RD Sharma Solutions for Class 12 Maths Exercise 21.1 Chapter 21 Areas of Bounded Regions are given here. The foremost objective is to help students understand and answer these questions. Students obtain more knowledge by referring to RD Sharma Solutions for Class 12 Maths Chapter 21. Download the PDF of Class 12 Chapter 21 from their respective links.
RD Sharma Solutions for Class 12 Maths Chapter 21 Exercise 1
Access RD Sharma Solutions for Class 12 Maths Chapter 21 Exercise 1
EXERCISE 21.1
Question. 1
Solution:
From the question, it is given that two equations,
x = 2 … [equation (i)]
y2 = 8x … [equation (ii)]
So, equation (i) represents a line parallel to the y-axis and equation (ii) represents a parabola with vertex at the origin and the x-axis as it axis, as shown in the rough sketch below,
Now, we have to find the area of OCBO,
Then, the area can be found by taking a small slice in each region of width Δx,
And length = (y – 0) = y
The area of the sliced part will be as it is a rectangle = y Δx
So, this rectangle can move horizontal from x = 0 to x = 2
The required area of the region bounded between the lines = Region OCBO
= 2 (region OABO)
Now, applying limits we get,
= 4√2[((2/3)2√2) – ((2/3)0√0)]
= 4√2(4√2/3)
Therefore, the required area = 32/3 square units.
Question. 2
Solution:
From the question, it is given that equation,
y – 1 = x … [equation (i)]
So, equation (i) represents a line that meets at (0, 1) and (-1, 0), is as shown in the rough sketch below,
Now, we have to find the area of the region bounded by the line y – 1 = x,
So, Required area = Region ABCA + Region ADEA
= [((9/2) + 3) – ((½) – 1)] + [((½) – 1) – (2 – 2)]
= [(15/2) + (½)] + [-½]
On simplification, we get,
= 8 + ½
A = 17/2 square units
Therefore, the area of the region bounded by the line y – 1 = x is 17/2 square units.
Question. 3
Solution:
From the question, it is given that two equations,
x = a … [equation (i)]
y2 = 4ax … [equation (ii)]
So, equation (i) represents a line parallel to the y-axis and equation (ii) represents a parabola with vertex at the origin and the x-axis as it axis, as shown in the rough sketch below,
Now, we have to find the area of OCBO,
Then, the area can be found by taking a small slice in each region of width Δx,
And length = (y – 0) = y
The area of the sliced part will be as it is a rectangle = y Δx
So, this rectangle can move horizontally from x = 0 to x = a
The required area of the region bounded between the lines = Region OCBO
Question. 4
Solution:
From the question, it is given that equation,
y = 4x – x2
Adding 4 on both sides,
y + 4 = 4x – x2 + 4
Transposing we get,
x2 – 4x + 4 = – y + 4
We know that (a – b)2 = a2 – 2ab + b2
(x – 2)2 = – (y – 4) … [equation (i)]
So, equation (i) represents a downward parabola with vertex (2, 4) and passing through (0, 0) and (0, 4), is as shown in the rough sketch below,
Then, the area can be found by taking a small slice in each region of width Δx,
And length = (y – 0) = y
The area of the sliced part will be as it is a rectangle = y Δx
So, this rectangle can move horizontally from x = 0 to x = a
The required area of the region bounded between the lines = Region OABO
Now we have to apply limits,
= [((4 × 16)/2) – (64/3)] – [0 – 0]
On simplification, we get,
= 64/6
Dividing both the numerator and denominator by 2 we get,
= 32/3
Therefore, the required area is 32/3 square units.
Question. 5
Solution:
From the question, it is given that two equations,
y2 = 4x … [equation (i)]
x = 3 … [equation (ii)]
So, equation (i) represents a parabola with vertex at the origin and the x-axis as its axis and equation (ii) represents a line parallel to the y-axis, as shown in the rough sketch below,
Now, we have to find the area of OCBO,
Then, the area can be found by taking a small slice in each region of width Δx,
And length = (y – 0) = y
The area of the sliced part will be as it is a rectangle = y Δx
So, this rectangle can move horizontal from x = 0 to x = 3
The required area of the region bounded between the lines = Region OCBO
= 2 (region OABO)
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