Question

the parallel sides of a trapezium are 40 cm and 20 CM if its non parallel sides are equal each being 26 cm find the area of a trapezium​

Answer 1

Given :

Parallel Sides are 40 cm and 20 cm

Non-parallel sides are 26 cm

To Find :

Area of the Trapezium

Solution :

Let ABCD be the Trapezium in which AB = 40 cm, DC = 20 cm, BC = 26 cm AD = 26 cm

Through C, draw CF || AD

Also, draw CE perpendicular to AB

Now,

$\star \sf \: FB= AB - AF$

$\sf FB = 40 - 20 = 20 \: cm$

In Triangle FBC :

$\sf \: FC = BC \: = 26 \: cm$

So, It is an isosceles Triangle.

Also, CE is perpendicular to FB. So, E is the mid-point of FB.

$\therefore\sf \: FE = \dfrac{1}{2} \: F B$

$\therefore\sf \: FE = \dfrac{1}{2} \times 20= \large \boxed{ \sf \: 10 \: cm}$

In a right angled triangle CEF,

$\star \sf \: {CF}^{2} = {FE}^{2} + {CE}^{2} \:$

$\sf \: {26}^{2} = {10}^{2} + {CE}^{2}$

$\sf {CE}^{2} = 676 - 100$

$\sf {CE}^{2} = 576$

$\sf \: {CE} = \sqrt{576}$

$\sf \: CE = \large \boxed{ \sf \: 24 \: cm}$

Area of Trapezium ABCD =

$\star \sf \: \dfrac{1}{2} (AB + DC) \times CE$

$\sf \: \dfrac{1}{2} (40 + 20) \times 24$

$\sf \: \dfrac{1}{2} \times 60 \times 24$

$\dfrac{1440}{2}$

$\large \boxed{ \sf \: 720 {cm}^{2}}$

Answer 2

$\sf \: the \: parallel \: sides \: of \: trapezium = 40cm \: and \: 20cm \\ \sf \: dc = 20 + x + x = 40 \\ \sf \: = 20 + 2x = 40 \\ \sf \: 2x = 40 - 20 \\ \sf \: x = \cancel20 \times \cancel \frac{1}{2} \\ \sf \: x= 10 \\ \\ \bf {\red\star{ \red{ \: by \:using \: pythagorean \: {th}^{m} : - 9 }}} \\ \sf \: \: ( {height})^{2} + ( {x)}^{2} = {(26)}^{2} \\ \sf \: ( {height})^{2} + ( {10})^{2} = 676 \\ \sf \: ( {height})^{2} = 676 - 100 \\ \sf \: ( {height})^{2} = 576 \\ \sf \: ( {height})^{2} = \sqrt{576} \\ \boxed{ \sf height = 24} \\ \\ \\ \bf \purple \star{ \purple{{area \: of \: trapezium : - }}} \\ \sf\: \frac{1}{2} \times (sum \: of \: parallel \: sides) \times height \\ \sf \: = \cancel\frac{1}{2} \times 60 \times \cancel 24 \\ \sf \: = 720 {cm}^{2}$

hope it helps you dear...