Jovany Clayton
2022-07-13
Answered

The area of a trapezium is $475c{m}^{2}$ and the height is 19 cm .Find the lengths of its two paallel sides if one side is 4 cm greater than the other.

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gozaderaradiox5

Answered 2022-07-14
Author has **19** answers

Step 1

We know that the area of a trapezium is

A $\frac{1}{2}\times (\text{sum of parallel lines})\times (\text{distance between parallel lines})$

Step 2

$A=475c{m}^{2}$

$A=19$

Let one parallel side length x

Another parallel side length $=x+4$

$A=\frac{1}{2}\times (x+x+4)\times 19$

$475=\frac{1}{2}\times (2x+4)\times 19$

$2x+4=\frac{2\times 475}{19}$

$2x+4=50$

$2x=46$

$x=23$

Second parallel side $x+4$

$=23+4$

$=27cm$

We know that the area of a trapezium is

A $\frac{1}{2}\times (\text{sum of parallel lines})\times (\text{distance between parallel lines})$

Step 2

$A=475c{m}^{2}$

$A=19$

Let one parallel side length x

Another parallel side length $=x+4$

$A=\frac{1}{2}\times (x+x+4)\times 19$

$475=\frac{1}{2}\times (2x+4)\times 19$

$2x+4=\frac{2\times 475}{19}$

$2x+4=50$

$2x=46$

$x=23$

Second parallel side $x+4$

$=23+4$

$=27cm$

delirija7z

Answered 2022-07-15
Author has **5** answers

Consider x cm as the smaller side of the trapeziumSo the larger parallel can be written as $(x+4)cm$We know that the area of trapezium $\frac{1}{2}\times $ sum of parallel sides x heightBy substituting the values$475=\frac{1}{2}\times (x+(x+4))\times 19$On further calculation$25=\frac{1}{2}\times (2x+4)$So we get$50=2x+4$By subtraction$2x=50-42$$x=46$

By division

$x=23cm$

So we get Larger parallel side $(x+4)=23+4=27cm$

Therefore, the length of two parallel sides is 23cm and 27cm.

By division

$x=23cm$

So we get Larger parallel side $(x+4)=23+4=27cm$

Therefore, the length of two parallel sides is 23cm and 27cm.

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