C
Step-by-step explanation:
Median of triangle: It is a line segment joining a vertex to the midpoint of the opposite side.
Consider ΔABC, point F id the midpoint of line segment AB and E is the midpoint of luine segment AC.
Draw line segments FC and BE(medians of triangle). G is the point where line segment FC and BE meet. Now, Join AG.
Let H be the point outside the ΔABC and AG passs through the point H such that AG intersects BC at D. BH and HC are dashed lines.
We need to show that D is the midpoint of BC. The correct logical order for proof will be:
III. GC is parallel to line segment BH and line segment BG is parallel to line segment HC.
IV. Line segment FG is parallel to line segment BH and line segment GE is parallel to line segment HC.
I. BGCH is a parallelogram as opposite sides are parallel (from III.)
II. Since, diagnols of a parallelogram bisect each other. Henc, we get BD=DC.
Therefore, D is mid pont of BC.
It implies that AD is also a median.
Hence, all the three medians that are: BE,FC and AD passes through a common vertex G.
II establishes the lines are parallel, then III renames the parallel lines to refer to specific segments. The order of these statements should be II, III.
IV establishes that BGCH is a parallelogram, then I makes use of the properties of a parallelogram. The order of these statements should be IV, I.
The appropriate choice is ...
... II, III, IV, I . . . . . . . the 3rd selection
IV, III, I, II
IV, III, I, II
Let's try to render the first part of the proof a bit more legibly.
Point F is a midpoint of Line segment AB
Point E is a midpoint of Line segment AC
Draw Line segment BE
Draw Line segment FCby Construction
Point G is the point of intersection between Line segment BE and Line segment FCIntersecting Lines Postulate
Draw Line segment AGby Construction
Point D is the point of intersection between Line segment AG and Line segment BCIntersecting Lines Postulate
Point H lies on Line segment AG such that Line segment AG ≅ Line segment GHby Construction
OK, now we continue. We need to prove some parallel lines; statement 4 lets us do so.
IVLine segment FG is parallel to line segment BH and Line segment GE is parallel to line segment HC Midsegment Theorem
Now that we've shown some segments parallel we extend that to collinear segments.
IIILine segment GC is parallel to line segment BH and Line segment BG is parallel to line segment HC Substitution
We have enough parallel lines to prove a parallelogram
IBGCH is a parallelogram Properties of a Parallelogram (opposite sides are parallel)
Now we draw conclusions from that.
IILine segment BD ≅ Line segment DC Properties of a Parallelogram (diagonals bisect each other)
IV III I II, second choice
B: II, IV, I, III
Step-by-step explanation:
We believe the proof statement — reason pairs need to be ordered as shown below
Point F is a midpoint of Line segment AB Point E is a midpoint of Line segment AC — given
Draw Line segment BE Draw Line segment FC — by Construction
Point G is the point of intersection between Line segment BE and Line segment FC — Intersecting Lines Postulate
Draw Line segment AG — by Construction
Point D is the point of intersection between Line segment AG and Line segment BC — Intersecting Lines Postulate
Point H lies on Line segment AG such that Line segment AG ≅ Line segment GH — by Construction
__
II Line segment FG is parallel to line segment BH and Line segment GE is parallel to line segment HC — Midsegment Theorem
IV Line segment GC is parallel to line segment BH and Line segment BG is parallel to line segment HC — Substitution
I BGCH is a parallelogram — Properties of a Parallelogram (opposite sides are parallel)
III Line segment BD ≅ Line segment DC — Properties of a Parallelogram (diagonals bisect each other)
__
Line segment AD is a median Definition of a Median
The answer is in the image
SI=(P*R*T)/100
P=2000
R=1.5
T=6
SI=(2000*1.5*6)/100
=(2000*9)/100
=180
Neil will earn interest of 180
It will provide an instant answer!