Question

pqrs is a trapezium as shown in the figure such that PQ parallel to SR and SP perpendicular to PQ find the area of a Trapezium pqrs if PQ= 15, SQ 17, and SR =11m.
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Answer 1

Answer:

Area of quadrilateral PQRS is

Step-by-step explanation:

it is an trapezium

hence area is equals to =

1/2 ( sum of parallel sides) ×height

By pythagoras thrm height is 8m

A (PQRS) = 1/2×(15+11)×8

= 1/2×26×8

= 104 m^2

Answer 2

Answer:

The area of the trapezium is 104 $m^2$.

Step-by-step explanation:

Given:-

PQRS is a trapezium such that PQ || SR and SP ⊥ PQ.

Also, PQ = 15 m, SQ = 17 m and SR = 11 m.

To find:-

The area of the trapezium.

Step 1 of 1

From the figure,

∠SPQ = 90°

⇒ ΔSPQ is a right triangle.

By the Pythagoras theorem - "In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides."

$SQ^2=SP^2+PQ^2$

$17^2=SP^2+15^2$

$289=SP^2+225$

$SP^2=289-225$

$SP=\sqrt{64}$

SP = 8 m

Thus, the height of the trapezium is 8 m.

So,

The area of the trapezium is,

= $\frac{1}{2}$ $\times$ sum of parallel sides $\times$ height

= $\frac{1}{2}$ × (15 + 11) × 8

= 26 × 4

= 104 $m^2$

Final answer: The area of the trapezium is 104 $m^2$.

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