Chapter List
Part 1
1. Rational and Irrational Numbers
2. Parallel Lines and Transversals
4. Altitudes and Medians of a Triangle
6. Factorisation of Algebraic Expressions
7. Variation
8. Quadrilateral : Constructions and Types
Part 2
11. Statistics
15. Area
In a circle with centre P, chord AB is drawn of length 13 cm, seg PQ ⊥ chord AB, then find l(QB).
We know that,
The perpendicular from the centre of a circle to a chord bisects the chord.
Therefore, it is given that,
AB=13 cm
PQ perpendicular to AB
|(QB) = AB/2
| (QB) = 13/2
| (QB) = 6.5 cm
Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if the length of the chord is 48 cm.
As we know that, the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore, OP perpendicular to CD and OP bisects the CD. Therefore, it makes a right angle triangle, which is ∆OPD. We have OD=25 cm and PD=48/2=24 cm.
By Pythagoras theorem,
OD2 = OP2 + PD2
OP2 = OD2 - PD2
OP2= (25)2 - (24)2
OP2 = 625 – 576
OP2 = 49
OP = 7 cm
Therefore, distance of the chord from the centre is 7 cm.
O is the centre of the circle. Find the length of the radius, if the chord of length 24 cm is at a distance of 9 cm from the centre of the circle.
As we know that, the perpendicular from the centre of a circle to a chord bisects the chord.
So let P is the point, which bisects chord AB. So OP is perpendicular, it makes a right angle triangle ∆OPA.
Now we have OP = 9cm and AP as 12 cm
So by Pythagoras theorem,
AO2 = AP2 + PO2
AO2 = (12)2 + (9)2
AO2 = 144+81
AO2 = 225
AO = 15 cm
Length of radius is 15 cm.
C is the centre of the circle whose radius is 10 cm. Find the distance of the chord from the centre if the length of the chord is 12 cm.
As we know that, the perpendicular from the centre of a circle to a chord bisects the chord.
So here we have C as a centre where CP is perpendicular on AB which bisects the chord AB and radius as CA = 10 cm and chord length = 12 cm, so AP=6cm.
It makes a right angle triangle ∆CPA.
Therefore, by using Pythagoras theorem, we have,
AC2 = CP2 + AP2
We have to find CP so
CP2 = AC2 - AP2
CP2 = (10)2 – (6)2
CP2 = 100 – 36
CP2= 64
CP = 8 cm
Therefore, a distance of the chord from the centre is 8 cm.
The diameters PQ and RS of the circle with centre C are perpendicular to each other at C. State, why arc PS and arc SQ are congruent. Write the other arcs, which are congruent to arc PS
As we know that, according to the theorem of the circle, two arcs are congruent, if their central angles are congruent, so arc PS and arc SQ are congruent because the angles between the chords are same and both are at 90° of the centre.
The other arcs, which are congruent to arcs PS, are
arc PS ≅ arc PR ≅ arc RQ because if two arcs of a circle are congruent, then their corresponding arcs are also congruent.
In the adjoining figure O is the centre of the circle whose diameter is MN. Measures of some central angles are given in the figure. Hence, find the following
(1) m∠AOB and m∠COD
(2) Show that arc AB ≅ arc CD
(3) Show that chord AB ≅ chord CD
(1) In given figure, we can see that
∠NOC + ∠COD +∠DOM = 180° (linear pair)
35° +∠COD+100° =180°
∠COD = 180° - 135°= 45°
So ∠COD and ∠AOB = 45°
(2) arc AB ≅ arc CD because the arcs are of equal measure 45° each angle and equal angle made equal sector.
(3) chord AB ≅ chord CD because corresponding chords of congruent arcs are congruent.
Find the volume of a box if its length, breadth, and height are 20 cm, 10.5 cm and 8 cm respectively.
Given:
Length = 20 cm
Breadth = 10.5 cm
Height = 8 cm
The box is nothing but a cuboid
Volumeof cuboid = l × b × h
= 20 × 10.5 × 8
= 1680 cm3
∴The volume of the box is 1680 cm3
A cuboid shape soap bar has volume 150 cc. Find its thickness if its length is 10 cm and breadth is 5 cm.
