Convert $65$ cm to $0.65$ m. Since the box is open at the top, required area $=lw+2lh+2wh=1.5\times1.25+2(1.5\times0.65)+2(1.25\times0.65)=5.375\text{ m}^2$. Cost $=5.375\times20=₹107.50$.
(i) $5.375\text{ m}^2$
(ii) $₹107.50$
Area of four walls $=2h(l+b)=2\times3(5+4)=54\text{ m}^2$. Area of ceiling $=5\times4=20\text{ m}^2$. Total area $=74\text{ m}^2$. Cost $=74\times7.50=₹555$.
$₹555$.
Cost $=₹15000$ at $₹10$ per $\text{m}^2$, so area of four walls $=1500\text{ m}^2$. If the floor perimeter is $250$ m, then $2(l+b)=250$. Wall area $=2h(l+b)=250h$. Hence $250h=1500$, so $h=6$ m.
$6$ m.
For one brick, total surface area $=2(lb+bh+hl)=2(22.5\times10+10\times7.5+22.5\times7.5)=937.5\text{ cm}^2=0.09375\text{ m}^2$. Number of bricks $=\dfrac{9.375}{0.09375}=100$.
$100$ bricks.
Cube LSA $=4a^2=4\times10^2=400\text{ cm}^2$. Cuboid LSA $=2h(l+b)=2\times8(12.5+10)=360\text{ cm}^2$. So the cube is greater by $40\text{ cm}^2$. Cube TSA $=6a^2=600\text{ cm}^2$. Cuboid TSA $=2(lb+bh+hl)=2(12.5\times10+10\times8+12.5\times8)=610\text{ cm}^2$. Thus the cube has the smaller TSA by $10\text{ cm}^2$.
(i) The cubical box has greater lateral surface area by $40\text{ cm}^2$.
(ii) The cubical box has smaller total surface area by $10\text{ cm}^2$.
The glass area is the total surface area of a cuboid: $2(lb+bh+hl)=2(30\times25+25\times25+30\times25)=4250\text{ cm}^2$. Tape is needed along all 12 edges, so total length $=4(l+b+h)=4(30+25+25)=320$ cm.
(i) $4250\text{ cm}^2$
(ii) $320$ cm
Big box TSA $=2(25\times20+20\times5+25\times5)=1450\text{ cm}^2$. With $5\%$ extra, area per big box $=1522.5\text{ cm}^2$. Small box TSA $=2(15\times12+12\times5+15\times5)=630\text{ cm}^2$. With $5\%$ extra, area per small box $=661.5\text{ cm}^2$. Total area for $250$ of each $=250(1522.5+661.5)=546000\text{ cm}^2$. Cost $=\dfrac{546000}{1000}\times4=₹2184$.
$₹2184$.
Tarpaulin is needed for four sides and the top, not the base. Area $=2lh+2bh+lb=2(4\times2.5)+2(3\times2.5)+4\times3=20+15+12=47\text{ m}^2$.
$47\text{ m}^2$.
CSA of a cylinder $=2\pi rh$. Thus $88=2\times\dfrac{22}{7}\times r\times14=88r$. Hence $r=1$ cm, so diameter $=2r=2$ cm.
$2$ cm.
Diameter $=140$ cm $=1.4$ m, so $r=0.7$ m and $h=1$ m. Sheet area for a closed cylinder is $2\pi r(h+r)=2\times\dfrac{22}{7}\times0.7(1+0.7)=7.48\text{ m}^2$.
$7.48\text{ m}^2$.
Inner radius $r=2$ cm, outer radius $R=2.2$ cm, and height $h=77$ cm. Inner CSA $=2\pi rh=2\times\dfrac{22}{7}\times2\times77=968\text{ cm}^2$. Outer CSA $=2\pi Rh=2\times\dfrac{22}{7}\times2.2\times77=1064.8\text{ cm}^2$. Area of the two ring-shaped ends $=2\pi(R^2-r^2)=2\times\dfrac{22}{7}(2.2^2-2^2)=5.28\text{ cm}^2$. Total surface area $=968+1064.8+5.28=2038.08\text{ cm}^2$.
