ABC is a triangle with AB=13cm,BC=15cm and CA=14cm,AD and BE are altitude from A and B to BC and AC respectively .H is the point of intersection of AD and BE then the ratio 5HD/HB is
ABC is a triangle with AB=13cm,BC=15cm and CA=14cm,AD and BE are altitude from A and B to BC and AC respectively .H is the point of intersection of AD and BE then the ratio 5HD/HB is
ABC is a triangle with AB = 13; BC =
There are two right-angled triangles ABE and BEC, with BE as the common side. Let AE = s, so the EC = (15-s). By Pythagoras theorem and from the two triangles BE^2 = AB^2 - AE^2, or
BE^2 = 13^2 - s^2 as also
BE^2 = 14^2 - (15-s)^2. Thus
13^2 - s^2 = 14^2 - (15-s)^2, or
169 - s^2 = 196 - (225 - 30s + s^2), or
169 - s^2 = 196 - 225 + 30s - s^2), or
30s = 169 - 196 + 225 = 198, or
s = 198/30 = 6.6.
So AE = 6.6 and EC = 15 - 6.6 - 8.4.
In right-angled triangle BHD, HD/HB = sin HBD. From triangle BEC, sin EBC = (same as sin HBD) = EC/BC = 8.4/14 = 1.2/2 = 0.6 = HD/HB
5HD/HB = 0.6*5 =3
ABC is a triangle with AB = 13; BC =
There are two right-angled triangles ABE and BEC, with BE as the common side. Let AE = s, so the EC = (15-s). By Pythagoras theorem and from the two triangles BE^2 = AB^2 - AE^2, or
BE^2 = 13^2 - s^2 as also
BE^2 = 14^2 - (15-s)^2. Thus
13^2 - s^2 = 14^2 - (15-s)^2, or
169 - s^2 = 196 - (225 - 30s + s^2), or
169 - s^2 = 196 - 225 + 30s - s^2), or
30s = 169 - 196 + 225 = 198, or
s = 198/30 = 6.6.
So AE = 6.6 and EC = 15 - 6.6 - 8.4.
In right-angled triangle BHD, HD/HB = sin HBD. From triangle BEC, sin EBC = (same as sin HBD) = EC/BC = 8.4/14 = 1.2/2 = 0.6 = HD/HB
5HD/HB = 0.6*5 =3
