Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 14 Proportion Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 14 Solutions Proportion
Proportion Class 9 Kerala Syllabus Questions and Answers
Kerala State Syllabus 9th Standard Maths Chapter 14 Proportion Solutions Questions and Answers
Class 9 Maths Chapter 14 Kerala Syllabus – Proportional Changes
Textual Questions And Answers
Question 1.
In each of the instances below, show that the second quantity changes proportionally with respect to the first. Also find the proportionally constant in each:
i) The length of sides of squares and their perimeters.
ii) The lengths of wires bent into squares and the length of the sides of the squares.
iii) The number of rotations of a circle rolling along a line, and the distance travelled along the line.
Answer:
i) Let s be the length of a side of a square. The perimeter P of a square is given by:
P = 4s
Now, we can express the ratio of the perimeter to the side length:
\(\frac{\mathrm{P}}{\mathrm{~s}}=\frac{4 \mathrm{~s}}{\mathrm{~s}}\) = 4
Since this ratio is constant (4), we conclude that the perimeter changes proportionally with respect to the length of the sides of the squares. The proportionality constant is 4.
ii) Let L be the total length of the wire, and let s be the length of a side of the square formed by bending the wire. The total length of the wire is given by:
L = 4s
Now, we express the ratio of the total length of the wire to the side length:
\(\frac{L}{s}=\frac{4 s}{s}\) = 4
Since the ratio is constant (4), we find that the lengths of wires bent into squares change proportionally with respect to the length of the sides of the squares. The proportionality constant is 4.
iii) Let r be the radius of the circle. The circumference C of the circle is given by:
C = 2πr
When a circle rolls along a line, the distance d travelled along the line is equal to the number of rotations n times the circumference of the circle:
d = nC
Substituting the circumference:
d = n (2πr)
Now, we express the ratio of the ratio of the distance travelled to the number of rotations:
\(\frac{\mathrm{d}}{\mathrm{n}}\) = 2πr
Since this ratio is constant (specifically, 27rr), we find that the distance travelled along the line changes proportionally with respect to the number of rotations of the circle. The proportionality constant is 2πr.
Question 2.
We have seen that in the picture, the height of a point on the slanted line from the horizontal line changes proportionally with respect to its distance from the corner. Calculate the proportionality constants 30°, 45° and 60°.
Answer:
When the angle is 30°
A base triangle ΔABC is drawn such that AC = 2cm, BC = 1cm and AB = √3 cm
Here triangle ABC and APQ are similar, we get
\(\frac{A C}{A Q}=\frac{B C}{P Q}=\frac{A B}{A P}\)
BC and PQ are heights and AC and AQ are distance.
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{2}{1}\)
AQ = 2 PQ
Therefore, the proportionality constant is 2
Similarly, when the angle is 60°
A base triangle ΔABC is drawn such that AC = 2cm, BC = √3 cm and AB = 1cm
Here triangle ABC and APQ are similar, we get
\(\frac{A C}{A Q}=\frac{B C}{P Q}=\frac{A B}{A P}\)
BC and PQ are heights and AC and AQ are distance.
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{2}{\sqrt{3}}\)
AQ = \(\frac{2}{\sqrt{3}}\)PQ
Therefore, the proportionality constant is \(\frac{2}{\sqrt{3}}\)
Similarly, when the angle is 45°
A base triangle A ABC is drawn such that AC = √2 cm, BC = 1 cm and AB = 1 cm
Here triangle ABC and APQ are similar, we get
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}=\frac{\mathrm{AB}}{\mathrm{AP}}\)
BC and PQ are heights and AC and AQ are distance.
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{\sqrt{2}}{1}\)
AQ = \(\frac{\sqrt{2}}{1}\) PQ
Therefore, the proportionality constant is √2.
Question 3.
Prove that in equilateral triangles, the perimeter changes proportionally with respect to the length of the sides. What is the proportionality constant? What can we say about other regular polygons?
Answer:
Let’s consider an equilateral triangle with side length’s’.
Perimeter of an equilateral triangle, P = 3s
The perimeter of an equilateral triangle changes proportionally with respect to the length of its sides, with a proportionality constant of 3.
Let’s consider a regular polygon with ‘n’ sides, each of length’s’.
Perimeter of a regular polygon, P = ns
The perimeter of any regular polygon changes proportionally with respect to the length of its sides,
with a proportionality constant equal to the number of sides.
Question 4.
Prove that the lengths of arcs of a fixed circle change proportionally with respect to their central angles. What is the proportionality constant? What about the relation between the area of a sector and its central angle?
Answer:
Let’s consider a circle with radius ‘r’ and central angle ‘θ’
Arc Length (L) = θ × r
The lengths of arcs of a fixed circle change proportionally with respect to their central angles, with a proportionality constant of r.
Area of a sector (A) = \(\frac{1}{2}\) r² θ
The area of a sector of a fixed circle changes proportionally with respect to its central angle with a proportionality constant of \(\frac{1}{2}\) r².
Class 9 Maths Kerala Syllabus Chapter 14 Solutions – Scale And Proportion
Textual Questions And Answers
Question 1.
i) What is the sum of the angles of a triangle? And the sum of the angles of a hexagon?
ii) Does the sum of the angles of polygons change proportionally with respect to the number of sides? Explain the reason.
Answer:
i) Sum of the interior angles of a triangle = (3 – 2) × 180° = 180°
Sum of the interior angles of a hexagon = (6 – 2) × 180° = 720°
ii) Yes, the sum of the angles of polygons changes proportionally with respect to the number of sides.
