That ones a hardddd one ill let someone else answer thatttttt
By comparing the given shape with easier ones like triangles and rectangles we will see that the area of the shape is 8 square units.
<h3>
How to simplify the shape.</h3>
So the given shape is a little bit complex, but you can actually see that it is a triangle with a base of 8 units with a height of 4 units, where a rectangle of 2 in by 3 in was removed, and also removed a triangle of height of 2 inches and base of 2 inches.
Remember that:
- A triangle of height H and base B has an area = B*H/2
- A rectangle of length L and width W has an area = L*W.
Then the area of the given shape is:
A = 8*4/2 - 3*2 - 2*2/2
A = 16 - 6 - 2
A = 8
So the given shape has an area of 8 square units.
If you want to learn more about areas, you can read:
brainly.com/question/14137384
<span>619.473970397 here is the square root</span>
Some basic formulas involving triangles
\ a^2 = b^2 + c^2 - 2bc \textrm{ cos } \alphaa 2 =b 2+2 + c 2
−2bc cos α
\ b^2 = a^2 + c^2 - 2ac \textrm{ cos } \betab 2=
m_b^2 = \frac{1}{4}( 2a^2 + 2c^2 - b^2 )m b2 = 41(2a 2 + 2c 2-b 2)
b
Bisector formulas
\ \frac{a}{b} = \frac{m}{n} ba =nm
\ l^2 = ab - mnl 2=ab-mm
A = \frac{1}{2}a\cdot b = \frac{1}{2}c\cdot hA=
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
\iits whatever A = prA=pr with r we denote the radius of the triangle inscribed circle
\ A = \frac{abc}{4R}A=
4R
abc
- R is the radius of the prescribed circle
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
