Question

If two parallel sides of a trapezium are 9 cm and 19 cm . and the other two sides are 6 cm and 8 cm then how much area it acquire

Answer 1

this is the correct answer

Answer 2

Answer:

Area=$67.2cm^{2}$

Step-by-step explanation:

Construction: Draw AE⊥DC and BF⊥Dc such that DE=xcm and FC=10-x

Solution: In ΔADE, $36=x^{2}+h^{2}$

⇒$h^{2}=36-x^{2}$                                                                         (1)

In ΔBFC,

$64=(10-x)^{2}+h^{2}$

⇒$h^{2}=64-(10-x)^{2}$                                                                    (2)

from (1) and (2),

$36-x^{2}=64-(10-x)^{2}$

⇒$36-x^{2}=64-100-x^{2}+20x$

⇒$72=20x$

⇒$x=\frac{36}{10}$

Now, from (1), $h^{2}=36-x^{2}$  

$36-(\frac{36}{10}) ^{2}$

⇒$h=\frac{48}{10}$

Hence, Area of the trapezium is : $\frac{1}{2}{\times}(a+b){\times}h$ where a and b are the two parallel sides of the trapezium.

Area= $\frac{1}{2}{\times}(9+19){\times}\frac{48}{10}$

       =$\frac{1}{2}{\times}(28)(4.8)$

Area=$67.2cm^{2}$