Question

A field is in the shape of a trapezium whose parallel sides are 25m and 10mm. The non parallel sides are 14m and 13mm. Find the area oftrapezium.

Answer 1

ans is 162.565 m sq. mark as brainliest

Answer 2

Answer:

From C, draw CE ll DA. Clearly, ADCE is a parallelogram having AD ll CE and DC ll AE such that AD = 13m and DC = 10m.

•°• AE = DC = 10m and CE = AD = 13m

$\implies$BE = AB - AE = (25-10) = 15m.

Thus in BCE, we have

BC = 14m, CE = 13m and BE = 15m.

Let s be the semi-perimeter of∆BCE. Then,

2s = BC + CE + BE = 14 + 13 + 15 = 42

$\rightarrow$ s = 21

•°• Area of ∆BCE = $\sf\sqrt{21\times(21-14)(21-13)(21-15)}$

$\rightarrow$ Area of ∆BCE = $\sf\sqrt{21\times7\times8\times6}$

$\rightarrow$ Area of ∆BCE = $\sf\sqrt{7^2\times3^2\times4^2}$ = 84 m².

Also, Area of ∆BCE = $\sf\frac{1}{2}(BE×CL)$

= 84 =$\sf\frac{1}{2}×15×CL$

CL = $\cancel\dfrac{168}{15}$

$\implies$ $\sf\frac{56}{5}$

$\implies$ height of llgm ADCE = CL = 56/5 m

$\therefore$ Area of llgm ADCE = Base $\times$ Height = AE $\times$ CL = 10 $\times$ $\sf\frac{56}{5}$ = 112 m²

Hence, Area of trapezium ADCD = Area of llgm ADCE + Area of ∆BCE

= (112+84)m² = 196 m².