Question

class 8 rs.agarwall book chapter number: Area of a trapeziume and a polygon....
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Answer 1

Answer:

1800 cm²

Step-by-step explanation:

Area of quadrilateral = ½ × ( one of the diagonal ) × ( sum of offsets corresponding to the diagonal)

Area = ½ × 60 × ( 38+22)

Area = ½ × 60 × 60

Area = 1800 cm²

Answer 2

Answer:

GIVEN :

The length of diagonal = 60 cm

perpendiculars drawn from the opposite vertices on this diagonal = 38 cm and 22 cm

TO FIND :

The area of the field

SOLUTION :

Let the quadrilateral be ABCD.

Thus, the length of each perpendicular becomes  BM and DN.

Let,

=> BM = 22 cm

=> DN = 38 cm

And, the diagonal mentioned = AC

=> AC = 60 cm

On observing the given figure carefully, we notice that the diagonal AC divides the quadrilateral into 2 triangles ABC and ADC.

=> area of quadrilateral ABCD = area of Δ ABC + area of Δ ADC

= ($\frac{1}{2}$ × 60 × 38) + ($\frac{1}{2}$ × 60 × 22)   .   .   .   . [area Δ = 1/ 2 × base × height]

= (30 × 38) + (30 × 22)

= 1140 + 660

= 1800 cm²

Thus, area of the quadrilateral is 1800 cm² (Ans

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METHOD II

area of a quadrilateral = 1/ 2 × diagonal × (sum of the perpendiculars from vertex on diagonal )

= 1/ 2 × 60 × (38 + 22)

= 30 × 60

= 1800 cm²