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primary 5 | Maths
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Need help again.thks
Date Posted: 4 years ago
Views: 205
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Thak you
Inappropriate approach for a Pri 5 question, since Pythagoras Theorem is only taught at Secondary 2.
The approach to take is
Area of 1 such triangle
= 1/2 * 3 * 4
= 6 cm^2
Area of 4 such triangles
= 4 * 6
= 24 cm^2
Observe that there is a small unshaded square in the middle. Since a length of 3 cm is placed against a length of 4 cm, the sides of the unshaded square are 1 cm each.
Area of unshaded square
= 1 * 1
= 1 cm^2
Total area of figure
= 24 + 1
= 25 cm^2
The approach to take is
Area of 1 such triangle
= 1/2 * 3 * 4
= 6 cm^2
Area of 4 such triangles
= 4 * 6
= 24 cm^2
Observe that there is a small unshaded square in the middle. Since a length of 3 cm is placed against a length of 4 cm, the sides of the unshaded square are 1 cm each.
Area of unshaded square
= 1 * 1
= 1 cm^2
Total area of figure
= 24 + 1
= 25 cm^2
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