Find the area of a hexagon ABCDEF as per the dimensions given in the figure alongside?
Answers
We have to find out Area of hexagon ABCDEF as per the dimensions given in the figure.
▶️ Area of hexagon ABCDEF = Area of ∆ABP + Area of ∆AQF + Area of ∆CDS + Area of ∆RDE + Area of trapezium BPSC + Area of trapezium QREF
➝ Area of ∆ABP = 1/2 × base × height
=> base = AP = 3.6 cm
=> height = QF = 10.5 cm
=> Area of ∆ABP = 1/2 × 3.6 × 4.5
=> Area of ∆ABP = 8.1 cm² .......(1)
➝ Area of ∆AQF = 1/2 × base × height
=> base = AQ = 3.6 + 4.3
= 7.9 cm
=> height = QF = 9 cm
=> Area of ∆AQF = 1/2 × 7.9 × 10.5
=> Area of ∆AQF = 41.47 cm²......(2)
➝ Area of ∆CDS = 1/2 × base × height
=> base = DS = 3.4 cm
=> height = CS = 9 cm
=> Area of ∆ABP = 1/2 × 3.4 × 9
=> Area of ∆ABP = 15.3 cm ² ......(3)
➝ Area of ∆RDE = 1/2 × base × height
=> base =RD = 3.4 + 3.4
= 6.8 cm
=> height = RE = 7.5 cm
=> Area of ∆ABP = 1/2 × 6.8 × 7.5
=> Area of ∆ABP = 25.5 cm² .......(4)
➝ Area of trapezium BPSC = 1/2 × ( sum of parallel sides ) × height
BP and CS are parallel sides.
=> Sum of parallel sides = 4.5+ 9 = 13.5 cm
=> height = PS = 4.3 + 4.8 + 3.4
= 12.4 cm
=> Area of trapezium BPSC = 1/2 ×13.5 × 12.4
=> Area of trapezium BPSC = 83.7 cm² ......(5)
➝ Area of trapezium QREF = 1/2 × ( sum of parallel sides ) × height
QF and RE are parallel sides.
=> Sum of parallel sides = 10.5 + 7.5= 18 cm
=> height = QR = 4.8
=> Area of trapezium QREF = 1/2 ×18 × 4.8
=> Area of trapezium QREF =43.2 cm² ......(6)
Now , Area of hexagon ABCDEF = (1)+ (2)+ (3)+ (4)+ (5)+ (6)
=> 8.1 +41.47+ 15.3 + 25.5 + 83.7 + 43.2
=>217.27 cm²