Find the area for these 2 shapes (b&c)
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The first shape is a right triangle composed with a semi circle.
The right triangle has a height of 50cm, and its base is the diameter of the circle, which is twice the radius, which is 30cm.
So, the area of the triangle is
[tex]\dfrac{bh}{2}=\dfrac{50\cdot 30}{2}=\dfrac{1500}{2}=750[/tex]
The area of the circle is simply
[tex]\pi r^2 = \pi\cdot 15^2 = 225\pi[/tex]
So, the area of the shape is
[tex]750+225\pi=75(10+3\pi)[/tex]
The second shape is the difference between an outer circle and an inner one. The area is thus the difference of the two areas: let [tex]r_o[/tex] be the outer radius and [tex]r_i[/tex] be the inner one. The area of the shape is
[tex]\pi r_o^2-\pi r_i^2=\pi(r_0^2-r_i^2) = \pi(90^2-40^2)=6500\pi[/tex]