Question

∆ABC is an isosceles right-angled triangle with ∠C = 90°. If D is any point on AB such that CD = 2√10 cm and BD = 8 cm. Find the value of AD?​

Answer 1

Answer:

AD = 8 cm

Step-by-step explanation:

By Congruency,

In $\triangle CDA$ and $\triangle CDB$,

∠CDA = ∠CDB = 90°      [$CD \perp AB$]

AC = BC                          [Equal sides of an isosceles triangle]

CD = CD                          [Common]

∴ $\triangle CDA$ ≅ $\triangle CDB$         [Right Angle - Hypotenuse - Side Congruence]

$AD = BD$                        [Corresponding Parts of Congruent Triangles are Congruent]

AD = 8 cm

Alternatively,

In $\triangle CDB$,

$BC^2 = CD^2+BD^2$           [Pythagoras theorem]

$BC = \sqrt{(2\sqrt{10})^2 + (8)^2}$

$BC = \sqrt{40+64}$

$AC = BC = \sqrt{104}$

In $\triangle CDA$,

$AC^2 = CD^2+AD^2$            [Pythagoras theorem]

$AD = \sqrt{(\sqrt{104})^2 - (2\sqrt{10})^2}$

$AD = \sqrt{104-40} = \sqrt{64}$

$AD = 8 cm$

Any corrections or suggestions for improvement are welcome :)

Below is a diagram of the question: