RD Sharma Solutions Chapter 8 Quadratic Equations Exercise 8.11 Class 10 Maths
Chapter Name  RD Sharma Chapter 8 Quadratic Equations 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 8.11 Solutions
1. The perimeter of a rectangular field is 82m and its area is 400 m^{2} . Find the breadth of the rectangle.
Solution
Let the breadth of the rectangle be x meters
Given that,
Perimeter = 82 m and Area = 400 m^{2}
We know that
Perimeter of a rectangle = 2(length + breadth)
⇒ 82 = 2(length + x)
⇒ 41 = length + x
⇒ length = (41  x)m
2. The length of a hail is 5m more than its breadth. If the area of the floor of the hail is 84 m^{2} , what are the length and breadth of the hail ?
Solution
Let the breadth of the rectangle (hall) be x meter.
Given that,
Length of the hall is 5 m more than its breadth i.e., length = (x + 5)m
Since, x cannot be negative. So, breadth of the hall = 7m
Hence, length of the hall = (x + 5)m = (7 + 5)m = 12m.
3. Two squares have sides x cm and (x + 4)cm. The sum of their areas is 656 cm^{2} . Find the sides of the squares.
Solution
Let S_{1} and S_{2} be two squares
Let x cm be the side of square S_{1} and (x + 4) cm be the side of square S_{2} .
We know that,
Area of a square = (side)^{2}
⇒ Area of square S_{1} = (x)^{2} = x^{2} cm^{2}
⇒ Area of square S_{2} = (x + 4)^{2} cm^{2}
Given that,
Area of square S_{1} + Area of square S_{2} = 656 cm^{2}
⇒ x^{2} cm^{2} + (x + 4)cm^{2} = 656 cm^{2}
4. The area of a right angled triangle is 165 m^{2} . Determine its base and altitude if the latter exceeds the former by 7 m.
Solution
Let the altitude of the right angled triangle be denoted by x meter
Given that altitude exceeds the base of the triangle by 7m.
⇒ Base = (x  7)m
Since, x cannot be negative. So, x = 22 m
∴ Altitude of the triangle ⇒ x = 22 m
And base of the triangle ⇒ (x  7)m = (22  7)m = 15m
5. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m^{2} ? If so, find its length and breadth.
Solution
Let the breadth of the rectangular mango grove be x meter.
Given that length is twice that of breadth ⇒ length = 2×x m = 2x m
Given that area of the grove is 800 m^{2} .
Since, x cannot be a negative value.
So, x = 20 m
∴ Breadth of the mango grove = 20 m and length of the mango grove
= 2x m = 2×20 = 40 m
Yes. It is possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m^{2}.
6. Is it possible to design a rectangular park of perimeter 80 m and area 400 m^{2} ? If so, find its length and breadth.
Solution
To prove the given condition, let us assume that the length of the rectangular park be denoted by x m.
Given that
Perimeter = 80m and
Area = 400 m^{2}
We know that,
Perimeter of a rectangle = 2(length + breadth)
∴ Length of the rectangular park ⇒ x = 20 m and breadth of the rectangular park ⇒ (40 x)m = (40  20)m = 20m
Yes, it is possible to design a rectangular park of perimeter 80m and area 400m^{2}.
7. Sum of the areas of two squares is 640 m^{2} . If the difference of their perimeters is 64 m, find the sides of the two squares.
Solution
Let the two squares be denoted as S_{1} and S_{2} and let side of squares S_{1} be denoted as x meter and that of square S_{2} by y m.
Given that,
Difference of their perimeter is 64m.
We know that
Perimeter of a square = 4×side
⇒ Perimeter of a square S_{1} = 4 × x m = 4x m
⇒ Perimeter of square S_{2} = 4 × y m = 4y m
Now, difference of perimeter = perimeter of square S_{1}  Perimeter of square S_{2}
⇒ 64m = (4x  4y)m
⇒ 64 = 4(x  y)
⇒ x  y = 16
⇒ x = y + 16
And also,
Given that sum of areas of two squares = 640 m^{2} .
We know that,
Area of a square = (Side)^{2}
⇒ Area of square S_{1} = x^{2} m^{2} .
⇒ Area of square S_{2} = y^{2} m^{2}
Now,
Sum of areas of two squares = Area of square S_{1} + Area of square S_{2}
Since, y cannot be a negative value, So, y = 8m
∴ Side of the square S_{2} is y = 8m
And side of the square S_{1} is x = (y + 16)m = (8 + 16)m = 24 m
Hence, sides of the two squares is 24m and 8m.