CBSE Class 9

Here we are providing Surface Areas and Volumes Class 9 Extra Questions Maths Chapter 13 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Surface Areas and Volumes with Answers Solutions

Extra Questions for Class 9 Maths Chapter 13 Surface Areas and Volumes with Solutions Answers

Surface Areas and Volumes Class 9 Extra Questions Very Short Answer Type

Question 1.
How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm?
Solution:
Here, radius (r) = 3.5 cm and height (h) = 12 cm
∴ Amount of ice-cream = \(\frac{1}{3}\) πr²
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 12
= 154 cm³

Question 2.
Calculate the edge of the cube if its volume is 1331 cm³.
Solution:
Volume of cube = 1331 cm³
(Side)³ = 1331
Side = (11 × 11 × 11)^{\(\frac{1}{3}\)} = 11 cm

Question 3.
The curved surface area of a cone is 12320 sq. cm, if the radius of its base is 56 cm, find its
height.
Solution:
Here, radius of base of a cone (r) = 56 cm
And, curved surface area = 12320 cm²
πrl = 12320
l = \(\frac{12320}{\pi r}\)
= \(\frac{12320 \times 7}{22 \times 56}\) = 70 cm
Again, we have
r² + h² = l²
h² = l² – r² = 70² – 56²
= 4900 – 3136 = 1764
h = √1764 = 42 cm
Hence, the height of the cone is 42 cm.

Question 4.
Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid.
Solution:
When two cubes are joined end to end, then
Length of the cuboid = 6 + 6 = 12 cm
Breadth of the cuboid = 6 cm
Height of the cuboid = 6 cm
Total surface area of the cuboid = 2 (lb + bh + hl)
= 2(12 × 6 + 6 × 6 + 6 × 12)
= 2(72 + 36 + 72) = 2(180)
= 360 cm²

Question 5.
A metallic sphere is of radius 4.9 cm. If the density of the metal is 7.8 g/cm², find the mass of the sphere (π = \(\frac{22}{7}\)).
Solution:
Here, radius of metallic sphere (r) = 4.9 cm

Question 6.
The volume of a solid hemisphere is 1152 π cm³. Find its curved surface area.
Solution:
Here, volume of hemisphere = 1152 π cm³
∴ \(\frac{2}{3}\)πr³ = 1152
⇒ r³ = (12)³ π
⇒ r\(\frac{1152 \times 3}{2}\) = 1728
⇒ r³ = (12)³
Now, curved surface area = 2πr²
= 2 × π × (12)² = 288π cm²

Question 7.
Find the diameter of a cylinder whose height is 5 cm and numerical value of volume is equal to

numerical value of curved surface area.
Solution:
Here, height of cylinder (h) = 5 cm
According to the statement of the question, we have
πr²h = 2πrh
r = 2 cm
Thus, diameter of the base of the cylinder is 2 × 2 i.e., 4 cm.

Question 8.
In a cylinder, if radius is halved and height is doubled, then find the volume with respect to original volume.
Solution:
Here, r = \(\frac{r}{2}\), h = 2h

Surface Areas and Volumes Class 9 Extra Questions Short Answer Type 1

Question 1.
A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm, find the volume of the spherical ball. [use π = 3.14]
Solution:
Since curved surface of half of the spherical ball = 56.57 cm²
2πr² = 56.57

= 113.04 cm³

Question 2.
Find the capacity in litres of a conical vessel having height 8 cm and slant height 10 cm.
Solution:
Height of conical vessel (h) = 8 cm
Slant height of conical vessel (l) = 10 cm
∴ r² + h² = l²
⇒ r² + 8² = 10²
⇒ r² = 100 – 64 = 36
⇒ r = 6 cm
Now, volume of conical vessel = \(\frac{1}{3}\)πr²h = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 8 = 301.71 cm³ = 0.30171 litre

Question 3.
Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside.
Solution:
Here, radius of hemispherical dome (r) = 14 m
Surface area of dome = 2πr²
= 2 × \(\frac{22}{7}\) × 14 × 14 = 1232 m²
Hence, total surface area to be whitewashed from outside is 1232 m².

Question 4.
A rectangular piece of paper is 22 cm long and 10 cm wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder.
Solution:
Since rectangular piece of paper is rolled along its length.
∴ 2πr = 22
r = \(\frac{22 \times 7}{2 \times 22}\) = 3.5 cm
Height of cylinder (h) = 10 cm
∴ Volume of cylinder = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 10 = 385 cm³

Question 5.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find it volume. If 1m³ wheat cost is ₹10, then find total cost.
Solution:
Diameter of cone = 10.5 m
Radius of cone (r) = 5.25 m
Height of cone (h) = 3 m
Volume of cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5.25 × 5.25 × 3
= 86.625 m³
Cost of 1m³ of wheat = ₹10
Cost of 86.625 m³ of wheat = ₹10 × 86.625
= ₹866.25

Question 6.
A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and1cm³ of water weighs 1 g, find the depth of water.
Solution:
Since 1 cm³ of water weighs 1 g.
∴ Volume of cylindrical vessel = 154 cm³
πr²h = 154
\(\frac{22}{7}\) × 3.5 × 3.5 × h = 154

h = \(\frac{154 \times 7}{22 \times 3.5 \times 3.5}\)
h = 4
cm Hence, the depth of water is 4 cm.

Surface Areas and Volumes Class 9 Extra Questions Short Answer Type 2

Question 1.
A wall of length 10 m is to be built across an open ground. The height of the wall is 5 m and thickness of the wall is 42 cm. If this wall is to be built with brick of dimensions 42 cm × 12 cm × 10 cm, then how many bricks would be required?
Solution:
Here, length of the wall (L) = 10 m = 1000 cm
Breadth of the wall (B) = 42 cm
Height of the wall (H) = 5 m = 500 cm
∴ Volume of the wall = L × B × H
= 1000 × 42 × 500 cm³
Volume of each brick = 42 × 12 × 10 cm³

= 4167
Hence, the required number of bricks is 4167.

Question 2.
The volume of cylindrical pipe is 748 cm. Its length is 0.14 m and its internal radius is 0.09 m. Find thickness of pipe.
Solution:
Internal radius (r) of cylindrical pipe = 0.09 m = 9 cm
Length (height) of cylindrical pipe (h) = 0.14 m = 14 cm
Let external radius of the cylindrical pipe be R cm.
Volume of cylindrical pipe = 748 cm³
⇒ π(R² – r²)h = 748
⇒ \(\frac{22}{7}\) (R² – 9²)14 = 748
⇒ R² – 81 = \(\frac{748 \times 7}{22 \times 14}\) = 17
⇒ R² = 81 + 17 = 98
⇒ R = √98 = 7√2 cm = 9.9 cm
Thus, thickness of the pipe = 9.9 -9 = 0.9 cm

Question 3.
The curved surface area of a cylinder is 154 cm. The total surface area of the cylinder is three times its curved surface area. Find the volume of the cylinder.
Solution:
Since curved surface area of cylinder = 154 cm² (given]
Total surface area of cylinder = 3 × curved surface area
2πrh + 2πr² = 3 × 154 3 154 + 2πr² = 462
2πr² = 462 – 154 = 308

= 539 cm³

Question 4.
A right-angled ∆ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. Find the volume of the solid generated. Also, find the total surface area of the solid.
Solution:
When rt. ∠ed ∆ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm. Slant height of the cone is 5 cm.

