## Tuesday, November 24, 2015

### [S1_20151124AESR] Slanted Rectangle does not need Pythagoras

Question

Introduction
This is another “Bonus Question” at a secondary level from somewhere that the question poser did not mention, but I guess it is most likely an Integrated Programme school in Singapore.  It is a beautifully crafted question.  The presence of a slant line seems to necessitate the usage of Pythagoras’ Theorem.  However, we have seen that Pythagoras’ Theorem can actually be avoided even in Primary (Elementary) School problems.  So a 10 year old kid with a rudimentary knowledge of algebra could do this.  Can you spot a short cut?

Making Observations
Stare at the diagram for a while.  What do you observe?

Solution
area of  DDBnCn =  ½  of the area of  ABnCnD.
area of  DDBnCn =  ½  of the area of  DBnPQ.
\  area of DBnPQ  =  area of ABnCnD = n cm2.

H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem

Suitable Levels
Lower Secondary Mathematics
* challenge for Primary school Olympiad
* other syllabuses that involve areas and a tiny bit of algebra

* anyone game itching for a challenge