Perimeter And Area - 2D Test

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Perimeter And Area - 2D Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

The figure shows the length and breadth of a rectangle in inches.

How much is the shaded area in square inches?

A
18

B
24

C
57

D
78

Solution

Area of shaded region = Area of rectangle – (Area of region A + Area of region B)
Area of rectangle = 8¢¢ 12¢¢ = 96'' square inches
Area of region A = 1/2 × 8¢¢ × 9¢¢ = 36 square inches
Area of region B = 3¢¢ 2¢¢ = 3'' square inches
Area of shaded region = (96'' – (36'' + 3'')) square inches = 57'' square inches
= 57 sq. inches
Option 3 is correct.
Question 2
1 / -0

What is the area of the shaded portion in the given figure?

A
196 cm²

B
224 cm²

C
240 cm²

D
254 cm²

Solution

Area of quadrant AEH = = Area of AEH = = cm²
Area of shaded region EH = Area of quadrant AEH - Area of AEH

=

=

= = Area of the required shaded region = 4 Area of shaded region EH

= 4 cm²

= 224 cm²
Question 3
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A punching machine is used to punch a circular hole of diameter two units from a square sheet of aluminium of width 2 units, as shown below. The hole is punched such that the circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in line with a diagonal of the square.

What is the proportion of the sheet area left after punching?

A

B

C

D

Solution
Question 4
1 / -0

One side of a rhombus is 10 cm and one of its diagonals is 12 cm. The area of the rhombus is

A
120 sq. cm

B
60 sq. cm

C
96 sq. cm

D
108 sq. cm

Solution

AD = 12 cm
AC = 10 cm
In AOC,
AC² = OC² + OA²
AC =
10 cm =
BC = 16 cm
Area of rhombus =
Hence, option (3) is correct.
Question 5
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In the diagram, ABC is equilateral, BC = 2CD, AF = 6, and DEF is perpendicular to AB. What is the area of quadrilateral FBCE?

A
144units

B
138 units

C
126units

D
108units

Solution

Let CD = x, BD = 2x + x = 3x
Then, BC = 2x
BF = 2x - 6
In BFD,
BD : BF : FD = 2 : 1 :

2(2x-6) = 3x ; 4x-12= 3x ; 4x-3x=12
x = 12
BC = 2x = 2 12 = = 24
BC = AB = AC = 24 (sides)
In AFE,
AF : AE : FE : 1 : 2 :
; FE =
Area of quad. FEBC = Area of equilateral triangle - Area of AFE
= (side)² - AF FE
= (24)² - square units
Option 3 is correct.
Question 6
1 / -0

A punching machine is used to punch a circular hole of diameter two units from a square sheet of aluminium of width 2 units, as shown below. The hole is punched such that the circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in line with a diagonal of the square.

Find the area of the part of the circle (round punch) falling outside the square sheet.

A

B

C

D

Solution

PG = 2 Units = BF (diameter of circle)
CD = 2 units = (side of a square sheet)
In BOP,
OB = 1 unit
BP = units (By Pythagoras theorem)
Similarly,

PF = units
Area of rt. PBF = = 1 square units
Area of semicircle BGF = = square units
Area of the part of the circle falling at side the square sheet = Area of circle - (Area of
PBF + semicircle BGF)
= (1)² -
=
Option 4 is correct.
Question 7
1 / -0

Based on the figure shown below, what is the value of x, if y = 10?

A
10

B
12

C
13

D
None of these

Solution

Let AC be equal to l.
In rt. ADC, By using Pythagoras theorem,
AC²= AD²+CD²
l²= (x+4)² + (x-3)²
l²= x²+16+8x+x²+9-6x = 2x²+2x+25â¦â¦(1)
Let CO = a
In rt. BOC, By using Pythagoras theorem
BC²= BO²+CO²
y²= (x - 3)²+a²
a²=100-(x²+ 9 - 6x) = 100 - x²- 9 + 6x=91 - x² + 6xâ¦..(2)
now AC = l = x + a
Squaring both sides,
l²= (x + a)²= x²+ a²+2xa
From (1) and (2),
2x²+ 2x + 25 = x²+ 91 - x² + 6x + 2x (91 - x² + 6x)^-1/2
2x²- 4x - 66 = 2x (91 - x² + 6x)^-1/2
x²- 2x - 33 = x (91 - x² + 6x)^-1/2
put x = 10, 11, 12 to check
by x=10 ; 47 ≠ 10
x=12 ; 87 ≠ 10
x=13 ; 100 ≠ 0
Question 8
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In the figure given below, AC + AB = 5 AD and AC – AD = 8 units. What is the area of the rectangle ABCD?

