In the diagram below of right triangle ACB, altitude CD intersects AB at D. If AD=3 and DB=4, find the length of CD in simplest radical form.
How do I go about figuring out this problem?
Please help. thanks
Which side is the hypotenuse? We can't see your "figure below".
drwls your answer seems very confusing isnt there a simpiler way to solve this????
The hypotenuse is side AB, or side ADB. The altitude goes from the vertex angle C down to point D, which is between A and B.
To solve the problem, we can use the concept of similarity of triangles.
First, let's label the length of CD as x. Since AD and CD are altitudes of triangles ADB and ADC respectively, we know that these two triangles are similar.
According to the similarity property for right triangles, the lengths of the corresponding sides of similar triangles are proportional. Therefore, we can set up the following proportion:
AD/DB = CD/AD
Substituting the given values,
3/4 = x/3
To solve for x, we can cross-multiply:
4 * x = 3 * 3
4x = 9
Now, divide both sides of the equation by 4:
x = 9/4
Since the problem asks for the length of CD in simplest radical form, note that 9/4 is already in this form. Thus, the length of CD is 9/4.
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let x be the angle ACD, part of the right angle at C. 90-x is then angle DCB.
You know that
CD tan x = 3 and
CD tan (90-x) = CD cot x = 4
Divide one equation by the other and the CD cancels out
tanx/cotx = tan^2 x = 3/4
tanx = (sqrt3)/2
CD = 3/tanx = 3*2/(sqrt3) = 2 sqrt3