Olympiad Mathematics - Mensuration
Exam Duration: 45 Mins Total Questions : 25
Two rectangles ABCD and DBEF are as shown in the figure. The area of rectangle DBEF (in square units) ________
- (a)
10
- (b)
12
- (c)
14
- (d)
15
In ΔDAB,DB2 = (3)2 + (4)2 ⇒ DB = 5 units
Area of ΔCBD= \(1\over 2\) x base x height
\(⇒{1\over 2}\times BC \times CD={1\over 2}\times DB\times CG\)
\(⇒\ {1\over 2}\times3\times4={1\over 2}\times5\times ⇒CG={12\over 5}\)
Also, CG = BE = \(12\over 5\)
∴ Area of rectangle DBEF = DB x BE = 5 x \({12\over 5}=12\)
The ratio of areas of two squares, if diagonal of one is double the diagonal of the other, is _____
- (a)
3 : 2
- (b)
4 : 1
- (c)
3 : 1
- (d)
4 : 3
Let side of one square be x
∴ Area = x2.
Now, its diagonal = √2 x x
So, diagonal of other square 2 x √2x = 2√2x
Now, required side of other square = 2x
∴ Area = (2X)2 = 4x2
so required ratio \(={4x^2\over x^2}={4\over 1}\) i.e., 4 : 1
Ratio of surface areas of two cubes is 25 : 36 Find the ratio of their volumes
- (a)
5 : 6
- (b)
125 : 216
- (c)
1 : 2
- (d)
64 : 216
Let I1 and l2 be the length of sides of the cubes.
According to question, \({6l_1^2\over 6l_2^2}={25\over36}\Rightarrow{l_1\over l_2}={5\over6}\)
∴ Ratio of volume of two cubes
\(=\left(l_1\over l_2\right)^3=\left(5\over6\right)^3={125\over216}i.e,125:216\)
The area of a quadrilateral is 342 sq.m. The perpendiculars from two of its opposite vertices to the diagonal are 12 m and 12 m. What is the length of the diagonal?
- (a)
28.6 cm
- (b)
25.3 cm
- (c)
28.5 cm
- (d)
22.5 cm
Let ABCD be a quadrilateral.
DE, BF are perpendiculars drawn to AC.
We have DE = BF = 12 m
Area of quadrilateral ABCD
= Area of (ΔACD + ΔABC)
⇒ 342 = \({1\over2}\) x AC x BF + \({1\over2}\) x AC x DE
⇒ 342 = \({1\over2}\) x AC(BE+DE)= \({1\over2}\) x AC x 24
\(⇒\ AC={2\times342\over 24}=28.5cm\)
The radii of the bases of two cylinders are in the ratio of 1 : 2 respectively and their heights are in the ratio 2 : 1 respectively. Find the ratio of their volumes.
- (a)
1 : 2
- (b)
1 : 4
- (c)
2 : 1
- (d)
4 : 1
Given that, \({r_1\over r_2}={1\over2}\) and \({h_1\over h_2}={2\over 1}\)
\(∴\ {V_1\over V_2}={\pi r_1^2h_1\over \pi r_2^2 h_2}=\left(r_1\over r_2\right)^2\times \left(h_1\over h_2\right)=\left(1\over2\right)^2\times\left(2\over1\right)={1\over 4}\times{2\over1}={1\over 2}\)
⇒ V1 : V2 = 1 : 2
Find the sum of the lengths of the parallel sides of a trapezium whose altitude is 11 cm and whose area is 0.55 m2
- (a)
25m
- (b)
15m
- (c)
12m
- (d)
10m
Height = 11 cm = \({11\over 100}m\)
Area of trapezium = 0.55 m2
⇒ \(1\over2\) x Height x (sum of parallel sides) = 0.55
⇒ Sum of parallel sides = \({0.55\times2\times100\over11}=10m\)
In the given figure, RSTV is a square inscribed in a circle with centre O and radius r. The total area of the shaded region is ________
- (a)
r2 (π - 2)
- (b)
2r2 (2-π)
- (c)
π (r2 - 2)
- (d)
8r2 - 8r
Radius of circle = r
∴ Area = πr2
Now, diameter of circle is equal to diagonal of square RSTV So, length of diagonal = 2r
⇒ Side of the square \(={2r\over \sqrt2}=\sqrt2r\)
∴ Area of square = (side)2 = (√2r)2 = 2r2
⇒ Area of shaded part = πr2 - 2r2 = (π - 2)r2
ABC is a right angled triangle with ㄥBAC = 90°. AH is drawn perpendicular to BC. If AB = 60 cm and AC = 80 cm, then BH = ________
- (a)
36cm
- (b)
32 cm
- (c)
24 cm
- (d)
30 cm
In ΔABC, BC2 = AB2 + AC2
= (60)2 + (80)2 = 3600 + 5400 = 10000
\(∴\ BC=\sqrt{10000}=100cm\)
Area of ΔABC = \(1\over2\) x base x height
⇒ \(1\over2\) x AB x AC = \(1\over2\)x BC x AH
⇒ \(1\over2\) x 60 x 80 = \(1\over2\)x 100 x AH ⇒ \(AH={60\times80\over100}=48cm\)
Now, in ΔABH, BH2 = AB2 - AH2 = (60)2 - (48)2 = 1296
\(∴\ BH=\sqrt{1296}=36cm\)
Three cubes whose edges are 3 cm, 4 cm and 5 cm respectively are melted without any loss of metal into a single cube. The edge of the new cube is ______
- (a)
6 cm
- (b)
12 cm
- (c)
9cm
- (d)
10cm
Let V1, V2 and V3 be the volumes of the three cubes.
