Let’s solve Quadrant by Quadrant , and please ignore the NEGATIVE Coordinates; Take all distance values to be POSITIVE
Scenario also requires Pythogras theorem & knowing properties of triangles
Triangle One[1] :
Base or Horizontal line [ A] =4 squares & Height or Vertical line [ B] =7 Squares. Those are easily obtained by counting . Remaining bit , the Slant is the Hypotenuse, let’s Lebel it C
In Pythogras for a right angled triangle,
A^2+B^2=C^2
substitute
4^ 2 +7^2=C^2
16 + 49 = C^2
65 =C2
√ 65 = 8.0622 Units
hence A[ Base ] =4 units , B [ Height] =7 units , C[ Hypotunese] =8.062 units
Triangle 2 :
Base [ Horizontal, A ] = 5 Units , Height [. reticle ] =7 Units , Hypotunese [ C] =?
A^2 + B ^2 =C^2
5^2+ 7^2 =C^2
25 + 49 = C^2
74 =C^2
√74 = 8.602 Units
There fore A = 5 units , B=7 units & C= 8.602 units
Triangle 3
Base[ A] =8 Units , Height [ B] = 2 units , Hypotunese C =? units
8^2 + 2^2=C^2
64 + 4 = C^2
68 =C^2
√68 =C =8.25Units
Triangle 4
Devide the base into two equal parts to get 2 Units each , Height is 2
Upon doing that , that , you get two equal small triangles with the Same Lengths Hypotenuse
Let’s only solve off the Hypotunese one small triangle
Base [ A]=2, Height =2, Hypotunese [ C] =?
Using A^2 + B ^2=C^2
2^2 +2^2=C^2
8 =C2
√8 =2.83 Units