#### Exercise 8.11

**Question 1:**

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

**Soln:**

Let the breadth of the rectangle be (x) m

Given,

Perimeter = 82 m

Area = 400 m^{2}

Perimeter of a rectangle = 2(length + breadth)

82=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}

^{Â Â }=^{Â Â }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?**

**Soln:**

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:Â Two squares have sides x and (x+4) cm. The sum of their area is 656 cm ^{2}** .

**Find the sides of the square.**

**Soln:**

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 }Â = 656 cm^{2}

^{Â }= x^{2}+ x^{2}+16+8x-656 =0

=2 x^{2}+16+8x-656 =0

= 2 (x^{2}+4x-320) =0

= x^{2}+4x-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:Â The area of the right-angled triangle is 165 cm ^{2}_{. }Determine the base and altitude if the latter exceeds the former by 7m.**

**Soln:**

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

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

We know

Area of the triangle = \(\frac{1}{2}\times base\times altitude\)

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

= 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:Â 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.**

**Soln:**

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

= 800Â Â Â = x(2x)

= 2x^{2}-800 =0

= x^{2}-400 =0 = x^{2}=400 = \(x=\sqrt{400}=20\)

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:Â Is it possible to design a rectangular park of perimeter 80 m and area 400 m ^{2}? If so find its length and breadth.**

**Soln:**

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 400m^{2}

**Question 7:Â 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.**

**Soln:**

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_{2Â }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 2Â = 8+16 = 24 m

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