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primary 6 | Maths
| Measurement
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Pls help!
Date Posted: 5 years ago
Views: 607
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See my comment for explanation.
Date Posted:
5 years ago
Let area of ∆ DEF be 1 unit.
∆CEF's base is twice that of ∆DEF,
and they have same height,
Area of ∆CEF = 2 units.
You can see from the diagram that the triangles on the other side are the same.
Area of ∆ ADF = 2 units + 1 unit + 1 unit
= 4 units
Area of ∆ ADF = 1/2 x base x height
= 1/2 x 6 x 18 = 54 cm²
So 4 units = 54 cm²
1 unit = 54 cm² ÷ 4 = 13.5 cm²
Area of shaded part
= Area of ∆ DBC - Area of ∆ DEF
= 1/2 x 18 x 18 - 1 unit
= 162 cm² - 13.5 cm²
= 148.5 cm²
∆CEF's base is twice that of ∆DEF,
and they have same height,
Area of ∆CEF = 2 units.
You can see from the diagram that the triangles on the other side are the same.
Area of ∆ ADF = 2 units + 1 unit + 1 unit
= 4 units
Area of ∆ ADF = 1/2 x base x height
= 1/2 x 6 x 18 = 54 cm²
So 4 units = 54 cm²
1 unit = 54 cm² ÷ 4 = 13.5 cm²
Area of shaded part
= Area of ∆ DBC - Area of ∆ DEF
= 1/2 x 18 x 18 - 1 unit
= 162 cm² - 13.5 cm²
= 148.5 cm²
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