Olympiad Mathematics - Perimeter and Area
Exam Duration: 45 Mins Total Questions : 30
The circumference of a circle is 44 m. What is its area?
- (a)
6084.5 m2
- (b)
276.5 m2
- (c)
154 m2
- (d)
44 m2
2\(\pi\)r =44 ⇒ r=\(\frac { 44 }{ 2\times \frac { 22 }{ 7 } } \)=7m
Area of a circle =\(\pi\)r2
=\(\frac { 22 }{ 7 } \) \(\times\)7\(\times\)7 =154 m2
The circular grass lawn of radius 35 m has a path of width 7 m around it on the outside.What is the area of the path?
- (a)
1496 m2
- (b)
1450 m2
- (c)
1576 m2
- (d)
1694 m2
r= 35 m; R = 35 + 7 = 42 m
Area of circular path
=\(\pi\)(R+r)(R-r)
= (42 + 35)(42 - 35) = 1694 m2
The difference between the circumference and radius of a circle is 37 m.What is the circumference of the circle
- (a)
7 m
- (b)
44 m
- (c)
154 m
- (d)
88 m
2\(\pi\)r-r = 37
r =\(\frac { 37 }{ 2\pi -1 } =\frac { 3 }{ 2\times \frac { 22 }{ 7 } -1 } \) =7m
Circumference
= 2\(\pi\)r-r =\(2\times \frac { 22 }{ 7 } \times 7\)= 44 m
How many plants (approximately) will be there in a circular bed whose outer edge measures 30 cm allowing 4 cm2 for each plant?
- (a)
18
- (b)
750
- (c)
24
- (d)
120
Circumference = 30 cm
Area =\(\frac { { C }^{ 2 } }{ 4\pi } =\frac { 30\times 30 }{ 4\times \frac { 22 }{ 7 } } \) =71.6 cm2
Number of plants =\(\frac { A }{ 4 } =\frac { 71.6 }{ 4 } \)
= 17.9 ≅ 18
If the ratio of circumferences of two circles is 4:9, what is the ratio of their areas?
- (a)
9:4
- (b)
16:81
- (c)
4:9
- (d)
2:3
The length and breadth of a rectangular hall in a model are 0.4 m and 30 cm respectively. What is the distance between the opposite corners of the wall in the model?
- (a)
34.16 m
- (b)
50 m
- (c)
34.16 m
- (d)
50 cm
Distance between two opposite corners = \(\sqrt { { l }^{ 2 }+b^{ 2 } } \)
=\(\sqrt { { 40 }^{ 2 }+30^{ 2 } } \)= 50 cm
The longer side of a parallelogram is 81 cm and the corresponding altitude is 16 cm. If the length of shorter side is 24 cm, what is the altitude corresponding to the shorter side?
- (a)
36 cm
- (b)
48 cm
- (c)
54 cm
- (d)
24 cm
One side of a parallelogram is 2.4 dam and its area is 576 m2. Find the corresponding altitude.
- (a)
24 m
- (b)
24 m
- (c)
24 dm
- (d)
24 dam
2.4 dam = 2.4\(\times\)10 = 24 m
(Since 10 m = 1 dam)
We have, area = 576 m2
⇒ 24\(\times\)altitude = 576
∴ altitude =\(\frac { 576 }{ 24 } \)=24 m
Two sides of a right triangle containing the right angle are 100 cm and 8.6 cm. Find its area.
- (a)
430 sq. cm
- (b)
43 sq. cm
- (c)
430 sq.m
- (d)
430 cm
If ABCD is a rectangle having length 30 cm and breadth 20 cm, E, F and G are midpoints of AB, CD and AD respectively, find the area of the unshaded part.
- (a)
400 cm2
- (b)
450 cm2
- (c)
24 cm
- (d)
26 cm
Area of rectangle = l\(\times\)b = 30\(\times\)20
= 600 cm2
Area of ∆DGF =\(\frac { 1 }{ 2 } \)\(\times\)b\(\times\)h
=75 cm2
Similarly area of ∆AGE = 75 cm2
Area of unshaded region
= 600 - (75 + 75) cm2
= 450 cm2
ABCD is a rectangle having length 30 cm and breadth 25 cm. P.Q. R and S are midpoints of AB,BC,
- (a)
375 m2
- (b)
375 cm2
- (c)
475 m2
- (d)
425 cm2
Consider PR = 25 cm as the base of two triangles, \(\triangle \) PQR and \(\triangle \)PSR.
