Answer to Question #287314 in Trigonometry for Mum

Question #287314

Explain the Pythagorean Theorem, its proofs and applications.


1
Expert's answer
2022-01-19T17:56:02-0500

Pythagoras Theorem

The Pythagoras Theorem states that the sum of the squares of the base and perpendicular is equal to the square of the hypotenuse in a right-angled triangle. Pythagoras' theorem states that if a triangle is right-angled (90 degrees), the square of the hypotenuse equals the sum of the squares of the other two sides.

Given a right angled triangle below,

the Pythagoras Theorem formula is given as,

"(Hypotenuse)^2=(Base)^2+(Perpendicular)^2"

Pythagoras Theorem proof

The proof of the Pythagoras Theorem involves the concept of similarity of the triangle. It states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given a right-angled triangle PQR, right angled at Q.



We proof that, "{PR^2}={PQ^2}+{QR^2}"

Draw a perpendicular line QD meeting PR at D.



We know that, "\\Delta RDQ\\sim\\Delta RQP"

So,

"{RD\\over QR}={QR\\over PR}" (Corresponding sides of similar triangles)

"\\implies{QR^2}=RD\\times PR.......(1)"

Also,

"\\Delta QDP\\sim\\Delta RQP"

So,

"{PD\\over PQ}={PR\\over PR}" (Corresponding sides of similar triangles)

So,

"PQ^2=PD\\times RP.......(2)"

Adding equation (1) and (2), we get,

"QR^2+PQ^2=(RD\\times PR)+(PD\\times PR)\\\\\n\\implies QR^2+PQ^2=PR(RD+PD)"

From the figure,

"RD+PD=PR"

Therefore,

"PR^2=PQ^2+QR^2" as required.

Applications of Pythagoras Theorem


  1. The Pythagoras Theorem is frequently used to determine the lengths of a right-angled triangle's sides.
  2. Architects employ the Pythagoras Theorem approach in the engineering and building professions.
  3. The Theorem is used to calculate the diagonal length of a rectangle, square, or other shape.
  4. It is also used in navigation to find the shortest route.
  5. Pythagoras Theorem is used in trigonometry to find the trigonometric ratios like "cot,sin,cosec,cos,tan,sec."

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS