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
In ΔABC, AC = BC, CD ⊥ AB with D ∈ AB , AB = 4 in, and CD = square root of 3 in. Find AC.
4 answers
tell us
The answer is the square root of 7.
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
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
1 answer