Question

ABC is a triangle right angled at B and BD is perpendicular to AC.if AD is 4cm and CD is 5cm,find BD and AB

Answer 1

Let AB = x, BD = y and BC = z.
From triangle ABC,
${x}^{2} + {z}^{2} = {(4 + 5)}^{2} = 81 - - (1)$
From triangle ADB,
${x}^{2} = {y}^{2} + 16 - - - (2)$
From triangle BDC,
${z}^{2} = {y}^{2} + 25 - - - (3)$
Substituting (2) and (3) in (1),
$( {y}^{2} + 16) + ( {y}^{2} + 25) = 81 \\ {y}^{2} = 20 - - - (4) \\ y = 2 \sqrt{5}$
Substituting (4) in (2),
${x}^{2} = 20 + 16 \\ x = 6$
Thus BD = 2√5; AB = 6

Answer 2

Answer:

Step-by-step explanation:

In triangle ABC & ADB

/_ A = /_A (common)

/_ABC = /_ ADB (each 90)

Therefore Triangle ABC similar to Triangle ADB

=> AB/AD = AC/AB

=> AB2= AD x AC

=> AB2= 4 x 9 (AC=AD+CD = 4+5 = 9 cm)

=> AB2= 36

=> AB = 6 cm

In Triangle ABD using Pythagoras Theorem

AB2= AD2+ BD2

62 = 42 + BD2

BD2= 36 - 16 = 20

BD = 2√5

Hope this helps you