Q. 44.1( 23 Votes )

# The sides of a re

Answer :

Let us consider a Rectangle **ABCD**,

According to the problem, it is given that

The sides are in the ratio **4:3**.

The area of Rectangle is **1728**m^{2}.

The cost of fencing is **Rs.2.50**/m.

Let us consider the length(**L**) of the Rectangle be **‘4x’**m and the breadth(**B**) of the rectangle be **‘3x’**m.

We know that Area(**A**) of a Rectangle is

**A** = **L × B**

From the given data we can write,

⇒ **4x** × **3x** = 1728

⇒ 12x^{2} = 1728

⇒

⇒ x^{2} = 144

⇒ x = √144

⇒ **x = 12**

Now we find the length and breadth of the triangle.

⇒ Length(**L**) = 4x

⇒ L = 4 × 12

⇒ **L = 48**m

⇒ Breadth(**B**) = 3x

⇒ B = 3 × 12

⇒ **B = 36**m

To find the total cost for fencing, we need to find the perimeter first.

As we know that the fencing the sides of the rectangle but not total internal part of the rectangle.

Now, let’s find the Perimeter (**P**) of the rectangle.

We know that Perimeter(**P**) of the Rectangle is

**P** = **2 × (L + B)**m

Now let’s substitute the values of length and breadth of the rectangle.

⇒ P = 2 × (48 + 36)

⇒ P = 2 × (84)

⇒ **P = 168**m

The total cost required for fencing(**F _{c}**) = Perimeter*cost per meter

⇒ **F _{c}** = 168 × 2.50

⇒ F_{c} = 420

The total fencing cost required is **Rs.420**.

Rate this question :

The sides of a reKarnataka Board - Mathematics Part II

| Construct squarKarnataka Board - Mathematics Part II

Construct rectangKarnataka Board - Mathematics Part II

All rectangles arKarnataka Board - Mathematics Part II

The side ofKarnataka Board - Mathematics Part II

In a rhombus ABCDKarnataka Board - Mathematics Part II

In a given squareKarnataka Board - Mathematics Part II

A rhombus is a paKarnataka Board - Mathematics Part II

| Construct squarKarnataka Board - Mathematics Part II

A rectangular yarKarnataka Board - Mathematics Part II