Question

In the given figure, if ar(ΔALM)/ar(trapeziumLMCB)=9/16, and LM||BC, Then AL:LB is equal to---- a. 3:5 b. 4:1 c. 3:4 d. 2:3

Answer 1

Answer:

Step-by-step explanation:

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

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Additional information ❤

What is a trapezium shape?

Trapezium, describing a geometric shape, has two contradictory meanings: Outside the US and Canada: a quadrilateral with at least one pair of parallel sides (known in the US as a trapezoid) In the US and Canada: a quadrilateral with no parallel sides (known elsewhere as a general irregular quadrilateral)

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:)

Answer 2

Triangle

"As figure is not attached with the question , I am attaching the figure as per the information given."

The ration of area of triangle to the area of trapezium is $\frac{9}{16}$ .

$\frac{ar. (\triangle ALM)}{ar.(trapezium LMCB)}=\frac{9}{16}$

and  $LM||BC$.

$\frac{ar. (\triangle ALM)}{ar.(\triangle ABC)}=\frac{9}{16}\\\\\frac{ar. (\triangle ALM)}{ar. (\triangle ALM)+ar.(trapezium LMCB)}=\frac{9}{16+9}=\frac{9}{25} ,$

We can use area theorem here ,

$\frac{ar. (\triangle ALM)}{ar. (\triangle ABC)} =\frac{AL^2}{AB^2}$

On equating,

$\frac{AL^2}{AB^2}=\frac{9}{25} \\\\\implies \frac{AL}{AB}=\frac{3}{5}$

$\implies \frac{AB}{AL}=\frac{5}{3}$

For getting the ration $\frac{AL}{LB}$, we will substract 1 from above ratio,

$\frac{AB}{AL}-1=\frac{AB-AL}{AL}=\frac{5}{3}-1=\frac{2}{3}$

$\implies\frac{AL}{LB}=\frac{3}{2}$

Hence the required ratio is 3:2.