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A short class with Sarah and her students. The other ways are under More.
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In a parallelogram, two opposite angles are (3x + 10)° and (5x − 30)°. Which property helps?
Opposite angles in a parallelogram are equal, so 3x + 10 = 5x − 30.
That gives 2x = 40, so x = 20 and both angles are 70°.
And the other two angles?
Adjacent angles add up to 180°, so they're 110° each.
Check with the angle sum.
70 + 110 + 70 + 110 = 360. It works.
A rhombus has diagonals 10 cm and 24 cm. How long is each side?
The diagonals bisect each other at right angles, so we get right triangles with legs 5 and 12. The side is 13 cm.
And its area?
Half the product of the diagonals: ½ × 10 × 24 = 120 cm².
Now an isosceles trapezium with parallel sides 10 and 16 and slant sides 5. How do we get its height?
Drop perpendiculars. They cut off 3 cm at each end, so the height is √(25 − 9) = 4 cm.
Then the area is ½(10 + 16) × 4 = 52 cm².
How can we prove ABCD is a parallelogram?
Show both pairs of opposite sides are parallel. AB and DC both have gradient ¼, and AD and BC both have gradient 4.
All four sides are √17. What does that make it?
A rhombus. But is it a square?
Test the angle at A.
¼ times 4 is 1, not −1, so the sides aren't perpendicular. It's a rhombus but not a square.
draw, mark the givens, identify the type, choose the properties, solve, then check.
angles sum to 360°; in a parallelogram opposite angles are equal and adjacent angles add to 180°.
diagonals often create right triangles, as in a rectangle or a rhombus.
parallelogram bh, trapezium ½(a + b)h, rhombus and kite ½d₁d₂.
equal gradients show parallel sides, the distance formula shows equal sides, and gradient products of −1 show right angles.
angle and length conditions turn into equations, sometimes quadratics, that pin down unknowns.
Parallelogram with opposite angles (3x + 10)° and (5x − 30)°: 3x + 10 = 5x − 30 gives x = 20, so the angles are 70°, 110°, 70° and 110°.
Rhombus with diagonals 10 cm and 24 cm: each side is √(5² + 12²) = 13 cm and the area is ½ × 10 × 24 = 120 cm².
Isosceles trapezium with parallel sides 10 cm and 16 cm and slant sides 5 cm: the height is √(5² − 3²) = 4 cm, so the area is ½(10 + 16) × 4 = 52 cm².
A(1, 1), B(5, 2), C(6, 6), D(2, 5): opposite sides have gradients ¼ and 4 and all sides are √17, so ABCD is a rhombus; ¼ × 4 = 1, not −1, so it is not a square.
Draw it, mark it, name its kind; list the facts that fit the find; solve it neatly, line by line, and check the angles total fine.
A detective called Quad visits a crime scene shaped like a four-sided room. She marks every clue (parallel walls, equal sides, square corners), names the room's type, and only then works out the missing angle.
"Draw, Mark, Name, Use, Solve, Check" - the six steps of the routine.
P.A.D.C. - Properties, Angles (360°), Diagonals (Pythagoras), Coordinates: the four toolkits for quadrilateral problems.
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In a parallelogram, what is true of adjacent angles?
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What is the area formula for a trapezium?
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How do you find the side of a rhombus from its diagonals?
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How can coordinates show two sides are parallel?
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What is the first step in any quadrilateral problem?
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A quadrilateral with both pairs of opposite sides parallel.
Two angles that add up to 180°.
Angles between two parallel lines on the same side of a transversal; they add up to 180°.
A quadrilateral with one pair of parallel sides.
An extra line added to a diagram, such as a perpendicular, to create shapes that are easier to work with.
The steepness of a line, (y₂ − y₁)/(x₂ − x₁); parallel lines have equal gradients and perpendicular lines have gradients that multiply to −1.
The distance between parallel sides measured at right angles to them.
This lesson gathers the techniques for solving quadrilateral problems: the 360° angle sum, the properties of special quadrilaterals, Pythagoras' theorem on diagonals, area formulas and coordinate checks, all used within a step-by-step problem-solving routine.
Quadrilateral problems are solved by spotting which facts apply and linking them. The angle sum of 360° finds missing angles; the properties of parallelograms, rectangles, rhombuses, kites and trapeziums give equal sides, equal or supplementary angles and special diagonals; Pythagoras' theorem turns diagonals into side lengths; area formulas handle parallelograms, trapeziums, rhombuses and kites; and gradients, lengths and midpoints classify a quadrilateral on a coordinate grid. This lesson works through an example of each technique, including setting up algebraic equations, and ends with a checklist for tackling unfamiliar problems: draw, mark, identify, choose properties, solve and check.
The earlier lessons in this section defined quadrilaterals, sorted them into types and listed their theorems. This lesson is about using those facts together.
A problem-solving routine.
Technique 1: the angle sum. Every quadrilateral's angles add up to 360°, so a missing angle is 360° minus the others. With algebraic angles, add them, set the total equal to 360° and solve.
Technique 2: angle properties of special quadrilaterals.
Technique 3: Pythagoras' theorem with diagonals.
Technique 4: area formulas.
Technique 5: coordinates. For A(1, 1), B(5, 2), C(6, 6) and D(2, 5):
Technique 6: combine with algebra. A rectangle has perimeter 34 cm and diagonal 13 cm. Then l + w = 17 and l² + w² = 169. Since (l + w)² = l² + w² + 2lw, we get 289 = 169 + 2lw, so lw = 60. The sides are the solutions of t² − 17t + 60 = 0, which are 12 and 5, so the rectangle is 12 cm by 5 cm.