Question 5.
A semicircular sheet of metal of radius 14 cm is bent to form an open conical cup. Find the capacity of the cup.
Solution:
Radius of semicircular sheet (r) = 14 cm
∴ Slant height (1) = 14 cm
Circumference of base = Circumference of semicircular sheet

Surface Areas and Volumes Class 9 Extra Questions Long Answer Type

Question 1.
It costs ₹3300 to paint the inner curved surface of a 10 m deep well. If the rate cost of
painting is of ₹30 per m², find :
(a) inner curved surface area
(b) diameter of the well
(c) capacity of the well.
Solution:
Depth of well (h) = 10 m
Cost of painting inner curved surface is ₹30 per m² and total cost is ₹3300

Hence, inner curved surface area is 110 m², diameter of the well is 2 × 1.75 i.e., 3.5 m and capacity of the well is 96.25 m³.

Question 2.
Using clay, Anant made a right circular cone of height 48 cm and base radius 12 cm. Versha reshapes it in the form of a sphere. Find the radius and curved surface area of the sphere so formed.
Solution:
Height of cone (h) = 48 cm
Radius of the base of cone = 12 cm
Let R be the radius of sphere so formed
∴ Volume of sphere = Volume of cone
\(\frac{4}{3}\)πR³ = \(\frac{1}{3}\)πr²h
4R³ = 12 × 12 × 48
R³ = 12 × 12 × 12
R = 12 cm
Now, curved surface area of sphere = 4πR²
= 4 × \(\frac{22}{7}\) × 12 × 12
= 1810.29 cm

Question 3.
A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of ₹498.96. If the rate of whitewashing is ₹4 per square metre, find the :
(i) Inside surface area of the dome
(ii) Volume of the air inside the dome.
Solution:
Here, dome of building is a hemisphere.
Total cost of whitewashing inside the dome = ₹498.96
Rate of whitewashing = ₹4 per m²

Question 4.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm. Find the volume of the solid so obtained. If it is now revolved about the side 12 cm, then what would be the ratio of the volumes of the two solids obtained in two cases ?
Solution:
Here, right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm.
∴ Radius of the base of cone = 12 cm
Height of the cone = 5 cm

= 12 : 5

Question 5.
A right triangle of hypotenuse 13 cm and one of its sides 12 cm is made to revolve taking side 12 cm as its axis. Find the volume and curved surface area of the solid so formed.
Solution:
Here, hypotenuse and one side of a right triangle are 13 cm and 12 cm respectively.

Now, given triangle is revolved, taking 12 cm as its axis
∴ Radius of the cone (r) = 5 cm
Height of the cone (h) = 12 cm
Slant height of the cone (1) = 13 cm
∴ Curved surface area = πrl = π(5)(13) = 65π cm²
Volume of the cone = \(\frac{1}{2}\)πr²h = \(\frac{1}{2}\)π × 5 × 5 × 12 = 100π cm³
Hence, the volume and curved surface area of the solid so formed are 100 π cm³ and 65 π cm² respectively.

Surface Areas and Volumes Class 9 Extra Questions HOTS

Question 1.
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Solution:
Let each edge of the cube be a cm.

Question 2.
A rectangular tank is 225 m × 162 m at base. With what speed should water flow into it through an aperture 60 cm × 45 cm so that the level of water is raised by 20 cm in 2.5 hours?
Solution:
Volume of water to be flown in 2.5 hour
= 225 m × 162 m × 20 cm

Hence, the speed of flow of water = 10.8 km/hour

Surface Areas and Volumes Class 9 Extra Questions Value Based (VBQs)

Question 1.
To maintain beauty of a monument, the students of the school cleaned and painted the dome of the monument. The monument is in the form of a hemisphere. From inside, it was white washed by the students whose area is 249.48 m².
(a) Find the volume of the air inside the dome. If white washing costs ₹2 per m², how much does it costs ?
(b) Which value is depicted by the students? (π = \(\frac{22}{7}\))
Solution:
(a) Here, dome of the monument is hemispherical in shape, which was whitewashed by the students.
Now, total area to be white washed = 249.48 m²
Cost of white washing = ₹2 per m²
∴ Total cost of white washing = ₹2 × 249.48
= ₹498.96
Also, 2πr² = 249.48

= 4191.264 m³
(b) Value: Cleanliness, beautification as well as preserving the heritage along with social values.

Question 2.
Salim provides water to a village, having a population of 4000 which requires 150 litres of water per head per day. He has storage tank measuring 20 m × 15 m × 6 m. For how many days will the water of his tank last? He increased the rate for providing water as the dependence of villagers increased on him. Which value is depicted by Salim?
Solution:
(i) Here, the population of the village = 4000
Requirement of water per head per day = 150 litres
∴ Total requirement of water per day = 4000 × 150 litres
= 600000 litres
Volume of water tank = 20 × 15 × 6
= 1800 m³
= 1800 × 1000 litres
Now, number of days for which water of the tank will last = \(\frac{1800 \times 1000}{600000}\) = 3 days
Hence, water tank can serve for 3 days.
(ii) Helping the needy.

Here we are providing Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Heron’s Formula with Answers Solutions

Extra Questions for Class 9 Maths Chapter 12 Heron’s Formula with Solutions Answers

Heron’s Formula Class 9 Extra Questions Very Short Answer Type

Question 1.
Find the area of an equilateral triangle having side 6 cm.
Solutioin:
Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) × (side)² = \(\frac{\sqrt{3}}{4}\) × 6 × 6 = 9√3 cm²

Question 2.
If the perimeter of an equilateral triangle is 90 m, then find its area.
Solutioin:

Question 3.
If every side of a triangle is doubled, then find the percent increase in area of triangle so formed.
Solutioin:
Let the sides of the given triangle be, a units, b units and c units.

Hence, percent increase = 300%

Question 4.
If the length of a median of an equilateral triangle is x cm, then find its area.
Solutioin:

Let each equal sides of given equilateral triangle be 2y². We know that median is also perpendicular bisector.
∴ y² + x² = 4y²
⇒ x² = 3y²
⇒ x = √3y
or
⇒ y = \(\frac{x}{\sqrt{3}}\)
Now, area of given triangle = \(\frac{1}{2}\)
× 2y × X = y × x = \(\frac{x}{\sqrt{3}}\)× x = \(\frac{x^{2}}{\sqrt{3}}\)

Heron’s Formula Class 9 Extra Questions Short Answer Type 2

Question 1.
Find the area of a triangle whose sides are 11 m, 60 m and 61 m.
Solutioin:
Let a = 11 m, b = 60 m and c = 61 m :

Question 2.
Suman has a piece of land, which is in the shape of a rhombus. She wants her two sons to work on the land and produce different crops. She divides the land in two equal parts by drawing a diagonal. If its perimeter is 400 m and one of the diagonals is of length 120 m, how much area each of them will get for his crops ?
Solutioin:
Here, perimeter of the rhombus is 400 m.
∴ Side of the rhombus = \(\frac{400}{4}\) = 100 m
Let diagonal BD = 120 m and this diagonal divides the rhombus ABCD into two equal parts.

