### Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

### At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

# Kite in a Square

##### Stage: 4 Challenge Level:

ABCD is a square. M is the midpoint of the side AB.
By constructing the lines AC, MC, BD and MD, the blue shaded quadrilateral is formed:

What fraction of the total area is shaded?

Below are three different methods for finding the shaded area. Unfortunately, the statements have been muddled up. Can you put them in the correct order?

Coordinates

one of which has vertices $(\frac{1}{3},\frac{2}{3}), (\frac{1}{2},\frac{1}{2}), (\frac{1}{2},1)$.

(B) The line joining $(0,0)$ to $(\frac{1}{2},1)$ has equation $y=2x$

(C) Area of the triangle $= \frac{1}{2}(\frac{1}{2} \times \frac{1}{6}) = \frac{1}{24}$

(D) The line joining $(0,1)$ to $(1,0)$ has equation $y=1-x$.

(E) Therefore the shaded area is $2 \times \frac{1}{24} = \frac{1}{12}$

(F) The point $(a,b)$ is at the intersection of the lines
$y=2x$ and $y=1-x$.

(G) Consider a unit square drawn on a coordinate grid.

(H) The perpendicular height of the triangle is $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.

(I) So $a = \frac{1}{3}, b=\frac{2}{3}$.

(J) The line joining $(0,0)$ to $(1,1)$ has equation $y=x$.

To help you reorder the statements above, here is a set of printable cards for you to cut out.

Similar Figures

(A) As line $AC$ intersects line $MD$ at point $E$,
the two opposite angles $\angle MEF$ and $\angle AED$ are equal.

(B) The line $MF$ is half the length of $AD$.

(C) Line $AD$ is parallel to line $MF$, so $\angle EDA$ and $\angle EMF$ are equal, and $\angle EAD$ and $\angle EFM$ are equal (alternate angles).

(D) Therefore, $\triangle AED$ and $\triangle FEM$ are similar.

(E) Therefore, the line $EH$ is half the length of $PE$.

(F) Let ABCD be a unit square.

(G) Therefore, the shaded area $MEFG = \frac{1}{24} \times 2 = \frac{1}{12}$ sq units.

(H) $PH$ has length $\frac{1}{2}$ units, so $PE$ has length $\frac{1}{3}$ units and $EH$ has length $\frac{1}{6}$ units.

(I) $\triangle MEF$ has area $\frac{1}{2}(\frac{1}{2}\times\frac{1}{6}) = \frac{1}{24}$ sq units.

To help you reorder the statements above, here is a set of Printable cards for you to cut out.

Pythagoras

(A) The area of $\triangle DMC = 2$ sq units.
The area of $\triangle DFC = 1$ sq unit.
Thus the combined area of $\triangle DFE$, $\triangle CFG$ and
shaded area $MEFG$ is $1$ sq unit.

(B) $(EH)^2+(HF)^2=(EF)^2$
$EH = HF$
$(EH)^2 = \frac{1}{2}(EF)^2$
$EH = \frac{EF}{\sqrt 2}$

(C) Areas of $\triangle DFE$, $\triangle CFG$ and shaded area $MEFG$ are equal
so each must have an area of $\frac{1}{3}$ sq units.

(D) Area of $\triangle MEF = \frac{1}{2}(1 \times EH) = \frac{1}{2}(\frac{EF}{\sqrt 2})$

(E) By Pythagoras, $DF$ has length $\sqrt 2$.

(F) The total area of the square is $4$ sq units, so the shaded area is
$\frac{1}{12}$ the area of the whole square.

(G) Area of $\triangle DFE = \frac{DF \times EF}{2}$
$= \frac{\sqrt 2 \times EF}{2} = \frac{EF}{\sqrt 2}$

(H) So the shaded area $MEFG$ is equal to the area of $\triangle DFE$.

(I) Assume that the sides of the square are each $2$ units long.
Thus, $DJ$ and $FJ$ are each $1$ unit long.

To help you reorder the statements above, here is a set of Printable cards for you to cut out.

Thanks to Jerome Foley for drawing our attention to this problem.