In the following figure, seg DH ⊥ seg EF and seg GK ⊥ seg EF. If DH = 18 cm, GK = 30 cm and `A(triangle DEF) = 450 cm^2`, then find:

1) EF

2) `A(triangle GFE)`

3) `A(square DFGE)`

#### Solution

1) Area of a triangle = 1/2 x base x height

`:. A(triangle DEF) = 1/2 xx EF xx DH`

`:. 450 = 1/2 xx EF xx 18` ......(Substituting the given values)

`:. (450xx2)/18 = EF`

`:. EF = 50 :. EF = 50 cm`

2) `triangle DEF` and `triangle GEF` have then common base EF

∴ their areas are proportional to their corresponding heights

`:. (A(triangle DEF))/(A(triangle GEF)) = "DH"/"GK"`

`:. 450/(A(triangle GEF)) = 18/30` ....(Substituting the given values).

`:. A(triangle GEF) = (450xx30)/18 = 750 cm^2`

`:. A(triangle GEF) = 750 cm^2`

3) `A(squareDFGE) = A(squareDEF) + A(squareGEF)` ....(Area addition postulate)

= 450 + 750 = 1200cm^{2}

`:. A(squareDFGE) = 1200 cm^2`