Question

In the given figure it AD 3. DE=4
AB=12 BF=& F G = 6, Be=16 Then
the value of his
(M is the area of the quadrilateral
FEDE and N is the area of the
triangle ABC)​

Answer 1

Given  : AD= 3. DE=4 AB=12 BF=2 FG = 6, BC=10

(M is the area of the quadrilateral FEDE and N is the area of the triangle ABC)​

To Find :  value of M/N

Solution:

AB = 12

BC = 10

Area of ΔABC  = N

Area of ΔADC  = (AD/AB)N  = (3/12)N  = N/4

Area of ΔBCD = N - N/4  = 3N/4

Area of ΔGCD =  (GC/BC) (3N/4) = (2/10) (3N/4)  = 3N/20

area of Quadrilateral ACGD = N/4 + 3N/20  = 8N/20  = 2N/5

area of ΔBCE = (BE/AB)N  = (5/12)N

area of Δ BFE = (BF/ BC) area of ΔBCE = (2/10) (5/12)N  =N/12

area of the quadrilateral FGDE   = Area of ΔABC  - area of Δ BFE - area of Quadrilateral ACGD

= N - N/12 - 2N/5

= N ( 60 - 5 - 24)/60

= 31N/60

area of the quadrilateral FGDE =   31N/60

=> M = 31N/60

=> M/N = 31/60

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