Question

In the given figure, ABC is a triangle and GHED is a rectangle in ABC. BC =12 cm, HE=6 cm, FC=BF and altitude AF=24 cm. Find the area of the rectangle.

Answer 1

See diagram.

There are two cases possible with given data, as per the rectangnle is concerned.

case 1:  
  As   AF = altitude,  CF = FB, The triangle ABC is isosceles.
  Then rectangle DEHG is symmetrically positioned around altitude.
   ΔAFC and EHC are similar.
       AF / FC = EH / HC   
       24 / 6 = 6 / HC
       HC = 1.50 cm
       So  FH = 6 - 1.50 = 4.50 cm
          =>GH = 9 cm

Area GHED = 9 * 6 = 54 cm²

==========

Case 2:
   As EH is symmetrically positioned around AF (altitude) EF = HF = 3 cm.
     =>  HC = 3 cm

 ΔGHC and ΔAFC are similar.
      GH / HC = AF / FC 
       GH = 3 * 24 / 6 = 12 cm
  
Area of rectangle = GHED =  6 * 12 = 72 cm²

Answer 2

Answer:

Step-by-step explanation:

As   AF = altitude,  CF = FB, The triangle ABC is isosceles.

Then rectangle DEHG is symmetrically positioned around altitude.

 ΔAFC and EHC are similar.

     AF / FC = EH / HC  

     24 / 6 = 6 / HC

     HC = 1.50 cm

     So  FH = 6 - 1.50 = 4.50 cm

        GH = 9 cm

Area GHED = 9 * 6 = 54 cm²

 As EH is symmetrically positioned around AF (altitude) EF = HF = 3 cm.

    HC = 3 cm

ΔGHC and ΔAFC are similar.

    GH / HC = AF / FC

     GH = 3 * 24 / 6 = 12 cm

Area of rectangle = GHED =  6 * 12 = 72 cm²