Day 20: A Right Trapezium Indeed

Drop a perpendicular from $C$ to $AB$ to give the point $X$ and in this way split the trapezium into a right angled triangle and a rectangle.

We know that $BX =3$ so by Pythagoras' Theorem, $CX=4$ . Therefore the area of the triangle is $6$ .

This tells us that the rectangle is of area $28$ . We know that the height is $4$ , so the length must be $7$ .

Finally we find that the perimeter is equal to $5+4+7+10=\boxed{26}$ as required.

The formula for the area of a trapezoid is $A=\dfrac{1}{2}(a+b)h$ where $a$ and $b$ are the length of the bases and $h$ is the height. We have

$34=\dfrac{1}{2}(AB+CD)(AD)$

Draw a line $CE$ perpendicular to line $AB$ , then

$BE=AB-CD$

However, $AB=3+CD$ , substituting, we get

$BE=3+CD-CD=3$

By pythagorean theorem on triangle $CEB$ , we have

$CE=AD=\sqrt{5^2-3^2}=4$

We substitute again in the formula of the area of a trapezoid, we have

$34=\dfrac{1}{2}(AB+CD)(AD)$

$68=(3+CD+CD)(4)$

$\dfrac{68}{4}=3+2CD$

$CD=7$

It follows that $AB=3+7=10$ .

So the desired perimeter is $10+5+7+4=\boxed{26}$

If we are to draw a straight line from point $C$ , parallel to line $AD$ , to a new point which we will call $Y$ , we can split the figure into a rectangle and right angled triangle. Since $AB$ is $3$ until longer than $CD$ and the line is being drawn from the end of $CD$ , from $C$ , we are subtracting the entire length of this line from $AB$ , so $BY$ equals $3$ units. We can then examine the figure ∆ $BCY$ , we can use Pythagorean Theorem to find the length of the missing side:

Through the theorem, we know that:

$A^2 + B^2 = C^2$

Therefore:

$A^2 = C^2 - B^2$

We then know that:

$YC^2 + BY^2 = BC^2$ and $YC^2 = BC^2 - BY^2$ $YC^2 = 5^2 - 3^2$ $=$ $YC^2 = 25 - 9$ $=$ $YC^2 = 16$

Finally:

$YC = 4$

Since $AYCD$ is a rectangle, this means that $YC = AD$ . Also, we know that the area of the trapezium is 34, so we can solve for the other side. First, we must solve for the area of the triangle:

If $A =$ $\frac{B*H}{2}$ , where $A$ is area, $B$ is the base and $H$ is the height. We can then see that:

$A =$ $\frac{B*H}{2}$ = $\frac{3*4}{2}$ = $6$

We can then examine the area of the rectangle, which is the total area minus the area of the triangle, $34 - 6$ or $28$ .

If $A = B*H$ , where $A$ is area, $B$ is the base and $H$ is the height. We can then see that:

$A = B*H$ $=$ $28 = B*4$ then $B =$ $\frac{28}{4}$ $= 7$

Finally, we can find the perimeter by adding together the sides of the trapezium, which are $5 + 7 + 10$ , (which is CD + 3) $+$ $4$ which is equal to $26$ .