[Highschool Algebra]Mensuration including sector area.

Question

The mark scheme says the answer is $5.68$. Can anyone explain me step by step how this was found?

Answer 1

• area of circle sector: $\frac{3}{4}\pi x^2$
• area of triangle: $\frac{1}{2}x^2$
• area of rectangle: $2x\sqrt{2x^2} = 2\sqrt{2}x^2$
• area of whole shape: $kx^2$ $$(\frac{3}{4}\pi + \frac{1}{2} + 2\sqrt{2})x^2 \approx 5.68x^2$$

Answer 2

For the circular part you get $$ (3/4) \pi x^2 $$

For the triangle you get $$ (1/2) x^2 $$ because it is a right triangle.

For the rectangle you get $$ (2\sqrt 2) x^2$$

So the total is $$(3/4 \pi+1/2+\frac 12+2\sqrt 2)x^2.$$

Answer 3

For me, the circular sector corresponds to an angle of $\theta=\frac{3\pi}2$ (in radians). The formula for the area of a circular sector with angle $\theta$ is $$\mathcal A_S=\frac12\theta \mkern2mu r^2.$$ (note that, for a full circle, $\theta=2\pi$, and you obtain what's expected: $\pi r^2$.)

As to the rectangle, you have to find its width. The triangle is an isosceles right triangle, its area is $\mathcal A_T=\frac12 x^2$, and its legs have length $x$, so by Pythagoras, its hypotenuse has length $\sqrt 2x$ . Finally the area of the rectangle is $$\mathcal A_R=2\sqrt2 x^2,$$ so that in all, the area of the shape is $$\mathcal A=\mathcal A_S+\mathcal A_T+\mathcal A_R=\frac{3\pi}4x^2+\frac12x^2+2\sqrt2 x^2$$ and $\quad k= \dfrac{3\pi}4+1+2\sqrt2\approx5.685$.