A
36 sq. units

B
50 sq. units

C
60 sq. units

D
48 sq. units

Solution

AC + AB = 5AD – (i)
AC – AD = 8 units – (ii)
AC = 8 units + AD – (iii)
Putting (iii) in (i), we get
AC + AB = 5AD
8 units + AD + AB = 5AD
AB = 4AD – 8 units
AB = 4 (AD – 2 units) ….(iv)
By using Pythagoras theorem, we get
AB² + BC² = AC²
From (iii) , (iv) and BC = AD,
(4(AD – 2 units))² + AD² = (8 units + AD)²
16 (AD² + 4 units² – 4AD units) + AD² = 64 units² + AD² +16AD units
16AD² + 64 units² – 64AD units + AD² = 64 units²+ AD² + 16AD units
16AD² = 80AD units
16AD = 80 units
AD = 5 units
Putting AD = 5 units in equation (iv),
AB = 4(AD - 2 units) = 4(5 units - 2 units) = 4(3 units) = 12 units
Area of rectangle = AD AB = 5 units 12 units = 60 sq. units
Hence, option (3) is correct.
Question 9
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In a circle of radius 49 cm, an arc subtends an angle of 72^o at the centre. Find the length of the arc and the area of the sector.

A
61.6 cm, 150.92 cm²

B
61.6 cm, 1509.2 cm²

C
61.6 cm, 2509 cm²

D
61.6 m, 150.92 m²

Solution

AOB = 72^o = radian
Arc AB = = 61.57 = 61.6 cm
Area of sector AOB:

×
Area of Sector AOB = = 1509.2 cm²
Option 2 is correct.
Question 10
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A cow is tethered at point A by a rope. Neither the rope nor the cow is allowed to enter the triangle ABC.
BAC = 30° and AB = AC = 10 m.

What is the area that can be grazed by the cow, if the length of the rope is 8 m?

A
134 m²

B
121 m²

C
132 m²

D
m²

Solution

We need to find the sector ACBO.
AOB = 330^o =

=
= Area of sector ACBO
m² = Area of sector ACBO
Option 4 is correct
Question 11
1 / -0

The lengths of sides of a triangle are 5 units, 12 units and 13 units. A rectangle is constructed which is equal in area to the triangle. The rectangle has a length of 10 units. What is the perimeter of the rectangle?

A
30 units

B
36 units

C
13 units

D
None of these

Solution

Option (4) is the correct answer.
Question 12
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The figure below shows two concentric circles with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at point B, C, D and A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD?

A

B

C

D

Solution

Let radius of the outer circle be x.
Circumference of the outer circle = 2x
Let length of the side of the square PORS be y.
By using Pythagoras theorem,
SQ² = QR² + RS²
= y² + y²
SQ² = 2 y²
4x² = 2 y²
Y =
CR = DR =
By using Pythagoras theorem,
CD² = CR² + DR² =
CD = x
Perimeter of ABCD = 4x
Ratio =
Question 13
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A rectangular pool 20 m wide and 60 m long is surrounded by a walkway of uniform width. If the total area of the walkway is 516 m², how wide (in metres) is the walkway?

A
43

B
4.3

C
3

D
3.5

Solution

Let x be the width of the walkway.
Then, rectangle formed including walkway has
Length = 60 + 2x
Width = 20 + 2x
Area of rectangle including walkway = 60 x 20 + 516
(60 + 2x) (20 + 2x) = 1716 m²
X² + 40x - 129 = 0
(x + 43) (x - 3) = 0
X = 43 m x = 3 m
X = 43 is not possible.
Therefore, correct option is 3.
Question 14
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PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR = OS. What is the ratio of the area of the circle to the area of the square?

