1. Let the breadth of the rectangle be \( x \) meters:
Since the length is 6 meters more than the breadth, the length is \( x + 6 \) meters.
2. Using the formula for the area of a rectangle:
The area is given by: \( \text{Area} = \text{Length} \times \text{Breadth} \).
We know the area is 112 m², so:
\( x(x + 6) = 112 \)
3. Simplifying the equation:
Expand the equation: \( x^2 + 6x = 112 \)
Rearrange it to form a quadratic equation:
\( x^2 + 6x - 112 = 0 \)
4. Solving the quadratic equation:
We can solve it using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here, \( a = 1 \), \( b = 6 \), and \( c = -112 \).
Substitute these values into the quadratic formula:
\( x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-112)}}{2(1)} \)
\( x = \frac{-6 \pm \sqrt{36 + 448}}{2} \)
\( x = \frac{-6 \pm \sqrt{484}}{2} \)
\( x = \frac{-6 \pm 22}{2} \)
Therefore, \( x = \frac{-6 + 22}{2} = \frac{16}{2} = 8 \) or \( x = \frac{-6 - 22}{2} = \frac{-28}{2} = -14 \).
Since the breadth cannot be negative, we have \( x = 8 \) meters.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
A carpenter needs to make a wooden cuboidal box, closed from all sides, which has a square base and fixed volume. Since he is short of the paint required to paint the box on completion, he wants the surface area to be minimum.
On the basis of the above information, answer the following questions :
Taking length = breadth = \( x \) m and height = \( y \) m, express the surface area \( S \) of the box in terms of \( x \) and its volume \( V \), which is constant.