Then V1 = (3)3 = 27 cm3, V2 = (4)3 = 64 cm3 and V3 = (5)3 = 125 cm3
∴ Total volume = 27 + 64 + 125 = 216 cm3
∴ Its edge = \(\sqrt[3]{216}=6cm\)
The volume of a cuboid is 440cm3 and the area of its base is 88 cm2. Find its height
- (a)
5 cm
- (b)
11 cm
- (c)
4 cm
- (d)
8 cm
Volume of cuboid = 440 cm3
Area of its base i.e., I x b = 88 cm2
As, I x b x h = 440
⇒ 88 x h = 440 ⇒ h = \({440\over88}=5cm\)
A hoop is resting vertically at staircase as shown in the diagram. AB = 12 cm and Be = 8 cm. The radius of the hoop is _________
- (a)
13cm
- (b)
12√2 cm
- (c)
14cm
- (d)
13√2 cm
ΔOEC is a right angle triangle, right angled at E.
Let OA = OC = x cm (radius)
∴ AE = 8 cm
Now in ΔOEC, OC2 = OE2 + EC2
⇒ x2 = (x - 8)2 + 122
⇒ x2 = x2 + 64 - 16x + 144
⇒ 16x = 208 ⇒ x = 13 cm
The capacity of a closed cylindrical water tank is 9.24 kilolitres. If the height of the cylinder is 1.5 m, what is its radius?
- (a)
1.4 m
- (b)
14 m
- (c)
7m
- (d)
0.7 m
Volume of cylindrical tank = 9.24 KL
= 9240 L = \(\left(9240\over1000\right)m^3= 9.24 m^3\)
Height of tank (h) = 1.5 m
Let r be the radius. Then, nr<h = 9.24
⇒ \(22\over 7\) x r2 x 1.5 = 9.24 ⇒ r2 = 1.96 ⇒ r = 1.4 m
The thickness of a hollow metallic cylinder is 2cm. It is 70 cm long with outer radius of 14 cm. Find the volume of the metal used in making the cylinder, assuming that it is open at both the ends. Also find its weight if the metal weighs 8 g per cm3
- (a)
10440 cm3 91250 g
- (b)
13440 cm3 90000 g
- (c)
13440 cm3 90000 g
- (d)
11440 cm3, 91520 g
- (e)
12440 cm3, 91550 g
Thickness of the hollow metallic cylinder = 2 cm
Height of the cylinder = 70 cm
Let Rand r be the outer and inner radii of the cylinder
∴ R = 14 cm and r = 14 - 2 = 12 cm
Volume of the metal used = πh(R2 - r2)
= \(22\over7\) x 70(142 -122 )= 220 x 52 = 11440 cm3
If the metal weighs 8 g per cm3, so 11440 cm3 metal weigh
= 11440 x 8 = 91520 g
A wooden box (including the lid) has external dimensions 40 cm by 34 cm by 30 cm. If the wood is 1 cm thick, then how many cm3 of wood is used in it?
