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The triangle inequality as an interval, the triangle centres, and the properties and tests that separate the parallelograms.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you find every length a third side of a triangle could have, and give the answer as an interval rather than as a number. You meet the four triangle centres and learn which kind of distance each one keeps equal. Then you sort the quadrilaterals into a family tree, use the properties of parallelograms, rectangles, rhombi and trapezoids, and tell a property that every such shape has from a test that proves a shape is one.
You know that the three angles of a triangle add to $180^\circ$, and you have just used the perpendicular bisector and the angle bisector as constructions. Here those two lines stop being drawings and become answers: each one is the set of points that keeps something equal.
Triangle inequality: any two sides of a triangle are together longer than the third.
Median: a segment from a vertex to the midpoint of the opposite side.
Centroid: where the three medians meet — the balance point.
Perpendicular bisector: the line at right angles to a segment through its midpoint. Circumcentre: where all three of them meet, the centre of the circle through the three vertices.
Angle bisector: the ray that cuts an angle in half. Incentre: where all three meet, the centre of the circle that touches all three sides.
Interval: a stretch of the number line, written $4 < x < 16$ when neither end counts and $4 \le x \le 16$ when both do.
The triangle inequality says any two sides are together longer than the third. Given two sides of $6$ and $10$, the third side $x$ has to satisfy $10 - 6 < x < 10 + 6$, that is $4 < x < 16$: an interval with both ends open, because at either end the triangle flattens onto a line. The longest side always faces the largest angle, so in a triangle with sides $5$, $7$ and $9$ the largest angle is opposite the $9$. The three medians meet at the centroid, two thirds of the way along each from the vertex, so a median of $18$ puts the centroid $12$ from the vertex and $6$ from the midpoint. The three perpendicular bisectors meet at the circumcentre, the one point equidistant from the three vertices; the three angle bisectors meet at the incentre, the one point equidistant from the three sides.
Another way: picture
A number line from $0$ to $20$ with the stretch from $4$ to $16$ shaded and hollow circles at both ends: every length in the shaded part closes up into a triangle with $6$ and $10$, and neither end does.
Another way: story
Walking from $A$ to $C$ the direct way must be shorter than walking from $A$ to $B$ and then $B$ to $C$. The triangle inequality is only that, said three times.
Checking only one pair of sides. With $4$, $5$ and $10$, the pairs $4 + 10$ and $5 + 10$ both beat the third side; it is $4 + 5$ that fails. Comparing the two shortest sides with the longest is the only comparison that can fail, so it is the only one worth making.
Including the ends of the interval. If the third side were exactly the sum, the two shorter sides would lie flat along it and enclose nothing. The ends are excluded, which is why the answer is $4 < x < 16$ and not $4 \le x \le 16$.
Treating all four centres as the same point. They coincide only in an equilateral triangle. The centroid balances, the circumcentre is level with the three vertices, the incentre is level with the three sides — and which one a question wants is decided by which kind of distance it mentions.
Two sides must beat the third: $6 + 10 > x$, so $x < 16$.
The upper end.
And $6 + x > 10$, so $x > 4$.
The lower end comes from the same rule read the other way.
$4 < x < 16$, both ends open.
At $4$ or $16$ the three lengths lie flat.
The centroid divides each median $2 : 1$ from the vertex.
Two thirds of $18$ is $12$, so the centroid is $12$ from the vertex.
The remaining $6$ reaches the midpoint of the opposite side.
The two pieces add back to $18$.
Upper end: $5 + 9 = 14$, so $x < 14$.
Lower end: $9 - 5 = 4$, so $4 < x < 14$.
Two sides of a triangle measure $10$ and $17$. Give the set of lengths $x$ the third side could have.
This task has no paper form; do it on a device.
Can a triangle have sides $4$, $7$ and $13$?
A median of a triangle is $24$ units long. How far along it, from the vertex, is the centroid?
answer
Which point of a triangle is the same distance from all three sides of it?
You have just used the parallel-line angle facts, and you know that opposite sides being parallel is what the word parallelogram means. Every property below is one of those angle facts, or one triangle congruence, applied to half of a quadrilateral.
