Exercise 16.1
Question 1
Given
(Diagram to be added)
ABCD is a square of side 24cm
AE = 15cm
(i) Perimeter of AEFD $=2(A E+A D)$
$=2(15+24)$
$=2(39)$
Perimeter of AEFD $=78 \mathrm{Cm}$
Perimeter of EBCF = $=2(E B+B C)$
$\begin{aligned} A B &=A E+E B \\ 24 &=15+E B \\ E B &=24-15 \\ E B &=9 C m \end{aligned}$
Perimeter of EBCF = $2(9+24)$
$2(33)$
Perimeter of EBCF = 66cm
Difference in perimeter = 78- 77= 12cm
∴ Perimeter of AEFD exceed the perimeter of EBCF by 12 cm
(ii) Area of AEFD(rectangle) = l$\times$ b
$=A E \times A D$
$=15 \times 24$
Area of AEFD $=360 \mathrm{c} \mathrm{m}^{2}$
Question 2
Dimensions of rectangular park ($l \times b $) = $180 \mathrm{~m} \times 120 \mathrm{~m}$
Perimeter of parks = $2(l+h)$
$=2(180+120)$
$=2(300)$
Perimeter of park = 600m
Distance covered by negma five rounds around park $=5 \times 660$
$=3000 \mathrm{~m}$
Speed of Negma = $7-5 \mathrm{~km} / \mathrm{hour}$
$\frac{7500}{3600} \mathrm{~m} / \mathrm{sec}$
Time $=\frac{\text { Distance }}{\text { speed }}$
$\frac{3000}{7500} \times 3600$
Time = 1440 sec = 24 min
Question 3
Area of rectangular plot = 540$m^{2}$
Length of rectangular plot (L) = 27m
Area of rectangle = $l \times b$
$540=27 \times b$
Breadth = $\frac{540}{27}$
Breadth = 20m
Perimeter of rectangular plot = $2(l+b$
$=2(27+20)$
$=2(47)$
Perimeter of rectangular plot = 94m
Question 4
Perimeter of rectangular plot = 151m
Breadth (b) = 32m
Length (l) = ?
Perimeter = 2(l+b)
151= 2(L+ 32)
L+ 32= 75.5
L = 75.5- 32
l = 43.5m
Area of rectangular plot = $\ell \times b$
$=43.5 \times 32$
Area of rectangular plot = 1522.5$m^{2}$
Area of EBCF (Rectangle)= l $\times$ b
$=E B \times B C$
$=9 \times 24$
Area of EBCF = $216 \mathrm{~cm}^{2}$
Difference in Area = 360- 216
$=144 \mathrm{~cm}^{2}$
Area of AEFD exceeds the area of EBCF by $=144 \mathrm{~cm}^{2}$
Question 5
Area of rectangular plot = $340 \mathrm{~m}^{2}$
Breadth (b) = 17m
Area = l $\times$ b
340= $l\times17$
$l=\frac{340}{17}$
$l=20 \mathrm{~m}$
Perimeter of rectangular plot = $2(+h)$
$=2(20+17)$
$=2(37)$
Perimeter of rectangular plot = 74m
Cost of fencing = Rs 5.70 meter
Total cost of fence around the plot = $74 \times 57$
= Rs $421.80$
Question 6
Let breadth of park = b
Length of park = l = 90m
Side of square park(s) = 60m
Area of square park = Area of rectangular park
$\begin{aligned} s^{2} &=l \times b \\ 60^{2} &=90 \times b \\ 3600 &=90 \times b \\ b &=\frac{3600}{90} \\ b &=40 \mathrm{~m} \end{aligned}$
∴ Breadth of rectangular park = 40m
Question 7
When wire is in the shape of rectangular
Length (l) = 40cm
Breadth (b) = 22cm
Perimeter (p)= $2(l+b)$
$=2(40+22)$
$=2(62)$
Perimeter (p)= 124cm
$\begin{aligned} \operatorname{Area}(A) &=\operatorname{l\times} b \\ &=40 \times 22 \end{aligned}$
Area (A) = 880$\mathrm{cm}^{2}$
When wire in the shape of square
Perimeter of square= Perimeter of rectangle
4s = p
4s = 124
$S=\frac{124}{4}$
$S=31 \mathrm{~cm}$
∴ Side of square = 31cm
Area of square = $s^{2}$
$=31^{2}$
Area of square = 961
Area of square > Area of rectangle
Square occupies more area than rectangle By (961-880) = $81 \mathrm{~cm}^{2}$
Question 8
Dimensions of wall ($l \times h$)= $4.5 \mathrm{~m} \times 3.6 \mathrm{~m}$
Dimensions of door $(l \times h)$ = $1 m \times 2 m$
Area of wall $\left(A_{1}\right)$ =$l\times h$
$=4.5 \times 3.6$
Area of wall $\left(A_{1}\right)=16.2 \mathrm{~m}^{2}$
$\begin{aligned} \text { Areaot door }\left(A_{2}\right)=& b \times h \\=& 1 \times 2 \\ \text { Areaot door }\left(A_{2}\right)=& 2 m^{2} \end{aligned}$
Area for white washing = $A_{1}-A_{2}$
= 16.2-2
Area for white washing = $14.2 \mathrm{~m}^{2}$
Cost for white washing = Rs 20 $m^{2}$
∴Cost of white washing = $14.2 \times 20$
=Rs 284
Question 9
(Diagram to be added)
Inner rectangle dimension
$(l \times b)=45 \times 30 \mathrm{~m}^{2}$
