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)
I belive the answer is 56
Answer:
A, C
Step-by-step explanation:
When you put the equation y = x -12 into a TI-84 calculator (because it has graph) and go to the table you can see that B,D,E don’t have the same y-axis even though their x-axis is the same but the y-axis isn’t, therefor A and C is the correct answer.
Sorry not the best explanation but I hope it helps
Answer:
B : 
Step-by-step explanation:
If you divide a rhombus using its diagonals, you get 4 right triangles, whose legs are both 1/2 the length of the diagonals.
This means that the legs of one of those 4 triangles have lengths of 2x/2, and 8x/2, so the legs of one of those triangles x and 4x. This makes the length of one side equal to 
. Because all 4 sides are the same length, you multiply this value by 4, and get , which is B.
Dang this is so hard bro but I think it’s 4
