In ΔABC, AC = BC, CD ⊥ AB with D ∈ AB , AB = 4 in, and CD = square root of 3 in. Find AC.
The answer is the square root of 7.
tell us
CB = sqrt 7
Explanation:
First, draw an iso triangle, since AC = BC
it should look like this: ( I did my best)
C
/ | \
/ - | - \
A D B
Since ACB is an iso triangle, and CD is perpendicular to the base, CD will be an angle bisector
so, AD = AB = 1/2 AB
AD = AB = 2
If we use the Pythagorean theorem, we'll get this equation:
(2)^2 + ((sqrt 3))^2 = (CB)^2
If we solve, we get
4+3=CB^2
7=CB^2
sqrt 7 = CB
To find AC in ΔABC, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
In this case, we have a right triangle ΔACD, where CD is the side opposite the right angle. We also know the lengths of AB and CD.
Given:
AB = 4 in
CD = √3 in
First, let's find AD by subtracting CD from AB:
AD = AB - CD
AD = 4 in - √3 in
Since AC = AD + CD (by the transitive property of equality), we can substitute the values of AD and CD into the equation:
AC = AD + CD
AC = (4 in - √3 in) + √3 in
To simplify, we can combine like terms:
AC = 4 in + (√3 in - √3 in)
Simplifying further:
AC = 4 in
Therefore, the length of AC in ΔABC is 4 inches.
Clearly ΔABC is isosceles, and CD bisects AB.
So AD = 2, and in ΔADC you now have 2 of the sides given
Use Pythagoras to find the hypotenuse AC
Let me know what you get