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Question

Assistance Needed with P5 Mathematics Question

Dear all,

I hope this email finds you well. I am a parent of a P5 student. I am reaching out to you for assistance with a P5 mathematics question that my child is struggling with. Thanks a lot!

Best rgds
Gennee

Answer

May I ask if the vertical marks (above the 6s and 10s, indicating
the segments are equal length) were originally in the question or
added by the student?

The question emphasized that it is not "not drawn to scale"
and there is no mention that the triangle is isosceles.
So if the vertical marks were not originally in the question,
then we cannot assume that the 2 horizontal segments are 6.
They sum up to 12 but are not necessarily equal to 6.

For example, this is a valid representation of the problem
in general:


In that case, we can solve it by noticing that the white portion
is in a square 12 x 12 (meshed portion in diagram below).



The white portion is made up of 4 triangles that are equal
in size to the 4 meshed triangles. So the white portion has
area 12 x 12 / 2 = 72.

So blue shaded area is 20 x 20 - 72 = 328cm2.

PS: Notice that we get the same answer as the case where
the triangle is isosceles such that the base segments are
indeed 6s and 10s. The reason is that white area does not
change when it is "sheared" horizontally because the height
of the white triangles are not changed.

 

0 Replies 2 Likes ✔Accepted Answer

Total area of square = 20 * 20 = 400 cm²

Area of 1 of the middle unshaded triangles
= (1/2) base * height
= (1/2) (16-4) * 6
= (1/2) 12 * 6
= 36

So 2 unshaded triangles = 36 *2 = 72

So area of shaded region = 400 – 72 = 328 cm² #

0 Replies 0 Likes