Hence, area of land allotted to two sons for their crops is 4800 m² each.

Question 3.
The perimeter of a triangular field is 144 m and its sides are in the ratio 3:4:5. Find the length of the perpendicular from the opposite vertex to the side whose length is 60 m.
Solutioin:
Let the sides of the triangle be 3x, 4x and 5x
∴ The perimeter of the triangular field = 144 m
⇒ 3x + 4x + 5x = 144
⇒ 12x = 144

Question 4.
Find the area of the triangle whose perimeter is 180 cm and two of its sides are of lengths 80 cm and 18 cm. Also, calculate the altitude of the triangle corresponding to the shortest side.
Solutioin:
Perimeter of given triangle = 180 cm
Two sides are 18 cm and 80 cm
∴ Third side = 180 – 18 – 80 = 82 cm

Hence, area of triangle is 720 cm² and altitude of the triangle corresponding to the shortest side is 80 cm.

Heron’s Formula Class 9 Extra Questions Long Answer Type

Question 1.
Calculate the area of the shaded region.

Solutioin:

= 2 × 2 × 3 × 7 = 84 cm²
Area of shaded region = Area of ∆ABC – Area of ∆AOB
= 84 cm² – 30 cm² = 54 cm²

Question 2.
The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses ? (use n = 3.14)
Solutioin:
The sides of the triangular park are 8 m, 10 m and 6 m.

Radius of the circle = \(\frac{2}{2}\) = 1 m
Area of the circle = πr² = 3.14 × 1 × 1 = 3.14 m²
∴ Area to be used for growing roses = Area of the park – area of the circle
= 24 – 3.14 = 20.86 m²

Heron’s Formula Class 9 Extra Questions HOTS

Question 1.
OPQR is a rhombus, whose three vertices P, Q and R lie on the circle with centre 0. If the radius of the circle is 12 cm, find the area of the rhombus.
Solutioin:
Since diagonals bisect each other at 90°.
∴ In right ∆QLR, (LR)² + (LQ)² = (QR)²

Question 2.
How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square with diagonal 60 cm?

Solutioin:
Since diagonals of a square are of equal length and bisect each other at right angles, therefore,
Area of ∆AOD = \(\frac{1}{2}\) × 30 × 30 = 450 cm²
Area of ∆AOD = Area of ∆DOC = Area of ∆BOC
= Area of ∆AOB = 450 cm²
[∵ ∆AOD = ∆AOB ≅ ∆BOC ≅ ∆COD, ∵ they² have equal area]
Now, area of ACEF (by Heron’s formula)
Here a = 20 cm, b = 20 cm and c = 30 cm

Now, area of orange shaded paper in kite
= Area of ∆AOD + Area of ∆CEF
= 450 cm² + 198.4 cm²
= 648.4 cm²
Area of blue shaded paper in kite
= Area of ∆AOB + Area of ∆COD
= 450 cm² + 450 cm² = 900 cm²
Area of black shaded paper in kite = Area of ∆BOC = 450 cm².

Heron’s Formula Class 9 Extra Questions Value Based (VBQs)

Question 1.
Sister Nivedita has trapezium shaped plot which she divided into three triangular portion for different purposes. I – for providing free education for orphan children, II – for providing dispensary for the needy villagers and III – for the library for villagers. Find the area of trapezium plot given in the figure. Which qualities of sister Nivedita are being depicted in question ?

Solutioin:
Here, ABCD is the trapezium with AB || DC.
Through C, Draw CF ⊥ AB
For – I: For area of ∆EBC

But, we know, Area of ∆EBC = \(\frac{1}{2}\)(base × height)
⇒ \(\frac{1}{2}\)1 × 30 × CF = 336
15 × CF = 336
⇒ CF = \(\frac{336}{15}\) = 22.4 m
Now, area of trapezium shaped plot = \(\frac{1}{2}\) (20 + 50)(22.4)
= 35 x 22.4 = 784 m²
Caring, kind, social, generous and visionary lady.

Question 2.
In an exhibition, an umbrella is made by stiching 10 triangular pieces of cloth with same message written on two triangular pieces. If each piece of cloth measures 60 cm, 60 cm and 20 cm, find how much cloth is required for each message.
Why we should respect women, educate women, save women and empower women ?

Solutioin:
Here, each triangular piece is an isosceles triangle with sides 60 cm, 60 cm and 20 cm.

Now, there are 2 triangular pieces with same message.
∴ Total area of cloth for each message = 2 × 591.61 = 1183.22 cm²
We should respect women, educate women, save women and empower women to serve humanity and give them equal opportunity so far they deprived.

Here we are providing Constructions Class 9 Extra Questions Maths Chapter 11 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Constructions with Answers Solutions

Extra Questions for Class 9 Maths Chapter 11 Constructions with Solutions Answers

Constructions Class 9 Extra Questions Very Short Answer Type

Question 1.
Draw a line segment AB = 8 cm. Draw \(\frac{1}{3}\) part of it. Measure the length of \(\frac{1}{3}\) part of AB.
Solution:

Steps of Construction :

1. Draw a line segment AB = 8 cm.
2. Draw its perpendicular bisector and let it intersect AB in M.
3. Draw the perpendicular bisector of MB and let it intersect AB in N. Thus, AN = \(\frac{1}{3}\) of AB = 6 cm.

Question 2.
Why we cannot construct a ∆ABC, if ∠A = 60°, AB = 6 cm and AC + BC = 5 cm but construction of ∆ABC is possible if ∠A = 60°, AB = 6 cm and AC – BC = 5 cm ?
Solution:
We know that, by triangle inequality property, construction of triangle is possible if sum of two sides of a triangle is greater than the third side. Here, AC + BC = 5 cm which is less than AB (6 cm) Thus, ∆ABC is not possible.
Also, by triangle inequality property, construction of triangle is possible, if difference of two sides of a triangle is less than the third side
Here, AC – BC = 5 cm, which is less than AB (6 cm)
Thus, ∆ABC is possible.

Question 3.
Construct an angle of 90° at the initial point of the given ray.
Solution:

Steps of Construction :

1. Draw a ray OA.
2. With O as centre and any convenient radius, draw an arc, cutting OA at P.
3. With P as centre and same radius, draw an arc cutting the arc drawn in step 2 at Q.
4. With Q as centre and the same radius as in steps 2 and 3, draw an arc, cutting the arc drawn in step 2 at R.
5. With Q and R as centres and same radius, draw two arcs, cutting each other in S.
6. Join OS and produce to B. Thus, ∠AOB is the required angle of 90°

Question 4.
Draw a straight angle. Using compass bisect it. Name the angles obtained.
Solution:

Steps of Construction :

1. Draw any straight angle (say ∠AOC).
2. Bisect ∠AOC and join BO.
3. ∠AOB is the required bisector of straight angle AOC.

Question 5.
Draw any reflex angle. Bisect it using compass. Name the angles so obtained.
Solution:

Steps of Construction :

1. Let ∠AOB be any reflex angle.
2. With O as centre and any convenient radius, draw an arc cutting OA in P and OB in Q.
3. With P and Q as centres, draw two arcs of radius little more than half of it and let they intersect each other in C. Join OC. Thus, OC is the required bisector. Angles so obtained are ∠AOC and ∠COB.

