Question

Let's say I have triangle ABC. If I draw a median line from A to the opposite side, lets call the new point D----what is the relationship of the area of the two new triangles, ABD and ACD? Are they equal and if so why? And will this always be true?

Answer 1

areas will be equal
the reason is that the median divides the base into two equal halves
so the resuling two triangles will have bases of equal length

Answer 2

also the height for both of them will be same

area of each triangle = 1/2 * base * height
since base and height for both is same, their areas will b equal....

Answer 3

Harkirat, yes the height would in this case be the same and the base would be BC is that correct?

Answer 4

no base will be half of BC , each triangle will have base = BC/2

Answer 5

A median is a line which joins the vertex to the mid-point of opposite side

Answer 6

Oh, that is right. Thanks for the help. I worked the original problem into this simplier problem. I hope that my work was correct. The original problem had a triangle ABC. And then with side AB they drew a square, (similar to how they do it for the Pyth. Th and then the problem drew another square below AC and then the two corners of the squares were connected call them D and E forming a new triangle ADE---- and this triangle is suppose to have the same area as the original triangle ABC.

Answer 7

if u cud put the picture here it wud make things easier to understand...

Answer 8

how do I draw a picture on this site?

Answer 9

if the problem is in a book, either scan the page and attache it sing the "Attach File" button right below
or take a good snap and attach it.....

Answer 10

Let me ask this, are you familiar with the standard proof of Pythagorean's Theorem where a right triangle is drawn, a and b are the legs and c is the hypotenuse. Then a square is formed below a (with each side being the length of a) and then a square is formed by using length c. Lets call length a PQ with Q being where the right angle would be in the triangle. Then on the hypotenuse side we can call it PR. The square formed, one would be PRST and the other PKMQ. If you followed all of this, then connect K and T, forming the triangle PKT. This is suppose to have the same area as the original triangle. Does this help in allowing you to draw this out so you can understand the problem?

Answer 11

ok i get the picture, but in your cse, is triangle ABC a right angled triangle???

Answer 12

Well, this is the problem. The original triangle is not necessarily right.

Answer 13

There is a hint given to me. HInt: what happens when you rotate one of the triangles by 90 degrees. -----after dong this, I noticed the triangle with the median, but the median is not necessarily perpendicular and thus when one applies the area formula for a triangle, the height will be.......I don't know

Answer 14

tes. they are equal

Answer 15

they have same height and BD=CD

Answer 16

Well, my problem asks me to show how they are equal. I'm attempting to come up with the steps and have yet to come up with a solid step-by-step 'proof'....

Answer 17

ok give me a minute and ill draw and send u a picture...

Answer 18

I think It has nothig else. what is the area of triangle?

Answer 19

(height * opposite side) /2

Answer 20

Here it is..
Pls see if it matches what u say...

Answer 21

(Height * opposite side)/ 2---------The triangles have to be equal ----I'm convinced that if I slide the two triangles together connecting one of the congruent sides that there is a base (the longest side) which could be called the base. Drawing an altitude from the vertex point to this base ---what a second, the longest side has the mid-point so does this 'median point', does it have to be perpendicular , that is it is a

Answer 22

have u seen the figure i sent??????

pls say if it is okay or not, then i can proceed with the answer..

Answer 23

Yes that is a good representative drawing....

Answer 24

I initially did not see your drawing as this is the first time I've ever done this.....

Answer 25

No problem,

so if we rotate triangle DAE by 90 deg to right, DA will coincide with BA as they r of same length

Answer 26

That is correct.

Answer 27

also EA which is equal to AC will also move by 90 degree

Answer 28

so EA will become an extension of CA towards A and so EC will b a straight line

Answer 29

Yes it does become a straight line,

Answer 30

I've attempted to follow this....I backtrack, in doing as you suggest, doesnt E and C occupy the same spot, segment BD has midpoint A

Answer 31

...and AC is the median

Answer 32

No, if u look at my diagram and rotate DA by 90 deg clockwise D coincides with B giving us the attached picture

Answer 33

then BA is the median...

Answer 34

is it clear now???

Answer 35

Okay, I slid the wrong side....I'm looking at your new diagram, BY must be the median and using the Area formula for a triangle, 1/2* BY * AC = BY* 1/2 * EA

Answer 36

no , no BA is the median and BY is the height/altitude !!!

Answer 37

see AE = AC as both equal
so when we rotate EA up by 90 deg to get EC then A is mid-point of EC and so BA is the median

Answer 38

Yes, I meant to say BY is the altitude and one needs that in order to do the formula for the area of a triangle and I did set that up correctly did I not.....and yes in looking at the original diagram you sent, EA = AC....I see that very clearly....

Answer 39

so if we find
1
area of triangle BEA = --- x EA (base) x BY (height)
2
1
we can also say area of triangle BEA = --- x AC x BY (becoz EA = AC)
2

Answer 40

also
1
area of triangle BAC = --- x AC (base) x BY (height)
2
1
we can also say area of triangle BAC = --- x AC x BY
2

so we find area trngBEA = area of trngBAC

Answer 41

I think we have got it!!!!!!! Wow, I can't believe how simply yet difficult (for me) this was. You are a Group Superhero!!!

Answer 42

Thank you, it was a pleasure doing it with you becoz u r INTERESTED in knowing how to do it rather than just wanting answer ( ; })

finally, trng BEA is actually trngDAE

so area of trngDAE = area of trng BAC

Answer 43

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Answer 44

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Answer 45

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Answer 46

I'm sorry, just above your name.... at the beginnig....I see your name Group Superhero then at the far right is a medal with a 0...

Answer 47

what browser r u using???

Answer 48

R u using I.Explorer???

Answer 49

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