If the perimeter of a rectangle is 10 cm and the area is 4 cm 2, then its length is
4 cm
Understanding the Rectangle Problem
The question asks us to find the length of a rectangle given its perimeter and area. We are provided with the perimeter being 10 cm and the area being 4 cm². We need to use the standard formulas for the perimeter and area of a rectangle to solve this problem.
Formulas for Rectangles
- The perimeter of a rectangle is given by the formula: \(P = 2 \times (\text{length} + \text{width})\)
- The area of a rectangle is given by the formula: \(A = \text{length} \times \text{width}\)
Setting up the Equations
Let's denote the length of the rectangle as \(l\) and the width as \(w\). Based on the given information, we can write two equations:
- Perimeter equation: \(10 = 2(l + w)\)
- Area equation: \(4 = lw\)
Solving the System of Equations
We have a system of two equations with two variables (\(l\) and \(w\)). We can solve this system to find the values of \(l\) and \(w\).
From the perimeter equation, we can simplify it:
\(10 = 2(l + w)\)
Divide both sides by 2:
\(\frac{10}{2} = l + w\)
\(5 = l + w\)
Now, we can express one variable in terms of the other. Let's express \(w\) in terms of \(l\):
\(w = 5 - l\)
Now, substitute this expression for \(w\) into the area equation:
\(4 = l \times w\)
\(4 = l(5 - l)\)
Expand the right side:
\(4 = 5l - l^2\)
Rearrange the terms to form a standard quadratic equation \(al^2 + bl + c = 0\):
\(l^2 - 5l + 4 = 0\)
Solving the Quadratic Equation
We need to solve the quadratic equation \(l^2 - 5l + 4 = 0\) for \(l\). We can solve this by factoring, using the quadratic formula, or completing the square. Factoring is often the simplest method if the quadratic can be factored easily.
We are looking for two numbers that multiply to +4 and add up to -5. These numbers are -1 and -4.
So, we can factor the quadratic equation as:
\((l - 1)(l - 4) = 0\)
For the product of two factors to be zero, at least one of the factors must be zero.
- Case 1: \(l - 1 = 0 \implies l = 1\)
- Case 2: \(l - 4 = 0 \implies l = 4\)
These are the two possible values for the length \(l\). For each value of \(l\), we can find the corresponding value of \(w\) using \(w = 5 - l\).
- If \(l = 1\) cm, then \(w = 5 - 1 = 4\) cm.
- If \(l = 4\) cm, then \(w = 5 - 4 = 1\) cm.
In geometry problems involving rectangles, the length is conventionally considered to be the longer side, and the width the shorter side. Comparing the dimensions (1 cm and 4 cm), the length is 4 cm and the width is 1 cm. However, mathematically, both (length=1, width=4) and (length=4, width=1) satisfy the given conditions for perimeter and area.
Let's verify both pairs of dimensions:
| Dimensions (l, w) | Perimeter \(2(l+w)\) | Area \(lw\) |
|---|---|---|
| (1 cm, 4 cm) | \(2(1+4) = 2(5) = 10\) cm | \(1 \times 4 = 4\) cm\(^2\) |
| (4 cm, 1 cm) | \(2(4+1) = 2(5) = 10\) cm | \(4 \times 1 = 4\) cm\(^2\) |
Both pairs of dimensions satisfy the given perimeter and area. The question asks for the length. Based on the convention that length is the longer side, the length is 4 cm.
Conclusion
The possible dimensions for the rectangle are 1 cm by 4 cm. The length is usually considered the longer side. Therefore, the length is 4 cm.
Revision Table: Rectangle Properties
| Property | Formula | Given Value |
|---|---|---|
| Perimeter (P) | \(2(l + w)\) | 10 cm |
| Area (A) | \(lw\) | 4 cm\(^2\) |
| Length (l) | To Find | ? |
| Width (w) | Derived (\(w = 5 - l\)) | ? |
Solving the system of equations \(10 = 2(l+w)\) and \(4 = lw\) leads to a quadratic equation for length, \(l^2 - 5l + 4 = 0\), which gives possible lengths of 1 cm and 4 cm. Conventionally, the length is the longer side, so the length is 4 cm.
Additional Information: Solving Quadratic Equations
A quadratic equation in the form \(ax^2 + bx + c = 0\) can be solved using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
In our case, the equation for length is \(l^2 - 5l + 4 = 0\). Here, \(a=1\), \(b=-5\), and \(c=4\). Plugging these values into the quadratic formula for \(l\):
\(l = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}\)
\(l = \frac{5 \pm \sqrt{25 - 16}}{2}\)
\(l = \frac{5 \pm \sqrt{9}}{2}\)
\(l = \frac{5 \pm 3}{2}\)
This gives two possible values for \(l\):
- \(l_1 = \frac{5 + 3}{2} = \frac{8}{2} = 4\)
- \(l_2 = \frac{5 - 3}{2} = \frac{2}{2} = 1\)
These solutions match the results obtained by factoring. This confirms that the possible dimensions are 1 cm and 4 cm. As discussed, the length is typically the larger value when solving such problems, which is 4 cm.
Similar Questions
ABCD is a trapezium in which AB is parallel to DC. Let E and F be the midpoints on AD and BC respectively. If EF = 10 cm and AB - DC = 4 cm, then what is the value of AB × DC?
The area of a sector of a circle of radius 4 cm is 25.6 cm 2. What is the radian measure of the arc of the sector?
- X, Y and Z are three equilateral triangles. The sum of the areas of X and Y is equal to the area of Z. If the side lengths of X and Y are 6 cm and 8 cm respectively, then what is the side length of Z?
If the perimeter of a right-angled triangle is 30 cm and the hypotenuse is 13 cm, then what is the area of the triangle?
The area of a rhombus is 336 square cm. If the length of one of its diagonals is 48 cm, then what is the perimeter of the rhombus ?
The lengths of the sides of a right-angled triangle are consecutive even integers (in cm). What is the product of these integers?
The arch of a bridge is in the form of an arc of a circle. If the span of the bridge is 40 m and height in the middle is 8 m, then what is the radius of curvature of the bridge?
ABC is a triangle with sides AB = 41 cm, BC = 28 cm and CA = 15 cm. If D, E and F are the mid-points of AB, BC and CA respectively, then what is the area of the triangle DEF?
What is the area of the circle (approximately) inscribed in a triangle with side lengths 12 cm, 16 cm and 20 cm ?
How long will a man take to walk around the boundary of a square field of area 25 hectares at the rate of 5 km/hr ?
Important Questions from Plane Figures
If the area of a square is 625 cm 2, then what is the perimeter of the square?
The area and the perimeter of a sheet of paper are 240 cm 2and 68 cm, respectively. What would be its length and breadth?
One side of rectangular field is 15 meters and one of its diagonals is 17 meters. Then find the area of the field.
The bisector of ∠B in ΔABC meets AC at D. If AB = 12 cm, BC = 18 cm and AC = 15 cm, then the length of AD (in cm) is:
The perimeter and the length of one of the diagonals of a rhombus is 26 cm and 5 cm respectively. Find the length of its other diagonal (in cm).
