15 x 3 = 45 Squares
15 x 2 = 30 Triangles
45 + 30 = 75 cm
Alternative method
----------------------
15 x 5 = 75 cm
Monday, January 3, 2011
Perimeter PSLE 2010 P1 Q28
The shaded figure below is formed using 3 squares and 3 equilateral triangles. The length of the straight line AB is 15 cm. Find the perimeter of the shaded figure.
15 x 3 = 45 Squares
15 x 2 = 30 Triangles
45 + 30 = 75 cm
Alternative method
----------------------
15 x 5 = 75 cm
15 x 3 = 45 Squares
15 x 2 = 30 Triangles
45 + 30 = 75 cm
Alternative method
----------------------
15 x 5 = 75 cm
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As appeared in PSLE 2010 exam:
ReplyDeleteThe shaded figure below is formed using 3 squares and 3 equilateral triangles. The length of the straight line AB is 15 cm.
Find the perimeter of the shaded figure.
Question has a shaded figure with three different sizes of squares and equilateral triangles. Line A is drawn from left to right and end at B.
On the left side, 2nd biggest square shared the same base as 2nd biggest equilateral triangle with the vertex facing downwards.
Centre is the largest equilateral triangle with the vertex facing upwards and sharing the same base with the largest square. On the right hand side, smallest square is sharing with the smallest equilateral triangle wit the vertex facing downnards.
From left to right, it's square, triangle and square on top.Then triangle, square and triangle facing downwards.All shared the same AB as the common base for this shaded figure.