All points at the same distance from a fixed point lie on a circle. Here the fixed point is $P$ and the distance is $4\text{ cm}$, so the circle has centre $P$ and radius $4\text{ cm}$.
They form a circle.
The central line $AB$ is $8\text{ cm}$. The first half-circle has diameter $AX$, which is half of $AB$, so $AX=4\text{ cm}$. The radius is half of $AX$, hence $2\text{ cm}$.
Radius $=2\text{ cm}$ and $AX=4\text{ cm}$.
- 1. PQSR
- 2. SPQR
- 3. RSPQ
- 4. QRSP
A valid name lists adjacent corners in order while moving around the square. $PQSR$ jumps across the square instead of following the boundary order.
PQSR.
On a dot grid, side lengths and right angles can be reasoned from equal horizontal and vertical moves, diagonal moves, and the relative positions of the corner dots.
Yes.
For each drawing, verify that a square has four equal sides and four right angles, and that a rectangle has opposite sides equal and four right angles.
Answers vary with the drawings.
Construct perpendicular sides of $4\text{ cm}$ and $6\text{ cm}$, then complete the rectangle. Check that opposite sides are equal and each angle is $90^\circ$.
The rectangle should have opposite sides $4\text{ cm}$ and $6\text{ cm}$, and all angles $90^\circ$.
Construct perpendicular sides of $10\text{ cm}$ and $2\text{ cm}$, complete the rectangle, and check opposite sides and right angles.
The rectangle should have opposite sides $10\text{ cm}$ and $2\text{ cm}$, and all angles $90^\circ$.
A four-sided figure with all angles $90^\circ$ is a rectangle, and in a rectangle opposite sides are equal.
No.
When $X$ and $Y$ are at the same distance from $A$ and $B$ on the two parallel sides, segment $XY$ remains parallel and equal to $AB$.
$XY$ equals $AB$.
The points $X$ and $Y$ are placed equally along opposite sides, so $XY$ is parallel and equal to $AB$, making $ABYX$ a rectangle.
i. $XY=AB$; ii. $ABYX$ is a rectangle.
The maximum separation occurs when $X$ and $Y$ are opposite corners, giving a diagonal of the rectangle.
The farthest distance between $X$ and $Y$ is equal to $AC$ or $BD$.
If each square has side $s$, three identical squares placed in a row form a rectangle of dimensions $3s\times s$.
Take the length of the rectangle to be three times its breadth.
For two identical squares in a row, the longer side must be twice the shorter side. For three identical squares in a row, the longer side must be three times the shorter side. The example dimensions do not satisfy those ratios.
Examples: $4\text{ cm}$ by $2.5\text{ cm}$ cannot be divided into two identical squares; $7\text{ cm}$ by $2\text{ cm}$ cannot be divided into three identical squares.
Join opposite vertices of the square. Their intersection is the square's centre; draw the circular hole with this point as centre.
The centre of the circle should be at the intersection of the two diagonals of the square.
In a square, adjacent sides are equal and the diagonal bisects the opposite right angles into $45^\circ$ and $45^\circ$.
It should be constructed as a square.
If a diagonal splits the right angle into $45^\circ$ and $45^\circ$, the two legs around that angle are equal, so the rectangle has all sides equal and is a square.
The rectangle becomes a square; adjacent sides are equal.
Draw two equal sides meeting at an angle that is not $90^\circ$. Then locate the fourth vertex using equal-radius arcs from the two open endpoints. The resulting four-sided figure has all sides equal but is not a square because its angles are not right angles.
Yes, a rhombus.