Given:
Volume of soap bar = 150 cc
Length = 10 cm
Breadth = 5 cm
Height = ?
The volume of cuboid = l × b × h
150 = 10 × 5 × h
h = 3 cm
The height of soap bar is 3 cm
How many bricks of length 25 cm, breadth 15 cm, and height 10 cm are required to build a wall of length 6 m, height 2.5 m, and breadth 0.5 m?
Given:
For one brick,
Length = 25 cm, breadth = 15 cm, height = 10 cm
For wall,
Length = 6 m = 6 × 100 cm = 600 cm
Breadth = 0.5 m = 0.5 × 100 = 50 cm
Height = 2.5 m 2.5 × 100 = 250 cm
Now, the number of bricks required to build a wall is given by,
Both wall and brick are cuboidal in shape.
Hence, the volume is given by,
The volume of wall = l × b × h
= 600 × 50 × 250
= 7500000 cm3
The volume of one brick = l × b × h
= 25 × 15 × 10
= 3750 cm3
= 2000 bricks
∴2000 bricks are required to build a wall of dimensions 6 × 0.5 × 2 m.
For rainwater harvesting, a tank of length 10 m, breadth 6 m, and depth 3m are built. What is the capacity of the tank? How many liters of water can it hold?
Given:
Length of tank = 10 m
Breadth of tank = 6 m
The height of tank = 3 m
Capacity is nothing but the volume of the tank.
As for length, breadth and height are given, the tank is cuboidal in shape.
The volume of tank = l × b × h
= 10 × 6 × 3
= 180 m3
The capacity of the tank is 180 m3
Now,
1 m3 = 1000 litre
∴180 m3 = 180 × 1000 = 180,000 litre
∴ The tank can hold 180,000 litres of water
In each example given below, the radius of the base of a cylinder and its height are given. Then find the curved surface area and total surface area.
(1) r = 7 cm, h = 10 cm
(2) r = 1.4 cm, h = 2.1 cm
(3) r = 2.5 cm, h = 7 cm
(4) r = 70 cm, h = 1.4 cm
(5) r = 4.2 cm, h = 14 cm
Curved surface area of cylinder(CSA) = 2πrh
Total surface area of cylinder(TSA) = 2πr(h+r)
1. r = 7 cm, h = 10 cm
CSA = 2πrh
= 2 × 3.14 × 7 × 10
= 440 cm2
TSA = 2πr(h+r)
= 2 × 3.14 × 7(10+7)
= 748 cm2
2. r = 1.4 cm, h = 2.1 cm
CSA = 2πrh
= 2 × 3.14 × 1.4 × 2.1
= 18.48 cm2
TSA = 2πr(h+r)
= 2 × 3.14 × 1.4(2.1+1.4)
= 30.8 cm2
3. r = 2.5 cm, h = 7 cm
CSA = 2πrh
= 2 × 3.14 × 2.5 × 7
= 110 cm2
TSA = 2πr(h+r)
= 2 × 3.14 × 2.5(7+2.5)
= 149.29 cm2
4. r = 70 cm, h = 1.4 cm
CSA = 2πrh
= 2 × 3.14 × 70 × 1.4
= 616 cm2
TSA = 2πr(h+r)
= 2 × 3.14 × 70(70+1.4)
= 31416 cm2
5. r = 4.2 cm, h = 14 cm
CSA = 2πrh
= 2 × 3.14 × 4.2 × 14
= 369.6 cm2
TSA = 2πr(h+r)
= 2 × 3.14 × 4.2(4.2+14)
= 480.48 cm2
Find the total surface area of a closed cylindrical drum if its diameter is 50 cm and height is 45 cm. (π = 3.14)
Total surface area of cylinder(TSA) = 2πr(h+r)
Here, 
h = 45 cm
Total Surface Area = 2 × 3.14 × 25(45+25)
= 10990 cm2
Total Surface Area of Cylinder is 10990 cm2
Find the area of base and radius of a cylinder if its curved surface area is 660 sqcm and height is 21 cm
Area of base of cylinder = π × r2
Curved surface area of cylinder(CSA) = 2π × r × h
Here, CSA = 660 sqcm, h = 21 cm, r = ?
