In triangle △abc, ∠abc=90°, bh is an altitude. find the missing lengths. ab=4 and bc=3, find ah, ch and bh.

Step-by-step explanation:

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

CH =

Now AH = AC-CH

= 5-1.8

=3.2

AH = 24 unit

Step-by-step explanation:

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

HC=

Step-by-step explanation:

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,

Hence, HC=.

  AH = 1 or 4

  CH = 4 or 1

Step-by-step explanation:

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

Step-by-step explanation:

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

Step-by-step explanation:

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

12

Step-by-step explanation:

In the right triangle ABC,

Thus,

In the right triangle BHC, by the Pythagorean theorem,

Then

24

Step-by-step explanation:

you can find the details in the attachment.

5.25

Step-by-step explanation:

H/L = L/S

Hypotenuse/Leg = Leg/Side

12/9 = 9/x

x=6.75

12-6.75 = 5.25

Step-by-step explanation:

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.