Outer rectangle dimensions
$\begin{aligned}(l \times b) &=45+2 \times 2.5 \times 30+2 \times 2.5 \\ &=45+5 \times 30+5 \\ &=50 \times 35 \mathrm{~m}^{\mathrm{2}} \end{aligned}$
Path area = Outer rectangle area - Inner rectangle area
=$50 \times 35-(45 \times 30)$
=$1750-1350$
Path area =$400 \mathrm{~m}^{2}$
Question 10
(Diagram to be added)
Outer Carpet size $(l\times b)=5 \mathrm{~m} \times 2 \mathrm{~m}$
Inner part of carpet size $=(5-2 \times 0.25) \times(2-2 \times 0.25)$
$=(5-0.5) \times(2-0.5)$
$=\quad 4.5 \times 1.5 \mathrm{~m}^{2}$
Red portion area = Outer part area - Inner part area
$5 \times 2-(4.5 \times 1.5)$
$10-6.75$
Red portion Area =$3.25 \mathrm{~m}^{2}$
Blue portion Area = $4.5 \times 1.5=6.75 \mathrm{~m}^{2}$
Ratio of Areas = $\frac{Red portion}{Blue portion area}$
$=\frac{3.25}{6.75}$
Ratio of Areas= $\frac{13}{27}$
Question 11
(Diagram to be added)
Width of verandah = 2.25m
Dimensions of Room =5.5mｘ4m
(i) Outside rectangle dimension
l = 5.5+ $2 \times 2.25=10 \mathrm{~m}$
b = $4+2 \times 2.25=8.5 \mathrm{~m}$
Area of verandah = Outside area - Inside area
= $10 \times 8.5-(5.5 \times 4)$
= 85-22
Area of Verandah = $63 m^{2}$
(ii) Cost of Cementing the floor of verandah = Rs $200 / \mathrm{ms}^{2}$
Total cost of cementing the floor of Verandah = $63 \times 200$
= Rs 12600
Question 12
(Diagram to be added)
Dimension of park = $70 \mathrm{~m} \times 45 \mathrm{~m}$
length path = l + b - w
=70+45-5
=70+40
$=110 \mathrm{~m}$
Area of path = Length of path ｘ Width of path
Area of path = $110 \times 5$
Area of path = $550 \mathrm{~m}^{2}$
Rate of Constructing the road = Rs 105
Total Cost of Constructing the road = $550 \times 105$
=₹ 57750
Question 13
Rectangle room dimension = $10 m \times 7.5 \mathrm{~m}$
Width of carpet = 1.25m
Area of room = $l \times b$
=$10 \times 1.5$
Area of Room = $75 \mathrm{~m}^{2}$
Length of carpet = $\frac{\text { Area of Room }}{\text { width of carpet }}$
=$\frac{75}{1.25}$
Length of carpet = 60m
Cost of carpet = Rs 250\m
Total cost of covering the floor with carpet = $=250 \times 60$
=Rs 30000
Question 14
Dimension of rectangular room = $6.5 \mathrm{~m} \times 5 \mathrm{~m}$
Dimension of square tile = $25 \mathrm{~cm} \times 25 \mathrm{~cm}$
$0.25 \mathrm{~m} \times 0.25 \mathrm{~m}$
$\begin{aligned} \text { Area of Room } &=l \times b \\ &=6.5 \times 5 \end{aligned}$
Area of Room $=532.5 \mathrm{~m}^{2}$
Area of tile $=S^{2}$
$=(0.25)^{2}$
Area of tile $=0.0625 \mathrm{~m}^{\mathrm{2}}$
No of tiles Required to cover the floor = $\frac{\text { Area of Room }}{\text { Area of tile }}$
=$\frac{32.5}{0.0625}$
No . of tiles = 520
Cost of one tile = Rs 9.40
Total cost of tiles = $9.4 \times 520$
Total cost of tiles = Rs 4888
Question 15
Side of square shape room = 4.8m
Perimeter of square tiles = 1.2m
Side of square tile = 4S = 1.2
$S=\frac{1.2}{4}$
Side of square tile = 0.3m
Area of square shape room = $(4.8)^{2}$
=2.3 .04$m^{2}$
No. of tiles required t cover the floor = $\frac{\text { Area of square shape room}}{\text { Area of tile }}$
$=\frac{23.04}{0.09}$
No. of tiles required = 256
Cost of one tile = Rs 27
Total Cost of tiles to cover the room = $27 \times 256$
=Rs 26912
Question 16
Width of rectangular plot land = 50m
Total cost of fencing = Rs 4680
Rate of cost of fencing = Rs 18\m
Total length of fencing = $\frac{\text { Total cost }}{\text { Rate of cost }}$
= $\frac{4686}{18}$
Total length of fencing = 260 m
Length of fencing = Perimeter of rectangle
$260=2(1+b)$
$260=2(1+50)$
$1+50=\frac{260}{2}$
$l+50=130$
$l=130-50$
$\ell=80 \mathrm{~m}$
Length of plot = 80m
(ii) Area of plot = $l \times b$
=$80 \times 50$
Area of plot = $4000 m^{2}$
Rate of cost for = Rs 7.6\$m^{2}$
Total cost of = $4000 \times 7.6=₹ 30400$
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