Question

PQRS is a parallelogram and ST$=$TR. What is the ratio of the area of triangle $QST$ to the area of the parallelogram?

• A. $1:2$
• B. $1:3$
• C. $1:4$
• D. $1:5$
• E. It cannot be determined

Answer 1

The correct option is C $1:4$

Given that $ST=TR$ So, T is the midpoint of SR.

Now, we know that a diagonal divider the parallelogram into two triangles of equal area.

So, Area of $△PQS=Areaof△QRS=\frac{1}{2}$ [area of parallelogram PQRS] -----------(1)

Now, in $△QRS,QT$ divides it into two $△$. Also height of $△QST=heightof△QTR.$

So, Area of $△QST=Areaof△QTR=\frac{1}{2}\left[areaof△QRS\right]$ -----------(2)

from (1),

Area of $△QRS=\frac{1}{2}$ [area of parallelogram PQRS] ---------(3)

From (2) Area of $QST=\frac{1}{2}\left[areaof△QRS\right]$

From (2),

Area of $△QST=\frac{1}{2}\left[areaof△QRS\right]$

$\Rightarrow$ Area of $△QRS=2\left[areaof△QST\right]$ --------(4)

LHS of (3) and (4) are some so , equating the RHS of (3) and (4) we get.

$\frac{1}{2}\left[areaofparallelogramPQRS\right]$ $=2$ [area of $△QST$]

$\Rightarrow$ Area of parallelogram $PQRS=4$ (area of \triangle QST)

$\Rightarrow \frac{Areaof△QST}{AreaofparallelogramPQRS}=\frac{1}{4}$

$\therefore$ The required ratio is $1:4$