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
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