The sum of interior angles (S) of a polygon with ‘n’ sides is given by:
S = (n- 2) × 180°
The sum of the angles of polygons changes proportionally with respect to the number of sides with proportionality constant 180°
Question 2.
Inside a triangle of base 6 centimeters and height 3 centimeters, lines are drawn parallel to the base. Prove that the lengths of these lines change proportionally with respect to the distance from the top vertex. Find the proportionality constant.
Answer:
Given, Base = 6 cm
Height = 3 cm
Consider parallel lines drawn from the top vertex.
Distance from top vertex = x
Length of parallel line = y
Using similar triangle property,
\(\frac{y}{x}=\frac{6}{3}\)
y = 2x
Thus we can say that lengths of these lines change proportionally with respect to the distance from the top vertex with proportionality constant 2
Question 3.
Within a semicircle of diameter 10 centimeters, lines are drawn parallel to the diameter:
(i) In each of the pictures below, calculate the length of the line parallel to the diameter:
ii) Does the length of the parallel line change proportionally with respect to the distance from the top of the semicircle. Explain the reason.
Answer:
i) Figure 1
Length of the parallel line = 2\(\sqrt{(5)^2-(4)^2}\) = 2 × 3 = 6 cm
Figure 2
Length of the parallel line = 2\(\sqrt{(5)^2-(3)^2}\) = 2 × 4 = 8 cm
ii) The length of the parallel line does not change proportionally with respect to the distance from the top of the semicircle.
SCERT Class 9 Maths Chapter 14 Solutions – Different Proportions
Textual Questions And Answers
Question 1.
i) Prove that the areas of equilateral triangles change proportionally with respect to the squares of the lengths of sides. What is the proportionality constant?
Answer:
Areas of equilateral triangles change proportionally with respect to the squares of the lengths of sides with proportionality constant of \(\frac{\sqrt{3}}{4}\).
ii) Are the areas of squares proportional to the squares of the lengths of sides? If so, what is the proportionality constant?
Answer:
Area of the squares are proportional to the squares of the lengths of side with proportionality constant of 1.
Question 2.
Consider all rectangles of area 1 square meter. The length of one side of such a rectangle depends on the length of the other side. Write this relation as an algebraic equation. How do we state it in terms of proportion?
Answer:
In rectangles of area one square meter
Let x be the length and y be the breadth of the rectangle
Then, area = length × breadth
1 = x × y
1 = x y
y = \(\frac{1}{x}\)
y is inversely proportional to x
Question 3.
Consider all triangles of a fixed area. How do we state in terms of proportion, the relation between the length of the longest side and the length of the perpendicular to it from the opposite vertex? What if we use the shortest side instead of the longest?
Answer:
Let ‘a’ be the longest side, ‘h’ be the length of perpendicular from opposite vertices, ‘A’ be the area then
A = \(\frac{1}{2}\)ah,
a = \(\frac{2A}{h}\)
Length of larger side is inversely proportional to the length of perpendicular from the opposite vertex.
That is, the length of small side is inversely proportional to the length of perpendicular line from the vertex of small side.
Question 4.
In regular polygons, can we say the relation between the number of sides and the measure of an outer angle, in terms of proportion? What is the proportionality constant?
Answer:
The sum of the exterior angles of all polygon is 360°.
If ‘n’ is the number of sides.
Measure of an exterior angle = \(\frac{\text { sum of exterior angle }}{\text { number of sides }}\)
If the measure of an outer angle is ‘x’
x = \(\frac{360^{\circ}}{n}\)
One outer angle and number of sides are inversely proportional.
The constant of proportionality is \(\frac{1}{n}\)
Proportion Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed by using different pipes. Write the relations between the following quantities as an algebraic equation and in terms of proportions.
i) The rate of water flow and the height of the water level.
ii) The rate of water flow and the time taken to fill the tank.
Answer:
i) Let x be the rate of water flowing, y be the height of water in the tank and A be the base area of the tank, then x = Ay
Height of the water level in the tank is proportional to the rate of the water flowing.
ii) If C is the volume of the tank, V be the volume of water flowing per second, the volume of water in’t’ second is given by C = V t
V = C × \(\frac{1}{t}\)
That is the rate of water flow and the time taken for filling the tank are inversely proportional.
C is the constant of proportionality.
Question 2.
Raghu invested Rs. 60000 and Nazar Rs. 100000 and started a business. Within one month a profit of Rs. 4800 was obtained. Raghu took 1800 and Nazar took Rs. 3000 out of the profit obtained. What is the ratio of the investment? Is the investment and the profit divided proportionally?
Answer:
Ratio of investments = 60000: 100000 = 6:10 = 3:5
Ratio of profit divided 1800: 3000 = 18:30 = 3:5
Ratio of investments and Ratio of profit divided are equal.
Hence they are proportional.
Question 3.
Are the length and breadth of a square having same perimeter inversely proportional?
Answer:
For a square the perimeter is 20 cm. so, length + breadth = 10.
Let x be the length and y be the breadth then possible values of x and y are
When x = 9 then y = 1
When x = 8 then y = 2
Here xy is not a constant term i.e. the changes in the x and y is not in the firm of xy = kept
Hence length and breadth are not in inversely proportional.
Question 4.
A car with 5L of petrol travels a distance of 75 km. What is the proportionality constant between the distance travelled and the quantity of petrol? How much petrol is needed for travelling 180 km?
Answer:
Taking the distance travelled as x and quantity of petrol as y, then the constant of proportionality 15
k = \(\frac{y}{x}=\frac{75}{5}\) = 15
k = \(\frac{y}{x}\) = 15 = \(\frac{100}{x}\)
x = \(\frac{180}{15}\) = 12
The quantity of petrol needed to travel 180 km = 12 litre