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

since it is a right angled triangle, the altitude CD is the same as the side CB.

NOW, BY PYTHAGORAS theorem,

AB^2 + BC^2 = AC^2
3^2 + 2^2 = AC^2
9+4= AC^2
11 = AC^2
$\sqrt{11} = ac$

Answer 2

BC² = BD x AB

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

x² = (3-x) . 3

x² = 9 - 3x

x² + 3x - 9 = 0

D = 3³ - 4 . (-9) = 9 + 36 = 45

√D = √45 = 3√5

x₁ = -3 - 3 √5/2 < 0

x₂ = -3 + 3 √5/2 > 0

Now if we take the value of these numbers positively.

AD = BC - 3 + 3 √5/ 2 cm.

2. AC² = AD. AB

AC² = -3 + 3 √5/2 . 3

= - 9 + 9√5/2

AC = √-9 + 9√5/2 cm

Answer: AC = √- 9 + 9√5/2 cm

If there is any confusion please leave a comment below.