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primary 5 | Maths
| Whole Numbers
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hi there please help me out with this question, I urgently need it
Date Posted: 2 years ago
Views: 204
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Let length of ST be x cm
Given that ST = TU = UR, and because it is a rectangle so PS = QR and PQ = 3 × ST. So,
Area of PST
= ½ × ST × PS
= ½PSx
Area of PTU
= ½ × TU × PS
= ½PSx
Area of trapezium PURQ
= ½ × (PQ + UR) × QR
= ½ × (3ST + ST) × PS
= 2PSx
Therefore,
fraction of area of shaded region over area of rectangle PQRS
= area of PTU / (area of PST + area of PTU + area of trapezium PURQ)
= ½PSx / (½PSx + ½PSx + 2PSx)
= ⅙
b) Area of shaded region = 37.5 cm²
Thus, area of rectangle = 37.5 × 6 = 225 cm²
Given that ST = TU = UR, and because it is a rectangle so PS = QR and PQ = 3 × ST. So,
Area of PST
= ½ × ST × PS
= ½PSx
Area of PTU
= ½ × TU × PS
= ½PSx
Area of trapezium PURQ
= ½ × (PQ + UR) × QR
= ½ × (3ST + ST) × PS
= 2PSx
Therefore,
fraction of area of shaded region over area of rectangle PQRS
= area of PTU / (area of PST + area of PTU + area of trapezium PURQ)
= ½PSx / (½PSx + ½PSx + 2PSx)
= ⅙
b) Area of shaded region = 37.5 cm²
Thus, area of rectangle = 37.5 × 6 = 225 cm²
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