Question

verify Pythagoras theorem......


please help me guys..... fast. I want to prepare my assignment

Answer 1

In the adjoining figure, ∆ PQR is a right angled triangle where QR is its hypotenuse and PR > PQ.

Square on QR is QRBA, square on PQ is PQST and the square on PR is PRUV.

The point of intersection of the diagonal of the square PRUV is O.

The straight line through the point O parallel to the QR intersects PV and RU at the point J and K respectively.

Again the straight line through the point O perpendicular to JK intersects PR and VU at the point L and  respectively.

As a result, the square PRUV is divided into four parts which is marked as 1, 2, 3, 4 and the square PQST is marked 5.

You can draw the same figure on a thick paper and cut it accordingly and now cut out the squares respectively from this figure. Cut the squares PRUV along JK and LM dividing it in four parts. Now, place the parts 1, 2, 3, 4and 5 properly on the square QRBA.

Note:

(i) these parts together exactly fit the square. Thus, we find that QR2 = PQ2 + PR2

(ii) Square drawn on side PQ, which means the area of a square of side PQ is denoted by PQ2.