ML Aggarwal Solution Class 9 Chapter 16 Mensuration Test
Test
Question 1
(a) Calculate the area of the shaded region.
(b) If the sides of a square are lengthened by 3 cm, the area becomes 121 cm2. Find the perimeter of the original square.
$=\frac{1}{2} \times 9 \times 12 \mathrm{~cm}^{2}=54 \mathrm{~cm}^{2}$
(b) Let the side of original square= x cm
Then length of given square=(x+3) cm
Area=side×side
⇒121=(x+3)(x+3)
⇒$(11)^{2}=(x+3)^{2}$
⇒11=x+3
⇒x+3=11
⇒x=11-3 cm
⇒x=8 cm
Question P.Q.
Length of each diagonal of the square ABCD
=32 cm
i.e. AC=BD=32 cm
Base EF of isosceles ΔBEF=8 cm
and each side=6 cm
Draw BL⟂EF
Area of square $=\frac{(\text { Diagonal })^{2}}{2}$
$=\frac{(32)^{2}}{2}=\frac{1024}{2} \mathrm{~cm}^{2}=512 \mathrm{~cm}^{2}$
In ΔBEL, ∠L=90°
$\mathrm{BL}^{2}=\mathrm{BE}^{2}-\mathrm{EL}^{2}$
(Pythagoras Theorem)
$=(6)^{2}-(4)^{2}$
=36-16=20
$\therefore \mathrm{BL}=\sqrt{20}=\sqrt{4 \times 5}=2 \sqrt{5} \mathrm{~cm}$
$\therefore$ Area of isosceles triangle $=\frac{1}{2}$ Base $\times$ alt.
$=\frac{1}{2} \times 8 \times 2 \sqrt{5} \mathrm{~cm}^{2}$
$=8 \sqrt{5} \mathrm{~cm}^{2}$
$\therefore$ Total area of the kite $=(512+8 \sqrt{5}) \mathrm{cm}^{2}$
$=512+8(2.236) \mathrm{cm}^{2}$
$=512+17.89=529.89 \mathrm{~cm}^{2}$
Question 2
(a) Find the area enclosed by the figure (i) given below. All measurements are in centimetres:
(b) Find the area of the quadrilateral ABCD shown in figure (ii) given below. All measurements are in centimetres.
(c) Calculate the area of the shaded region shown in figure (iii) given below. All measurements are in metres.
$\Rightarrow \mathrm{BD}^{2}=36+64 \Rightarrow \mathrm{BD}^{2}=100 \Rightarrow \mathrm{BD}=10 \mathrm{~cm}$
In ΔBCD
By Pythagoras theorem
$\mathrm{BC}^{2}=\mathrm{BD}^{2}+\mathrm{CD}^{2}$
$\Rightarrow(26)^{2}=(10)^{2}+(\mathrm{CD})^{2}$
$\Rightarrow 676=100+\mathrm{CD}^{2}$
$\Rightarrow \mathrm{CD}^{2}+100=676 \Rightarrow \mathrm{CD}^{2}=676-100$
$\Rightarrow \mathrm{CD}^{2}=576 \Rightarrow \mathrm{CD}=\sqrt{576 \mathrm{~cm}} \Rightarrow \mathrm{CD}=24 \mathrm{~cm}$
Area of given fig. Area of ΔABD+Area of ΔBCD
$=\frac{1}{2} \times$ Base $\times$ height $+\frac{1}{2} \times$ Base $\times$ height
$=\frac{1}{2} \times \mathrm{AB} \times \mathrm{AD}+\frac{1}{2} \times \mathrm{CD} \times \mathrm{BD}$
$=\left[\frac{1}{2} \times 6 \times 8+\frac{1}{2} \times 24 \times 10\right] \mathrm{cm}^{2}=[3 \times 8+12 \times 10] \mathrm{cm}^{2}$
$=(24+120) \mathrm{cm}^{2}=144 \mathrm{~cm}^{2}$
(c) Area of figure (iii) =Area of ABCD-(Area of 1st part + Area of 2nd part + Area of 3rd part)
$=(\mathrm{AB} \times \mathrm{BC})-\left[\left(\frac{1}{2} \times\right.\right.$ Base $\times$ height $)$ $+\left(\frac{1}{2} \times\right.$ base $\times$ height $)+\frac{1}{2}$ (sum of $\|$ side $\times$ height $]$
$=(12 \times 12) \mathrm{m}^{2}-\left[\frac{1}{2} \times 5 \times 5+\frac{1}{2} \times 5 \times 7+\frac{1}{2}(7+3) \times 12\right] \mathrm{m}^{2}$
$=144 \mathrm{~m}^{2}-\left[\frac{25}{2}+\frac{35}{2}+10 \times 6\right] \mathrm{m}^{2}$
$=144 \mathrm{~m}^{2}-\left(\frac{60}{2}+60\right) \mathrm{m}^{2}=144 \mathrm{~m}^{2}-(30+60) \mathrm{m}^{2}$
$=144 \mathrm{~m}^{2}-90 \mathrm{~m}^{2}=54 \mathrm{~m}^{2}$
Hence, required area of given figure $=54 \mathrm{~m}^{2}$
Question 3
Asifa cut an aeroplane from a coloured chart paper (as shown in the adjoining figure). Find the total area of the chart paper used, correct to 1 decimal place.
$s=\frac{5+5+1}{2}=\frac{11}{2}$
∴area $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{\frac{11}{2}\left(\frac{11}{2}-5\right)\left(\frac{11}{2}-5\right)\left(\frac{11}{2}-1\right)}$
$=\sqrt{\frac{11}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{9}{2}}=\sqrt{\frac{11 \times 9}{16}}=\frac{3}{4} \sqrt{11} \mathrm{~cm}^{2}$
$=\frac{3}{4} \times 3.316=3 \times 0.829=2.487$
$=2.48 \mathrm{~cm}^{2}$
Area of rectangle II (MCFH)=l×b
$=6.5 \times 1=6.5 \mathrm{~cm}^{2}$
area of $\Delta \mathrm{III}+\mathrm{IV}=2 \times \frac{1}{2} \times 6 \times 1.5$
$=9.0 \mathrm{~cm}^{2}$
area of 3 equilateral $\Delta \mathrm{s}$ formed trapezium III
$=3 \times \frac{\sqrt{3}}{4} \times(1)^{2} \mathrm{~cm}^{2}$
$=\frac{3}{4} \times 1.732 \mathrm{~cm}^{2}$
$=3 \times 0.433=1.299=1.3 \mathrm{~cm}^{2}$
$\therefore$ Total area $=2.48+6.50+9.00+1.30 \mathrm{~cm}^{2}$
$=19.28 \mathrm{~cm}^{2}=19.3 \mathrm{~cm}^{2}$
Question 4
If the area of a circle is 78.5 cm2, find its circumference. (Take π = 3.14)
Let r be the radius
$\therefore r^{2}=\frac{\text { Area }}{\pi}=\frac{78.50}{3.14}=25=(5)^{2}$
$\therefore r=5 \mathrm{~cm}$
Now circumference =2𝛑r
$=2 \times 3.14 \times 5 \mathrm{~cm}=31.4 \mathrm{~cm}$
Question 5
From a square cardboard, a circle of biggest area was cut out. If the area of the circle is 154 cm2, calculate the original area of the cardboard.
Let r be the radius
$\therefore \pi r^{2}=154$
$\Rightarrow \frac{22}{7} r^{2}=154$
$\Rightarrow r^{2}=\frac{154 \times 7}{22}=49=(7)^{2}$
$\Rightarrow r=7 \mathrm{~cm}$
Now side of the square$=7 \times 2=14 \mathrm{~cm}$
$\therefore$ Area of the original cardboard
$=a^{2}=(14)^{2}=196 \mathrm{~cm}^{2}$
Question 6
(a) From a sheet of paper of dimensions = 2m x 1.5m, how many circles can you cut of radius 5cm. Also find the area of the paper wasted. Take π = 3.14.
(b) If the diameter of a semicircular protractor is 14cm, then find its perimeter.
Radius of circle $=5 \mathrm{~cm}$
$\therefore$ No. of circles in lengthwise
$=\frac{200}{5 \times 2}=20$
and widthwise $=\frac{150}{10}=15$
$\therefore$ No. of circles $=20 \times 15=300$ -
Area of one circle $=\pi r^{2}$
$=3 \cdot 14 \times 5 \times 5 \mathrm{~cm}^{2}$
Area of 300 circles
$=300 \times \frac{314}{100} \times 25 \mathrm{~cm}^{2}=23550 \mathrm{~cm}^{2}$
$\therefore$ Area of the remaining portion
= Area of square - area of 300 circles
$=(30000-23550) \mathrm{cm}^{2}$
$=6450 \mathrm{~cm}^{2}$
(b) Diameter of semicircular protractor =14 cm
$\therefore$ Its perimeter $=\frac{1}{2} \pi d+d$
$=\frac{1}{2} \times \frac{22}{7} \times 14+14=22+14=36 \mathrm{~cm}$
Question 7
A road 3.5 m wide surrounds a circular park whose circumference is 88 m. Find the cost of paving the road at the rate of Rs. 60 per square metre.
$\Rightarrow 2 \times \frac{22}{7} \quad r=88$
$\Rightarrow r=\frac{88 \times 7}{2 \times 22}=14 \mathrm{~m}$
Outer radius (R) =14+3.5=17.5m
Now area of the path
$=\frac{22}{7} \times(17.5+14)(17.5-14)$
$=\pi\left(\mathrm{R}^{2}-r^{2}\right)=\frac{22}{7}\left[(17.5)^{2}-(14)^{2}\right] \mathrm{m}^{2}$
$=\frac{22}{7}(17.5+14)(17.5-14)$
$=\frac{22}{7} \times 31.5 \times 3.5 \mathrm{~m}^{2}=346.5 \mathrm{~m}^{2}$
Rate of paving the road $=\mathrm{Rs} .60$ per $\mathrm{m}^{2}$
$\therefore$ Total cost $=\mathrm{Rs} .60 \times 346.5$
=20790
Question 8
The adjoining sketch shows a running tract 3.5 m wide all around which consists of two straight paths and two semicircular rings. Find the area of the track.
Width of track $=3.5 \mathrm{~m}$
Inner length of rectangular base $=140 \mathrm{~m}$ and width $=42 \mathrm{~m}$
Outer length $=140+2 \times 3.5=140+7$
$=147 \mathrm{~m}$
and width $=42+2 \times 3.5=42+7=49 \mathrm{~m}$
Radius of inner semicircle $(r)=\frac{42}{2}=2 \operatorname{lm}$
and outer radius $(R)=21+3.5=24.5 \mathrm{~m}$
Now area of track $=2(140 \times 3.5)+2 \times \frac{1}{2} \pi$
$\left(\mathrm{R}^{2}-r^{2}\right)$
$=2(490)+\frac{22}{7}\left[(24.5)^{2}-(21)^{2}\right]$
$=980+\frac{22}{7}(24.5+21)(24.5-21)$
$=980+\frac{22}{7} \times 45.5 \times 3.5 \mathrm{~m}^{2}$
$=980+500.5=1480.5 \mathrm{~m}^{2}$
Question 9
In the adjoining figure, O is the centre of a circular arc and AOB is a line segment.Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)
$\therefore \Delta \mathrm{ABC}$ is a right angled triangle
Now $A B^{2}=A C^{2}+B C^{2}$
(Pythagoras Theorem)
$=12^{2}+16^{2}$
$=144+256=400$
$=(20)^{2}$
$\therefore A B=20 \mathrm{~cm}$
$\therefore$ Radius of semicircle $=\frac{20}{2}$
=10 cm
(i) Area of shaded portion
= Area of semicircle $-$ area of $\Delta \mathrm{ABC}$
$=\frac{1}{2} \pi r^{2}-\frac{A C \times B C}{2}$
$=\frac{1}{2} \times 3.142(10)^{2}-\frac{12 \times 16}{2}$
$=\frac{314.2}{2}-96=157.1-96.0$
$=61.1 \mathrm{~cm}^{2}$
(ii) Perimeter of shaded portion
=circumference of semicircle+AC+BC
$=\pi r+12+16=3.142 \times 10+28$
=31.42+28=59.42 cm
=59.4 cm
Question 10
(a) In the figure (1) given below, the radius is 3.5 cm. Find the perimeter of the quarter of the circle.
(b) In the figure (ii) given below, there are five squares each of side 2 cm.
(i) Find the radius of the circle.
(ii) Find the area of the shaded region. (Take π= 3.14).
