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primary 6 | Maths
| Geometry
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Date Posted: 5 years ago
Views: 384
Triangle ADC is half of Rectangle ACYZ.
Area of triangle ADC = 1/2 X 8cm (AB = DC) X 12cm (AD = BC) = 48cm2
Area of Rectangle ACYZ = 48cm2 X 2 = 96cm2
Area of triangle ADC = 1/2 X 8cm (AB = DC) X 12cm (AD = BC) = 48cm2
Area of Rectangle ACYZ = 48cm2 X 2 = 96cm2
The question is asking for the breadth of
ACYZ.
ADC = 48 cm² (refer to above comment).
Notice that the vertical height of ADC is the same as the breadth of ACYZ. And it's base is just the length of ACYZ. So that's why it's ½ of ACYZ.
Area of ADC = ½ x vertical height x base
Area of ADC = ½ x breadth of ACYZ x length of ACYZ
48 cm² = ½ x breadth of ACYZ x 15 cm
Breadth of ACYZ
= 48 cm² ÷ ½ ÷ 15 cm
= 6.4 cm
ACYZ.
ADC = 48 cm² (refer to above comment).
Notice that the vertical height of ADC is the same as the breadth of ACYZ. And it's base is just the length of ACYZ. So that's why it's ½ of ACYZ.
Area of ADC = ½ x vertical height x base
Area of ADC = ½ x breadth of ACYZ x length of ACYZ
48 cm² = ½ x breadth of ACYZ x 15 cm
Breadth of ACYZ
= 48 cm² ÷ ½ ÷ 15 cm
= 6.4 cm
Alternatively, knowing that ADC is half of
ACYZ,
area of ACYZ = 48 x 2 = 96 cm²
Breadth of ACYZ = area ÷ length
= 96 cm² ÷ 15 cm
= 6.4 cm
ACYZ,
area of ACYZ = 48 x 2 = 96 cm²
Breadth of ACYZ = area ÷ length
= 96 cm² ÷ 15 cm
= 6.4 cm
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Ans: 6.4cm
Date Posted:
5 years ago
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