$P Q R S$ is an isosceles trapezium. $l (P Q) = 7$ cm. Segment $P M$ is perpendicular to segment $S R$ , $l (S M) = 3$ cm, Distance between two parallel sides is $4$ cm, Find the area of $P Q R S .$

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Solution

Length of $P Q = 7$ cm
Length of $P M =$ distance between two parallel sides $= 4$ cm
Length of $S M = 3$ cm
Length of $S R = S M + M N + N R$
because $P Q R S$ is isosceles trapezium,
so, $S M = N R = 3$ cm and $M N = P Q = 7$ cm
so, Length of $S R = 3$ cm $+ 7$ cm $+ 3$ cm $= 13$ cm
Now, area of trapezium $P Q R S$
$= \frac{1}{2} (P Q + S R) \times P M$ sq.cm
$= \frac{1}{2} (7 + 13) \times 4$ sq.cm
$= 20 \times 2$ sq.cm
$= 40$ sq.cm

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PQRS is an isosceles trapezium. l(PQ)=7 cm. Segment PM is perpendicular to segment SR, l(SM)=3 cm, Distance between two parallel sides is 4 cm, Find the area of PQRS.

Length of PQ=7 cmLength of PM= distance between two parallel sides =4 cm Length of SM=3cm Length of SR=SM+MN+NR because PQRS is isosceles trapezium, so, SM=NR=3 cm and MN=PQ=7cm so, Length of SR=3cm +7cm +3cm =13cmNow, area of trapeziumPQRS=12(PQ+SR)×PM sq.cm=12(7+13)×4 sq.cm=20×2 sq.cm=40sq.cm

$P Q R S$ is an isosceles trapezium. $l (P Q) = 7$ cm. Segment $P M$ is perpendicular to segment $S R$ , $l (S M) = 3$ cm, Distance between two parallel sides is $4$ cm, Find the area of $P Q R S .$