(i) $968\text{ cm}^2$
(ii) $1064.8\text{ cm}^2$
(iii) $2038.08\text{ cm}^2$
In one revolution the roller covers its curved surface area: $\pi dh=\dfrac{22}{7}\times84\times120=31680\text{ cm}^2$. In $500$ revolutions, area $=31680\times500=15840000\text{ cm}^2=1584\text{ m}^2$.
$1584\text{ m}^2$.
Diameter $=50$ cm $=0.5$ m, so $r=0.25$ m and $h=3.5$ m. Curved surface area $=2\pi rh=2\times\dfrac{22}{7}\times0.25\times3.5=5.5\text{ m}^2$. Cost $=5.5\times12.50=₹68.75$.
$₹68.75$.
$2\pi rh=4.4$. With $r=0.7$ m, $2\times\dfrac{22}{7}\times0.7\times h=4.4h=4.4$. Hence $h=1$ m.
$1$ m.
Inner radius $r=\dfrac{3.5}{2}=1.75$ m and depth $h=10$ m. Inner curved surface area $=2\pi rh=2\times\dfrac{22}{7}\times1.75\times10=110\text{ m}^2$. Cost $=110\times40=₹4400$.
(i) $110\text{ m}^2$
(ii) $₹4400$
Diameter $=5$ cm $=0.05$ m, so $r=0.025$ m and $h=28$ m. Radiating surface is the curved surface area: $2\pi rh=2\times\dfrac{22}{7}\times0.025\times28=4.4\text{ m}^2$.
$4.4\text{ m}^2$.
Here $r=2.1$ m and $h=4.5$ m. Curved surface area $=2\pi rh=2\times\dfrac{22}{7}\times2.1\times4.5=59.4\text{ m}^2$. The tank is closed, so finished surface area $=2\pi r(h+r)=2\times\dfrac{22}{7}\times2.1(4.5+2.1)=87.12\text{ m}^2$. If $\dfrac1{12}$ of the steel used was wasted, the finished tank used $\dfrac{11}{12}$ of the steel. Thus steel actually used $=87.12\times\dfrac{12}{11}=95.04\text{ m}^2$.
(i) $59.4\text{ m}^2$
(ii) $95.04\text{ m}^2$ of steel
The cloth forms a rectangle whose breadth is the circumference of the base, $\pi d=\dfrac{22}{7}\times20=\dfrac{440}{7}$ cm. The effective height is $30+2.5+2.5=35$ cm. Required cloth area $=\dfrac{440}{7}\times35=2200\text{ cm}^2$.
$2200\text{ cm}^2$.
A penholder has one circular base and curved surface, so area per penholder $=2\pi rh+\pi r^2=2\times\dfrac{22}{7}\times3\times10.5+\dfrac{22}{7}\times3^2=198+\dfrac{198}{7}=\dfrac{1584}{7}\text{ cm}^2$. For $35$ competitors, total cardboard $=35\times\dfrac{1584}{7}=7920\text{ cm}^2$.
$7920\text{ cm}^2$.
Radius $r=\dfrac{10.5}{2}=5.25$ cm and slant height $l=10$ cm. CSA $=\pi rl=\dfrac{22}{7}\times5.25\times10=165\text{ cm}^2$.
$165\text{ cm}^2$.
Radius $r=12$ m and $l=21$ m. TSA $=\pi r(l+r)=\dfrac{22}{7}\times12\times(21+12)=\dfrac{8712}{7}\text{ m}^2$.
$\dfrac{8712}{7}\text{ m}^2$, approximately $1244.6\text{ m}^2$.
$\pi rl=308$. With $l=14$, $\dfrac{22}{7}\times r\times14=308$, so $44r=308$ and $r=7$ cm. TSA $=\pi r(l+r)=\dfrac{22}{7}\times7\times(14+7)=462\text{ cm}^2$.
(i) $7$ cm
(ii) $462\text{ cm}^2$
Slant height $l=\sqrt{24^2+10^2}=\sqrt{676}=26$ m. Canvas area is the curved surface area: $\pi rl=\dfrac{22}{7}\times24\times26=\dfrac{13728}{7}\text{ m}^2$. Cost $=\dfrac{13728}{7}\times70=₹1,37,280$.