Constructions Class 9 Extra Questions Short Answer Type 1 and 2

Question 1.
Construct a triangle whose sides are in the ratio 2 : 3 : 4 and whose perimeter is 18 cm.
Solution:

Steps of Construction :

1. Draw a line segment AB =18 cm.
2. At A, construct an acute angle ∠BAX (< 90°).
3. Mark 9 points on AX, such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆
   = A₆A₇ = A₇A₈ = A₈A₉.
4. Join A₉B.
5. From A₂ and A₅, draw A₂M || A₅N || A₉B, intersecting AB in M and N respectively.
6. With M as centre and radius AM, draw an arc.
7. With N as centre and radius NB, draw another arc intersecting the previous arc at L.
8. Join LM and LN. Thus, ∆LMN is the required triangle.

Question 2.
Construct a ∆ABC with BC = 8 cm, ∠B = 45° and AB – AC = 3.1 cm.
Solution:

Steps of Construction :

1. Draw any line segment BC = 8 cm.
2. At B, construct an angle ∠CBX = 45°.
3. From BX, cut off BD = 3.1 cm.
4. Join DC.
5. Draw the perpendicular bisector ‘p’ of DC and let it intersect BX in A.
6. Join AC. Thus, ∆ABC is the required triangle.

Question 3.
Construct a ∆ABC such that BC = 3.2 cm, ∠B = 45° and AC – AB = 2.1 cm.
Solution:

Steps of Construction :

1. Draw a line segment BC = 3.2 cm.
2. At B, construct an angle ∠CBX = 45° and produce it to point X’.
3. Cut-off BD = 2.1 cm and join CD.
4. Draw the perpendicular bisector of CD and let it intersect X’BX in A.
5. Join AC. Thus, ∆ABC is the required triangle.

Question 4.
Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. Name them as QX and RY respectively. Are they both parallel ?
Solution:

Steps of Construction :

1. Draw a line segment QR = 5 cm.
2. With Q as centre, construct an angle of 90° and let this line through Q is QX.
3. With R as centre, construct an angle of 90° and let this line through R is RY. Yes, the perpendicular lines QX and RY are parallel.

Question 5.
Construct an isosceles triangle whose two equal sides measure 6 cm each and whose base is 5 cm. Draw the perpendicular bisector of its base and show that it passes through the opposite vertex.
Solution:

Steps of Construction :

1. Draw a line segment AB = 5 cm.
2. With A and B as centres, draw two arcs of radius 6 cm and let they intersect each other in C.
3. Join AC and BC to get ∆ABC.
4. With A and B as centres, draw two arcs of radius little more than half of AB. Let they intersect each other in P and Q. Join PQ and produce, to pass through C.

Constructions Class 9 Extra Questions Long Answer Type

Question 1.
Construct a triangle ABC in which BC = 4.7 cm, AB + AC = 8.2 cm and ∠C = 60°.
Solution:

Given : In ∆ABC, BC = 4.7 cm, AB + AC = 8.2 cm and ∠C = 60°.
Required : To construct ∆ABC.
Steps of Construction :

1. Draw BC = 4.7 cm.
2. Draw
3. From ray CX, cut off CD = 8.2 cm.
4.  Join BD.
5.  Draw the perpendicular bisector of BD meeting CD at A.
6. Join AB to obtain the required triangle ABC.

Justification :
∵ A lies on the perpendicular bisector of BD, therefore, AB = AD
Now, CD = 8.2 cm
⇒ AC + AD = 8.2 cm
⇒ AC + AB = 8.2 cm.

Question 2.
Construct ∆XYZ, if its perimeter is 14 cm, one side of length 5 cm and ∠X = 45°.
Solution:

Here, perimeter of ∆XYZ = 14 cm and one side XY = 5 cm
∴  YZ + XZ = 14 – 5 = 9 cm and ∠X = 45°.
Steps of Construction :

1. Draw a line segment XY = 5 cm.
2. Construct an ∠YXA = 45° with the help of compass and ruler.
3. From ray XA, cut off XB = 9 cm.
4. Join BY.
5. Draw perpendicular bisector of BY and let it intersect XB in Z.
6. Join ZY. Thus, ∆XYZ is the required triangle.

Question 3.
To construct a triangle, with perimeter 10 cm and base angles 60° and 45°.
Solution:

Given : In ∆ABC,
AB + BC + CA = 10 cm, ∠B = 60° and ∠C = 45°.
Required : To construct ∆ABC.
Steps of Construction :

1. Draw DE = 10 cm.
2. At D, construct ∠EDP= 5 of 60°= 30° and at E, construct DEQ = 1 of 45o = 22°
3. Let DP and EQ meet at A.
4. Draw perpendicular bisector of AD to meet DE at B.
5. Draw perpendicular bisector of AE to meet DE at C.
6. Join AB and AC. Thus, ABC is the required triangle.

Constructions Class 9 Extra Questions HOTS

Question 1.
Construct an equilateral triangle whose altitude is 6 cm long.
Solution:

Steps of Construction :

1. Draw a line PQ and take any point S on it.
2. Construct the perpendicular SR on PQ.
3. From SR, cut a line segment SA = 6 cm.
4. At the initial point A of the line segment AS, construct ∠SAB = 30° and ∠SAC = 30°.
5. The arms AB and AC of the angles ∠SAB and ∠SAC meet PQ in B and C respectively. Then, ∆ABC is the required equilateral triangle with altitude of length 6 cm.

Question 2.
Construct a rhombus whose diagonals are 8 cm and 6 cm long. Measure the length of each side of the rhombus.
Solution:

Steps of Construction :

1. Draw a line segment PR = 8 cm.
2. Draw the perpendicular bisector XY of the line segment PR. Let O be the point of intersection of PR and XY, so that O is the 8 cm mid-point of PR.
3. From OX, cut a line segment OS = 3 cm and from OY, cut a line segment OQ = 3 cm.
4. Join PS, SR, RQ and QP, then PQRS is the required rhombus.
5. Measure the length of segments PQ, QR, RS and SP, each is found to be 5 cm long.

Here we are providing Circles Class 9 Extra Questions Maths Chapter 10 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Circles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 10 Circles with Solutions Answers

Circles Class 9 Extra Questions Very Short Answer Type

Question 1.
In the figure, O is the centre of a circle passing through points A, B, C and D and ∠ADC = 120°. Find the value of x.

Solution:
Since ABCD is a cyclic quadrilateral
∠ADC + ∠ABC = 180°
[∴ opp. ∠s of a cyclic quad. are supplementary]
120° + ∠ABC = 180°
∠ABC = 180° – 120° = 60°
Now, ∠ACB = 90° [angle in a semicircle]
In rt. ∠ed ∆CB, ∠ACB = 90°
∠CAB + ∠ABC = 90°
x + 60° = 90°
x = 90° -60°
x = 30°

Question 2.
In the given figure, O is the centre of the circle, ∠AOB = 60° and CDB = 90°. Find ∠OBC.

