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HELP NEEDED ASAP... 20 POINTS. WILL MARK BRAINLIEST!

1). Points B, D, and, F are midpoints of the sides of ACE. EC= 33 and DF= 20. Find AC. The diagram is not to scale. (The first photo).
A). 40
B). 33
C). 10
D). 66

2). Find the length of the midsegment. The diagram is not to scale. (The second photo).
A). 26
B). 17
C). 14
D). 52

3). A rope is tied between the edge of the top of the building and a stake in the ground just above the top of a tree. What angle is formed by the rope and the building? (The third photo).
A). 138
B). 52
C). 42
D). 48

4). DF bisects <EDG. Find the value of x. The diagram is not to scale. (The fourth photo).
A). 25
B). 3/16
C). 120
D). 8

5). E is equidistant from the sides of <HGF. Find m< FGH. The diagram is not to scale. (The fifth photo).
 A). 52
B). 26
C). 22
D). 6

HELP NEEDED ASAP 20 POINTS WILL MARK BRAINLIEST 1 Points B D and F are midpoints of the sides of ACE EC 33 and DF 20 Find AC The diagram is not to scale The fir class=
HELP NEEDED ASAP 20 POINTS WILL MARK BRAINLIEST 1 Points B D and F are midpoints of the sides of ACE EC 33 and DF 20 Find AC The diagram is not to scale The fir class=
HELP NEEDED ASAP 20 POINTS WILL MARK BRAINLIEST 1 Points B D and F are midpoints of the sides of ACE EC 33 and DF 20 Find AC The diagram is not to scale The fir class=
HELP NEEDED ASAP 20 POINTS WILL MARK BRAINLIEST 1 Points B D and F are midpoints of the sides of ACE EC 33 and DF 20 Find AC The diagram is not to scale The fir class=
HELP NEEDED ASAP 20 POINTS WILL MARK BRAINLIEST 1 Points B D and F are midpoints of the sides of ACE EC 33 and DF 20 Find AC The diagram is not to scale The fir class=

Respuesta :

1 is C
2 is b
3 is B
4 idrk
5 idk sorry hope i helped

Answer:

1). Option A

2). Option A

3). Option D

4). Option D

5). Option A

Step-by-step explanation:

1). Points B, D,are the midpoints of the sides of Δ ACE.

Side EC = 33 and DF = 20, then we have to find the length of AC.

Since ΔAEC and ΔEFD are similar,

so [tex]\frac{EC}{ED}=\frac{AC}{FD}[/tex]

[tex]\frac{2\timesED}{ED}=\frac{AC}{20}[/tex] [Since D is the midpoint of EC]

AC = 2×20

AC = 40 units

Option A. is correct

2). In the give picture two triangles shown are similar.

Therefore, their respective sides will be in the same ratio.

[tex]\frac{6x+2}{4x+36}=\frac{45}{90}[/tex]

[tex]\frac{6x+2}{4x+36}=\frac{1}{2}[/tex]

2(6x + 2) = 4x + 36 [By cross multiplication]

12x + 4 = 4x + 36

12x - 4x = 36 - 4

8x = 32

x = (4×6 + 2)

x = 24 + 2

x = 26

Option A. is the answer

3). In the given picture, tree and building are parallel to each other.

Rope is a transverse to both the parallel lines.

Therefore, angle formed between the tree and rope is 48° will be equal to the angle between rope and the building because they are internal alternate angles.

Therefore, Option D. 48° will be the answer.

4). DF bisects ∠EDG. So by definition, opposite sides of these equal angles will be equal.

3x + 96 = 15x

15x - 3x = 96

12x = 96

x = 8

Option D. is correct.

5). E is equidistant from the sides of ∠HGF. We have to find the measure of ∠FGH.

Since EG is the bisector of ∠HGF, so ∠HGE ≅ ∠FGE

4x + 2 = 6x - 10

6x - 4x = 10 + 2

2x = 12

x = 6

And ∠HGF = (4x + 2) + (6x - 10)

                  = 10x - 8

By putting the value of x = 6

∠HGF = 10×6 - 8

          = 60 - 8

          = 52°

Option A. is correct.