All categories

3 years ago

The diagram above shows a rectangle inscribed in a circle AB=10 and AC =12 caculate the total surface area of the shaded part

Mathematics

1 answer:

erastova [34]3 years ago

5 0

Answer:

$71.63 \: \: \mathrm{cm^2 }$

Step-by-step explanation:

Once we know the diameter of the circle, we can figure out the problem.

The diameter of the circle = The diagonal of the rectangle inscribed in the circle

To find the diagonal of the rectangle, we can use a formula.

$d=\sqrt{w^2 + l^2}$

The width is 10 cm and the length is 12 cm.

$d=\sqrt{10^2 + 12^2}$

$d \approx 15.62$

The diagonal of the rectangle inscribed in the circle is 15.62 cm.

The diameter of the circle is 15.62 cm.

Find the area of the whole circle.

$A=\pi r^2$

The $r$ is the radius of the circle, to find radius from diameter we can divide the value by 2.

$r = \frac{d}{2}$

$r=\frac{15.62}{2}$

$r=7.81$

Let’s find the area now.

$A=\pi (7.81)^2$

$A \approx 191.625$

Find the area of rectangle.

$A=lw$

Length × Width.

$A = 12 \times 10$

$A=120$

Subtract the area of the whole circle with the area of rectangle to find area of shaded part.

$191.625-120$

$71.625 \approx 71.63$

You might be interested in

Use cross products to find the area of the triangle in the xy-plane defined by (1, 2), (3, 4), and (−7, 7).

Free_Kalibri [48]

I love these. It's often called the Shoelace Formula. It actually works for the area of any 2D polygon.

We can derive it by first imagining our triangle in the first quadrant, one vertex at the origin, one at (a,b), one at (c,d), with (0,0),(a,b),(c,d) in counterclockwise order.

Our triangle is inscribed in the $a \times d$ rectangle. There are three right triangles in that rectangle that aren't part of our triangle. When we subtract the area of the right triangles from the area of the rectangle we're left with the area S of our triangle.

$S = ad - \frac 1 2 ab - \frac 1 2 cd - \frac 1 2 (a-c)(d-b) = \frac 1 2(2 ad - ab -cd - ad +ab +cd -bc) = \frac 1 2(ad -bc)$

That's the cross product in the purest form. When we're away from the origin, a arbitrary triangle with vertices $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ will have the same area as one whose vertex C is translated to the origin.