There is a right angle triangle ABC with angle B = 90 deg. AB = 6, BC = 8. D is the midpoint of AC. Find BD.?

Answer 1

I realized an error in Pascal's and farha_heusen's solutions.
Note that angle BDA or angle BDC does not equal to 90 degrees.
To assume this requires the fact that AB = BC, which is not as given in the question.
So, we cannot assume BD is the height and AC is the base for the triangle.

Instead, I believe the solution requires one to rotate the right angle triangle by 180 degrees clockwise to form a rectangle. This rectangle is of side 6 and 8, with diagonal 10 (AC = 10). So then BD is half of the length of the other diagonal = 10/2 = 5

Answer 2

Basically I have generalized to three methods to prove AD = BC = BC, thus rendering the soln BD = 5.
1) Using rotation to form a rectangle of side 6 and 8, so that BD is half the length of a diagonal, which is 10/2 = 5
2) Using congruent triangles. Let P be a perpendicular from D to AB. Then you realize that triangle APD is congruent to BPD, from which the result is rendered as AD = BD = 5.
3) Using circumcentre. The circumcentre of the triangle is at the point D. Thus we can draw a circle with centre D passing through the points A, B, C. This would mean that AD = BD = CD, which means that BD = 5.

Answer 3

By Pythogoras theorem,

AC^2 = AB^2 + BC^2
AC^2 = (6)^2 + (8)^2
AC^2= 36 + 64
AC^2= 100
AC = Square Root of 100
AC = 10

Since D is the midpoint of AC , AD=DC=5cm

Now,
BD^2 = AD*DC
BD^2= 5cm * 5cm
BD^2=25cm^2
BD=Square root of 25cm^2
BD=5cm

Answer 4

Note that BD divides ABC into two triangles - ADB and CDB, with ∠D=90°. Thus, AD² + BD² = AB² ⇒BD² = AB² - AD². But AD = AC/2 and AC² = AB² + BC² ⇒ BD² = AB² - (AB² + BC²)/4 ⇒ BD = √(36 - (36 + 64)/4) ⇒ BD = √11

BD = √11 ≈ 3.3166

Edit: Oops... TT's right - the assumption that ADB would form a right triangle was completely unjustified. Ignore this answer.

Answer 5

First you have to find the length of AC.

Line AB ^2 + Line BC ^2 = Line AC ^2

SO...

6^2 + 8^2 = AC^2
36 + 64 = AC^2
100=AC^2
10=AC

Then to find the distance to the midpoint (MIDDLE) divide by two.
10/2=5.

So BD = 5

Answer 6

We can easily apply Pythagoras to learn that AC = 10. Hence, AD = DC = 5. Drop a perpendicular from D to AB, and call the point of intersection E. By similarity, AE = EB = 4. Since triangles ADE and BDE are congruent, AD = BD = 5.
-- Robert A. Saunders, Lake Stevens, WA.

Answer 7

hey, this looks like a maths enrichment stage (gauss) question!!!!!!!

however, if we draw a vertical line from D to make it perpendicular to line BC (let's call that point X) then triangle DXC is similar to triangle ABC
so CD/CA = 5/10 = CX/CB so it has half the dimensions
so DX is half of AB which is 4 cm.
triangle DXB (X is right angle) has sides 4 and 5 cm and to find hypoteneuse (BD) is 4^2+5^2=25
BD is route of 25
=5

Answer 8

(AB)^2 + (BC)^2 = (AC)^2
6^2 + 8^2 = (AC)^2
36 + 64 = (AC)^2
100 = (AC)^2
AC = 10

half of 10 is 5

now find half of the other 2 sides, which is 3 and 4

If you go from point D and go straight down, you will notice that from D to the midpoint of BC, lets call it E, forms a right triangle, if you use

(BE)^2 + (DE)^2 = (BD)^2
4^2 + 3^2 = (BD)^2
16 + 9 = (BD)^2
25 = (BD)^2
BD = 5

ANS : 5

So although BD doesn't from a right triangle with BDC, it still forms a right triangle with BED

Answer 9

bda and bdc are neither rt triangles, bda is not a rt angle .. ad=5, ba=6,cd=5, bc=8, however the rt angle from the mid point of bc, and the rt angle from the midpoint of ba each bisect ac.. bc/2=4,ba/2=3 ,; you have 4 rt triangles with 3,4,5 side ratios ,; bd=5.... bs,bs,bs

Answer 10

6,8,10 as sides is a standard right angled traingle.by the given information AC=10 ,AD=DC=10/2=5.D is mid point of hypotenuse AC=CIRCUM CENTER of traingle ABC
according to def of circum center it is eqidistant to all
vertices A B C there fore AD=CD=BD=5