Area of \(\triangle \)PQR= \(\frac { 1 }{ 2 } \times \)25\(\times\)15
=\(\frac { 375 }{ 2 } \) cm2
Similarly area of
\(\triangle \)PSR= \(\frac { 1 }{ 2 } \times \)25\(\times\)15
=\(\frac { 375 }{ 2 } \) cm2
Area of shaded region = Area of rectangle - (area of \(\triangle \)PQR + area of \(\triangle \)PSR)
= (25\(\times\)15)-\(\left( \frac { 375 }{ 2 } +\frac { 375 }{ 2 } \right) \)
= 750-375 cm2 =375 cm2
The playground of school is as shown.What is the perimeter of the playground?
- (a)
420 m
- (b)
200 m
- (c)
220 m
- (d)
840 m
The measurements in the given figure are in cm
- (a)
200 cm2
- (b)
186 cm2
- (c)
210 cm2
- (d)
196 cm2
Clearly, the figure is a rectangle
∴ 2(x + 3) = x + 8
⇒ 2x+6=x+8 ⇒ x=2cm
∴ Area = (x + 8) (6x + 9)
= (2 + 8) (6\(\times\)2 + 9)
=10\(\times\)21 = 210 cm2
The diameter of a wheel of a cycle is 70 cm.lt moves slowly along a road. What distance will it cover in 24 complete revolutions?
- (a)
5820 cm
- (b)
5280 cm
- (c)
5028 cm
- (d)
5082 cm
Radius of a wheel =\(\frac { 70 }{ 2 } \)= 35 cm
In one revolution, the wheel covers a distance equal to its circumference
∴ \(2\pi { r }=2\times \frac { 22 }{ 7 } \times 35\) = 220 cm
In 24 complete revolutions, distance covered
= 24\(\times\)220 = 5280 cm
The length around a rectangular mat is 21 m. If its length is 5 m 60 cm, what is its width?
- (a)
4 m 90 cm
- (b)
490 m
- (c)
49 m
- (d)
49 cm
Perimeter = 21 m = 2100 cm
Length = 5 m 60 cm = 560 cm
Width = \(\frac { P }{ 2 } -l\)
\(=\left( \frac { 2100 }{ 2 } -560 \right) cm\)
= 490 cm = 4 m 90 cm
The floor of a room measures 12 m by 10 m. A carpet is placed on the floor from wall to wall.What isthe area of the carpet?
- (a)
120 m2
- (b)
1200 m
- (c)
120 m
- (d)
1200 m2
Length of the floor = 12 m
Breadth = 10 m
\(\therefore\) Area = 12 \(\times\) 10 = 120 m2
\(\therefore\) Area ofthe carpet = 120 m2
Find the unshaded area of the given figure.
- (a)
111 cm2
- (b)
100 cm2
- (c)
183 cm2
- (d)
294 cm2
We have, AB = 24.5 cm, BC = 12 cm
\(\therefore\) Area of rectangle ABCD = 24.5 \(\times\) 12 = 294 cm2
Now, GH = [24.5 - (3 + 3)] cm
= (24.5-6)cm= 18.5cm
GF= [12-(3 + 3)] cm = (12-6)cm = 6 cm
\(\therefore\) Area of rectangle EFGH
= 18.5 \(\times\) 6 = 111 cm2
So, unshaded area = (294 - 111) cm2= 183 cm2
In the given figure (not drawn to scale), B, D, F and H are the midpoints of AC, CE, EG and GA respectively. Find the area of the shaded portion, if AGEC is a square of area 100 cm2.
- (a)
25 cm2
- (b)
50 cm2
- (c)
35 cm2
- (d)
75 cm2
Area of square AGEC = 100 cm2
Area of shaded portion
\(=\frac { 1 }{ 4 } \times (Area\quad of\quad square\quad AGEC)\)
\(=\frac { 1 }{ 4 } \times (100){ cm }^{ 2 }=25{ cm }^{ 2 }\)
The rectangle is made of 12 identical squares. It is divided into 4 parts. Which of the following 2 parts will be removed to form \(\frac { 3 }{ 12 } \) of the rectangle?
- (a)
L and N
- (b)
M and O
- (c)
L and M
- (d)
M and N
Total number of squares = 12
Required number of squares to form \(\frac { 3 }{ 12 } \) of rectangle \(=\frac { 3 }{ 12 } of\quad 12=\frac { 3 }{ 12 } \times 12=3\)
L and O have 3 squares together.
So, parts m and N must be removed to form \(\frac { 3 }{ 12 } \) of the rectangle.