- (a)
6752 cm2
- (b)
6750 cm3
- (c)
5752 cm3
- (d)
5750 cm3
Thickness of wood = 1 cm
Internal length of wooden box = 40 - (1 + 1) = 38 cm
Internal breadth of the box = 34 - (1 + 1) = 32 cm
Internal height of wooden box = 30 - (1 + 1) = 28 cm
∴ Internal volume of box = (38 x 32 x 28) cm3 = 34,048 cm2
External volume of box = (40 x 34 x 30) cm3 = 40,800 cm3
∴ Volume of wood used
= External volume - Internal volume
= 40,800 - 34,048 = 6752 cm2
How many bricks of size 22 cm x 10 cm x 7 cm are required to construct a wall 11m long, 3.5 m high and 40 cm thick, if the cement and sand used in the construction occupy (1/10)th part of the wall?
- (a)
8000
- (b)
9000
- (c)
7000
- (d)
10000
Volume of a brick = (22 x 10 x 7) cm3 = 1540 cm3
Now, length of the wall = 11 m = 1100 cm
Height of the wall = 3.5 m = 350 cm
Width of the wall = 40 cm
∴ Volume of the wail = (1100 x 350 x 40) cm3 = 15400000 cm3
It is given that cement and sand occupy \(1\over10\)th part of 15400000, i.e., 1540000 cm3
Volume of the wall occupied by the bricks
= 15400000 - 1540000 = 13860000
∴ Number of bricks
\(={volume\ of\ wall\ in\ which\ bricks\ are\ occupied\over volume\ of\ a\ brick}\)
\(={13860000\over1540}=9000\)
Hence, 9000 bricks are required to construct the wall.
A rectangular block of ice measures 40cm by 25 cm by 15 cm. Calculate its weight (in kg), if ice weighs \(9\over 10\) of the weight of the same volume of water and 1 cm3 of water weighs 1 gm
- (a)
9
- (b)
13.5
- (c)
8
- (d)
9.5
Volume of rectangular block of ice
= (40 x 25 x 15) cm3 = 15000 cm3
Now, weight of 1 cm3 of water = 1 gm
Weight of 1 cm3 of ice = \(\left(9\over10\right)^{th}\) of weight of 1 cm3 of water
∴ Weight of 1 cm3 of ice =\(\left(9\over10\right)gm\)
∴ Weight of rectangular block of ice
\(=\left({9\over 10}\times15000\right)gm=13.5kg\)
A cylindrical tower is 5 metres in diameter and 14 metres high. The cost of white washing its curved surface at 50 paise per m2 is ______
- (a)
Rs.90
- (b)
Rs.97
- (c)
Rs.110
- (d)
Rs.95
Radius of tower \(={5\over 2}m\)
Height of tower = 14 m
∴ Curved surface area = \(=2\pi rh={2\times{22\over7}\times{5\over2}\times14}=220m^2\)
∴ Cost of white washing the curved surface area
\(=220\times{50\over 100}=Rs.110\)
A square garden measuring 8 m on a side is surrounded by a 1 m wide path. What is the area of the path?
- (a)
8m2
- (b)
9m2
- (c)
28m2
- (d)
36m2
Side of garden = 8 m
∴ Its area = (side)2 = (8)2 = 64 m2
Now, side of square garden including path
= 8 + 1 + 1 = 10 m
∴ Its area = (10)2 = 100 m2
So, area of path = 100 - 64 = 36 m2
A well 12 m deep with a diameter 3.5 m is dug up and earth from it is evenly spread to form a platform 10.5 m long and 8.8 m wide. Find the height of the platform
- (a)
2.5 m
- (b)
12.5 m
- (c)
1.25 m
- (d)
1.5 m
Let r and h be the radius and depth of well
Volume of earth dug out = πr2h
\(={22\over7}\times{3.5\over2}\times{3.5\over2}\times12=115.5m^2\)
Let x be the height of platform
Now, volume of platform = volume of earth dug out
\(⇒\ 10.5\times8.8\times x = 115.5 ⇒ x ={115.5\over10.5\times8.8}=1.25m\)
A room is 9 m long, 6 m wide and 4 m high. Find the cost of plastering its walls and ceiling at the rate of Rs.2.50 per square metre
- (a)
Rs.435
- (b)
Rs.600
- (c)
Rs.502
- (d)
Rs.534
Surface area of walls and ceiling
= 2h(l + b) + Ib = 2 x 4(9 + 6) + 9 x 6
= 120 + 54 = 174m2
Cost of plastering walls and ceiling
= Rs.(174 x 2.5) = Rs. 435
A solid iron rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.