Solution:
Since angle subtended at the centre by an arc is double the angle
subtended at the remaining part of the circle.
∴ ∠ACB = \(\frac{1}{3}\) ∠AOB = \(\frac{1}{3}\) x 60° = 30°
Now, in ACBD, by using angle sum property, we have
∠CBD + ∠BDC + ∠DCB = 180°
∠CBO + 90° + ∠ACB = 180°
[∵ ∠CBO = ∠CBD and ∠ACB = ∠DCB are the same ∠s]
∠CBO + 90° + 30° = 180°
∠CBO = 180o – 90° – 30° = 60°
or ∠OBC = 60°

Question 3.
In the given figure, O is the centre of the circle with chords AP and BP being produced to R and Q respectively. If ∠QPR = 35°, find the measure of ∠AOB.

Solution:
∠APB = ∠RPQ = 35° [vert. opp. ∠s]
Now, ∠AOB and ∠APB are angles subtended by an arc AB at centre and at the remaining part of the circle.
∴ ∠AOB = 2∠APB = 2 × 35° = 70°

Question 4.
In the figure, PQRS is a cyclic quadrilateral. Find the value of x.

Solution:
In ∆PRS, by using angle sum property, we have
∠PSR + ∠SRP + ∠RPS = 180°
∠PSR + 50° + 35o = 180°
∠PSR = 180° – 85o = 95°
Since PQRS is a cyclic quadrilateral
∴ ∠PSR + ∠PQR = 180°
[∵ opp. ∠s of a cyclic quad. are supplementary]
95° + x = 180°
x = 180° – 95°
x = 85°

Question 5.
In the given figure, ∠ACP = 40° and BPD = 120°, then find ∠CBD.

Solution:
∠BDP = ∠ACP = 40° [angle in same segment]
Now, in ∆BPD, we have
∠PBD + ∠BPD + ∠BDP = 180°
⇒ ∠PBD + 120° + 40° = 180°
⇒ ∠PBD = 180° – 160o = 20°
or ∠CBD = 20°

Question 6.
In the given figure, if ∠BEC = 120°, ∠DCE = 25°, then find ∠BAC.

Solution:
∠BEC is exterior angle of ∆CDE.
∴ ∠CDE + ∠DCE = ∠BEC
⇒ ∠CDE + 25° = 120°
⇒ ∠CDE = 95°
Now, ∠BAC = ∠CDE [∵ angle in same segment are equal]
⇒ ∠BAC = 95°

Circles Class 9 Extra Questions Short Answer Type 1

Question 1.
In the given figure, PQR = 100°, where P, Q and R are points on a circle with centre O. Find LOPR.

Solution:
Take any point A on the circumcircle of the circle.
Join AP and AR.
∵ APQR is a cyclic quadrilateral.
∴ ∠PAR + ∠PQR = 180° [sum of opposite angles of a cyclic quad. is 180°]
∠PAR + 100° = 180°
⇒ Since ∠POR and ∠PAR are the angles subtended by an arc PR at the centre of the circle and circumcircle of the circle.
∠POR = 2∠PAR = 2 x 80° = 160°
∴ In APOR, we have OP = OR [radii of same circle]
∠OPR = ∠ORP [angles opposite to equal sides]
Now, ∠POR + ∠OPR + ∠ORP = 180°
⇒ 160° + ∠OPR + ∠OPR = 180°
⇒ 2∠OPR = 20°
⇒ ∠OPR = 10°

Question 2.
In figure, ABCD is a cyclic quadrilateral in which AB is extended to F and BE || DC. If ∠FBE = 20° and DAB = 95°, then find ∠ADC.

Solution:
Sum of opposite angles of a cyclic quadrilateral is 180°
∴ ∠DAB + ∠BCD = 180°
⇒ 95° + ∠BCD = 180°
⇒ ∠BCD = 180° – 95° = 85°
∵ BE || DC
∴ ∠CBE = ∠BCD = 85°[alternate interior angles]
∴ ∠CBF = CBE + ∠FBE = 85° + 20° = 105°
Now, ∠ABC + 2CBF = 180° [linear pair]
and ∠ABC + ∠ADC = 180° [opposite angles of cyclic quad.]
Thus, ∠ABC + ∠ADC = ∠ABC + 2CBF
⇒ ∠ADC = CBF
⇒ ∠ADC = 105° [∵ CBF = 105°]

Question 3.
If the diagonals of a cyclic quadrilateral are diameters of the circle through the opposite vertices of the quadrilateral. Prove that the quadrilateral is a rectangle.
Solution:
Here, ABCD is a cyclic quadrilateral in which AC and BD are diameters.

Since AC is a diameter.
∴ ∠ABC = ∠ADC = 90°
[∵ angle of a semicircle = 90°]
Also, BD is a diameter
∴ ∠BAD = ∠BCD = 90° [∵ angle of a semicircle = 90°]
Now, all the angles of a cyclic quadrilateral ABCD are 90 each.
Hence, ABCD is a rectangle.

Question 4.
Equal chords of a circle subtends equal angles at the centre.

Solution:
Given : In a circle C(O, r), chord AB = chord CD
To Prove : ∠AOB = ∠COD.
Proof : In ∆AOB and ∆COD
AO = CO (radii of same circle]
BO = DO [radii of same circle]
Chord AB = Chord CD (given]
⇒ ∆AOB = ACOD [by SSS congruence axiom]
⇒ ∠AOB = COD (c.p.c.t.]

Question 5.
In the figure, chord AB of circle with centre O, is produced to Csuch that BC = OB. CO is joined and produced to meet the circle in D. If ∠ACD = y and ∠AOD = x, show that x = 3y.

Solution:
In AOBC, OB = BC
⇒ ∠BOC = ∠BCO = y [angles opp. to equal sides are equal]

∠OBA is the exterior angle of ∆BOC
So, ∠ABO = 2y [ext. angle is equal to the sum of int. opp. angles]
Similarly, ∠AOD is the exterior angle of ∆AOC
∴ x = 2y + y = 3y

Question 6.
In the given figure, P is the centre of the circle. Prove that : ∠XPZ = 2(∠X∠Y + ∠YXZ).

Solution:
Arc XY subtends ∠XPY at the centre P and ∠XZY in the remaining part of the circle.
∴ ∠XPY = 2 (∠X∠Y)
Similarly, arc YZ subtends ∠YPZ at the centre P and ∠YXZ in the remaining part of the circle.
∴ ∠YPZ = 2(∠YXZ) ….(ii)
Adding (i) and (ii), we have
∠XPY + ∠YPZ = 2 (∠XZY + ∠YXZ)
∠XP2 = 2 (∠XZY + ∠YXZ)

Circles Class 9 Extra Questions Short Answer Type 2

Question 1.
In the given figure, AB and CD are two equal chords of a circle with centre O. OP and OQ are perpendiculars on chords AB and CD respectively. If ∠POQ = 120°, find ∠ APQ.

