Given :

• AD = 16 cm
• CD = 9cm
• ∆ADB & ∆CDB are two right TRIANGLE included in right triangle ∆ABC
• BD is common base between ∆ADB & ∆CDB

To find :

• AB, BC & BD

Solution :

• According to the PYTHAGORAS THEOREM

★(Hypotenuse)²=(perpendicular)²+(base)²

• In ∆ABC

→ (AC)² = (AB)² + (BC)²

→ (16 + 9)² = AB² + BC²

→ (25)² = AB² + BC²

→ AB² + BC² = 625 ---(i)

• Now, In ∆ADB

→ (AB)² = (BD)² + (AD)²

→ AB² = BD² + (16)²

→ AB² - BD² = 256 -----(ii)

• In ∆CDB

→ (BC)² = (DC)² + (DB)²

→ BC² = (9)² + DB²

→ BC² - DB² = 81 -----(iii)

Subtract equations (ii) & (iii) the equations

→ AB² - BD² - (BC² - DB²) = 256 - 81

→ AB² - BD² - BC² + DB² = 175

→ AB² - BC² = 175 ------(iv)

Add equations (i) and (iv)

→ AB² + BC² + AB² - BC² = 625 + 175

→ 2AB² = 800

→ AB² = 800/2

→ AB = √400 = 20 cm

Put the value of AB in eqⁿ (i)

→ AB² + BC² = 625

→ (20)² + BC² = 625

→ 400 + BC² = 625

→ BC² = 625 - 400

→ BC = √225 = 15cm

Put the value AB in eqⁿ (ii)

→ AB² - BD² = 256

→ (20)² - BD² = 256

→ 400 - BD² = 256

→ BD² = 400 - 256

→ BD = √144 = 12 cm

•°• AB = 20cm, BC = 15cm & BD = 12cm