Find the area of the trapezium PQRS given in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To find the area of the trapezium $PQRS$,draw a line $ST$ perpendicular to $PQ$ at $T$. Since $PQ \parallel SR$ and $QR \perp PQ$,$QR \perp SR$ as

Vedclass · Accepted Answer

$114$

Vedclass · Answer

(A) To find the area of the trapezium $PQRS$,draw a line $ST$ perpendicular to $PQ$ at $T$. Since $PQ \parallel SR$ and $QR \perp PQ$,$QR \perp SR$ as well. Thus,$TQRS$ is a rectangle. Given $PQ = 12 \, m$ and $SR = 7 \, m$,we have $TQ = SR = 7 \, m$. Therefore,$PT = PQ - TQ = 12 \, m - 7 \, m = 5 \, m$. In the right-angled triangle $	riangle PTS$,by Pythagoras theorem: $TS^2 = PS^2 - PT^2 = 13^2 - 5^2 = 169 - 25 = 144$. So,$TS = \sqrt{144} = 12 \, m$. Since $TQRS$ is a rectangle,$QR = TS = 12 \, m$. The area of the trapezium is given by the formula: $	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$ $	ext{Area} = \frac{1}{2} 	imes (PQ + SR) 	imes QR$ $	ext{Area} = \frac{1}{2} 	imes (12 + 7) 	imes 12 = 19 	imes 6 = 114 \, m^2$.

Find the area of the trapezium $PQRS$ given in the figure.

Similar Questions

From a point in the interior of an equilateral triangle,perpendiculars are drawn on the three sides. The lengths of the perpendiculars are $14 \, cm$,$10 \, cm$,and $6 \, cm$. Find the area of the triangle.

The perimeter of a rhombus is $100\,cm$ and one of its diagonals measures $48\,cm$. Then,the length of the other diagonal is ....... $cm.$

Write True or False and justify your answer.
If the side of a rhombus is $10 \, cm$ and one diagonal is $16 \, cm$,the area of the rhombus is $96 \, cm^2$.

Find the area of a triangle whose sides are $13\, cm$,$14\, cm$,and $15\, cm$. (in $, cm^2$)

Find the area of the triangle with the side lengths $51 \, cm$,$52 \, cm$,and $101 \, cm$. (in $, cm^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module