Question

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Answer 1

Answer:

In ⊥ΔACB,∠C=90

In ⊥ΔACB,∠C=90 ∘

In ⊥ΔACB,∠C=90 ∘ ,BC=?

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8)

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10)

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2 =100

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2 =100CB

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2 =100CB 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2 =100CB 2 =100−64

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2 =100CB 2 =100−64CB

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2 =100CB 2 =100−64CB 2

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2 =100CB 2 =100−64CB 2 =36

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2 =100CB 2 =100−64CB 2 =36∴CB=6

In ⊥ΔACB,∠C=90 ∘ ,BC=?AC 2 +CB 2 =AB 2 (8) 2 +CB 2 =(10) 2 64+CB 2 =100CB 2 =100−64CB 2 =36∴CB=6∴ Ladder is at a distance of 6m from the base of the wall.

Answer 2

Answer:

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