Question

In the adjoining figure, DE is parallel to BC and AD = 1 cm, BD = 2 cm. What is the ratio of the area of $\triangle ABC$ to the area of $\triangle ADE$?

Answer 1

Answer:

The ratio of ar(ΔABC) : ar( ΔADE) is 9 : 1.

Step-by-step explanation:

Given:

DE || BC, AD = 1 cm, BD = 2 cm.

 

In ΔABC, DE || BC.

According to BASIC PROPORTIONALITY THEOREM (BPT) :  

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

AD/AB = AE/AC  

AD/(AD + DB) = AE/AC

1/(2+1) = AE/AC

AE/AC = ⅓

ΔABC ~ ΔADE

[Two triangles are said to be similar, if their corresponding sides are proportional]

 

ar(ΔABC)/ar( ΔADE) = (AC/AE)²

[The ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.]

ar(ΔABC)/ar( ΔADE) = (3/1)² = 9/1

ar(ΔABC)/ar( ΔADE) = 9/1

ar(ΔABC) : ar( ΔADE) = 9 : 1

Hence, the ratio of ar(ΔABC) : ar( ΔADE) is 9 : 1.

 HOPE THIS ANSWER WILL HELP YOU ..

Answer 2

since de is parrallel to bc.

so by thales theorem AD / DB = AE / EC.

SO BOTH , AD / DB and , AE / EC = 1 :2.

and , AD / AB = 1:3.

or AB / AD = 3:1

the triangles are similar by the SAS axiom.

hence, the areas of triangle are in ratio of squares of corresponding sides..

so ratio of area of trianglr ABC / TO THE AREA OF TRIANGLE ADE = 3 ^2 : 1 ^2.

that is , 9:1... (9 ratio 1)