Window Area Calculation: Perimeter and Composite Shape
The problem requires finding the area of a window composed of a square and an equilateral triangle.
Perimeter Determination
Let 's' represent the side length of the square. Since the equilateral triangle sits atop the square, sharing a side, the triangle's side length is also 's'.
The window's perimeter comprises the bottom edge of the square and the two vertical sides of the square, plus the two exposed vertical sides of the triangle.
- Bottom side of the square: s
- Left side (square + triangle): s + s = 2s
- Right side (square + triangle): s + s = 2s
Wait, let's reconsider the perimeter. The perimeter is the outer boundary. This includes:
- The bottom side of the square (length s).
- The left side of the square (length s).
- The right side of the square (length s).
- The left side of the triangle (length s).
- The right side of the triangle (length s).
Therefore, the total perimeter is the sum of these 5 sides:
Perimeter = s + s + s + s + s = 5s.
We are given that the perimeter is 6 m:
$5s = 6 \text{ m}$
Solving for the side length 's':
$s = \frac{6}{5} \text{ m} = 1.2 \text{ m}$
Area Calculation Steps
The total area of the window is the sum of the area of the square portion and the area of the equilateral triangle portion.
Square Area Calculation
The formula for the area of a square is $A_{square} = s^2$.
Using $s = 1.2$ m:
$A_{square} = (1.2 \text{ m})^2 = 1.44 \text{ m}^2$
Equilateral Triangle Area Calculation
The formula for the area of an equilateral triangle is $A_{triangle} = \frac{\sqrt{3}}{4} s^2$.
Using $s = 1.2$ m:
$A_{triangle} = \frac{\sqrt{3}}{4} (1.2 \text{ m})^2 = \frac{\sqrt{3}}{4} \times 1.44 \text{ m}^2$
$A_{triangle} = \sqrt{3} \times 0.36 \text{ m}^2$
Using the approximation $\sqrt{3} \approx 1.732$:
$A_{triangle} \approx 1.732 \times 0.36 \text{ m}^2 \approx 0.6235 \text{ m}^2$
Total Window Area
Sum the areas of the square and the triangle:
$A_{total} = A_{square} + A_{triangle}$
$A_{total} \approx 1.44 \text{ m}^2 + 0.6235 \text{ m}^2$
$A_{total} \approx 2.0635 \text{ m}^2$
Rounding to two decimal places gives the final area.
The area of the window is approximately 2.06 $m^2$.
Important Questions from Mensuration and Geometry
The city of Atlantis was crafted by the God of the seas, Poseidon. It was made of alternating concentric circular rings of land (shaded) and water (not shaded) as represented in the figure (not to scale). The radius of Inner Island was 2.5 stades (a unit of length used in ancient Greece). The water surrounding Inner Island was one stade wide (length AB). This was surrounded by two pairs of alternating rings of land and water. The first pair of land and water was two stades wide each (lengths BC and CD), and the outer pair is three stades wide each (lengths DE and EF).
The ratio of the surface area of the land to that of the water in the city of Atlantis is _________ (round off to two decimal places).
In the given figure, $P, Q$, and $R$ are three points on a circle of radius 10 cm with $O$ as its center, $\overline{PQ} = \overline{RQ}$, and $\angle PQR = 45^\circ$. The figure is representative.
The area of the shaded region $PQRO$ is ______________ cm$^2$.
- A straight line $y = x - 1$ intersects a circle with center at $x = 1, y = 1$ and radius of magnitude 1 at two points. The length of the chord formed by this intersection is _______. (rounded off to three decimal places)
- The shell of a hollow spherical nanoparticle has a uniform thickness of 3 nanometers (nm). The outer radius of the nanoparticle is 5 nm. The ratio of the volume of the shell to the volume of the hollow core is ________
(Round off to one decimal place) - The volume of a sphere of diameter 1 unit is ______ than the volume of a cube of side 1 unit.
