Q. 95.0( 2 Votes )

In Fig. 4.178,

(i) If DE = 4 cm, BC = 6 cm and area () = 16 cm2, find the area of .

(ii) If DE = 4 cm, BC = 8 cm and area () = 25 cm2, find the area of .

(iii) If DE : BC = 3 : 5. Calculate the ratio of the areas of and the trapezium BCED.


Answer :

(i) We have , DE||BC, DE = 4cm, BC = 6cm and area (ΔADE) = 16cm2


In ΔADE and ΔABC


<A = <A (Common)


<ADE = <ABC (Corresponding angles)


Then, ΔADE ~ ΔABC (BY AA similarity)


So, By area of similar triangle theorem


Area of ΔADE/Area of ΔABC = DE2 /BC2


16/Area of ΔABC = 42/62


Or, Area (ΔABC) = 16 x 36/16


= 36cm2


(ii) We have , DE||BC, DE = 4cm, BC = 8cm and area (ΔADE) = 25cm2


In ΔADE and ΔABC


<A = <A (Common)


<ADE = <ABC (Corresponding angles)


Then, ΔADE ~ ΔABC (BY AA similarity)


So, By area of similar triangle theorem


Area of ΔADE/Area of ΔABC = DE2 /BC2


25/Area of ΔABC = 42/82


Or, Area (ΔABC) = 25 x 64/16


= 100 cm2


(iii) We have DE||BC, And DE/BC = 3/5 ……………(i)


In ΔADE and ΔABC


<A = <A (Common)


<ADE = <ABC (Corresponding angles)


Then, ΔADE ~ ΔABC (BY AA similarity)


So, By area of similar triangle theorem


Area of ΔADE/Area of ΔABC = DE2 /BC2


Area of ΔADE/Area of ΔADE + Area of trap. DECB = 32/52


Or, 25 area ΔADE = 9 Area of ΔADE +9 Area of trap. DECB


Or 25 area ΔADE - 9 Area of ΔADE = 9 Area of trap. DECB


Or, 16 area ΔADE = 9 Area of trap. DECB


Or, area ΔADE / Area of trap. DECB = 9/16


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