Question

abc is a triangle and ghed is a rectangle.bc=12cm,he=6cm,fc=bf and altitude af =24cm.find the area of the rectangle

Answer 1

we have 
BC = 12 cm and FC = BF 

FC = BC/2 = 12/2 = 6 cm

in triangle AFC and triangle EHC

AF and EH are parallel  lines and AC is transversal

∠ FAC = ∠ HEC

∠F = ∠H = 90 degrees

therefore by A - A congruency 

triangle AFC - triangle EHC 

= AF/FC = EH/HC

= 24/6 = 6/HC

= HC = 36/24 = 1.5 cm

FH = FC - HC = 6 - 1.5 = 4.5

GH = 2 x FH = 2 x 4.5 = 9 cm

Area of rectangle DGHE = GH x HE

= 9 x 6

= 54 cm square

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