Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Additional Questions
Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Additional Questions
Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Additional Questions
Additional Questions And Answers
Exercise 2.1
Very Short Answers [2 Marks]
Question 1.
Find the length of arc if the perimeter of sector is 45 cm and radius is 10 cm.
Solution:
Given Radius of the sector = 10 cm
Perimeter of the sector P = 45 cm
l + 2r = 45
l + 2(10) = 45
l + 20 = 45
l = 45 – 20
l = 25 cm
Length of the arc l = 25 cm
Question 2.
Find the radius of sector whose perimeter and length of arc are 30 cm and 16 cm respectively.
Solution:
Given length of the arc = 16 cm
Perimeter of the arc = 30 cm
l + 2r = 30
16 + 2 r = 30
2 r = 30 – 16
2 r = 14
r = 14/2
r = 7 cm
Radius of the sector = 7 cm
Question 3.
Find the length of arc whose radius is 7 cm and central angle 90°.
Solution:
∴ Length of the arc = 6.6 cm
Long Answers [5 Marks]
Question 1.
A sector is cut from a circle of radius 21 cm. The angle of the sector is 150°. Find the length of its arc and area of the sector.
Solution:
Radius of the sector = 21 cm
Length of the arc
∴ Length of the arc = 55 cm
Area of the sector = 577.5 cm2
Question 2.
Find the perimeter of sector whose area is 324 sq. cm and radius is 27 cm.
Solution:
Radius of the sector = 27 cm
Area of the sector =324 cm2
Perimeter of the sector P = (l + 2r) units = 24 + 2(27) cm = (24 + 54) cm = 78 cm
Exercise 2.2
Question 1.
PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal semi-circles drawn on PQ and Question as diameters. Find the p perimeter and area of the shaded region.
Solution:
PS = Diameter of a circle of radius 6 cm = 12 cm
PQ = QR = RS = 12/3 = 4 cm Question = QR + RS = 4 + 4 = 8 cm
∴ Perimeter of the shaded part = Arc length of semi-circle of radius 6 cm + Arc length of semicircle of radius 4 cm + Arc length of semi-circle of radius 2 cm.
= (π × 6) + (π × 4) + (π × 2) cm
P = 12 π cm
Area required = Area of semicircle with PS as diameter + Area of semi circle with PQ as diameter – Area of semi-circle with Question as diameter.
Question 2.
In the figure AOBCA represents a quadrant of a circle of radius 3.5cm with center ‘O’ calculate the area of the shaded portion (π = 22/7)
Solution:
∴ Area of shaded region = Area of the quadrant – Area of triangle
= 9.625 – 3.5 cm2 = 6.125 cm2
Question 3.
Find the area of the shaded region in the figure
Solution:
Radius of the big semicircle = 14 cm
∴ Required area = 308 + 154 cm2 = 462 cm2
Exercise 2.3
Very Short Answers [2 Marks]
Question 1.
What is the least number of planes that can enclose a solid? What is the name of the solid?
Solution:
Least number of planes = 4, the solid is tetrahedron.
Question 2.
Can a polyhedron have for its faces = 12 edges = 16 and vertices = 6.
Solution:
Verifying Euler’s formula
F + V – E = 12 + 6 – 16 = 18 – 16 = 2
Yes, the polyhedron can have F = 12, E = 16 and V = 6
Short Answers [3 Marks]
Question 1.
Verify Euler’s formula for a pyramid.
Solution:
A pyramid has faces = 5, Vertices = 5, Edges = 8
By Euler’s formula F + V – E = 5 + 5 – 8 = 10 – 8 = 2
Question 2.
Verify Eulers formula for a triangular prism.
Solution:
For a triangular prism
Faces = 5, Edges = 9, Vertices = 6
By Euler’s formula F + V – E = 5 + 6 – 9 = 11 – 9 = 2
Long Answers [5 Marks]
Question 1.
(a) Dice are cubes where the numbers on the opposite faces must total 7. Is the following a die.
(b) The following shows a net with areas of faces. What can be the shape?
Solution:
(a) 2 + 5 = 6 + 1 = 3 + 4 = 7
∴ It can be a die.
(b) It is a cuboid
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