(a) Radius of quadrant=3.5 cm
$\therefore$ Perimeter $=2 r+\frac{1}{4} \times 2 \pi r$
$=2 r+\frac{1}{2} \times \pi r=2 \times 3 \cdot 5+\frac{1}{2} \times \frac{22}{7} \times 3 \cdot 5 \mathrm{~cm}$
=7+5.5=12.5 cm
(b) In the figure,
OB=2+1=3 cm
AB=1 cm
$\therefore \mathrm{OA}=\sqrt{\mathrm{OB}^{2}+\mathrm{AB}^{2}}$
$=\sqrt{(3)^{2}+(1)^{2}}=\sqrt{9+1}=\sqrt{10}$
$\therefore$ Radius of the circle $=\sqrt{10} \mathrm{~cm}$
$\therefore$ Area of the circle $=\pi r^{2}$
$=3.14 \times(\sqrt{10})^{2} \mathrm{~cm}^{2}$
$=3.14 \times 10 \mathrm{~cm}^{2}=31.4 \mathrm{~cm}^{2}$
Area of 5 square of side $2 \mathrm{~cm}$ each $=(2)^{2} \times 5=4 \times 5=20 \mathrm{~cm}^{2}$
$\therefore$ Area of shaded portion $=31.4-20$ $=11.4 \mathrm{~cm}^{2}$
Question 11
(a) In the figure (i) given below, a piece of cardboard in the shape of a quadrant of a circle of radius 7 cm is bounded by perpendicular radii OX and OY. Points A and B lie on OX and OY respectively such that OA = 3 cm and OB = 4 cm. The triangular part OAB is removed. Calculate the area and the perimeter of the remaining piece.
Now (i) Area of shaded portion
$=\frac{1}{4} \pi r^{2}-\frac{1}{2} \mathrm{OA} \times \mathrm{OB}$
$=\frac{1}{4} \times \frac{22}{7} \times(7)^{2}-\frac{1}{2} \times 3 \times 4$
$=\frac{1}{4} \times \frac{22}{7} \times 49-6 \mathrm{~cm}^{2}$
$=\frac{77}{2}-6=\frac{65}{2} \mathrm{~cm}^{2}=32 \cdot 5 \mathrm{~cm}^{2}$
(ii) Perimeter of shaded portion
(b) In the figure (ii) given below , ABCD is a square . Points A,B,C and D are centres of quadrants of circles of the same radius . If the area of the shaded portion is $21\frac{3}{7}$ cm2, find the radius of the quadrants. Take π = $\frac{22}{7}$
Sol :
$=\frac{1}{4}(2 \pi r)+\mathrm{AX}+\mathrm{BY}+\mathrm{AB}$
$=\frac{1}{2} \times \frac{22}{7} \times 7+4+3+5$
=11+12=23 cm
(b) ABCD is a square and with centres A, B , C and D quadrants are drawn
Let side of square=a
$\therefore$ Radius of each quadrant $=\frac{a}{2}$
$\therefore$ Area of shaded portion
$=a^{2}-4 \times\left[\frac{1}{4} \pi\left(\frac{a}{2}\right)^{2}\right]$
$=a^{2}-\frac{11 a^{2}}{14}=\frac{3 a^{2}}{14}$
But area of shaded portion
$=21 \frac{3}{7}=\frac{150}{7} \mathrm{~cm}^{2}$
$\therefore \frac{3}{14} a^{2}=\frac{150}{7} \Rightarrow a^{2}=\frac{150}{7} \times \frac{14}{3}$
$\Rightarrow \quad a^{2}=100=(10)^{2}$
$\therefore \quad a=10$
$\therefore$ Radius of each quadrant
$=\frac{a}{2}=\frac{10}{2}=5 \mathrm{~cm}$
Question 12
In the adjoining figure, ABC is a right angled triangle right angled at B. Semicircle are drawn on AB, BC and CA as diameter. Show that the sum of areas of semi circles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.