(i) $26$ m
(ii) $₹1,37,280$
The slant height is $l=\sqrt{6^2+8^2}=10$ m. Curved surface area $=\pi rl=3.14\times6\times10=188.4\text{ m}^2$. For tarpaulin $3$ m wide, required length $=\dfrac{188.4}{3}=62.8$ m. Adding $20$ cm $=0.2$ m gives $63$ m.
$63$ m.
Radius $r=7$ m and $l=25$ m. Curved surface area $=\pi rl=\dfrac{22}{7}\times7\times25=550\text{ m}^2$. Cost at $₹210$ per $100\text{ m}^2$ is $\dfrac{550}{100}\times210=₹1155$.
$₹1155$.
Slant height $l=\sqrt{7^2+24^2}=25$ cm. Sheet for one cap is its curved surface area: $\pi rl=\dfrac{22}{7}\times7\times25=550\text{ cm}^2$. For $10$ caps, area $=5500\text{ cm}^2$.
$5500\text{ cm}^2$.
Radius $r=20$ cm $=0.2$ m and height $h=1$ m. Slant height $l=\sqrt{1^2+0.2^2}=\sqrt{1.04}=1.02$ m. CSA of one cone $=3.14\times0.2\times1.02=0.64056\text{ m}^2$. For $50$ cones, area $=32.028\text{ m}^2$. Cost $=32.028\times12=₹384.336$, approximately $₹384.34$.
Approximately $₹384.34$.
Surface area of a sphere is $4\pi r^2$. For $r=10.5$ cm, area $=4\times\dfrac{22}{7}\times10.5^2=1386\text{ cm}^2$. For $r=5.6$ cm, area $=4\times\dfrac{22}{7}\times5.6^2=394.24\text{ cm}^2$. For $r=14$ cm, area $=4\times\dfrac{22}{7}\times14^2=2464\text{ cm}^2$.
(i) $1386\text{ cm}^2$
(ii) $394.24\text{ cm}^2$
(iii) $2464\text{ cm}^2$
Use $r=\dfrac d2$ and surface area $=4\pi r^2$. For $d=14$ cm, $r=7$ cm and area $=616\text{ cm}^2$. For $d=21$ cm, $r=10.5$ cm and area $=1386\text{ cm}^2$. For $d=3.5$ m, $r=1.75$ m and area $=38.5\text{ m}^2$.
(i) $616\text{ cm}^2$
(ii) $1386\text{ cm}^2$
(iii) $38.5\text{ m}^2$
Total surface area of a hemisphere $=3\pi r^2=3\times3.14\times10^2=942\text{ cm}^2$.
$942\text{ cm}^2$.
Surface area of a sphere is proportional to $r^2$. Therefore the ratio is $7^2:14^2=49:196=1:4$.
$1:4$.
Inner radius $r=5.25$ cm. Inner curved surface area of the hemisphere $=2\pi r^2=2\times\dfrac{22}{7}\times5.25^2=173.25\text{ cm}^2$. Cost $=\dfrac{173.25}{100}\times16=₹27.72$.
$₹27.72$.
$4\pi r^2=154$. Using $\pi=\dfrac{22}{7}$, $r^2=\dfrac{154\times7}{88}=12.25$, so $r=3.5$ cm.
$3.5$ cm.
The ratio of diameters, and hence radii, is $1:4$. Surface areas of spheres are proportional to the squares of their radii. Therefore the ratio of surface areas is $1^2:4^2=1:16$.
$1:16$.
Outer radius $R=5+0.25=5.25$ cm. Outer curved surface area of the hemisphere $=2\pi R^2=2\times\dfrac{22}{7}\times5.25^2=173.25\text{ cm}^2$.
$173.25\text{ cm}^2$.
The enclosing cylinder has radius $r$ and height $2r$. Surface area of the sphere $=4\pi r^2$. Curved surface area of the cylinder $=2\pi rh=2\pi r(2r)=4\pi r^2$. Therefore the ratio is $4\pi r^2:4\pi r^2=1:1$.
(i) $4\pi r^2$
(ii) $4\pi r^2$
(iii) $1:1$