Parallelogram: a quadrilateral with both pairs of opposite sides parallel.
Rectangle: a parallelogram with four right angles. Rhombus: a parallelogram with four equal sides. Square: both at once.
Trapezoid: a quadrilateral with exactly one pair of parallel sides, its bases. Kite: two pairs of equal adjacent sides.
Diagonal: a segment joining opposite corners.
Midsegment of a trapezoid: the segment joining the midpoints of the two non-parallel sides.
Test: a condition that is enough to prove a shape is a parallelogram, as opposed to a property every parallelogram happens to have.
In a parallelogram, opposite sides are equal, opposite angles are equal, consecutive angles are supplementary — so an angle of $70^\circ$ has a neighbour of $110^\circ$ — and the diagonals bisect each other, so a diagonal of $24$ is cut into two pieces of $12$. A rectangle adds one thing: its diagonals are equal in length. A rhombus adds a different thing: its diagonals are perpendicular. A square is both, which is why it sits under both in the family tree, and everything under that tree is a quadrilateral. One pair of sides that is both parallel and equal is already enough to force a parallelogram; that is a test rather than a property. A trapezoid's midsegment averages its bases, so bases of $8$ and $14$ give $11$.
Another way: picture
A family tree with quadrilateral at the top, parallelogram below it, rhombus and rectangle side by side under that, and square joined up to both. Every arrow points from the more special shape to the more general one.
Another way: story
Think of the shapes as job titles. A square holds every job below it in the tree; a parallelogram holds only its own. Asking whether a rectangle is a rhombus is asking whether the general job includes the special one, and it does not.
Reading the family tree backwards. Every square is a rectangle; not every rectangle is a square. When a question says always, follow the arrows from the special shape to the general one and never against them.
Confusing a property with a test. The diagonals of a parallelogram bisect each other is a property. A quadrilateral whose diagonals bisect each other is a parallelogram is the test. Which one you need depends on whether you are given the shape or asked to prove it.
Expecting a rectangle's diagonals to be perpendicular. Equal diagonals belong to the rectangle; perpendicular diagonals belong to the rhombus; a square has both, which is exactly why it sits under both of them in the family tree.
Consecutive angles are supplementary: $180 - 70 = 110^\circ$.
The parallel sides make the two angles same-side interior.
The angle opposite the $70^\circ$ is also $70^\circ$, and the fourth is $110^\circ$.
The four add to $360^\circ$.
Every parallelogram has diagonals that bisect each other, so that cannot separate them.
A property they all share settles nothing.
Four equal sides force the diagonals to cross at right angles, so the answer is the rhombus; the rectangle is the one with equal diagonals.
The midsegment averages the bases.
$(7 + 19) \div 2 = 13$.
Join each shape to the shape it is always an example of. Draw the shortest link in each case: a square is a rhombus and a rhombus is a parallelogram, so a square needs no arrow straight to the parallelogram.
This task has no paper form; do it on a device.
One angle of a parallelogram measures $113^\circ$. What does the angle next to it measure?
Answer:
A trapezoid has parallel bases of $13$ and $17$. How long is the midsegment joining the midpoints of the two slanted sides?
Answer:
Join each shape to the shape it is always an example of. Draw the shortest link in each case: a square is a rhombus and a rhombus is a parallelogram, so a square needs no arrow straight to the parallelogram.
This task has no paper form; do it on a device.
A joiner builds a four-sided frame with opposite sides equal: $30$ cm one way and $40$ cm the other. He cannot get a set square inside it, so he measures the two diagonals instead and gets $50$ cm and $50$ cm. Is the frame a rectangle?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Two sides of a triangle measure $3$ and $14$. Give the set of lengths $x$ the third side could have.
This task has no paper form; do it on a device.
One angle of a parallelogram measures $87^\circ$. What does the angle next to it measure?
Answer:
You can decide whether three lengths make a triangle, give the possible third sides as an interval, place the quadrilaterals in their family, and use their properties. Explain why the ends of that interval are excluded, and which measurement tells a rectangle from any other parallelogram.
8. Your turn: two sides are $5$ and $9$, step 2
19. Your turn: a trapezoid with bases $7$ and $19$, step 2