ΔABC is a right angled triangle right angled at B
$\therefore \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$...(i)
(Pythagoras theorem)
Now area of semicircle on AC as diameter
$=\frac{1}{2} \pi\left(\frac{\mathrm{AC}}{2}\right)^{2}$
$=\frac{1}{2} \pi \times \frac{\mathrm{AC}^{2}}{4}=\frac{\pi \mathrm{AC}^{2}}{8}$
Area of semicircle on AB as diameter
$=\frac{1}{2} \pi\left(-\frac{\mathrm{AB}}{2}\right)^{2}=\frac{1}{2} \pi \frac{\mathrm{AB}^{2}}{4}$
$=\frac{\pi \mathrm{AB}^{2}}{8}$
and area of semicircle on BC as diameter
$=\frac{1}{2} \pi\left(\frac{\mathrm{BC}}{2}\right)^{2}=\frac{1}{2} \pi \frac{\mathrm{BC}^{2}}{4}$
$=\frac{\pi \mathrm{BC}^{2}}{8}$
$\therefore \frac{\pi \mathrm{AB}^{2}}{8}+\frac{\pi \mathrm{BC}^{2}}{8}=\frac{\pi}{8}\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}\right)$
$=\frac{\pi}{8}\left(\mathrm{AC}^{2}\right)$ [From (i)]
$=\frac{\pi \mathrm{AC}^{2}}{8}$
Hence proved
Question 13
The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 15 minutes.
$=\pi r^{2} \times \frac{15}{60}=\frac{22}{7} \times 14 \times 14 \times \frac{1}{4} \mathrm{~cm}^{2}$
$=154 \mathrm{~cm}^{2}$
Question 14
Find the radius of a circle if a 90° arc has a length of 3.5 n cm. Hence, find the area of sector formed by this arc.
$\therefore$ Radius of the arc $=\frac{3 \cdot 5 \pi}{2 \pi} \times \frac{360}{90}$
$=\frac{3 \cdot 5 \times 4}{2}=7 \mathrm{~cm}$
and area of the sector $=\pi r^{2} \times \frac{90^{\circ}}{360^{\circ}}$
$=\frac{22}{7} \times 7 \times 7 \times \frac{1}{4} \mathrm{~cm}^{2}=\frac{77}{2}=38 \cdot 5 \mathrm{~cm}^{2}$
Question 15
A cube whose each edge is 28 cm long has a circle of maximum radius on each of its face painted red. Find the total area of the unpainted surface of the cube.
$=6 \times 28 \times 28=4704 \mathrm{~cm}^{2}$
Now diameter of each circle $=28 \mathrm{~cm}$
Radius $=\frac{28}{2}=14 \mathrm{~cm}$
$\therefore$ Area of each circle
$=\pi r^{2}=\frac{22}{7} \times 14 \times 14 \mathrm{~cm}^{2}=616 \mathrm{~cm}^{2}$
and area of such 6 circles drawn on 6 faces of cube
$=616 \times 6=3696 \mathrm{~cm}^{2}$
$\therefore$ Area of remaining portion of the cube
$=4704-3696=1008 \mathrm{~cm}^{2}$
Question 16
Can a pole 6.5 m long fit into the body of a truck with internal dimensions of 3.5m, 3 m and 4m?
Question 17
$=(40 \times 28 \times 25) \mathrm{cm}^{3}=\frac{40 \times 28 \times 25}{1000}$ litre
$\left(\because 1000 \mathrm{~cm}^{3}=1\right.$ litre $)$
Average of car=13.5 km per litre
Then, distance traveled by car
$=\frac{40 \times 28 \times 25}{1000} \times 13.5 \mathrm{~km}$
$=\frac{(40 \times 25) \times 28}{1000} \times \frac{135}{10} \mathrm{~km}$
$=\frac{1 \times 28}{1} \times \frac{135}{10} \mathrm{~km}=\frac{14 \times 135}{5} \mathrm{~km}$
$=14 \times 27 \mathrm{~km}=378 \mathrm{~km}$
Hence, The car can travel 378 km with a full tank of petrol
Question 18
An aquarium took 96 minutes to completely fill with water. Water was filling the aquarium at a rate of 25 litres every 2 minutes. Given that the aquarium was 2 m long and 80 cm wide, compute the height of the aquarium.
