol: (IMAGE TO BE ADDED)

Area of cross roads 

=Area of ABCD+ EFGH -IJKL 

$75 \times 2+62 \times 2-2 \times 2$

$=150+124-4$

$274-4=270 \mathrm{~m}^{2}$ Answer