Find the length of DC to the nearest first decimal place.

Question

In the trapezium ABCD shown in the figure, AB = (x + 3) cm, DC = (2x − 3) cm and BE = EC. If the area of the trapezium is $15 cm^2$, find the length of DC to the nearest first decimal place. (Take$ \sqrt{19}$ = 4.36)

Any Ideas on how to begin?

Many Thanks

score 1 · Answer 1 · answered Aug 22 '16 at 04:06

1

To begin, note that $EC=BE=x-6$ Now use the area you are given to find $x$.

answered Aug 22 '16 at 04:06

Ross Millikan

374,822

EC=BE=x−6 How did this happen? – user360471 Aug 22 '16 at 04:26
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Compare $DC$ to $AB$ and note that $DE=AB$ – Ross Millikan Aug 22 '16 at 04:32
$15$=$\frac{3 (x-6) x}{2}$ Correct – user360471 Aug 22 '16 at 04:40
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Yes. I believe it would be clearer to write $15=(x+3)(x-6)+\frac 12(x-6)^2$, which shows you added the rectangle and triangle, but your equation is equivalent. Now when you solve for $x$ you will see where $\sqrt{19}$ comes in. – Ross Millikan Aug 22 '16 at 04:45
DC=7.36, Correct? :) – user360471 Aug 22 '16 at 08:42
No, you dropped a factor $2$ on the $\sqrt{19}$ – Ross Millikan Aug 22 '16 at 14:37

score 1 · Answer 2 · answered Aug 22 '16 at 04:06

1

hint: Height $h = BE = EC = (2x-3) -(x+3) = ..$, and use the area formula for a trapezoid: $15 = \dfrac{(b+B)h}{2} = \dfrac{((x+3)+(2x-3))h}{2}$ to solve for $x$.

answered Aug 22 '16 at 04:06

DeepSea

77,651

$15 = \dfrac{(b+B)h}{2} = \dfrac{((x+3)+(2x-3))h}{2}$= $\dfrac{(3x)h}{2}$= $\dfrac{3hxh}{2}$= $\dfrac{4hx}{2}$ = $2hx $ Correct? – user360471 Aug 22 '16 at 04:23
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You need to redo it because I spotted a mistake in your comment... – DeepSea Aug 22 '16 at 04:25
can please show me where? – user360471 Aug 22 '16 at 04:26

Find the length of DC to the nearest first decimal place.

2 Answers2