Question

15. Polygon ABCDE is divided into parts as shown in
adjoining figure. Find its area if AD = 8 cm, AH =
6 cm, AĞ = 4 cm, AF = 3 cm and perpendiculars
BF = 2 cm, CH = 3 cm, EG = 2.5 cm.​

Answer 1

In the given figure, ABCDE, AD = 8 cm, AH = 6 cm, AG = 4 cm,

AF = 3cm ⊥ BF = 2 cm CH = 3 cm and ⊥ EG = 2.5 cm

The given figure, consists of 3 triangles and one trapezium.

Now,

Area of ∆AED = ½ × AD × GE

= ½ × 8 × 2.5

= 10 cm²

Area of ∆ABF = ½ × AF × BF

= ½ × 3 × 2

= 3 cm²

Area of ∆CDH = ½ × HD × CH

= ½ × (AD – AH) × 3

= ½ × (8 – 6) × 3

= ½ × 2 × 3

= 3 cm²

Area of trapezium BFHC = ½ × (BF + CH) × FH

= ½ × (2 + 3) × (AH – AF)

= ½ × 5 × (6 – 3)

= ½ × 5 × 3

= 7.5 cm²

Hence,

Total area of the figure = Area of ∆AED + Area of ∆ABF + Area of ∆CDH + Area of trapezium BFHC

= 10 + 3 + 3 + 7.5

= 23.5 cm²

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Answer 2

Answer:

this is it

answer is 23.5 cm2