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Answers
Answer:
61 cm²
Step-by-step explanation:
VX = PX - PV = 8 - 3 = 5cm
- ΔPQV
Consider ΔPQV, Base = PV = 3 cm and Height = QV = 5 cm
Area of ΔPQV = 1/2 * Base * Height = 1/2 * 3 * 5 = 7.5 cm²
- Quadrilateral QVXR
Join line QA to meeet RX at A, such that QA ║ PS.
So, QA = VX = 5cm and AX = QV = 5cm and RA = RY - AX = 6 - 5 = 1cm
Consider ΔQRA, Base = RA = 1 cm and Height = QA = 5 cm
Area of ΔPQV = 1/2 * Base * Height = 1/2 * 1 * 5 = 2.5 cm²
Consider square QAXV, Length = QA = 5 cm
Area of square QAXV = 4 * Length = 4 * 5 = 20 cm²
- ΔRYS
YS = PS - PY = 11 - 9 = 2cm
Consider ΔRYS, Base = YS = 2 cm and Height = RX = 6 cm
Area of ΔRYS = 1/2 * Base * Height = 1/2 * 2 * 6 = 6 cm²
- ΔTYS
YS = PS - PY = 11 - 9 = 2cm
Consider ΔTYS, Base = YS = 2 cm and Height = TX = 2 cm
Area of ΔTYS = 1/2 * Base * Height = 1/2 * 2 * 2 = 2 cm²
- Quadrilateral UWXT
Join line TB to meeet UW at B, such that TB ║ WX.
WX = PX - PW = 8 - 5 = 3 cm
So, TB = WX = 3 cm and WB = TX = 2 cm and UB = WU - WB = 4 - 2 = 2cm
Consider ΔTBU, Base = UB = 2 cm and Height = TB = 3 cm
Area of ΔTBU = 1/2 * Base * Height = 1/2 * 2 * 3 = 3 cm²
Consider rectangle WXTB, Length = WX = 3 cm and Breadth = TX = 2cm
Area of rectangle WXTB = 2 (Length + Breadth) =2 (3 + 2) = 2 * 5= 10 cm²
- ΔPWU
Consider ΔPWU, Base = PW = 5 cm and Height = WU = 4 cm
Area of ΔPWU = 1/2 * Base * Height = 1/2 * 5 * 4 = 10 cm²
So, Area of polygon PQRSTU = Area of ΔPQV + Area of Quadrilateral QVXR + Area of ΔRYS + Area of ΔTYS + Area of Quadrilateral UWXT + Area of ΔPWU
Area of polygon PQRSTU = 7.5 +2.5 + 20 + 6 + 2 + 3 + 10 + 10 = 61 cm²
