Wong Wei Jie's answer to JJ's Primary 5 Maths Whole Numbers Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
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
Date Posted:
2 years ago
360 Tutoring Program