Solution: Since
AB = CD
∴ OP = OQ [∵ equal chords are equidistant from the centre]
∠OPQ = ∠OQP:
[by using isosceles triangle property, angles opp. to equal sides of a ∆]
In APOQ, by using angle sum property, we have
∠OPQ + ∠OQP + ∠POQ = 180°
⇒ ∠OPQ + ∠OPQ + 120° = 180°
⇒ 2∠0PQ = 60°
⇒ ∠OPO = 30°
Now, ∠APQ + ∠OPQ = 90°
⇒ ∠APQ + 30° = 90°
⇒ ∠APQ = 90° – 30o = 60°
Hence, ∠APQ = 60°

Question 2.
Two circles whose centres are O and O’ intersect at P. Through P, a line parallel to OO’, intersecting the circles at C and D is drawn as shown in the figure. Prove that CD = 2OO’.

Solution:
Draw OA ⊥ CD and O’B ⊥ CD
Now, OA ⊥ CD
OA ⊥ CP
CA = AP = \(\frac{1}{2}\)CP
CP = 2AP ….(i)

Similarly, O’B ⊥ CD
O’B ⊥ PD
⇒ PB = BD = \(\frac{1}{2}\)PD
⇒ PD = 2PB
Also, CD = CP + PD
= 2AP + 2PB = 2(AP + PB) = 2AB
CD = 2OO’ [∵ OABO’ is a rectangle]

Question 3.
ABCD is a parallelogram. The circle through A, B and C intersects (produce if necessary) at E. Prove that AE = AD.
Solution:

Given : ABCD is a parallelogram. Circle through A, B and C intersects CD produced in E.
To Prove: AE = AD
Proof : ABCE is a cyclic quadrilateral.
∴ ∠B + ∠E = 180° …(i)
ABCD is a parallelogram.
∴ ∠B = ∠1 … (ii)
Also, ∠1 + ∠2 = 180° [linear pair]
∠B + ∠2 = 180° …(iii) [using (ii)]
Now, from (i) and (iii), we have
∠B + ∠E = ∠B + ∠2
∠E = ∠2 In ∆DE, we have
∠E = ∠2
⇒ AD = AE [side opposite to equal angles of a A]

Question 4.
If two equal chords of a circle intersect within a circle, prove that the line segment joining the point of intersection to the centre makes equal angles with the chords.
Solution:

Join OP, draw OL ⊥ AB and OM ⊥ CD.
Thus, L and M are the mid-points of AB and CD respectively. Also, equal chords are equidistant from the centre.
∴ OL = OM
Now, in right-angled As OLP and OMP
OL = OM
OP = OP [common]
∠OLP = ∠OMP [each = 90°]
So, by RHS congruence axiom, we have
∆OLP ≅ ∆OMP
Hence, ∠OPL = ∠OPM [c.p.c.t.]

Question 5.
If two circles intersect in two points, prove that the line through their centres is the perpendicular bisector of the common chord.
Solution:

Given : Two circles Clo, r) and C(O’, s)intersect at P and Q.
To Prove : OO’ is perpendicular bisector of the chord PQ.
Const. : Join OP, OQ, O’P and O’Q
Proof : In ∆OPO’ and ∆OQO’
OP = OQ [radii of same circle]
OᏢ = QQ [radii of same circle]
OO’ = OO [common]
⇒ ∆OPO’ ≅ ∆OQO’ [by SSS congruence axiom]
⇒ ∠POM = ∠QOM [c.p.c.t.]
Now, in ∆POM and ∆QOM
OP = OQ
(radii of same circle]
∠POM = ∠OOM [proved above]
OM = OM [common]
∆POM ≅ ∆QOM [by SAS congruence axiom]
PM = QM and ∠PMO = ∠QMO [c.p.c.t.]
Also, ∠PMO+ ∠QMO = 180° [linear pair]
⇒ ∠PMO = ∠QMO = 90°
Hence, OO’ is the perpendicular bisector of the chord PQ.

Circles Class 9 Extra Questions Long Answer Type

Question 1.
In the given figure, O is the centre of a circle of radius r сm, OP and OQ are perpendiculars to AB and CD respectively and PQ = 1cm. If AB || CD, AB = 6cm and CD = 8cm, determine r.

Solution:
Since the perpendicular drawn from the centre of the circle to a chord bisects the chord. Therefore, P and Q are mid-points of AB and CD respectively.
Consequently, AP = BP = \(\frac{1}{2}\)AB = 3 cm
and CQ = QD = \(\frac{1}{2}\)CD = 4 cm
In right-angled AQAP, we have
OA² = OP² + AP²
r² = OP² + 3²
r² = OP² + 9
In right-angled ∆OCQ, we have
OC² = OQ² + CQ²
r² = OQ² + 4²
p² = OQ² + 16 … (ii)
From (i) and (ii), we have
OP² + 9 = OQ² + 16
OP² – OQ² = 16 – 9
x² – (x – 1)² = 16 -9 [where OP = x and PQ = 1 cm given]
x² – y² – 1 + 2x = 7
2x = 7 + 1
x = 4
⇒ OP = 4 cm
From (i), we have
r² = (4)² + 9
r² = 16 + 9 = 25
r = 5 cm

Question 2.
In a circle of radius 5 cm, AB and AC are two chords such that AB = AC = 6 cm, as shown in the figure. Find the length of the chord BC.

Solution:
Here, OA = OB = 5 cm [radii]
AB = AC = 6 cm
∴ B and C are equidistant from A.
∴ AO is the perpendicular bisector of chord BC and it intersect BC in M.
Now, in rt. ∠ed ∆AMB, M = 90° ….(i)
∴ By using Pythagoras Theorem, we have
BM² = AB² – AM²
= 36 – AM²
Also, in rt. ∠ed ∆BMO, ∠M = 90°
∴ By using Pythagoras Theorem, we have
BM² = BO² – MO² = 25 – (AO – AM)²
From (i) and (ii), we obtain
25 – (AO – AM)² = 36 – AM\(\frac{1}{2}\)
25 – AOC – AM² + 240 × AM = 36 – AM\(\frac{1}{2}\)
25 – 25 + 2 × 5 × AM = 36
10 AM = 36
AM = 3.6 cm
From (i), we have
BM² = 36 – (3.6)² = 36 – 12.96 = 23.04
BM = √23.04 = 4.8 cm
Thus, BC = 2 × BM
= 2 × 4.8 = 9.6 cm
Hence, the length of the chord BC is 9.6 cm.

Question 3.
In the given figure, AC is a diameter of the circle with centre O. Chord BD is perpendicular to AC. Write down the measures of angles a, b, c and d in terms of x.

Solution:
Here, AC is a diameter of the circle.
∴ ∠ADC = 90°
⇒ ∠a + ∠d = 90°
In right-angled ∆AED, ∠E = 90°
∴ ∠a + 2b = 90°
From (i) and (ii), we obtain
∠b = ∠d ….(iii)
Also, ∠a = ∠c … (iv)
[∠s subtended by the same segment are equal]
Now, ∠AOB and ∠ADB are angles subtended by an arc AB at the centre and at the remaining part of the circle.