CSA = 2π × r × h
660 = 2π × r × 21 
r = 5 cm
Area of base = π × r2
= 3.14 × 25 × 25
= 78.5 cm2
Area of the base is 78.5 cm2 and radius is 5 cm
Find the area of the sheet required to make a cylindrical container which is open at one side and whose diameter is 28 cm and height is 20 cm. Find the approximate area of the sheet required to make a lid of height 2 cm for this container.
Given:
Diameter = 28 cm
Radius 
height = 2 cm
As the cylindrical container is open at one side, Total area of a cylinder is given as,
Area of Cylinder = area of the base + curved surface area
Area of base = π × r2
Curved surface area = 2π × r × h
∴Area of Cylinder =π × r2 + 2π × r × h
= 3.14 × 142 + 2 × 3.14 × 14 × 20
= 615.44 + 1759.3
= 2376 cm2
Now, the area of the sheet required to make a cylindrical container is nothing but an area of the cylinder.
∴ Area of Sheet = 2376 cm2
Now, we need to make a lid for the open cylinder. Given the height of the lid is 2 cm.
As the lid is for the cylinder, it’s radius will be the radius of the cylinder.
Hence, For lid,
Radius = 14 cm
Height = 2 cm
Area of lid = area of the base of the lead + curved surface area
= π × r2 + 2π × r × h
= 3.14 × 142 + 2 × 3.14 × 14 × 2
= 615.44 + 175.84
= 792 cm2
∴ Area of Sheet = 2376 cm2
∴ Area of Lid = 792 cm2
Find the volume of the cylinder if height (h) and radius of the base (r) are as given below.
(1) r = 10.5 cm, h = 8 cm
(2) r = 2.5 m, h = 7 m
(3) r = 4.2 cm, h = 5 cm
(4) r = 5.6 cm, h = 5 cm
Volume of cylinder=π × r2 × h
1. r = 10.5 cm, h = 8 cm
Volume = π × r2 × h
= 3.14 × 10.52 × 8
= 2772 cm3
2. r = 2.5 m, h = 7 m
Volume = π × r2 × h
= 3.14 × 2.52 × 7
= 137.5 cm3
3. r = 4.2 cm, h = 5 cm
Volume = π × r2 × h
= 3.14 × 4.22 × 5
= 277.2 cm3
4. r = 5.6 cm, h = 5 cm
Volume = π × r2 × h
= 3.14 × 5.62 × 5
= 492.8 cm3
How much iron is needed to make a rod of length 90 cm and diameter 1.4 cm?
Given,
length/height of the cylindrical rod = 90 cm
The radius of rod 
Here, we need to calculate the amount of iron required to make a rod.
That mean, we need to calculate the volume of the rod.
Volume of rod = π × r2 × h
= 3.14 × 0.72 × 90
= 138.6 cm3
∴ Amount of iron required is 138.6 cm3
How much water will a tank hold if the interior diameter of the tank is 1.6 m and its depth is 0.7 m?
Given,
Radius 
Height = 0.7 m
The volume of tank = π × r2 × h
= 3.14 × 0.82 × 0.7
= 1.408 m3
Now, 1m3 = 1000 litre
1.408 m3 = 1408 litre
∴ The tank can hold 1408 liter of water
Find the volume of the cylinder if the circumference of the cylinder is 132 cm and height is 25 cm.
Given,
Circumference = 132 cm
Height = 25 cm
Volume = ?
The circumference of cylinder = 2 × π × r
132 = 2 × π × r
The volume of cylinder = π × r2 × h
= 3.14 × 212 × 25
= 34650 cm3
∴ The volume of the cylinder is 34650 cm3