Here the figure is made up of a quadrilateral and a semi circle.
ABCD is the quadrilateral here. We will find the sides of the quadrilateral by using the distance formula.
If (xâ, yâ) and (xâ, yâ) are two points given, then the distance between two points by using distance formula is,
[tex] d =\sqrt({ x_{1}-x_{2})^2 +( y_{1} -y_{2} )^2} [/tex]
The co-ordinate of A is (-1,2) and co-ordinate of B is (-2,-1).
So the length of side AB = [tex] \sqrt{(-1-(-2))^2+(2-(-1))^2} [/tex]
=[tex] \sqrt{(-1+2)^2+(2+1)^2} [/tex](As negative times negative is positive)
= [tex] \sqrt{(1)^2+(3)^2} =\sqrt{1+9} =\sqrt{10} [/tex]
The co-ordinate of C is (4,-3) and D is (5,0)
The length of side CD
= [tex] \sqrt(4-5)^2+(-3-0)^2} [/tex]
= [tex] \sqrt{(-1)^2+(-3)^2} [/tex]
= [tex] \sqrt{1+9} =\sqrt{10} [/tex]
So the sides AB and CD are equal.
The length of side AD
= [tex] \sqrt{(-1-5)^2+(2-0)^2} [/tex]
= [tex] \sqrt{(-6)^2+(2)^2} =\sqrt{36+4} =\sqrt{40} [/tex]
The length of side BC
= [tex] \sqrt{(-2-4)^2+(-1-(-3))^2} [/tex]
= [tex] \sqrt{(-2-4)^2+(-1+3)^2} [/tex]
= [tex] \sqrt{(-6)^2+(2)^2}=\sqrt{36+4} =\sqrt{40} [/tex]
So the lengths of the sides AD and BC are equal.
So the quadrilateral is a rectangle whose length is [tex] \sqrt{40} [/tex] and width is [tex] \sqrt{10} [/tex].
Area of a rectangle = length Ă width
= [tex] (\sqrt{40}) (\sqrt{10}) [/tex]
= [tex] \sqrt{(40)(10)}=\sqrt{400} [/tex]
= [tex] 20 unit^2 [/tex]
Now the diameter of the semicircle is the side AD = [tex] \sqrt{40} [/tex]
So, the radius of the semi-circle = [tex] \frac{\sqrt{40}}{2} [/tex]
= [tex] \frac{\sqrt{(4)(10)}}{2} [/tex]
= [tex] \frac{(\sqrt{4})(\sqrt{10})}{2} [/tex]
= [tex] \frac{2\sqrt{10}}{2} [/tex] = [tex] \sqrt{10} [/tex]
Area of semi-circle = [tex] \frac{1}{2} \pi r^2 [/tex], where r is the radius.
= [tex] \frac{1}{2} \pi (\sqrt{10})^2 [/tex]
= [tex] \frac{1}{2} \pi (10) [/tex]
= [tex] \frac{(\pi)(10)}{2} [/tex]
= [tex] \frac{10\pi}{2} = 5\pi [/tex] = [tex] 15.7 unit^2 [/tex] ( Approximately taken to the nearest tenth)
Total area of the figure = [tex] (20+15.7) unit^2 [/tex] = [tex] 35.7 unit^2 [/tex]
We have got the required answer.
Option a is correct here.