A
/3

B
11/7

C
3/

D
7/11

Solution

SOR is a right angled triangle.
Let SO = R (Radius of circle)
According to the question,
SO = TR = R
OR = (OT + TR) = 2R (OT = TR =SO = radius)
Applying Pythagoras theorem,
OR² = OS² + SR²
= SR² = 3R²
Area of square = SR² (side of square PQRS)² = 3 R²
Area of circle = R²
Required ratio =
Option 1 is correct.
Question 15
1 / -0

A cow is tethered at point A by a rope. Neither the rope nor the cow is allowed to enter the triangle ABC.
BAC = 30° and AB = AC = 10 m.

What is the area that can be grazed by the cow, if the length of the rope is 12 m?

A
134 m²

B
121 m²

C
132 m²

D
m²

Solution

We need to find the sector ACBO.
AOB = 330^o =

=
= Area of sector ACBO
m² = Area of sector ACBO
m² = Area of sector ACBO

Option 3 is correct.
Question 16
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Four identical coins are placed in a square. For each coin, the ratio of area to circumference is same as the ratio of circumference to area. Find the area of the square that is not covered by the coins.

A
16 ( – 1) sq.units

B
16 (8 – ) sq.units

C
16 (4 – ) sq.units

D
16 sq.units

Solution

We have four equal circles.
Let radius of one circle be R.
According to the question,

According to the given figure,
4 R = side of the square
Side of square = 8
Area of square = (side of square)² = (8)² = 64
Area of four circles = 4 x Area of one circle = 4 x (2)² = 16
Area of the square that is not covered by the coin = Area of square - Area of four circles
= 64 - 16
= 16 (14 - ) sq. units
Option 3 is correct.
Question 17
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Four friends start from four towns, which are at the four corners of an imaginary rectangle. They meet at a point which falls inside the rectangle, after three of them have travelled the distances of 40 m, 50 m and 60 m. The maximum distance that the fourth could have travelled is approximately

A
47 m

B
52 m

C
22.5 m

D
67 m

Solution

For any point inside rectangle:
OB² + OC² = OA² + OD²
(40)² + OC² = (60)² + (50)²
OC = 67.08 m 67 m
Option 4 is correct.
Question 18
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AB is the diameter of the given circle, while points C and D lie on the circumference as shown. If AB is 15 cm, AC is 12 cm and BD is 9 cm, find the area of quadrilateral ACBD.

A
54 cm²

B
216 cm²

C
162 cm²

D
None of these

Solution

C and D = 90^o

In right triangle ABC:
AB² = BC² + AC²
(15)² = BC² + (12)²
BC = 9 cm
In right triangle ADB and ABC:
BC = BD
AB = AB (Reflexive property)
ABC ADB
BC = BD = 9 cm
AD = AC = 12 cm
Area of triangle ABC = Area of ADB = x 9 x 12
= 54 cm²
Area of two right triangles = 2 x 54 = 108 cm²
Area of two right triangles = Area of quadrilateral ACBD = 108 cm²
Option 4 is correct.
Question 19
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The sum of the areas of two circles, which touch each other externally, is 153m². If the sum of their radii is 15 m, find the ratio of the larger to the smaller radius.

A
4 : 1

B
2 : 1

C
3 : 1

D
1 : 1

Solution

According to the question,
Area of circle O + Area of circle A = 153 m²
r² + R² = 153
r² + R² = 153 â¦â¦. (1)
r + R = 15 m
r = 15 - R
Put value of r in eq (1).
(15 - R)² + R² = 153
225 + R² - 30 R + R² = 153
2R² - 30 R + 72 = 0
R² - 15R + 36=0
R² - 12R - 3R + 36=0
(R - 12) (R - 3) = 0
Radius of larger circle = 12 m
Radius of smallest circle = 3 m
Required Ratio = 12 : 3
= 4 : 1
Option 1 is correct.
Question 20
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Three identical cones with base radius r are placed on their bases such that each is touching the other two. What is the radius of the circle drawn through their vertices?

A
Smaller than r

B
Equal to r

C
Larger than r

D
Depends on the height of the cones

Solution

(Circle Radius) R > r
It is clear from the figure that radius of the circle drawn through the vertices is larger than the radius of base circle.
Hence, option (3) is the correct answer.