Question

Construct a trapezium PQRS in which vertices P, Q , R and S are P(3,0),  Q(7,9),R(-6,9) and S(-2,0). Also find area.​

Answer 1

Answer:

81 unit ^2

step by step explanation

mark this guy as brainlest

Answer 2

Answer:

Here we constructed a trapezium and the area of this trapezium is 81 square unit .

Step-by-step explanation:

Explanation:

PQRS is a trapezium  .

P(3 , 0 ) , Q(7 , 9) , R(-6,9) and S(-2 , 0).

Plot the all given points in the graph so we get a trapezium .

Let height of trapezium be h  .

Step 1:

Area of trapezium =  $\frac{a+ b}{2}$× h

Therefore  , from the graph we have

$a = 5 units \ and \ b = 13 units$

and h =  9units

Where , $a \ and \ b$ are the base of the trapezium  and h is the height .

Put the value of a = 5 unit , b = 13  and h = 9 units in the above formula

⇒ Area of trapezium = $\frac{a+ b}{2} h$ = $\frac{5 + 13}{2}. 9$

⇒Area of trapezium = $\frac{18}{2}.9 =$ 9× 9 = 81 square unit.

Final answer:

Hence , area of trapezium is 81 square unit.

#SPJ3