The following shape is drawn ** to scale **:

Give the most specific name for the shape.

We start by counting the number of sides. There are four sides in this figure and so it is either just a quadrilateral or one of the special types of quadrilateral.

Next we ask ourselves if there are any parallel lines in the figure. You can look at the figure to see if any of the lines look parallel or make a quick sketch of the image and see if any pairs of opposite lines meet at a point.

Both pairs of opposite sides are parallel. This means that the figure can only be one of the following: parallelogram, rectangle, rhombus or square.

Next we ask ourselves if all the interior angles are 90°. All the interior angles are 90° and so this must be a square or a rectangle. Finally we check to see if all the sides are equal in length. In this figure the sides are not equal in length and so it is a rectangle.

Therefore this is a rectangle.

The shape is also a parallelogram and a quadrilateral. This question, however, asked for the most specific name for the shape.

The following shape is drawn ** to scale **:

Give the most specific name for the shape.

We start by counting the number of sides. There are four sides in this figure and so it is either just a quadrilateral or one of the special types of quadrilateral.

Next we ask ourselves if there are any parallel lines in the figure. You can look at the figure to see if any of the lines look parallel or make a quick sketch of the image and see if any pairs of opposite lines meet at a point.

Both pairs of opposite sides are parallel. This means that the figure can only be one of the following: parallelogram, rectangle, rhombus or square.

Next we ask ourselves if all the interior angles are 90°. All the interior angles are not 90° and so this must be a parallelogram or a rhombus. Finally we check to see if all the sides are equal in length. In this figure the sides are equal in length and so it is a rhombus.

Therefore this is a rhombus.

The shape is also a parallelogram and a quadrilateral. This question, however, asked for the most specific name for the shape.

Based on the shape that you see list the all the names of the shape. The figure is drawn to scale

Both pairs of opposite sides are not parallel. This means that the figure can only be some combination of the following: trapezium, kite, or quadrilateral.

The shape is definitely a quadrilateral because it has four sides. It does not have any special properties: it does not have parallel sides, or right angles, or sides which are equal in length. Therefore it is only a quadrilateral.

Based on the shape that you see list the all the names of the shape. The figure is drawn to scale

Both pairs of opposite sides are not parallel. This means that the figure can only be some combination of the following: trapezium, kite, or quadrilateral.

The shape is definitely a quadrilateral because it has four sides. It is also a kite because it has two pairs of adjacent sides which are the same lengths. It cannot be a square or a rectangle because it does not have right angles. It cannot be a parallelogram or a trapezium because it does not have any parallel sides. And it is not a rhombus because the four sides are not all the same length.

Therefore the correct answer is: kite and quadrilateral.

Based on the shape that you see list the all the names of the shape. The figure is drawn to scale

Both pairs of opposite sides are parallel. That means that this shape can belong to one or more of these groups: square, rhombus, rectangle or parallelogram.

The given shape is a square. However, it is also a rectangle. A square is also a parallelogram, because it has parallel sides; and it is a rhombus as well, it just happens to have right angles. A square is also a kite, a trapezium and of course a quadrilateral.

Therefore the correct answer is: square, rectangle, rhombus, parallelogram, kite, trapezium and quadrilateral.

Find the area of \(ACDF\) if \(AB = 8,~ BF = 17,~ FE = EC,~ BE = ED,~ \hat{A} = 90^{\circ} ,~ C\hat{E}D = 90^{\circ}\)

Construct \(G\) such that \(AC = FG\)

\(BCDF\) is a rhombus (diagonals bisect at right angles)

Since \(BCDF\) is a rhombus \(BC = DF\). We constructed \(G\) such that \(AC = FG\). Therefore \(AB = DG\).

In \(\triangle ABF\) and \(\triangle CGD\):

\begin{align*}
B\hat{A}F & = C\hat{G}D = 90^{\circ} \quad \text{(given and by construction)} \\
AB & = DG \quad \text{(by construction)} \\
BF & = CD \quad \text{(} BCDF \text{ is a rhombus)}
\end{align*}

Therefore \(ABF \equiv CGD\) (RHS)

Therefore \(AF = CG\) and so \(ACGF\) is a rectangle (both pairs of opposite sides equal in length and all interior angles are 90°).

We are given the length of \(AB\) and \(BF\). Since \(\triangle ABF\) is right-angled we can use the theorem of Pythagoras to find the length of \(AF\):

\begin{align*}
BF^{2} & = AB^{2} + AF^{2} \\
(17)^{2} & = (8)^{2} + AF^{2} \\
AF^{2} & = 225 \\
AF & = 15
\end{align*}

We also know that \(FD = BF = 17\) and so \(AC = 17 + 8 = 25\).

Therefore the area of rectangle \(ACGF\) is:

\begin{align*}
A_{\text{rectangle}} & = l \times b \\
& = (25)(15) \\
& = 375
\end{align*}

We are almost there. We now need to calculate the area of triangle \(CDG\) and subtract this from the area of the rectangle to get the area of \(ACDF\).

The area of triangle \(CDG\) is:

\begin{align*}
A_{\text{triangle}} &= \frac{1}{2} DG \times CG \\
& = \frac{1}{2} (8 \times 15) \\
&= 60
\end{align*}

Therefore the area of \(ACDF\) is \(375 - 60 = 315\).