Ask Singapore Homework?

Upload a photo of a Singapore homework and someone will email you the solution for free.



Question

primary 6 | Maths
One Answer Below

Anyone can contribute an answer, even non-tutors.

Answer This Question
JK Reddy
JK Reddy

primary 6 chevron_right Maths chevron_right Singapore

Need help on P6-Maths-SA2-2016-Singapore-Chinese-Girls question 12

Date Posted: 4 years ago
Views: 387
J
J
4 years ago
E is the point of intersection of the line drawn from F and the line CD. It might have been left out by accident.


a)


Draw a line from F to D to form triangle CDF.

Triangle CDF and triangle CEF have the same height. Since CE is 3/4 of DC, the base of CEF is 3/4 of CDF. This tells you that the area of CEF is also 3/4 of CDF.

So area of triangle CDF = 24cm² ÷ 3 x 4
= 32cm²


b)

Area of ABF = twice of CEF
= 2 x 24cm²
= 48cm²

Notice that triangle ADF and triangle CDF have the same base length (which is the breadth of the triangle). Their combined height is the length of the rectangle ABCD

So, this means that their combined area = ½ x breadth x length of rectangle ABCD
= ½ area of ABCD

Combined area = 48cm² + 32cm²
= 90cm²

Area of ABCD = 2 x 90cm² = 180cm²
JK Reddy
JK Reddy
4 years ago
Thank you very much, It is helpful
J
J
4 years ago
Welcome. Have edited a small typo, it should be 'Their combined height is the length of rectangle ABCD' instead

See 1 Answer

done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
JK Reddy
Jk Reddy's answer
1 answers (A Helpful Person)
1st