Answer:
[tex] \rm \displaystyle A_{ \text{triangle}} = \frac{147 \sqrt{7} }{2 } \quad \text{or} \quad 73.5 \sqrt{7} [/tex]
Step-by-step explanation:
we have a right angle triangle composited in a semi-circle we want to figure out the area of the triangle
in order to do so we need to figure out the adjacent side (height AB)
since the given triangle is a right angle triangle we can consider Pythagoras theorem given by
[tex] \displaystyle {a}^{2} + {b}^{2} = {c}^{2} [/tex]
given that,a=21 and c=28
substitute:
[tex] \displaystyle {21}^{2} + {b}^{2} = {28}^{2} [/tex]
move left hand side expression to right hand side and change its sign:
[tex] \displaystyle {b}^{2} = {28}^{2} - {21}^{2} [/tex]
square root both sides:
[tex] \displaystyle {b}^{} = \sqrt{{28}^{2} - {21}^{2} }[/tex]
use algebraic indentity:
[tex] \rm \displaystyle {b}^{} = \sqrt{({28}^{} + 21) (28 - {21}^{})}[/tex]
simplify:
[tex] \rm \displaystyle {b}^{} = \sqrt{({49}) ( 7)}[/tex]
use redical property:
[tex] \rm \displaystyle {b}^{} = 7 \sqrt{ 7}[/tex]
now to figure out the area we can consider the following formula:
[tex] \displaystyle A_{ \text{triangle}} = \frac{1}{2} bh[/tex]
remember that,h=AB=7√7 and b=BC=21 thus substitute:
[tex] \displaystyle A_{ \text{triangle}} = \frac{1}{2} \times 21 \times 7 \sqrt{7} [/tex]
simplify multiplication:
[tex] \rm \displaystyle A_{ \text{triangle}} = \frac{147 \sqrt{7} }{2 } \quad \text{or} \quad 73.5 \sqrt{7} [/tex]
hence,
the area of the triangle is 194 (approximately)
note: you can also use horon's formula but I like to keep things simple
