What is the length of DC in this trapezium?

[mathlearn]

View attachment 5910

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 cm2, find the length of DC to the nearest first decimal place.

Take $\sqrt{19}$ = $4.36$

Have I done correctly,

Okay first we know that area of a trapezium

$15 = \dfrac{((x+3)+(2x-3))h}{2}$
$h=(2x-3) -(x+3)$ $h=2x-3 -1(x+3)$
$h=2x-3 -x-3$
$h=x-3 -3$
$h=x-6$

, Correct ?

Many Thanks :)

 

[Greg]

Start with

$$\dfrac{(2x-3+x+3)h}{2}=15\implies hx=10$$

and

$$2x-3-h=x+3$$

Can you continue?

 

[mathlearn]

$2x−3−h=x+3$

As I have found earlier,

$$2x-3-h=x+3$$

Can you continue?

$h=x−6$

$$\therefore$$ $2x−3-x-3=h$, $x-6=h$

$15 = \dfrac{((x+3)+(2x-3))h}{2}$

$15 = \dfrac{((x+3)+(2x-3))*(x-6)}{2}$

$30= {(3x)*(x-6)}$
$30= {(3x^2-18x)}$

$10= {(x^2-6x)}$

Okay from there on I will factor using complete the square method,

$10+ \frac{b}{2}^2= {(x^2-6x)+\frac{b}{2}^2}$

$10+ \frac{-6}{2}^2= {(x^2-6x)+\frac{-6}{2}^2}$

$10+ 9= {(x^2-6x)+9}$

$19 = {(x^2-6x)+9}$

Now factorising

$$\sqrt{19}$$= $\sqrt{(x-3)^2}$

The problem has given the value of $$\sqrt{19}$$= 4.36;

$$4.36$$= $(x-3)$
$$4.36$$= $(x-3)$
$$7.36$$= $x$

Now have I done correct ? :)

 

[Greg]

$$hx=10\implies h=\dfrac{10}{x}$$

$$2x-3-h=x+3$$

$$2x-3-\dfrac{10}{x}=x+3$$

$$2x^2-3x-10=x^2+3x$$

$$\implies x^2-6x-10=0$$

$$\implies x=\dfrac{6+\sqrt{76}}{2}=3+\sqrt{19}$$

$$2x-3=2(3+\sqrt{19})-3=3+2\sqrt{19}\approx11.7\text{ cm.}$$

 

[mathlearn]

$2x-3=2(3+\sqrt{19})-3=3+2\sqrt{19}\approx11.7\text{ cm.}$. A Little explanation here please? :)

Many Thanks :)

 

[MarkFL]

Another approach would be to let:

$$\overline{DC}=u$$

thus:

$$\overline{AB}=\frac{u+9}{2}$$

Now compute the area two ways:

$$A=\frac{h}{2}(u+9)+\frac{1}{2}h^2=\frac{h}{2}(u+h+9)$$

$$A=\frac{h}{2}\left(u+\frac{u+9}{2}\right)=\frac{h}{2}\left(\frac{3u+9}{2}\right)$$

Hence:

$$3u+9=2u+2h+18$$

$$h=\frac{u-9}{2}$$

Thus:

$$\frac{u-9}{8}(3u+9)=15$$

$$(u-9)(u+3)=40$$

$$u^2-6u-67=0$$

Taking the positive root, we obtain:

$$u=3+2\sqrt{19}$$

 

[mathlearn]

:),
Hi but I cannot still understand $\displaystyle u=3+2\sqrt{19}$ .

:):)

 

[soroban]

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 cm2, find the length of DC to the nearest first decimal place.

Take $\sqrt{19}$ = $4.36$

Have I done correctly,

Okay first we know that area of a trapezium

$15 = \dfrac{((x+3)+(2x-3))h}{2}$
$h=(2x-3) -(x+3)$ $h=2x-3 -1(x+3)$
$h=2x-3 -x-3$
$h=x-3 -3$
$h=x-6$

, Correct ?

Many Thanks :)

There is a simpler way to find h = BE.

Note that DE = AB = x+3.

Also: \underbrace{DC}_{2x-3} \:=\: \underbrace{DE}_{x+3} + EC

Hence, EC = BE= x-6

 

[MarkFL]

:),
Hi but I cannot still understand $\displaystyle u=3+2\sqrt{19}$ .

:):)

Where did I lose you? :)

 

[mathlearn]

:) So far I have found the height or 'h' but after factoring using the complete the square method i get the square root of 19 or 4.36 as x-3 so I find that x is 7.36 But I know I have missed a factor 2 on the of $\sqrt{19}$.

I want to know where does that 2 come from and why I am missing it?

Many Thanks :)

 

[Greg]

$2x−3−h=x+3$

As I have found earlier,

$h=x−6$

$$\therefore$$ $2x−3-x-3=h$, $x-6=h$

$15 = \dfrac{((x+3)+(2x-3))h}{2}$

$15 = \dfrac{((x+3)+(2x-3))*(x-6)}{2}$

$30= {(3x)*(x-6)}$
$30= {(3x^2-18x)}$

$10= {(x^2-6x)}$

Okay from there on I will factor using complete the square method,

$10+ \frac{b}{2}^2= {(x^2-6x)+\frac{b}{2}^2}$

$10+ \frac{-6}{2}^2= {(x^2-6x)+\frac{-6}{2}^2}$

$10+ 9= {(x^2-6x)+9}$

$19 = {(x^2-6x)+9}$

Now factorising

$$\sqrt{19}$$= $\sqrt{(x-3)^2}$

The problem has given the value of $$\sqrt{19}$$= 4.36;

$$4.36$$= $(x-3)$
$$4.36$$= $(x-3)$
$$7.36$$= $x$

Now have I done correct ? :)

You have correctly found $x$. $\overline{DC}=2x-3=2(7.36)-3=14.72-3=11.72$.

:) So far I have found the height or 'h' but after factoring using the complete the square method i get the square root of 19 or 4.36 as x-3 so I find that x is 7.36 But I know I have missed a factor 2 on the of $\sqrt{19}$.

I want to know where does that 2 come from and why I am missing it?

Many Thanks :)

The '2' comes from $\overline{DC}=2(3+\sqrt{19})-3=6+2\sqrt{19}-3=3+2\sqrt{19}$. You're confusing $x$ and $\overline{DC}$.

 

[mathlearn]

:) Yes ,You explained it clearly :)

View attachment 5913

:) Many Thanks