Question

to BC 1. In the figure 12.109 (not drawn to scale), LMis parallel AB= 6 cm, AL = 2 cm and AC = 9 cm. Calculate: (i) The length of CM (ii) The value of the ratio Area of triangle ALM (ICSE 1996) Area of trapezium LBCM M M B Fig. 12.109​

Answer 1

Answer:

Area of ΔALM / Area of Trap LBCM   = 1/8

Step-by-step explanation:

In the adjoining figure, LM is parallel to BC. AB =

6 cm, AL = 2 cm and AC = 9 cm.

LM ║ BC

=> ΔALM ≅ ΔABC

=> AL/AB  = AM/AC

=> 2/6  = AM/9  = 1/3

=> AM = 3 cm

Area of ΔALM / Area of ΔABC  =  (1/3)²

=> Area of ΔALM / Area of ΔABC = 1/9

=> Area of ΔABC = 9 * Area of ΔALM

Area of Trap LBCM = Area of ΔABC  - Area of ΔALM

=> Area of Trap LBCM = 9 * Area of ΔALM  - Area of ΔALM

=> Area of Trap LBCM = 8 * Area of ΔALM

=> Area of ΔALM / Area of Trap LBCM   = 1/8