Question

In △ABC, CD is an altitude, such that AD = BC. Find AC, if AB = 3 cm, and CD = √3

Answer 1

Answer:

$AC=\sqrt{7}$ cm

Step-by-step explanation:

As shown in the figure

Let AC = x and BD = y

Therefore,

AD = 3-y

and BC = 3-y   (∵ AD = BC)

In ΔBDC by Pythagoras theorem

$BC^2=CD^2+BD^2$

$(3-y)^2=(\sqrt{3})^2+y^2$

$\implies 9-6y+y^2=3+y^2$

$\implies 6y=6$

$\implies y=1$

Therefore,

$AD=3-1$

$\implies AD=2$

In ΔADC by Pythagoras theorem

$AC^2=CD^2+AD^2$

$\implies x^2=(\sqrt{3})^2+2^2$

$\implies x^2=3+4$

$\implies x^2=7$

$\implies x=\sqrt{7}$ cm

Hope this helps.

Answer 2

Answer:

AC = √7 cm

Step-by-step explanation:

Let say AD = x  cm

then BC = x cm  

BD = AB - AD = 3 - x   cm

CD = √3 cm

CD ⊥ AB

in Δ BDC

BC² = BD² + CD²

=> x² = (3 - x)² + (√3)²

=> x² = 9 + x² - 6x + 3

=> 6x = 12

=> x = 2

in Δ ADC

AC² = AD² + CD²

=> AC² = x² + (√3)²

=> AC² = 2² + 3

=> AC² = 7

=> AC = √7 cm