Question

In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.

Answer 1

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

$BC^2=BD\cdot AB.$

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

$x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.$

Take positive value x. You get

$AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.$

2. According to the previous theorem,

$AC^2=AD\cdot AB.$

Then

$AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

Answer: $AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

$CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D$

This means that you cannot find solutions of this equation. Then CD≠2 cm.

Answer 2

Answer:

it is sqrt(7)

Step-by-step explanation:

when the altitude hits the triangles edge at D, it splits AB in to 2 segments which are ad and db. ad = (3-x) and db =x. and we have 2 right triangles. DCB and ACD are right triangles. so-> looking at BCD (3-x)^2+sqrt(3)^2=x^2, using pythagorean thereom, we get x=2. so then we look and triangle ADC and we see 2^2 +sqrt(3)^2 =7. we take the square root of that to get ac which is sqrt(7)