Question

In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths. AC=5 and BH=2, find AH and CH.

Answer 1

Answer:

AH= 1 or 4.

CH = 4 or 1.

Step-by-step explanation:

Please find the attachment.

We have been given that in  triangle △ABC, ∠ABC=90°, BH is an altitude. We are also told that AC equals 5 and BH equals 2.

Altitude geometric mean theorem states that altitude drawn to hypotenuse of a right triangle separates the hypotenuse into two segments and the length of this altitude is geometric mean between the lengths of these segments.

Using Altitude geometric mean theorem we can set an equation as:

$\frac{AH}{BH}=\frac{BH}{CH}$

Upon substituting BH=2 in our equation we will get,

$\frac{AH}{2}=\frac{2}{CH}$

Let us cross multiply our equation.

$2*2=AH*CH$

$4=AH*CH...(1)$

We can see from our attachment that $AC=AH+CH$.

Substituting AC equals 5 in this equation we will get,

$5=AH+CH...(2)$

We have two equations and two unknowns, so we will use substitution method to solve system of equations.

From equation (2) we will get,

$5-AH=CH$

Substituting this value in equation (1) we will get,

$4=AH*(5-AH)$

$4=5AH-AH^2$

$AH^2-5AH+4=0$

Let us factor out our quadratic equation by splitting the middle term.

$AH^2-4AH-AH+4=0$

$AH(AH-4)-1(AH-4)=0$

$(AH-1)(AH-4)=0$

$(AH-1)=0\text{ or }(AH-4)=0$

$AH=1\text{ or }AH=4$

1st case: If AH is 1.

Upon substituting AH=1 in equation (2) we will get,

$5=1+CH$

$5-1=1-1+CH$

$4=CH$

When AH is 1, then CH will be 4.

2nd case: If AH is 4.

Upon substituting AH=4 in equation (2) we will get,

$5=4+CH$

$5-4=4-4+CH$

$1=CH$

When AH is 4, then CH will be 1.