From the given diagram, answer the following questions.

‘AC’ is 23 cm in the adjoining figure. Consider the figure and answer the questions

The total area of figure is

$175.5 cm$

Here, we have, 

ΔDBC : where base (DB) = 10 + 3 = 13 cm, height (CG) = 5cm

ΔAGB : where base (BG) = 3 cm, height (AG) = AC - CG = 23 - 5 = 18 cm

ΔEFA : where base (EF) = 9cm, heigh (AF) = 11cm

Trapezium (DGFE) : where two parallel sides (DG & EF) = 10cm(a) ,9cm(b) and distance b/w them (GF) = AC - CG - AF = 23 - 5 - 11 = 7cm(H)

Now,

⇒ Area of figure = ar ΔDBC + ar ΔAGB + ar ΔEFA + ar of DGFE 

ar ΔDBC = $\frac{1}{2}×B×H⇒\frac{1}{2}×13×5=32.5cm^2$

ar ΔAGB = $\frac{1}{2}×B×H⇒\frac{1}{2}×18×3=27cm^2$

ar ΔEFA = $\frac{1}{2}×B×H⇒\frac{1}{2}×9×11=49.5cm^2$

ar DGFE = $\frac{1}{2}(a+b)×H⇒\frac{1}{2}(10+9)×7=66.5cm^2$

So total area is

$⇒ 32.5 + 27 + 49.5 + 66.5 = 175.5$