Question

9. A field is in the shape of a quadrilateral ABCD
in which side AB = 18 m, side AD = 24 m,side BC= 40 m,DC = 50m and angle A = 90 degree . Find the area of the field.


the answers is
$816m {}^{2}$
I have given the answers above the chapter is mensuration area of a Trapezium and a polygon please do in class 8th standard explain the method and solve it. only solve if you know otherwise don't solve thank you .​ don't give me any reply If you are going to write rubbish ​
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Answer 1

Answer:

816 m²

Step-by-step explanation:

We have,

AB = 18 m, AD = 24 m, BC = 40 m, DC = 50 m, ∠A = 90°.

BD is joined.

In Δ ABD

By applying Pythagoras theorem,

⇒ BD² = AB² + AD²

⇒ BD² = 18² + 24²

⇒ BD² = 900

⇒ BD = 30 m.

(i)

Area of ΔABD = (1/2) * base * height

                       = (1/2) * 24 * 18

                       = 216 m²

(ii)

Now,

Semi-perimeter of BCD = (50 + 40 + 30)/2

                                       = 120/2

                                       = 60 m

Using heron's formula:

Area of ΔBCD = √s(s - a)(s - b)(s - c)

                        = √60(60 - 30)(60 - 40)(60 - 50)

                        = √60 * 30 * 20 * 10

                        = √360000

                        = 600 m²

∴ Area of Quadrilateral ABCD = Area of ΔABD + Area of ΔBCD

                                                  = 216 + 600

                                                  = 816 m²

Therefore, Area of the field = 816 m².