Water fill in 96 minutes $=\frac{25}{2} \times 96$ litres
$=25 \times 48$ litres $=1200$ litres
i.e. Capacity of aquarium=1200 litres...(1)
But, Length of aquarium $=2 \mathrm{~m}=2 \times 100 \mathrm{~cm}$
=200 cm
Breadth of aquarium=80 cm
Let height of aquarium $=h \mathrm{~cm}$
Then, capacity of aquarium $=200 \times 80 \times h \mathrm{~cm}^{3}$
$=\frac{200 \times 80 \times h}{1000}$ litre $=\frac{1}{5} \times 80 \times h$ litre
=16 h litre ....(2)
From (1) and (2)
$16 h=1200 \Rightarrow h=\frac{1200}{16} \mathrm{~cm} \Rightarrow h=75 cm$
Hence , height of aquarium=75 cm
Question 19
The lateral surface area of a cuboid is 224 cm2. Its height is 7 cm and the base is a square. Find :
(i) a side of the square, and
(ii) the volume of the cuboid.
Lateral Surface Area $=2(l+b) \times h$
$\Rightarrow \quad 224=2(x+x) \times 7$
$\Rightarrow \quad 224=2 \times 2 x \times 7 \Rightarrow 224=28 x$
$\Rightarrow \quad 28 x=224 \Rightarrow x=\frac{224}{28} \mathrm{~cm}=8 \mathrm{~cm}$
(i) Hence, side of the square=8 cm
(ii) Volume of the cuboid$=\ell \times b \times h$
$=8 \times 8 \times 7 \mathrm{~cm}^{3}=448 \mathrm{~cm}^{3}$
Question 20
If the volume of a cube is V m3, its surface area is S m2 and the length of a diagonal is d metres, prove that 6√3 V = S d.
Diagonal $(d)=\sqrt{3} \cdot a$
Now $\mathrm{Sd}=6 a^{2} \times \sqrt{3} a=6 \sqrt{3} a^{3}$
$=6 \sqrt{3} \mathrm{~V} \quad\left(\because \mathrm{V}=a^{3}\right)$
Hence $6 \sqrt{3} \mathrm{~V}=\mathrm{Sd}$
Question 21
The adjoining figure shows a victory stand, each face is rectangular. All measurement are in centimetres. Find its volume and surface area (the bottom of the stand is open).
In the figure, it has three parts as indicated by 3,1 and 2 .
$\therefore$ Volume of part $(3)=50 \times 40 \times 12 \mathrm{~cm}^{3}$
$=24000 \mathrm{~cm}^{3}$
Volume of part $(1)=50 \times 40$ $\times(16+24) \mathrm{cm}^{3}$
$=50 \times 40 \times 40 \mathrm{~cm}^{3}=80000 \mathrm{~cm}^{3}$
and volume of part $(2)=50 \times 40 \times 24 \mathrm{~cm}^{3}$
$=48000 \mathrm{~cm}^{3}$
$\therefore$ Total volume $=(24000+8000+48000)$ $\mathrm{cm}^{3}=153000 \mathrm{~cm}^{3}$
Now total surface area = Area of front and back+area of vertical faces + area of top faces
$=2(50 \times 12+50 \times 40+50 \times 24) \mathrm{cm}^{2}$ $+(12 \times 40+28 \times 40+16 \times 40+24 \times 40) \mathrm{cm}^{2}$ $+3(50 \times 40) \mathrm{cm}^{2}$
$=2(600+2000+1200) \mathrm{cm}^{2}+(480+$ $1120+640+960) \mathrm{cm}^{2}+3 \times 2000 \mathrm{~cm}^{2}$
$=2(3800)+3200+6000 \mathrm{~cm}^{2}$
$=7600+3200+6000=16800 \mathrm{~cm}^{2}$
Question 22
The external dimensions of an open rectangular wooden box are 98 cm by 84 cm by 77 cm. If the wood is 2 cm thick all around, find :
(i) the capacity of the box
(ii) the volume of the wood used in making the box, and
(iii) the weight of the box in kilograms correct to one decimal place, given that 1 cm3 of wood weighs 0.8 gm.
