This blog is managed by Song Hock Chye, author of Improve Your Thinking Skills in Maths (P1-P3 series), which is published and distributed by EPH.

Thursday, September 17, 2009

Raffles Girls Pri Sch 2007 PSLE Math Prelim Q45

The figure below shows a rectangle PQRS. The lines are extended from point P, Q, R and S and they meet at point Y. The length of QR is 20 cm and the length of XY is 4 cm. Given that QS is a straight line, the area of Triangle PQY is 72 square cm and the area of Triangle SRY is 84 square cm, find the shaded area of Triangle QSY.



Solution

Area of Triangle PQY + Area of Triangle SRY
72 sq cm + 84 sq cm = 156 square cm

Area covered by Triangles PQY and SRY is also 1/2 of area of Rectangle PQRS. This also means that Area of Triangle PQS is also 156 sq cm.

Area of Triangle PSY = 1/2 x 20 cm x 4 cm = 40 sq cm

Area of Triangle PQY = 72 sq cm

Shaded area
Area of PQS – Area of PQY – Area of PSY
=156 sq cm - 72 sq cm - 40 sq cm= 44 square cm

Answer: 44 square cm

2 comments:

Dolphin said...

Area covered by Triangles PQY and SRY is also 1/2 of area of Rectangle PQRS.

Hi Mr Song
Could you explain why?

Excel Eduservice said...

If you draw a triangle in a rectangle such that:

1)The base of the triangle is one of the breadths of the rectangle,

2)Extend the triangle's height to be equal to the length of the triangle,

..you will notice that the area of the triangle is half the area of the rectangle.

In the diagram, we have 2 triangles, both of which have bases that are the breadth of the rectangle. At the same time the sum of heights of both triangles is equal to the length of the rectangle.

Hence, both triangles cover half the area of the rectangle.