let's start with the triangle, since that seems very obvious per se, the triangle has a base of 20 and a height of 14.
[tex]\textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b= base\\ h= height\\[-0.5em] \hrulefill\\ b=20\\ h=14 \end{cases}\implies A=\cfrac{1}{2}(20)(14)\implies A=140[/tex]
now let's do the sector.
[tex]\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=20\\ \theta =45 \end{cases}\implies \begin{array}{llll} A=\cfrac{(45)\pi (20)^2}{360} \\\\\\ A=50\pi \implies A\approx 157.08 \end{array}[/tex]
now let's do the segment.
[tex]\textit{area of a segment of a circle}\\\\ A=\cfrac{r^2}{2}\left( \cfrac{\pi \theta }{180}~~ - ~~sin(\theta ) \right)~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=20\\ \theta =45 \end{cases} \\\\\\ A=\cfrac{20^2}{2}\left( \cfrac{\pi (45)}{180}~~ - ~~sin(45^o) \right)\implies A=200\left(\cfrac{\pi }{4}~~ - ~~\cfrac{1}{\sqrt{2}} \right) \\\\\\ A=50\pi - 100\sqrt{2}\implies A\approx 15.66[/tex]
