Question

Abc is triangle ghed is rectangle bc=12cm he=6cm fc=bf altitude af=24cm find area of rectangle

Answer 1

Given: In ΔABC : BC = 12 cm
             ⇒EH = DG = 6cm (∵In Rectangle Opp Sides are ║)
             ⇒ BC = 12 cm ⇔ BF = FC = 12/2 =6cm
             ⇒AF= 24 cm ( Altitude of ΔABC)
             ⇒DE = GH (∵Opposite sides of Rectangle are Equal)
Ar. of Δ ABCD = 1/2 x BC x AF ⇒1/2 x 12 x 24 = 144 cm
From The Figure We can say that DG = IF = EH = 6 cm
So AI = AF - IF = 24 - 12 = 18 cm
Ar. Δ ADE = 1/2 x DE x AI = 1/2 x GH x 18 = (9 GH) cm
∵ BF = FC And AF is Altitude ⇒ Altitude Bisects the Side Which Is Property of and Isoceles Triangle.
∴ ∠ABC = ∠ACB
∴ΔBDG ≈ ΔHEC (Side Angle Side Angle Congruency)
⇒BG = HC (∵ΔBDG ≈ ΔHEC)
Ar. of ΔBDG = 1/2 x DG x BG ⇒1/2 x 6 x BG = (3 BG) cm
∴Ar. of ΔBDG = (3 BG) cm
Area of DEGH = length x breadth = EF x GH = 6 GH
From The Figure:
Area of ΔABC = Ar.ΔADE + Ar.ΔBDE + Ar.ΔHEC + Ar. DEGH
                       = 9 GH + 2(Ar.ΔBDE) + 6GH
                144 = 15 GH + 2(3BG)
15(BC - (BG+HC)) +6BG = 144
15(12 - 2BG) + 6 BG = 144 (∵BG = HC)
180 - 30 BG + 6 BG = 144
24 BG = 36
BG = 1.5

⇒ BG + HC + GH = BC = 12 cm
⇒GH = 12 - 2 BG (∵BG=HC)
⇒GH = 12 - 2(1.5)
⇒GH = 9
 
Area Of DGEH = 6 GH = 6 x 9 = 54 cm²
∴The Area of Given Rectangle = 54cm²

Answer 2

see the image
hope it helps.................