Question

Q12 A field in the shape of a trapezium whose parallel sides are 25m and 10m. The non parallel sides are 14m and 13m. Find the area of the field.
OR
A park, in the shape of a quadrilateral ABCD has ∠C=90⁰, AB=9m, BC= 12m, CD=5m and AD=8m. How much area does it occupy?

Answer 1

Answer:

The trapezium field is shown below in figure:

Drawing line CF parallel to AD and a line perpendicular to AB, we obtain

field

Then in quadrilateral ADCF,

CF || AD ............................ \left [ \because\ by\ construction \right ]

CD || AF ............................ \left [ \because ABCD\ is\ a\ trapezium \right ]

Therefore, ADCF is a parallelogram.

So, AD = CF = 13\ m and CD = AF = 10\ m

\left ( \because Opposite\ sides\ of\ a\ parallelogram \right )

Therefore, BF = AB-AF = 25-10 = 15\ m

Now, the sides of the triangle;

a = 13\ m,\ b =14\ m\ and\ c = 15\ m.

So, the semi-perimeter of the triangle will be:

s= \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21\ m

Therefore, the area of the triangle can be found by using Heron's Formula:

Area = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{21(21-13)(21-14)(21-15)}

= \sqrt{21(8)(7)(6)}

= \sqrt{7056} = 84\ m^2.

Also, the area of the triangle is given by,

Area = \frac{1}{2}\times BF\times CG

\Rightarrow \frac{1}{2}\times BF\times CG = 84\ m^2

\Rightarrow \frac{1}{2}\times 15\times CG = 84\ m^2

Or,

\Rightarrow CG = \frac{84\times2}{15} = 11.2\ m

Therefore, the area of the trapezium ABCD is:

= \frac{1}{2} \times (AB+CD)\times CG

= \frac{1}{2} \times (25+10)\times 11.2

= 35\times5.6

= 196\ m^2

Hence, the area of the trapezium field is 196\ m^2.

OR

A park in the shape of a quadrilateral ABCD , has angle C =90 , AB=9m , CD=5m , BC=12m and AD=8

❤❤❤PLEASE,PLEASE,PLEASE MARK THIS AS A BRAINLIST.❤❤❤

Answer 2

Answer:

the answer to your question is in the attachment

pls mark as brainliest