Question

If two triangles are on the same base and between the same parallels, then the ratio of
their areas is :

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $1:4$

Answer 1

The correct option is A $1:1$

$△ADB$ and $△ABC$ are triangles on same base $AB$ and between the same parallels $AB$ and $DC.$

We have to prove that

$ar\left(ADB\right)=ar\left(ABC\right)\Rightarrow 1:1$

When we construct a line through $A$ parallel to $BC$ meeting $DC$ at $E$ i.e, $AB\parallel BC$ and construct a line through $B$ parallele to $AD$ meeting $DC$ at $F$ i.e, $BF\parallel AD$

From $ABFD,$ From $ABCE,$

$DF\parallel AB$ (Given as $AB\parallel CD$) $EC\parallel AB$ (Given as $AB\parallel CD$)

and $BF\parallel AD$ ( By construction ) and $AE\parallel BC$ ( By construction )

Thus, both pair of opposite sides are parallel. Thus, both pair of opposite sides are parallel.

Thus, $ABFD$ and $ABCE$ are two parallograme with the same base $AB$ and between two parallel $AB$and$EF$

$\therefore$ $ar\left(ABFD\right)=ar\left(ABCE\right)$

$\therefore ar\left(\Delta ADB\right)=ar\left(\Delta ABC\right)$

$\therefore$ The ratio of their area will be equal to $1:1$

Hence, the answer is $1:1.$