Find the area of the given figure (not drawn to scale).
- (a)
200 cm2
- (b)
94 cm2
- (c)
84 cm2
- (d)
100 cm2
Area of square ABCD = (6)2 = 36 cm2
Area of rectangle CEFG = EF \(\times\) EC
= (4 \(\times\)10) cm2 = 40 cm2
Area of rectangle FHIJ = JI \(\times\) HI
= 6 \(\times\) 2 = 12 cm2
Area of rectangle GKLM = GM \(\times\) LM = (6 \(\times\) 2) cm2
= 12 cm2
\(\therefore\) Required area = (36 + 40 + 12 + 12)cm2 = 100 cm2
In the given figure, all triangles are equilateral and PQ = 12 units. Other triangles have been formed by taking the mid points of the sides. What is the perimeter of the figure?
- (a)
62.3 units
- (b)
64.5 units
- (c)
65.8 units
- (d)
67.5 units
perimeter of the given figure
=(9\(\times\)1.5 + 6 \(\times\)3 + 6 \(\times\) 6) units
= (13.5 + 18 + 36) units = 67.5 units
The area of two circles are in the ratio 25 : 36. Then the ratio of their circumference is ___________
- (a)
6: 5
- (b)
3: 4
- (c)
4: 3
- (d)
5: 6
Ratio of areas = \(\frac { 25 }{ 36 } \Rightarrow \frac { { \pi r }_{ 1 }^{ 2 } }{ \pi { r }_{ 2 }^{ 2 } } =\frac { 25 }{ 36 } \)
[where r1 and r2 are the radii of two circles]
\(\Rightarrow \quad \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 5 }{ 6 } \)
So, ratio of circumference of the circles \(=\frac { 2\pi { r }_{ 1 } }{ 2\pi { r }_{ 2 } } =\frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 5 }{ 6 } i.e.,\quad 5:6\)
The lengths of the sides of a triangle are 5 m, 1.2 decametre and 130 dm. Then its area is ______________
- (a)
24 m2
- (b)
30 m2
- (c)
48 m2
- (d)
40 m2
Let BC = 1.2 decameter = 12 m
AC = 130 dm = \(\frac { 130 }{ 10 } m\)= 13 m
and AB = 5 m
Now, it is clear that,
AB2 + BC2 = AC2
So, it is a right angled triangled.
\(\therefore Area=\frac { 1 }{ 2 } \times BC\times AB=\left( \frac { 1 }{ 2 } \times 12\times 5 \right) { m }^{ 2 }=30{ m }^{ 2 }\)
A rectangular field is 112 m long and 100 m wide. Two roads, each 3 m wide are to be constructed at the centre of this field; one is parallel to the length of the field and the other is parallel to the breadth of the field.
Find:
(i) Area of the part of the field, not covered by the roads.
(ii) Cost of constructing the road at Rs 17 per m2
- (a)
(i) (ii) 3248 m2 Rs 2890 - (b)
(i) (ii) 10573 m2 Rs 10659 - (c)
(i) (ii) 16582 m2 Rs 20480 - (d)
(i) (ii) 10573 m2 Rs 1250
(i) Area of path BSTV = (100 \(\times\) 3) m2 = 300 m2
Area of path MNDP = (112 \(\times\) 3) m2 = 336 m2
\(\therefore\) Total area of two paths
= (300 + 336) - Area of square HGFE
= 636 - 3 \(\times\) 3 = 627 m2
\(\therefore\) Area of field not covered by roads
Total area of field - Area of two paths
(112 \(\times\) 100) - 627 = 10573 m2
(ii) Cost of constructing 1 m2 area = Rs 17
\(\therefore\)Cost of constructing the roads
= Rs (17 \(\times\) 627) = Rs 10659
In the given diagram, PQU is an equilateral triangle, QRVU is a square and RSTU is a rhombus. Find the perimeter, (in cm) of the whole diagram.
- (a)
17
- (b)
21
- (c)
26
- (d)
27
Since PQU is an equilateral triangle.
\(\therefore\) PQ = PU = QU = 4 cm
Also, QUVR is a square.
\(\therefore\) QU = UV = VR = QR = 4 cm
Also, RUTS is a rhombus.
So, UT = TS = SR = RU = 5 cm
Now, required perimeter
= PU + UT + TS + SR + OR + PO
= (4 + 5 + 5 + 5 + 4 + 4) cm = 27 cm
A field is 225 m long and 175 m wide. It has two roads in its centre of uniform width of 5 m, one parallel to its length and the other parallel to its breadth. Find the cost of levelling the roads at Rs 3 per square metre.