Given that external dimensions of open rectangular wooden box $=98 \mathrm{~cm}, 84 \mathrm{~cm}$, and $77 \mathrm{~cm}$
Thickness = 2 cm
Then internal dimensions of open rectangular wooden box $(98-2 \times 2) \mathrm{cm},(84-2 \times 2) \mathrm{cm}$ and
(77-2) cm
=(98-4) cm, (84-4) cm , 75 cm=94 cm, 80 cm , 75 cm
(i) Capacity of the box $=94 \mathrm{~cm} \times 80 \mathrm{~cm} \times 75 \mathrm{~cm}$
$=564000 \mathrm{~cm}^{3}$
(ii) Internal volume of box $=564000 \mathrm{~cm}^{3}$
External volume of $\mathrm{box}=98 \mathrm{~cm} \times 84 \mathrm{~cm} \times 77 \mathrm{~cm}$
$=633864 \mathrm{~cm}^{3}$
Volume of wood used in making the
Box $=633864 \mathrm{~cm}^{3}-564000 \mathrm{~cm}^{3}=69864 \mathrm{~cm}^{3}$
(iii) Weight of $1 \mathrm{~cm}^{3}$ wood $=0.8 \mathrm{gm}$ Weight of $69864 \mathrm{~cm}^{3}$ wood $=0.8 \times 69864 \mathrm{gm}$
$=\frac{0.8 \times 69864}{1000} \mathrm{~kg}=\frac{55891.2}{1000} \mathrm{~kg}$
=55.9 kg (correct to one decimal)
Question 23
A cuboidal block of metal has dimensions 36 cm by 32 cm by 0.25 m. It is melted and recast into cubes with an edge of 4 cm.
(i) How many such cubes can be made ?
(ii) What is the cost of silver coating the surfaces of the cubes at the rate of Rs. 1.25 per square centimetre ?
Volume of cuboidal block $=36 \mathrm{~cm} \times 32 \mathrm{~cm} \times$ $0.25 \mathrm{~m}$
$=36 \mathrm{~cm} \times 32 \mathrm{~cm} \times(0.25 \times 100) \mathrm{cm}=(36 \times 32$ $\times 25) \mathrm{cm}^{3}$
Volume of cube having edge is $4 \mathrm{~cm}$
$=4 \mathrm{~cm} \times 4 \mathrm{~cm} \times 4 \mathrm{~cm}=64 \mathrm{~cm}^{3}$
(i) Number of cubes
$=\frac{\text { volume of cuboidal block }}{\text { volume of one cube }}$
$=\frac{36 \times 32 \times 25}{64}=\frac{36 \times 25}{2}=18 \times 25=450$
(ii) Total surface area of one cube
$=6(a)^{2}=6(4)^{2} \mathrm{~cm}^{2}=6 \times 4 \times 4 \mathrm{~cm}^{2}=96 \mathrm{~cm}^{2}$
Total surface area of 450 cube $=450 \times 96 \mathrm{~cm}^{2}=$ $43200 \mathrm{~cm}^{2}$
Cost of silver coating the surface for $1 \mathrm{~cm}^{2}$ $=$ Rs. 1.25
cost of silver coating the surface for 43200 $\mathrm{cm}^{2}$
$=43200 \times 1.25=\mathrm{Rs} 54000$
Question 24
Three cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and recast into a single cube. Find the cost of coating the surface of the new cube with gold at the rate of Rs. 3.50 per square centimetre.
$=(3 \mathrm{~cm})^{3}=3 \times 3 \times 3 \mathrm{~cm}^{3}=27 \mathrm{~cm}^{3}$
Volume of second cube $=(\text { edge })^{3}$
$=(4 \mathrm{~cm})^{3}=4 \times 4 \times 4 \mathrm{~cm}^{3}=64 \mathrm{~cm}^{3}$
Volume of third cube $=(\text { edge })^{3}$
$=(5 \mathrm{~cm})^{3}=5 \times 5 \times 5 \mathrm{~cm}^{3}=125 \mathrm{~cm}^{3}$
Total volume $=(27+64+125) \mathrm{cm}^{3}=216 \mathrm{~cm}^{3}$
Made new cube whose volume $=216 \mathrm{~cm}^{3}$
$(\text { edge })^{3}=216 \mathrm{~cm}^{3} \Rightarrow(\text { edge })^{3}=(6 \mathrm{~cm})^{3}$
$\Rightarrow \quad$ edge $=6 \mathrm{~cm}$
Surface area of new cube $=6$ (edge) $^{2}$ $=6(6)^{2} \mathrm{~cm}^{2}=6 \times 6 \times 6 \mathrm{~cm}^{2}=216 \mathrm{~cm}^{2}$
Cost of coating the surface for $1 \mathrm{~cm}^{2}$
$=\mathrm{Rs} .3 .50$
Cost of coating the surface for $216 \mathrm{~cm}^{2}$
= Rs. $3.50 \times 216=$ Rs. 756
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