or 6.8 In a right triangle, the square of the hypotenuse is equal to t

1.	or 6.8 In a right triangle, the square of the hypotenuse is equal to theum of the squares of the other two sides.
Answer» Given: A right triangle ABC right angled at B. To Prove: AC^2= AB^2+ BC^2 Construction: Draw BD ⊥ AC Proof: In Δ ADB and Δ ABC, ∠ADB = ∠ ABC (each 90°) ∠BAD = ∠ CAB (common) Δ ADB ~ Δ ABC (By AA similarity criterion) Now, AD/AB =AB/AC(corresponding sides are proportional) AB2= AD × AC … (i) Similarly, Δ BDC ~ Δ ABC BC2= CD × AC … (ii) Adding (i) and (ii) AB^2+ BC^2= (AD × AC) + (CD × AC) AB^2+ BC^2= AC × (AD + CD) AB^2+ BC^2= AC^2 Hence Proved.

or 6.8 In a right triangle, the square of the hypotenuse is equal to theum of the squares of the other two sides.

Answer»

Given: A right triangle ABC right angled at B.

To Prove: AC^2= AB^2+ BC^2

Construction: Draw BD ⊥ AC

Proof:

In Δ ADB and Δ ABC,

∠ADB = ∠ ABC (each 90°)

∠BAD = ∠ CAB (common)

Δ ADB ~ Δ ABC (By AA similarity criterion)

Now, AD/AB =AB/AC(corresponding sides are proportional)

AB2= AD × AC … (i)

Similarly, Δ BDC ~ Δ ABC

BC2= CD × AC … (ii)

Adding (i) and (ii)

AB^2+ BC^2= (AD × AC) + (CD × AC)

AB^2+ BC^2= AC × (AD + CD)

AB^2+ BC^2= AC^2

Hence Proved.