- (a)
112.5 m
- (b)
112 m
- (c)
212 m
- (d)
312 m
Volume of block = (4.4 x 2.6 x 1)
= (440 x 260 x 100) cm2
Let r = internal radius of pipe = 30 cm
and R = external radius of pipe = (30 + 5 ) cm = 35 cm
∴ Volume of iron used = nh (R2 - r2)
= πh (352 - 302) = πh (325)
Volume of iron pipe = Volume of block
⇒ 325πh = 440 x 260 x 100
\(⇒\ h = 440 \times 260 \times 100 \times{7\over22}\times{1\over325}\)
⇒ h = 11200 cm = 112 m
Find the area of the given field. All dimensions are in m.
- (a)
9000 m2
- (b)
11700 m2
- (c)
11000 m2
- (d)
12000 m2
Area of given field
Area (ΔEHF) + Area ( FHJG) + Area(ΔAJG) + Area (ΔABK) + Area ( KBCI) + Area (☐ICDE)
\(={1\over2}\times 40 \times 20+{1\over2}\times20+40)\times80+{1\over2}\times40\times80+{1\over2}\times30\times60+{1\over2}(50+30)\times60+80\times50\)
= 400 + 2400 + 1600 + 900 + 2400 + 4000 = 11700 m2
Fill in the blanks.
(i) The perimeter of a rectangle becomes ______ times its original perimeter, if its length and breadth are doubled.
(ii) The curved surface area of a cylinder is reduced by _____ percent if the height is half of the original height.
(iii)Opposite faces of a cuboid are ______ in area.
(iv) If the diagonal d of a quadrilateral is doubled and the heights h1 and h2 falling on d are halved, then the new area of quadrilateral is __________
- (a)
(i) (ii) (iii) (iv) 2 50 equal \({1\over2}\)(h1+h2)d - (b)
(i) (ii) (iii) (iv) 2 35 unequal \({1\over2}\)(h1+h2)d - (c)
(i) (ii) (iii) (iv) 3 40 equal (h1+h2)d - (d)
(i) (ii) (iii) (iv) 3 35 unequal (h1+h2)d
Match the figure in column I with their total surface area in column II
- (a)
P ⟶ (iii); Q ⟶ (ii); R ⟶ (iv); S ⟶ (i)
- (b)
P ⟶ (ii); Q ⟶ (iv); R ⟶ (i); S ⟶ (iii)
- (c)
P ⟶ (ii); Q ⟶ (iii); R ⟶ (i); S ⟶ (iv)
- (d)
P ⟶ (ii); Q ⟶ (iv); R ⟶ (iii); S ⟶ (i)
P. Total surface area = 2(lb + bh + hl)
= 2(10 x 5 + 5 x 8 + 8 x 10)
= 2(50 + 40 + 80) = 2 x 170 = 340 cm2
Q. Total surface area = 2πr (r + h)
\(=2\times{22\over 7}\times{21\over 2}\left({21\over 2}+9\right)=1287cm^2\)
R. We have, n« = 154
\(⇒{22\over 7}\times r^2=154⇒ r^2=49⇒r=7cm\)
∴ Surface area = \(2\pi r(r+h)=2\times{22\over7}\times7(7+12)=836cm^2\)
S. Let I be the edge of cube.
⇒ l3 = 2197 ⇒ l= 13 cm
Total surface Area = 612= 6 (13)2 = 1014 cm2
The area of a trapezium with equal nonparallel sides is 168 m2. If the lengths of the parallel sides are 36 m and 20 m, find the length of the non-parallel sides
- (a)
8m
- (b)
10m
- (c)
15 m
- (d)
4 m
Let the length of non-parallel sides AD and BC be x m. Here
AB II CD and AB = 20 m, CD = 36 m
Area of trapezium ABCD = 168 m2
Let the distance between two parallel sides be h.
\(∴\ 168={1\over2}\times(Ab+CD)\times h\)
\(168={1\over2}\times56\times h⇒ h=6m\)
Now draw BE II AD
Given, AB II CD ⇒ AB II DE
∴ ABED is a parallelogram.
⇒ BE = AD = x and AB = DE = 20 m
So, EC = 36 - 20 = 16 m
In ΔBEC, we have BE = BC = x m
⇒ ΔBEC is an isosceles triangle.
\(∴\ Em=MC={1\over2}EC={1\over2}\times165=8m\)
By Pythagoras theorem in ΔBEM,
BE2 = BM2 + EM2
⇒ x2= 62 + 82 = 100
⇒ x = 10 m
Hence, the length of non-parallel side is 10m.