- (a)
Rs 1975
- (b)
Rs 1125
- (c)
Rs 5925
- (d)
Rs 6125
Area of roads
=[(175 \(\times\)5) + (225 \(\times\)5) - (5)2]m2
=(875 + 1125 - 25)m2 = 1975 m2
\(\therefore\) Cost of levelling the roads = Rs (1975 \(\times\) 3) = Rs 5925
A rectangular piece of canvas measures 25 cm by 16 cm. A triangular piece, with base 14 cm and height 10 cm is cut off from the canvas. Find the area of the remaining piece.
- (a)
400 cm2
- (b)
120 cm2
- (c)
140 cm2
- (d)
330 cm2
Let ABCE be the rectanglar piece of canvas.
\(\Delta\)LBM is the triangular piece to be cut off.
Required area = area of rectangle ABCE - area of \(\Delta\)LBM
\(=\left[ (25\times 16)-\left( \frac { 1 }{ 2 } \times 14\times 10 \right) \right] { cm }^{ 2 }\)
= ( 400 - 70) cm2 = 330 cm2
If the radius of a circle is tripled, the area becomes
- (a)
9 times
- (b)
3 times
- (c)
6 times
- (d)
30 times
Let the radius of original circle be r.
\(\therefore\) Area of the circle = \(\pi\)r2
Now, radius becomes tripled i.e., 3r
Thus, area of the new circle = \(\pi\)(3r)2
= 9\(\pi\)r2 = 9 \(\times\) (original area)
State 'T' for true and 'F' for false.
(P) Length of ribbon required to cover the semicircular disc of radius 10 cm is 51.4 cm.
(Q) Ratio of circumference of a circle to its radius is always 2\(\pi\) : 1.
(R) 500 m2 = 5 hectares
(S) If 1 m2 = \(\times\) mm2 , then the value \(\times\) of is 100000.
- (a)
P Q R S T F F F - (b)
P Q R S F F T F - (c)
P Q R S T F F T - (d)
P Q R S T T F F
(P) We have, Radius of disc = 10 cm
\(\therefore\) Length of ribbon required = \(\left( \frac { 1 }{ 2 } \times 2\pi r+2r \right) \)
\(=\left( \frac { 22 }{ 7 } \times 10+2\times 10 \right) cm=51.42\quad cm\)
(Q) \(\frac { Circumference\quad of\quad circle }{ Radius\quad of\quad circle } =\frac { 2\pi r }{ r } =\frac { 2\pi }{ 1 } \)
(R) 1 hectare = 10000 m2
\(\therefore \quad 1{ m }^{ 2 }=\frac { 1 }{ 10000 } hectare\)
\(\Rightarrow 500{ m }^{ 2 }=\frac { 500 }{ 10000 } hectare=\frac { 1 }{ 20 } hectare\)
(S) 1 m2 = (1000\(\times\)1000) mm2 = \(\times\)mm2
\(\therefore\) x = 1000000
Which of the following statements is true?
Statement-1: A saree 5.5 m long and 1.25 m wide has a zari border 2.5 cm broad along its length on either side. Along the width, on one edge the border was 5 cm wide and on the other side the zari was 25 cm wide. The area of the zari is 5855 cm2.
Statement-2: The moon is about 384000 km from earth and its path around the earth is nearly circular. The length of path described by moon in one complete revolution will be 2,411,520 km. (\(\pi\) = 3.14)
- (a)
Only Statement-1
- (b)
Only Statement-2
- (c)
Both Statement-1 and Statement-2
- (d)
Neither Statement-1 nor Statement-2
Statement-1:
Length of saree = 5.5 m = (5.5 \(\times\) 100) cm = 550 cm
Width of saree = 1.25 m = (1.25 \(\times\) 100) cm
= 125 cm
Shaded region shows zari border.
Area of zari along the lengths of saree
= (2 \(\times\) 2.5 \(\times\) 550) cm2 = 2750 cm2
Area of zari along the width of saree
= (120 \(\times\) 5 + 120 \(\times\) 25) cm2
= 600 + 3000 = 3600 cm2
Hence, total area covered with zari
= (2750 + 3600) cm2 = 6350 cm2
Statement-2: Distance of moon from earth = 384000 km
Length of path is equal to the circumference of the circular path.
So, 2\(\pi\)r= 2 \(\times\) 3.14 \(\times\) 384000 km = 2,411,520 km
Hence, the length of path described by moon in one complete revolution is 2,411,520 km.