Question

Question 10
Find the area of the trapezium PQRS with height PQ given in the figure given below:

Answer 1

We have, trapezium PQRS, in which draw a line RT perpendicular to PS.
Where side, ST = PS – TP = 12 – 7 = 5m $[\because TP=PQ=7m]$

$Inrightangled\Delta STR(SR{)}^2=(ST{)}^2+\left(TR{)}^2\right[byusingPythagoras]$
$\Rightarrow (13{)}^2=(5{)}^2+(TR{)}^2$
$\Rightarrow (TR{)}^2=169-25$
$\Rightarrow (TR{)}^2=144$
$\therefore TR=12m$
[taking positive square root because length is always positive]
$Now,areaof\Delta STR=\frac{1}{2}\times TR\times TS=\frac{1}{2}\times 12\times 5=30m^2$
$[\because areaoftriangle=\frac{1}{2}(base\times height\left)\right]$

Now, area of rectangle $PQRT=PQ\times RQ=12\times 7=84m^2$
$[\because areaofrectangle=\frac{1}{2}(length\times breadth\left)\right]$
$[\because PQ=TR=12m]$

$\therefore$ Area of trapezium = Area of DSTR + Area of rectangle PQRT = 30 + 84 = 114 $m^2$

Hence, the area of trapezium is 114 $m^2$.

Alternate Method
Find TR as in the above method
$\therefore$ Area of trapezium =$\frac{1}{2}$ (Sum of parallel lines) $\times$ Distance between two points
$=\frac{1}{2}(PS+QR)\times TR=\frac{1}{2}\times (12+7)\times 12$
$=\frac{1}{2}\times 19\times 12=114m^2$
Hence, the area of trapezium is $114m^2$.