Please see the picture attached also.
In triangle ABC, since two legs are given using Pythagorus theorem
we get AC = 5
We have triangle ABC and CHB having congruent angles
Hence they are similar and sides are proportional
In right angled triangle CHB, we have two sides.
Hence third side
Now AH = AC-CH
AH = 24 unit
Since, in triangles ABC and BHC,
( Right angles)
( Reflexive )
Hence, By AA similarity postulate,
By the property of similar triangle,
Here, BC = 9 unit and HC = 3 unit,
Since, by the below diagram,
AH = AC - HC
AH = 27 - 3
AH = 24 unit
Let HC=x, it is given that AH=3HC, then AH=3x.
Since, from the given figure, ΔABC is similar to ΔBHC and ΔABC is similar to ΔABH.
Therefore, ΔABH is similar to ΔBHC, hence using the similarity conditions,
AH = 1 or 4
CH = 4 or 1
An altitude divides a right triangle into similar triangles. That means the sides are in proportion, so ...
AH/BH = BH/CH
AH·CH = BH²
The problem statement tells us AH + CH = AC = 5, so we can write
AH·(5 -AH) = BH²
AH·(5 -AH) = 2² = 4
This gives us the quadratic ...
AH² -5AH +4 = 0 . . . . in standard form
(AH -4)(AH -1) = 0 . . . . factored
This equation has solutions AH = 1 or 4, the values of AH that make the factors be zero. Then CH = 5-AH = 4 or 1.
BH = 2
The altitude divides the triangle into similar triangles. The ratio of the long side to the short side is the same for all, so ...
AH/BH = BH/HC
4/BH = BH/1
4 = BH² . . . . multiply by BH
2 = BH . . . . . take the square root
HC = 1
AC = HC +AH = HC +(HC+2) = 2·HC +2
The altitude divides the triangle into similar triangles, so the ratio of hypotenuse to short side is the same for all. That is ...
BC/HC = AC/BC
2/HC = (2HC +2)/2
4 = 2(HC)(HC +1) . . . . . cross multiply
0 = HC² +HC -2 . . . . . . divide by 2, subtract 2
0 = (HC -1)(HC +2) . . . . factor. Solutions are those values of HC that make the factors be zero.
The useful solution is ...
HC = 1
In the right triangle ABC,
In the right triangle BHC, by the Pythagorean theorem,
you can find the details in the attachment.
H/L = L/S
Hypotenuse/Leg = Leg/Side
12/9 = 9/x
12-6.75 = 5.25
Please find the attachment.
We have been given that triangle ABC is a right triangle, having a right angle at point B and BH is the altitude.
We can see from our attachment that the altitude BH is drawn to hypotenuse AC.
Altitude geometric mean theorem states that the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. Each leg of the right triangle is the mean proportional of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.
Using the above theorem we can set proportions for our given side lengths as:
Upon substituting our given values we will get,
Upon cross multiplying our equation we will get,
Taking square root of both sides of our equation we will get,
Therefore, BH is equals to 2 units.