RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.11 (Updated for 2021-22)

RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11: The utilization of quadratic equations to solve problems with mensuration is highlighted in this exercise. Students who are having difficulty solving problems can use the RD Sharma Class 10 Solutions. The RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 PDF can also be used to clear up any conceptual doubts about this exercise.

Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 Free PDF

RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11

Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.11- Important Question with Answers

Question 1.
The perimeter of a rectangular field is 82 m and its area is 400 m². Find the breadth of the rectangle.
Solution:
Perimeter of a rectangle field = 82 m
Length + Breadth = 822 = 41 m
Let breadth = x m
Length = (41 – x) m
According to the condition,
Area = Length x breadth
400 = x (41 – x)
⇒ 400 = 4x – x²
⇒ x² – 41x + 400 = 0
⇒ x² – 16x – 25x + 400 = 0
⇒ x (x – 16) – 25 (x – 16) = 0
⇒ (x – 16) (x – 25) = 0
Either x – 16 = 0, then x = 16
or x – 25 = 0 then x = 25
25 > 16 and length > breadth
Breadth = 16 m

Question 2.
The length of a hall is 5 m more than its breadth. If the area of the floor of the hall is 84 m², what are the length and breadth of the hall?
Solution:
Let breadth of the hall = x m
Then length = x + 5
Area of the floor = 84 m2
Now according to the condition,
x (x + 5) = 84
⇒ x² + 5x – 84 = 0
⇒ x² + 12x – 7x – 84 = 0
⇒ x (x + 12) – 7 (x + 12) = 0
⇒ (x + 12) (x – 7) = 0
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 7 = 0, then x = 7
Breadth of the hall = 7 m and length = 7 + 5 = 12 m

Question 3.
Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm². Find the sides of the squares.
Solution:
Side of first square = x cm
and side of the second square = (x + 4) cm
According to the condition,
x² + (x + 4)² = 656
⇒ x² + x² + 8x + 16 = 656
⇒ 2x² + 8x + 16 – 656 = 0
⇒ 2x² + 8x – 640 = 0
⇒ x² + 4x – 320 = 0 (Dividing by 2)
⇒ x² + 20x – 16x – 320 = 0
⇒ x (x + 20) – 16 (x + 20) 0
⇒ (x + 20) (x – 16) = 0
Either x + 20 = 0, then x = -20 which is not possible being negative
or x – 16 = 0, then x = 16
Side of first square = 16 cm
and side of second square = 16 + 4 = 20 cm

Question 4.
The area of a right-angled triangle is 165 m². Determine its base and altitude if the latter exceeds the former by 7 m.
Solution:
Area of a right angled triangle = 165 m²
Let its base = x m
Then altitude = (x + 7) m
According to the condition,
12 x (x + 7) = 165
⇒ 12 (x² + 7x) = 165
⇒ x² + 7x = 330
⇒ x² + 7x – 330 = 0
⇒ x² + 22x – 15x – 330 = 0
⇒ x (x + 22) – 15 (x + 22) = 0
⇒ (x + 22) (x – 15) = 0
Either x + 22 = 0, then x = -22 which is not possible being negative
or x – 15 = 0, then x = 15
Base = 15 m
and altitude = 15 + 7 = 22 m

Question 5.
Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m²? If so, find its length and breadth.
Solution:
Area of rectangular mango grove = 800 m²
Let breadth = x m
Then length = 2x m
According to the condition,
2x x x = 800
⇒ 2x² = 800
⇒ x² = 400 = (±20)²
Yes, it is possible,
x = 20, -20
But x = -20 is not possible being negative
Breadth = 20 m
and length = 20 x 2 = 40 m

Question 6.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadth:
Solution:
Perimeter of rectangular park = 80 m
Length + Breadth = 802 = 40 m
Let length = x m
Them breadth = 40 – x
According to the condition,
Area = Length x Breadth
x (40 – x) = 400
⇒ 40x – x² = 400
⇒ x² – 40x + 400 = 0
⇒ (x – 20)² = 0
⇒ x – 20 = 0
⇒ x = 20
Yes, it is possible
Length = 20 m
and breadth = 40 – x = 40 – 20 = 20 m