Question 4.
Show that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

Solution:
Given : A cyclic quadrilateral ABCD in which AP, BP, CR and DR are the angle bisectors of ∠A, ∠B, 2C and ∠D respectively such that a quadrilateral PQRS is formed. To Prove: PQRS is a cyclic quadrilateral.
Proof : Since ABCD is a cyclic quadrilateral.
∴ ∠A + 2C = 180° and ∠B + ∠D = 180°
Also, AP, BP, CR and DR are the angle bisectors of ∠A, ∠B, ∠C and ∠D respectively.

or ∠1 + ∠3 = 90°
and ∠2 + ∠4 = 90°
Now, in ∆APB, by angle sum property of a ∆
∠1 + ∠2 + ∠P = 180° … (iii)
Again, in ∆CRD, by angle sum property of a ∆
∠3 + ∠4 + ∠R = 180° …(iv)
Adding (iii) and (iv), we have
∠1 + ∠2 + ∠3 + ∠4 + ∠P + ∠R = 180° + 180°
90° + 90° + ∠P + ∠R = 360° [using (ii)]
∠P + ∠R= 360° – 180° = 180°
i.e., the sum of one pair of the opposite angles of quadrilateral PQRS is 180°.
Hence, the quadrilateral PQRS is a cyclic quadrilateral.

Circles Class 9 Extra Questions HOTS

Question 1.
PQ and PR are the two chords of a circle of radius r. If the perpendiculars drawn from the centre of the circle to these chords are of lengths a and b, PQ = 2PR, then prove that:
\(b^{2}=\frac{a^{2}}{4}+\frac{3}{4} r^{2}\)
Solution:
In circle Clo, r), PQ and PR are two chords, draw OM I PQ, OL I PR, such that OM = a
and OL = b. Join OP. Since the perpendicular from the centre of the circle to the chord of the circle, bisects the chord.

Question 2.
Bisectors of angles A, B and C of triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the ∠DEF are 90° – \(\frac{\angle A}{2}\), 90° – \(\frac{\angle B}{2}\) and 90° – \(\frac{\angle C}{2}\) respectively.
Solution:

Let ∠BAD = x, ∠ABE = y
and ∠ACF = 2, then
∠CAD = x, ∠CBE = y
and ∠BCF = 2 [AD, BE and CF is bisector of ∠A, ∠B and ∠C]
In ∆BC,
∠A + ∠B + ∠C = 180°
⇒ 2x + 2y + 2Z = 180°
or x + y + Z = 90° …(i)
Now, ∠ADE = ∠ABE
and ∠ADF = ∠ACF [angles in the same segment of a circle]
⇒ ∠ADE = y and ∠ADF = Z
⇒ ∠ADE + ∠ADF = y + Z
or ∠D = y + Z …(ii)
From (i) and (ii), we have
x + 2D = 90°
⇒ ∠D = 90° – x

Circles Class 9 Extra Questions Value Based (VBQs)

Question 1.
A small cottage industry employing people from a nearby slum area prepares round table cloths having six equal designs in the six segment formed by equal chords AB, BC, CD, DE, EF and FA. If O is the centre of round table cloth (see figure). Find ∠AOB, ∠AEB and ∠AFB. What value is depicted through this question ?

Solution:
Since six equal designs in the six segment formed by equal chords AB, BC, CD, DE, EF and FA.
Therefore, we have six equilateral triangles as shown in the figure. Since ∆AOB, ∆BOC, ∆COD, ∆DOE, ∆EOF
∴ Each angle is equal to 60°.

∠AOB = 60°
∠AOB, ∠AEB and ∠AFB are angles subtended by an arc AB at the FK centre and at the remaining part of the circle.
∴ ∠AEB = ∠AFB = \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) × 60° = 30°
Thus, ∠AEB = ∠AFB = 30°
Value depicted : By employing people from a slum area to prepare round table clothes reali∠e their social responsibility to work for helping the ones in need.

Question 2.
A circular park of radius 10 m is situated in a colony. Three students Ashok, Raman and Kanaihya are standing at equal distances on its circumference each having a toy telephone in his hands to talk each other about Honesty, Peace and Discipline.
(i) Find the length of the string of each phone.
(ii) Write the role of discipline in students’ life.
Solution:
(i) Let us assume A, B and C be the positions of three students Ashok, Raman and Kanaihya
respectively on the circumference of the circular park with centre O and radius 10 m. Since the centre of circle coincides with the centroid of the equilateral ∆ABC.

Thus, the length of each string is 10√3 m.
(ii) In students’ life, discipline is necessary. It motivates as well as nurture the students to make him a responsible citizen.

On this page, you will find why Do We Fall ill Class 9 Notes Science Chapter 13 Pdf free download. CBSE NCERT Class 9 Science Notes Chapter 13 Why Do We Fall ill will seemingly help them to revise the important concepts in less time.

CBSE Class 9 Science Chapter 13 Notes Why Do We Fall ill

Why Do We Fall ill Class 9 Notes Understanding the Lesson

1. Cells are basic unit of living organisms and a lot of activities and chemical reactions are taking place in the cells, so even though they look static they are in a dynamic state as a lot of activities are continuously going inside them.

2. Growth and repair of cells is a continuous process which occurs in living organisms.

3. Activities of all the organs in our body are occurring in a coordinated and interconnected way. For example, poisonous substances would accumulate in the body if the urine is not filtered properly by the kidneys. This in turn would affect the activities of the brain and it will not be able to think properly.

4. Food and energy is needed by all the tissues of a living body in order to function properly.

5. If the functioning of cells and tissues gets adversely affected it would result in lack of proper activity of the body.

6. ‘Health’ is defined as a state of physical, emotional, mental and social well being of a person.

7. The health of an individual depends on the surroundings or the environment of an individual. The environment includes the physical environment.

8. Social environment is also an important factor for health of an individual. Even our physical environment is decided by our social environment.

9. Personal hygiene, public cleanliness, social equality, harmony and good economic conditions are necessary for individual health.

10. Disease refers to any condition that disturbs the normal functioning of the living organism. It is a condition of disturbed ease when an individual feels uncomfortable. Being disease-free is a concept which applies to individual sufferer.

11. Even a disease free person can be considered as having a poor health if the person is unfit in terms of social and mental health. So, health is a concept which applies to societies and communities.

12. The functioning or the appearance of one or more systems of the body gets adversely affected when a person is suffering from a disease. These changes result in symptoms and signs of the disease.

13. Signs and Symptoms

• Symptoms are the physical disturbances that indicate that something is wrong with the functioning of the body and that there may be a disease, but they don’t indicate what the disease is. For example, headache, cough, pus in wound, etc., are symptoms of a disease.
• On the basis of the symptoms, physicians look for the signs of a disease. Signs give a more definite indication of the presence of a particular disease.

14. Types of Disease – Acute disease and chronic disease

• The disease which lasts for only very short periods of time is called acute disease.
  For example: common cold
• The disease which can last for a long time, even as much as a lifetime is called as chronic disease. They have a long-term effect on the health of a person.
  For example: elephantiasis

15. Causes of disease – Immediate and contributory

• Immediate causes: These are the first level causes or the primary cause of the disease.
  Example: The various bacteria, virus, fungi, protozoans, etc., which cause infection in an organism.
• Contributory causes: These are the second level of causes or the internal factors which result in a disease.
  For example: Pollutants in the atmosphere, inadequate diet, poor health, improper nourishment, lack of public services, poor economic condition, genetic abnormalities, etc.
• The immediate cause of disease can be of two types: Infectious diseases and non-infectious dis¬eases.
• Infectious diseases: These diseases are caused by infectious agents. These can spread from one person to another through some medium or by direct contact.
  Example: Pneumonia, common cold, tuberculosis, etc.
• Non-infectious diseases: These do not spread from one person to another person in the community. They are caused due to internal factors.
  Example: Cancer, high blood pressure, diabetes, etc.
• Infectious agents: The organisms (unicellular or multicellular) which cause infection are called as the infective agents.
  Example: virus, bacteria, worms, fungi, etc.

