True. The only area where a line parallel to one side would be able to intersect both other sides would be in a place that proportionally divides the other sides.
Answer:
The image in the attached figure
Step-by-step explanation:
we know that
The dimensions of the image of the given triangle are equal to the original dimensions of the pre-image multiplied by the scale factor
In this problem the scale factor is 1/4
The height of the pre-image is 4 units
The base of the pre-image is 8 units
Find out the dimensions of the image
The height of the image is 
The base of the image is 
The image in the attached figure
Answer:
Q= 6x+1
R= -7x+1
Step-by-step explanation:
Answer:
Ix = Iy =
Radius of gyration x = y = 
Step-by-step explanation:
Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.
Mass of disk = ρπR2
Moment of inertia about its perpendicular axis is
. Moment of inertia of quarter disk about its perpendicular is
.
Now using perpendicular axis theorem, Ix = Iy =
=
.
For Radius of gyration K, equate MK2 = MR2/16, K= R/4.