## RD Sharma class 10 solutions Chapter 8 Quadratic Equations Ex 8.11

### RD Sharma Class 10 Solutions Quadratic Equations Exercise 8.11

**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 = \(\frac { 82 }{ 2 }\) = 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,

\(\frac { 1 }{ 2 }\) x (x + 7) = 165

⇒ \(\frac { 1 }{ 2 }\) (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 = \(\frac { 80 }{ 2 }\) = 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 = \(\frac { x }{ 4 }\) cm

and side of second square = (\(\frac { x }{ 4 }\) + 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 = \(\frac { -33 }{ 2 }\) 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 centre 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 centre of a rectangular lawn of dimensions 50 m x 40 m. So, the distance between pond and lawn would be same around the pond. Say x m.

Now, length of rectangular lawn (l_{1}) = 50 m

and breadth of rectangular lawn (b_{1}) = 40 m

Length of rectangular pond (l_{2}) = 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

l_{1} x b_{1} – l_{2} x b_{2}= 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.

**Exercise 8.11**

#### RD Sharma Class 10 Solutions

- Chapter 8 Quadratic Equations Ex 8.1
- Chapter 8 Quadratic Equations Ex 8.2
- Chapter 8 Quadratic Equations Ex 8.3
- Chapter 8 Quadratic Equations Ex 8.4
- Chapter 8 Quadratic Equations Ex 8.5
- Chapter 8 Quadratic Equations Ex 8.6
- Chapter 8 Quadratic Equations Ex 8.7
- Chapter 8 Quadratic Equations Ex 8.8
- Chapter 8 Quadratic Equations Ex 8.9
- Chapter 8 Quadratic Equations Ex 8.10
- Chapter 8 Quadratic Equations Ex 8.11
- Chapter 8 Quadratic Equations Ex 8.12
- Chapter 8 Quadratic Equations Ex 8.13