Question

In figure ABC , B = 90°, BD AC, ifAB = 10 cm. BC = 20 cm. Find BD​

Answer 1

Solution :

From attachment,

In ∆ABC, ∠B = 90°, AB = 10 cm, BC = 20 cm.

So using Pythagoras theorem,

→ AC² = AB² + BC²

→ AC² = 10² + 20²

→ AC² = 500

→ AC = √500

→ AC = √(5 × 100)

→ AC = 10√5 cm.

→ Area of ∆ABC = 1/2 × AB × BC

Now as BD ⊥ AC,

So, Area of ∆ABC = 1/2 × BD × AC

Therefore,

→ 1/2 × AB × BC = 1/2 × BD × AC

→ AB × BC = BD × AC

→ 10 × 20 = BD × 10√5

→ 200 = BD × 10√5

→ BD = 200/(10√5)

→ BD = 20/√5

→ BD = (20√5)/(√5)²

→ BD = 20√5/5

→ BD = 4√5 cm

Therefore,

Required answer : (2) 4√5 cm.

Answer 2

EXPLANATION.

In the figure ABC.

⇒ ∠B = 90°.

⇒ AB = 10 cm.

⇒ BC = 20 cm.

As we know that,

Pythagoras Theorem.

⇒ H² = P² + B².

Hypotenuse > Perpendicular > Base.

⇒ (AC)² = (AB)² + (BC)².

⇒ (AC)² = (10)² + (20)².

⇒ (AC)² = 100 + 400.

⇒ (AC)² = 500.

⇒ (AC) = 10√5 cm.

To find : BD.

Area of triangle = 1/2 x base x height.

⇒ 1/2 x BC x AB = 1/2 x AC x BD.

⇒ BC x AB = AC x BD.

⇒ 20 x 10 = 10√5 x BD.

⇒ 200 = 10√5 x BD.

⇒ 20 = √5 x BD.

⇒ 20/√5 = BD.

⇒ (20/√5) x (√5/√5) = BD.

⇒ (20√5)/(5) = BD.

⇒ 4√5 = BD.

∴ BD = 4√5 cm.

Option [2] is correct answer.