The problem involves finding the uniform width of a gravel path inside a rectangular grassy plot given the plot dimensions and the path's area.
The path is inside the plot, so the inner dimensions (the grassy area excluding the path) are reduced by $2x$ (width $x$ on both sides).
For the inner dimensions to be valid, $38 - 2x > 0$ and $32 - 2x > 0$, which means $2x < 32$, or $x < 16$ m.
The area of the outer rectangle ($A_{outer}$) is $L \times W$. The area of the inner rectangle ($A_{inner}$) is $l \times w$. The area of the path is the difference between these two areas.
Outer Area:
$A_{outer} = 38 \times 32 = 1216 \text{ m}^2$Inner Area:
$A_{inner} = (38 - 2x)(32 - 2x)$Path Area Equation:
$A_{path} = A_{outer} - A_{inner}$ $600 = 1216 - (38 - 2x)(32 - 2x)$Substitute the expression for $A_{inner}$:
$600 = 1216 - (1216 - 76x - 64x + 4x^2)$ $600 = 1216 - 1216 + 76x + 64x - 4x^2$ $600 = 140x - 4x^2$Rearrange into a standard quadratic equation ($ax^2 + bx + c = 0$):
$4x^2 - 140x + 600 = 0$Divide the entire equation by 4 to simplify:
$x^2 - 35x + 150 = 0$Factor the quadratic equation:
$(x - 5)(x - 30) = 0$This gives two possible values for $x$:
$x = 5$ or $x = 30$We must consider the physical constraint that the width of the path cannot be larger than half of the plot's width. As determined earlier, $x < 16$ m.
Therefore, the width of the path is 5 m.
A regular hexagonal garden with each side measuring 14 metres is surrounded by a circular jogging track that has a radius of 21 metres from the centre of the hexagon. Find the sum of the perimeters of both shapes.
(Use $\pi = \frac{22}{7}$)
What is the area of this parallelogram in sq.cm.? The measurements are in cms.
