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Zepler [3.9K]
3 years ago
7

URGENT!!!!!

Mathematics
1 answer:
Marysya12 [62]3 years ago
8 0
Remember the pythagorean theorem?
this is like that
the distance is the hypotnuse

D=\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}

D=\sqrt{(8-(-3))^{2}+(-7-4)^{2}}
D=\sqrt{(11)^{2}+(-11)^{2}}
D=\sqrt{169+169}
D=\sqrt{338}
D=11\sqrt{2}
aprox 15.55 units



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In triangle abc shown below side ab is 6 and side ac is 4
kaheart [24]

Question is Incomplete, Complete question is given below:

In Triangle ABC shown below, side AB is 6 and side AC is 4.

Which statement is needed to prove that segment DE is parallel to segment BC and half its length?

Answer

Segment AD is 3 and segment AE is 2.

Segment AD is 3 and segment AE is 4.

Segment AD is 12 and segment AE is 4.

Segment AD is 12 and segment AE is 8.

Answer:

Segment AD is 3 and segment AE is 2.

Step-by-step explanation:

Given:

side AB = 6

side AC = 4

Now we need to prove  that segment DE is parallel to segment BC and half its length.

Solution:

Now AD + DB = AB also AE + EC = AC

DB = AB - AD also EC = AC - AE

Now we take first option Segment AD is 3 and segment AE is 2.

Substituting we get;

DB = 6-3 = 3 also EC = 4-2 =2

From above we can say that;

AD = DB and EC = AE

So we can say that segment DE bisects Segment AB and AC equally.

Hence From Midpoint theorem which states that;

"The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side."

Hence Proved.

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Answer:

Step-by-step explanation:

The description is too ambiguous to reconstruct the diagram. You need to post the actual diagram.

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That doesn’t help you answer this particular question, though.

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