Question

A rectangle plank √2 m wide is placed symmetrical on the diagonal of a square of side 8 metres as shown. What is the area of the plank ?​

Answer 1

Answer:

Let AF=AE=a

⇒AF

2

+AE

2

=EF

2

...( By Pythagoras theorem)

⇒a

2

+a

2

=(

2

)

2

⇒2a

2

=2

⇒a=1

∴AF=CG=1M

⇒FB=BG=8m−1m=7m

In ΔFGB,

FG

2

=FB

2

+BG

2

...( Pythagoras Theorem)

⇒FG

2

=7

2

+7

2

=98

⇒FG=7

2

∴ Area of the plank =FG×FE

=(7

2

×

2

)m

2

=14m

2

Answer 2

Let AF = AE = a

AF^2+AE^2=EF^2(By Pythagoras theorem)

a^2+a^2=(√2)^2

a=1

AF=CG=IM

FB=BG = 8m-1m = 7m

In ∆FGB,

FG^2=FB^2+BG^2(By Pythagoras theorem)

FG^2=7^2+7^2

FG^2=98

FG=7√2

Area of the plank=FG×FE

=(7√2×√2)m^2 =14m^2