Question

In triangle ABC, angle B = 90° and BD is perpendicular to AC.
(1.) If CD = 10cm and BD = 8cm; find AD.
(2.)If AC = 18cm and AD = 6cm; find BD.
(3.)If AC = 9 cm and AB = 7 cm; find AD.

Answer 1

I was able to solve the first and second part, not the third one. Sorry.

Answer 2

Value for AD & BD is 6.4 cm and 8.5 cm Respectively

Step-by-step explanation:

We have,     ∠ABC = 90°

⇒∠ABD + ∠DBC = 90°      .....(1)

In ΔBDC,

∠BDC + ∠DCB + ∠DBC = 180°   (Angle sum property)

⇒90° + ∠DCB + ∠DBC = 180°  

⇒∠DCB + ∠DBC = 90°    .....(2)

Now, from (1) and (2), we get

∠ABD + ∠DBC = ∠DCB + ∠DBC

⇒∠ABD = ∠DCB    ......(3)

In ΔADB and ΔBDC,

• ∠ABD = ∠DCB   [Using (3)]
• ∠ADB = ∠BDC  [90° each]
• ΔADB ~ ΔBDC [AA similarity]

⇒   [Corresponding sides of similar Δ's are proportional]

⇒$\frac {AD}{BD} = \frac {BD}{CD}$

⇒  $BD^2 = AD \times CD$ ........(4)

Now, put CD = 10 cm; BD = 8 cm in (4), we get    

$(8)^2 = AD \times 10$

⇒ AD = 6.4 cm

In ΔADB and ΔABC,      ∠A = ∠A   [Common]

• ∠ADB = ∠ABC   [90° each]

• ΔADB ~ ΔABC   (AA similarity)

⇒   (Corresponding sides of similar Δ's are proportional)

⇒$\frac {AD}{AB} =\frac {AB}{AC}$

⇒ $AB^2 = AD \times AC$   .....(5)

Put AD = 6 cm; AC = 18 cm in (5), we get

$AB^2 = 6 \times 18$

⇒ $AB^2 = 108$

⇒ $AB = 6\sqrt 3 cm$

In ΔADB,        [Pythagoras theorem]

⇒$BD^2 = 108-36$

⇒$BD^2 = 72$

⇒BD = $6\sqrt 2\ cm$ = 8.5 cm

Now, put AC = 9 cm; AB = 7 cm in (5), we get    

$(7)^2 = AD \times 9$

⇒AD = $\frac {49}{9}$

⇒AD = $5\frac {4}{9}\ cm$ = 5.4 cm