A dilation of figure A by a factor of 1/2 is the last figure with sides 1in, 2 in, 0.5 in and 2.5 inch. Since dilation means becoming larger, smaller or wider from one figure, this is also same with similar figures where the ratios of corresponding sides should be the same. In this problem, the ratio is 0.5 or 1/2. Therefore figure A is similar to the last figure.
Answer:
-x/6+0
Step-by-step explanation:
You know that the two triangles are congruent because two sides and one angle are congruent.
According to CPCTC, corresponding parts of congruent triangles are congruent, which means that PR and SU are equal.
3y-2=y+4
3y=y+6
2y=6
y=3
3(3)-2=9-2=7
7+4+6=17
The perimeter is 17 feet.
Hope this helps!
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
