triangle problem?

Question

equilateral triangle ABC has a side that forms the height of equilateral triangle DBE. What is the ratio of the areas of these two triangles? can't paste the picture, but you might be able to imagine it. please tell me the trick of it.

Answer 1

1 : how is the height of the first triangle ABC ?

we note F the middle of AC

then :
angle f = angle AFB = 90°

and :
AF = AC/2 = AB/2

BF = sqrt( AB² - AF²) = sqrt(AB² - AB²/4) = sqrt(3.AB²/4)
BF = AB sqrt(3/4)

then, the area of ABC = AB * BF/2 = AB² . sqrt(3/4) / 2

area ABC = AB² . sqrt(3/16)

2. area of the second triangle DBE.

we know that the height of DBE is equal to AB ... but we also know that the ratio between the side and the height is :

height = side * sqrt(3/4)

or :

side = height / sqrt(3/4)

the ratio between the two area is : (sqrt(3/4))² = 3/4

area BDE = (area ABC)*4/3

Answer 2

I think I know what you mean. Immagine a triangle with sides of 5cm, the area is then apprpox 10.825cm2, calculated by the formula:

Area = (Side Length Squared x Root 3) / 4

If and equilateral triangle has a height of 5 cm, its other two lengths can be calculated using simple trigonometry. This is made easier if an equilateral triangle is assumed to be two identical isoscelese (right angle) triangles. Then using the same formula as above to can calculate the area of the second triangle and thus the ratio.

Do this for 3 different size triangle and you will notice the ratio remain unchanged. I'm not going to give you the answer as you will invariably need this in an exam and will do better to understand how you came up with the answer, good luck.

Answer 3

The height of the equilateral triangle DBE cuts the side by the half. So, imagine a side of DBE (which I call b ) is the hypothenuse of a rectangle triangle with sides a side from ABC (which I call a) and hakf a sideof DBE (since in an equilateral triangle the hight cuts a side in two

so you can write b^2 = (b2/2)^4+a^2= b^2/4+ a^2

or 3b^2/4 = a^2 b= a (4/3)

As the area is proportionnal to the square of a side

the ratio areaDBE/area ABC = 4/3= 1.3333

Answer 4

triangle ABC has sides=x
triangle DBE has height =x

for an equilateral triangle, the ratio of the height to a side is √3 /2

so the ratio of heights is 1:√3/2 the ratio of areas is this ratio squared.

1^2:(√3 /2)^2
1:3/4
Area of DBE is 4/3 the area of ABC

Answer 5

let s=length of a side of ABC
then area of ABC is (s^2)*sqrt(3)/4 {just look up the formula for area of an equilateral or derive it}

now s is the height of DBE, so side is
2s/sqrt(3), now use this side length with the formula for area

then form and simplify the ratio of areas
area ratio DBE/ABC = 4/3

Answer 6

if i take the side of triangle ABC to be 1 unit, it would have an area of ABC would be √3/4 * 1/2 = √3/16
(you can use the pythagoras theorem to get to this point)
thus 1 would be the side of triangle DBE
the side would be √4/3(using pythagoras again)
the area would be √4/3*1=√4/3

ratio of area of ABC is to DBE=
3/16:4/3= 9:64

i hope i didn mess up some things!(:

Answer 7

A 2nd triangle interior a triangle potential that's ambiguous. basically meaning 2 triangles (in calculus). to comprehend no count if that's ambiguous, you label A on the backside left nook of the triangle, B on the backside appropriate nook of the triangle and C on the suitable of the triangle. Label the values you have and if A is acute AND its length is below b's length, then you definately likely have an ambiguous triangle. (additionally, on exams, if it states to locate all "achieveable" measures for besides the fact that, they're would desire to probably giving it away that it would be ambiguous.) enable us to sparkling up the triangle via utilising sine regulation and cosine regulation. Sine regulation for fixing lengths: a/SinA = b/SinB = c/SinC Sine regulation for fixing angles: SinA/a = SinB/b = SinC/c all of us comprehend a's length and degree, so we are able to apply it to sparkling up for b's perspective. SinA/a = SinB/b Sin50/22 = SinB/26 Isolate SinB via multiplying the two factors via 26. 26Sin50/22 = SinB Simplify. 0.905 = SinB Inverse sine for the two factors. Sin-a million(0.905) = (Sin-a million)SinB Simplify. sixty 4.86 = B sixty 5 = B for this reason, B = sixty 5 tiers. we could upload this to 50 then subtract a hundred and eighty to get C's perspective degree. a hundred and eighty - (50 + sixty 5) a hundred and eighty - one hundred fifteen sixty 5 for this reason, C = sixty 5 tiers. provided that I certainly have given you the occasion of utilising the sine regulation, you may end this triangle via utilising the sine regulation and sparkling up for c's length. As for the ambiguous"ness", you haven't any longer given me sufficient training on what to precisely sparkling up, via fact there are 2 angles in contact in an abiguous case. yet you may easily sparkling up it via utilising reference angles.

Answer 8

Area 1 = (BC^2√3)/4
Area 2 = BC^2/√3
Area 2/Area 1 = (BC^2/√3)/((BC^2√3)/4) = 4/3

Answer 9

If there are no measurements given then it is 1:1

Answer 10

it sounds lyk they're both the same... 1:1