Question

Prove that in a right triangle the square of the hypotenuse is equal to sum of square of other two using the above that in rhombus abcd for a b square is equal to ac square + b square

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other sides.

Proof: Construct a right-angled triangle ABC, Draw a perpendicular from a A intersecting AC at D.

Given:

In ΔABC,

∠ABC = 90°

∠ADB = ∠CDB = 90°

To Prove:

AC² = AB² + BC²

Proof:

In ΔABC and ΔBDC,

∠ABC = ∠BDC = 90° (Given + construction)

∠ACB = ∠BCD (Common angle)

∴ ΔABC ΔBDC by AA similarity.

We know that corresponding parts of similar triangles are proportional. Therefore,

Cross multiplying BC/DC and AC/BC we get,

In ΔABC and ΔADB,

∠ABC = ∠ADB = 90° (Given + construction)

∠BAC = ∠BAD (Common angle)

∴ ΔABC ΔADB by AA similarity.

We know that corresponding parts of similar triangles are proportional, ∴

Cross-multiplying AB/AD and AC/AB we get,

Adding equations 1 and 2 we get,

BC² + AB² = (AC × DC) + (AC × AD)

AB² + BC² = AC (DC + AD)

AB² + BC² = AC (AC)

AB² + BC² = AC²

Hence Proved.

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