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secondary 2 | Maths
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Hi, I am stuck at part e of the question. Can anyone help? The answer is 4.2 cm², but I am not sure why and how.
Date Posted: 4 years ago
Views: 266
Yup.
Draw a line perpendicular to BK which hits point D.
That is the height of Triangle BDK.
You can extend the that line to the left until it hits the the line parallel to BK.
Note that the two parallelograms are identical to each other.
Draw a line perpendicular to BK which hits point D.
That is the height of Triangle BDK.
You can extend the that line to the left until it hits the the line parallel to BK.
Note that the two parallelograms are identical to each other.
Hello Adwin.
If you have the solution to the question, could you post it so i can see it?
Thanx.
Yuan.
If you have the solution to the question, could you post it so i can see it?
Thanx.
Yuan.
Sorry, I do not have the solution as the paper only has the final answer of 4.8 cm².
My final working, based on your help is as follows:
Area of ΔJFK / Area of ΔBDK = (3 / 9)²
2.4 / Area of ΔBDK = 1 / 9
Area of ΔBDK = 21.6 cm²
Height of ΔBDK = 21.6 cm² / 9cm × 2
= 4.8 cm
Height of ΔBDK = Height of ΔGFK = 4.8 cm
Area of ΔGFK = ½ × 4.8 cm × 3 cm
= 7.2 cm²
Area of ΔGJF = Area of ΔGFK - Area of ΔJFK
= 7.2 cm² - 2.4 cm²
= 4.8 cm²
My final working, based on your help is as follows:
Area of ΔJFK / Area of ΔBDK = (3 / 9)²
2.4 / Area of ΔBDK = 1 / 9
Area of ΔBDK = 21.6 cm²
Height of ΔBDK = 21.6 cm² / 9cm × 2
= 4.8 cm
Height of ΔBDK = Height of ΔGFK = 4.8 cm
Area of ΔGFK = ½ × 4.8 cm × 3 cm
= 7.2 cm²
Area of ΔGJF = Area of ΔGFK - Area of ΔJFK
= 7.2 cm² - 2.4 cm²
= 4.8 cm²
Thanx, Adwin.
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Part 1 of 2.
Date Posted:
4 years ago
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Part 2 of 2.
Date Posted:
4 years ago
Thanks. Can you explain the part about triangles BDK and HGF having similar heights? I don't see it.
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Basically how I come to the conclusion the heights are the same.
Date Posted:
4 years ago
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