Examples of diseases caused by infectious agents

• By viruses: Common cold, influenza, dengue fever and AIDS.
• By bacteria: Typhoid, cholera, tuberculosis, anthrax, etc.
• By fungi: Many common skin infections like ringworm disease.
• By Protozoans: Malaria and kala-azar.
• By worms: Intestinal worm infections, elephantiasis.

Antibiotic: Chemicals produced by certain microorganisms like bacteria and fungi which inhibit the growth of the other microorganisms are called as antibiotics.

Mode of action of antibiotic: Antibiotics usually block the important biochemical pathways of bacteria. Example: The antibiotic penicillin blocks the processes that help in synthesis of a cell wall in the bacteria. Due to this, the bacteria become unable to make cell walls to protect them and die easily. Human cells don’t make a cell wall, so penicillin does not affect them.

16. Antibiotics do not work against viruses:
Reason: Viruses do not use their own biochemical pathways for the synthesis of their cell wall. They have very few biochemical mechanisms. So, antibiotics do not work against viral infections.

17. Means of spread of diseases
Infectious diseases can spread from an infected person to a healthy person by various transmitting agents, so they are called as communicable diseases. These diseases can be spread through air, water, vectors and physical contact.

(i) By air: The little droplets thrown out by an infected person who sneezes or coughs can result in a new infection in a healthy person if the person breathes in the droplets released by the infected person. Example: common cold, pneumonia and tuberculosis.

(ii) By water: If the excreta from someone suffering from an infectious gut disease, such as cholera, get mixed with the drinking water used by people living nearby. The microbes will enter new hosts through the water they drink and cause disease in them.
Example: cholera.

(iii) By Sexual contact: Some diseases are transmitted by sexual contact from one partner to the other. Example: syphilis, AIDS, etc.
It is important to note that: AIDS does not spread by casual physical contacts like handshakes or hugs or sports, like wrestling. AIDS can spread in four ways, sexual contact, transfusion of AIDS infected blood, from an infected mother to her baby during pregnancy or through breast feeding.

(iv) By vectors: Many diseases are transmitted by animals from an infected person to a healthy person. These animals act as intermediaries and are called vectors. The most common example of vectors are mosquitoes. The females of many species of mosquitoes need highly nutritious food in the form of blood in order to be able to lay mature eggs.

18. Different species of microbes have evolved to affect different parts of the body. The selection is generally connected to their point of entry.

• They usually go to the lungs if they enter from the air via the nose.
  Example: bacteria causing tuberculosis.
• They can stay in the gut lining or go to the liver if they enter through the mouth. Example: bacteria which causes typhoid remains in the gut lining whereas the viruses that cause jaundice go to the liver.

19. Other ways are:

• The AIDS causing virus HIV comes into the body via the sexual organs and then spreads to lymph nodes all over the body.
• Malaria-causing microbes which enter through a mosquito bite go to the liver, and then to the red blood cells.
• The virus causing Japanese encephalitis or brain fever enters through a mosquito bite and goes on to infect the brain.

20. The signs and symptoms of a disease depend on the tissue or organ which the microbe targets.

• If the lungs are the targets, then symptoms will be cough and breathlessness.
• If the liver is the target, there will be jaundice.
• If the brain is the target, it will result in headache, vomiting, fits or unconsciousness.

21. Inflammation: An active immune system recruits many cells to the affected tissue to kill off the disease- causing microbes. This recruitment process is called inflammation. Local effects such as swelling and pain, and general effects such as fever are caused as a part of this process.

In AIDS, the virus called HIV goes to the immune system and damages its function. As a result, the body can no longer fight off even the minor infections. For example, a very small cold can become pneumonia or minor gut infection can produce major diarrhoea with blood loss. Ultimately, the immune system of the person becomes very weak and these and other infections kill people suffering from HIV-AIDS.

22. Principles of Treatment
There are two ways for treatment of infectious diseases:

•  to reduce the effects of the disease: The main focus is to reduce the symptoms which are usually due to inflammation. For example, rest is taken to conserve energy, medicines are taken to bring down fever or to reduce pain or loose motions, etc.
• to kill the cause of the disease: The main focus on this is to get rid of the disease causing microbe from the body. For example, medicines like antibiotics are taken to kill microbes.

23. Principles of Prevention
Prevention is better than cure because the methods which deal with getting rid of an infection have three limitations.

• During treatment, the body functions of the person get damaged and may never recover completely.
• The person suffering from a disease is likely to be dridden for some time as the treatment of a disease takes time.
• The person suffering from an infectious disease can serve as the source of spread of infection to other people.

23. Ways of Prevention of disease are
(i) General ways: These focus on preventing exposure to the disease causing organisms. Public hygiene is important for the prevention of infectious diseases.
For example,

• By providing living conditions that are not overcrowded, exposure to air-borne infections can be avoided.
• By providing safe drinking water and treating water to kill microbes in it, the water-borne infections can be prevented.
• By providing clean environment which do not allow spread of microbes/vectors, the vector-borne infections can be prevented.
• For vector-borne infections, we can provide clean environments. This would not, for example, allow mosquito breeding.

(ii) Specific ways: These methods are adopted to make the immune system of the person strong by vaccination to create immunisation and taking proper and sufficient food for keeping the body healthy.

24. Vaccine: A modified suspension of disease causing microorganisms which have been weakened so that they are not able to cause disease but help to activate the immune system of the organism to produce antibodies against the disease causing organism. Edward Jenner developed the first vaccine against the smallpox disease. The process of inoculation of a vaccine in a person to develop immunity against a disease is called vaccination.

Class 9 Science Chapter 13 Notes Important Terms

Health: The state of physical, emotional, mental and social well being of an individual.

Disease: Any condition that causes disturbed ease in a person or disturbs the normal functioning of the body.

Acute disease: The disease which lasts for a short time.

Chronic disease: The disease which persists for a long time.

Infectious disease: A disease which can spread from one person to other. E.g., common cold

Non-infectious disease: A disease which cannot spread from one person to other. E.g., diabetes

Antibiotics: The chemicals secreted by some microbes like fungi and bacteria which are used to control the growth of bacteria or kill the bacteria.

Immunity: The ability of the body to fight against the disease causing organisms.

Vaccine: A biological preparation that provides immunity to a person against a disease.

Vector: The organism or the agent which helps in the transmission of a disease.

Inflammation: It is the process by which the white blood cells of our blood protect us from the infections by disease-causing organisms.

Symptom: The physical disturbances which indicate the presence of a disease.

Syndrome: Group of symptoms

AIDS: Acquired immunodeficiency syndrome

HIV: Human immuno-deficiency virus

Immunisation: The process by which a person is made immune to an infectious disease . administering a vaccine.