In this chapter, we provide RD Sharma Class 10 Ex 8.11 Solutions Chapter 8 Quadratic Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 8.11 Solutions Chapter 8 Quadratic Equations pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 8 |

Chapter Name | Quadratic Equations |

Exercise | 8.11 |

Category | RD Sharma Solutions |

**RD Sharma Solutions for Class 10 Chapter** **8**** Quadratic Equations** Ex 8.11 Download PDF

**Quadratic Equations**Ex 8.11 Download PDF

**Chapter 8: Quadratic Equations Exercise – 8.11**

**Question: 1**

The perimeter of the rectangular field is 82 m and its area is 400 m^{2}. Find the breadth of the rectangle?

**Solution:**

Let the breadth of the rectangle be (x) m

Given,

Perimeter = 82 m

Area = 400 m^{2}

Perimeter of a rectangle = 2(length + breadth)

80 = 2(length + x)

41 = (length + x)

Length = (41 – x)m

We know,

Area of the rectangle = length * breadth

400 = (41 – x)(x)

400 = 41x – x^{2}

= x^{2 }– 41x +400 = 0

= x^{2 }– 25x – 16x + 400 = 0

= x(x – 25) – 16(x – 25) = 0

= (x – 16)(x – 25) = 0

Either x – 16 = 0 therefore x = 16

Or, x – 25 = 0 therefore x = 25

Hence the breadth of the above mentioned rectangle is either 16 m or 25 m respectively.

**Question: 2**

The length of the hall is 5 m more than its breadth. If the area of the floor of the hall is 84 m^{2}, what is the length and breadth of the hall?

**Solution:**

Le the breadth of the rectangle be x m

Let the length of the hall is 5 m more than its breadth = (x + 5) m

Also given that,

Area of the hall is = 84 m^{2}

The shape of the hall is rectangular

Area of the rectangular hall = length * breadth

84 = x(x + 5)

= x^{2 }+ 5x – 84 = 0

= x^{2 }+ 12x – 7x – 84 = 0

= x(x + 12) – 7(x + 12) = 0

= (x + 12)(x – 7) = 0

Either x + 12 = 0 therefore x = – 12

Or, x – 7 = 0 therefore x = 7

Since the value of x cannot be negative

So x = 7

= x + 5 = 12

The length and breadth of the rectangle is 7 and 12 respectively.

**Question: 3**

Two squares have sides x and (x + 4) cm. The sum of their area is 656 cm^{2}. Find the sides of the square.

**Solution:**

Let S_{1} and S_{2} be the two square

Let x cm be the side square S_{1} and (x + 4) cm be the side of the square S_{2}.

Area of the square S_{1} = x^{2} cm^{2}

Area of the square S_{2} =(x + 4)^{2} cm^{2}

According to the question,

Area of the square S_{1 }+ Area of the square S_{2 }= 656 cm^{2}

= x^{2} cm^{2 }+ (x + 4)^{2} cm^{2 }= 656 cm^{2}

= x^{2}+ x^{2 }+ 16 + 8x – 656 = 0

= 2x^{2 }+ 16 + 8x – 656 = 0

= 2(x^{2 }+ 4x – 320) = 0

= x^{2 }+4 x – 320 = 0

= x^{2 }+ 20x – 16x – 320 = 0

= x(x + 20) – 16(x + 20) = 0

= (x + 20)(x – 16) = 0

Either x + 20 = 0 therefore x = -20

Or, x – 16 = 0 therefore x = 16

Since the value of x cannot be negative so the value of x = 16

The side of the square S_{1}= 16 cm

The side of the square S_{2} = 20 cm

**Question: 4**

The area of the right-angled triangle is 165 cm^{2}. Determine the base and altitude if the latter exceeds the former by 7m.

**Solution:**

Let the altitude of the right angles triangle be denoted by x m

Given that the altitude exceeds the base by 7m = x – 7m

We know

Area of the triangle = 1/2 × base × altitude

= 165 = 1/2 × (x − 7) × x

= x(x – 7) = 330

= x^{2 }– 7x – 330 = 0

= x^{2 }– 22x + 15x – 330 = 0

= x(x – 22) + 15(x – 22) = 0

= (x – 22)(x + 15) = 0

Either x – 22 = 0 therefore x = 22

Or, x + 15 = 0 therefore x = -15

Since the value of x cannot be negative so the value of x = 22

= x – 7 = 15

The base and altitude of the right angled triangle are 15 cm and 22 cm respectively.

**Question: 5**

Is it possible to design a rectangular mango grove whose length is twice its breadth and area is 800 m^{2}.find its length and breadth.

**Solution:**

Let the breadth of the rectangular mango grove be x m

Given that length of rectangle is twice of its breadth

Length = 2x

Area of the grove = 800 m^{2}

We know,

Area of the rectangle = length * breadth

= 800 = x(2x)

= 2x^{2 }– 800 = 0

= x^{2 }– 400 = 0

Breadth of the rectangular groove is 20 m

Length of the rectangular groove is 40 m

Yes, it is possible to design a rectangular groove whose length is twice of its breadth.

**Question: 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:**

In order to prove the given condition let us assume that the length of the rectangular park is denoted by x m

Given that,

Perimeter = 8 cm

Area = 400 cm^{2}

Perimeter of the rectangle = 2(length + breadth)

80 = 2(x + breadth)

Breadth = (40 – x)m

We know,

Area of the rectangle = (length) (breadth)

= 400 = x(40 – x)

= 40x – x^{2 }= 400

= x^{2 }– 40x + 400 = 0

= x^{2 }– 20x – 20x + 400 = 0

= x(x – 20) – 20(x – 20) = 0

= (x – 20)(x – 20) = 0

= (x – 20)^{2} = 0

= x – 20 = 0 therefore x =20

Length of the rectangular park is = 20 m

Breadth of the rectangular park = (40 – x) = 20 m

Yes, it is possible to design a rectangular Park of perimeter 80 m and area 400 m^{2}

**Question: 7**

Sum of the area of the square is 640 m^{2}.if the difference of their perimeter is 64 m, find the sides of the two squares.

**Solution:**

Let the two squares be S_{1} and S_{2 }respectively.

Let he sides of the square S_{1 }be x m and the sides of the square S_{2 }be y m

Given that the difference of their perimeter is 64 m

We know that the

Perimeter of the square = 4(side)

Perimeter of the square S_{1} = 4x m

Perimeter of the square S_{2} = 4y m

Now, difference of their perimeter is 64 m

= 4x – 4y = 64

x – y = 16

x = y + 16

Also, given that the sum of their two areas

= area of the square 1 +area of the square 2

= 640 = x^{2 }+ y^{2}

= 640 = (y + 16)^{2 }+ y^{2}

= 2y^{2 }+ 32y + 256 – 640 = 0

= 2y^{2 }+ 32y – 384 = 0

= 2(y^{2 }+ 16y – 192) = 0

= y^{2 }+ 16y – 192 = 0

= y^{2 }+ 24y – 8y – 192 = 0

= y(y + 24) – 8(y + 24) = 0

= (y + 24)(y – 8) = 0

Either y + 24 = 0 therefore y = -24

Or, y – 8 = 0 therefore y = 8

Since the value of y cannot be negative so y = 8

Side of the square 1 = 8 m

Side of the square 2 = 8 + 16 = 24 m

The sides of the squares 1 and 2 are 8 and 24 respectively.

**All Chapter RD Sharma Solutions For Class10 Maths**

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