Question 7.
Sum of the areas of two squares is 640 m². If the difference of their perimeters is 64 m, find the sides of the two squares. [CBSE 2008]
Solution:
Let side of first square = x m
and of second squares = y m
According to the given conditions,
4x – 4y = 64
⇒ x – y = 16 ….(i)
and x² + y² = 640 ….(ii)
From (i), x = 16 + y
In (ii)
(16 + y)² + y² = 640
⇒ 256 + 32y + y² + y² = 640
⇒ 2y² + 32y + 256 – 640 = 0
⇒ y² + 16y – 192 = 0 (Dividing by 2)
⇒ y² + 24y – 8y – 192 = 0
⇒ y (y + 24) – 8 (y + 24) = 0
⇒ (y + 24)(y – 8) = 0
Either y + 24 = 0, then y = -24, which is not possible as it is negative
or y – 8 = 0, then y = 8
x = 16 + y = 16 + 8 = 24
Side of first square = 24 m
and side of second square = 8m

Question 8.
Sum of the areas of two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of two squares. [CBSE 2013]
Solution:
Let perimeter of the first square = x cm
Then perimeter of second square = (x + 16) cm
Side of first square = x4 cm
and side of second square = (x4 + 4) cm
Sum of areas of these two squares = 400 cm²

Question 9.
The area of a rectangular plot is 528 m². The length of the plot (in metres) is one metre more then twice its breadth. Find the length and the breadth of the plot. [CBSE 2014]
Solution:
Area of a rectangular plot = 528 m²
Let breadth = x m
Then length = (2x + 1) m
x (2x + 1) = 528 (∴ Area = l x b)
2x² + x – 528 = 0
⇒ 2x² + 33x – 32x² – 528 = 0
⇒ x (2x + 33) – 16 (2x + 33) = 0
⇒ (2x + 33) (x – 16) = 0
Either 2x + 33 = 0 then 2x = – 33 ⇒ x = −332 but it is not possible being negative
or x – 16 = 0, then x = 16
Length = 2x + 1 = 16 x 2 + 1 = 33 m
and breadth = x = 16 m

Question 10.
In the center of a rectangular lawn of dimensions 50 m x 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m². Find the length and breadth of the pond. [NCERT Exemplar]
Solution:
Given that a rectangular pond has to be constructed in the center of a rectangular lawn of dimensions 50 m x 40 m. So, the distance between the pond and the lawn would be the same around the pond. Say x m.

Now, length of rectangular lawn (l₁) = 50 m
and breadth of rectangular lawn (b₁) = 40 m
Length of rectangular pond (l₂) = 50 – (x + x) = 50 – 2x
Also, area of the grass surrounding the pond = 1184 m²
Area of rectangular lawn – Area of rectangular pond = Area of grass surrounding the pond
l₁ x b₁ – l₂ x b₂= 1184 [∵ area of rectangle = length x breadth]
⇒ 50 x 40 – (50 – 2x) (40 – 2x) = 1184
⇒ 2000 – (2000 – 80x – 100x + 4x²) = 1184
⇒ 80x + 100x – 4x² = 1184
⇒ 4x² – 180x + 1184 = 0
⇒ x² – 45x + 296 = 0
⇒ x² – 21x – 8x + 296 = 0 [by splitting the middle term]
⇒ x (x – 37) – 8 (x – 37) = 0
⇒ (x – 37) (x – 8) = 0
∴ x = 8
[At x = 37, length and breadth, of pond are -24 and -34, respectively but length and breadth cannot be negative. So, x = 37 cannot be possible]
Length of pond = 50 – 2x = 50 – 2(8) = 50 – 16 = 34 m
and breadth of pond = 40 – 2x = 40 – 2(8) = 40 – 16 = 24 m
Hence, required length and .breadth of pond are 34 m and 24 m, respectively.

We have provided complete details of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.

FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11