Question

pls tell it's urgent​

Answer 1

Step-by-step explanation:

In ∆ FED , base is 10cm , l DE and l FE = 13cm , Angle Y is 90°

So , (13)^2 = (YE)^2 + (5)^2 [Half of base]. {Pythagoras theorem}

So YE = 12cm

So area of ∆FED is 1/2 × base × height

Area of ∆ FED = 60cm^2

In ∆BAF , by pythegoras theorem

Base BF is 2√2

Area of ∆BAF = 4√2cm^2

So , In rectangle BCDF , Length is 10cm and breadth is 4√2

Area of rectangle = 40√2 cm^2

Now add the areas , 60 + 44√2cm^2