Step-by-step explanation:
Solve b part yourself in the same way.
1. M is the midpoint of LN and O is the midpoint of NP. This makes the triangle MNO equal to half of LNP. Then you can get this equation
MO= (1/2) LP
If you insert MO = 2x +6 and LP = 8x – 20 the calculation would be:
2x+6= (1/2)( 8x-20)
2x+6= 4x-10
2x-4x= -10 - 6
-2x= -16
x=8
2. Centroid is the point that intersects with three median lines of the triangle. The centroid should divide the median lines into 1:2 ratio. In AC lines, A located in the base so A.F:FC would be 1:2
Then, the answer would be:
A.F= 1/(1+2) * AC
A.F= 1/3 * 12= 4
FC= 2/(1+2) * AC
FC= 2/3 * 12= 8
3. Since
∠BAD=∠DAC
∠ABD=∠ACD
AD=AD
The triangle ABD and ACD are similar. You can get this equation
BD=DC
x+8= 3x+12
x-3x= 12-8
-2x=4
x=-2
DC=3x+12= 3(-2) +12= 6
4. Orthocenter made by intersection of triangle altitude
A
BC lines slope would be (-4)-(-1)/1-4= -3/-3= 1. The altitude line slope would be -1, the function would be:
y=-x +a
0= 1+a
a=-1
y=-x-1
B
AC lines slope would be (-4)-(-1)/1-0= -3. The altitude line slope would be 1/3, the function would be:
y=1/3x+a
-1=1/3(4)+a
a=-7/3
y=1/3x - 7/3
C
BC lines slope would be (-1)-(-1)/4 = 0/4.
The line would be
0=x+a
a=-1
0=x-1
x=1
y=-x-1 = 1/3x-7/3
-x-(1/3x)=-7/3 +1
-4/3x= -4/3
x=1
y=-x-1
y=-1-1= -2
The orthocenter would be (1,-2)
5.
a. Circumcenter: the intersection of perpendicular bisector lines<span>
b. Incenter: the intersection of bisector lines
c. Centroid: </span>the intersection of median lines<span>
d. Orthocenter: </span>the intersection of altitude lines
A^2 + b^2 = c^2
3^2 + 8^2 = c^2
9 + 64 = c^2
73 =c^2
take the square root of each side
c= 8.544
Answer:
57 units^2
Step-by-step explanation:
First find the area of the triangle on the left
ABC
It has a base AC which is 9 units and a height of 3 units
A = 1/2 bh = 1/2 ( 9) *3 = 27/2 = 13.5
Then find the area of the triangle on the right
DE
It has a base AC which is 6 units and a height of 1 units
A = 1/2 bh = 1/2 ( 6) *1 = 3
Then find the area of the triangle on the top
It has a base AC which is 3 units and a height of 3 units
A = 1/2 bh = 1/2 ( 3) *3 = 9/2 = 4.5
Then find the area of the rectangular region
A = lw = 6*6 = 36
Add them together
13.5+3+4.5+36 =57 units^2
6 students – 3 rolls of clay divided by each portion given to a student (0.5) equals 6, meaning only six students where able to receive some clay
