PSA_LEVEL-UP_QNR19. In the figure below, the radii of the circles with centres at A and B are 5 cm and 3 cm respectively. If ∠FAG = ∠DBC = 90°. What is the perimeter of ΔCEG?
(i) 38.63 cm
(ii) 30.14 cm
(iii) 42.18 cm
(iv) 27.31 cm
Ans: Let us join the point F and the “meeting point” of the two circles, P. Also, join the points D and P.
∠FAG = ∠DBC = 90°. This indicates that GP and PC are the two diameters and GAPBC is a straight line. [FA and DB are two parallel lines intersected by line GAPBC].
Perimeter of ΔCEG
=GC + CE + EG
=(GP+PC) + (CD + DE) + (GF + FE)
GP = 10 cm.
PC = 6 cm.
CD = √(BC² + BD²) = 3√2 cm.
GF = √(FA² + GA²) = 5√2 cm.
Also, the figure EFPD is a rectangle [∠GFP and ∠PDC are two opposite angles and their measure is 90° each as they lie in respective semicircles.]
Thus, DE= FP = GF =5√2 cm.
and FE=PD=CD =3√2 cm.
Perimeter of ΔCEG =(GP+PC) + (CD + DE) + (GF + FE) = (10 + 6) + (3√2 + 5√2) + (5√2 + 3√2) = 16 + 16√2 = 16*2.414 = 38.64. Option (i).