What does a gradient of 3 mean?

For every 1 unit you move to the right, the line goes up by 3 . The bigger the gradient, the steeper the line, so a gradient of 3 is steeper than a gradient of 2.

How do you get the two lengths needed for a gradient?

Find a right-angled triangle that has the line itself as its hypotenuse. The horizontal side of that triangle is the horizontal distance and the vertical side is the vertical height, and the gradient is the second divided by the first.

True or False? The vertical height and the horizontal distance must be in the same unit before you divide.

True. A gradient compares two lengths, so they have to be measured on the same scale for the comparison to mean anything. Leaving one in metres and the other in centimetres would make the answer a hundred times too big or too small.

True or False? The gradient of a straight line is the same wherever along the line you measure it.

True. A straight line has the same steepness all the way along, so any right-angled triangle drawn on it gives the same ratio. That is why you are free to choose whichever two points make the arithmetic easiest.

The hypotenuse is the longest side of a right-angled triangle. It is always the side opposite the right angle, which is the reliable way to spot it however the triangle is turned round.

What is the first step in a Pythagoras problem that is not just a bare triangle?

Find a right-angled triangle in which you already know two of the sides. You may have to draw a line yourself to create one, for instance splitting an isosceles triangle down the middle into two right-angled halves.

What sorts of problem might need Pythagoras before you can finish them?

Any problem where a length is missing before the real calculation can start. Common cases are: perimeters and areas lengths in compound shapes made of two right-angled triangles diameters of circles the perpendicular height of an isosceles triangle In each of these the theorem supplies a length that the question does not give you directly.

National 5MathsSQAApplications of MathematicsFlashcardsGeometric SkillsGradients & Pythagoras

Gradients & Pythagoras (SQA National 5 Applications of Mathematics): Flashcards

Exam code: X844 75

1/14

0Still learning

Know0

FrontGradient
What does a gradient of 3 mean?

Was this flashcard helpful?

Cards in this collection (14)

What does a gradient of 3 mean?
For every 1 unit you move to the right, the line goes up by 3.
The bigger the gradient, the steeper the line, so a gradient of 3 is steeper than a gradient of 2.
How do you get the two lengths needed for a gradient?
Find a right-angled triangle that has the line itself as its hypotenuse.
The horizontal side of that triangle is the horizontal distance and the vertical side is the vertical height, and the gradient is the second divided by the first.
A slide starts 2 m above the ground and ends 20 cm above it, over a horizontal distance of 1.2 m. What is the gradient?
The gradient is 1.5.
Working in centimetres, the vertical height is the difference between the two ends, $200 - 20 = 180$ cm, not the 200 cm start height.
Dividing by the horizontal distance gives $180 \div 120 = 1.5$ .
True or False?
The vertical height and the horizontal distance must be in the same unit before you divide.
True.
A gradient compares two lengths, so they have to be measured on the same scale for the comparison to mean anything.
Leaving one in metres and the other in centimetres would make the answer a hundred times too big or too small.
Complete the method for a gradient from two points:
The horizontal distance is the difference between the two $______$ coordinates, and the vertical height is the difference between the two $______$ coordinates.
The completed method is:
The horizontal distance is the difference between the two x coordinates, and the vertical height is the difference between the two y coordinates.
The same given formula is then used, so nothing new has to be remembered for coordinates.
What is the gradient of the line joining $(4, 11)$ and $(12, 17)$ ?
The gradient is 0.75.
The horizontal distance is $12 - 4 = 8$ and the vertical height is $17 - 11 = 6$ .
Dividing the vertical height by the horizontal distance gives $6 \div 8 = 0.75$ .
True or False?
The gradient of a straight line is the same wherever along the line you measure it.
True.
A straight line has the same steepness all the way along, so any right-angled triangle drawn on it gives the same ratio.
That is why you are free to choose whichever two points make the arithmetic easiest.
Define hypotenuse.
The hypotenuse is the longest side of a right-angled triangle.
It is always the side opposite the right angle, which is the reliable way to spot it however the triangle is turned round.
Complete the two uses of Pythagoras' theorem:
To find the hypotenuse you square the two shorter sides and $______$ them, but to find a shorter side you square the other two and $______$ them instead.
The completed sentence is:
To find the hypotenuse you square the two shorter sides and add them, but to find a shorter side you square the other two and subtract them instead.
Either way you finish by taking the positive square root, and this choice is the part the formula sheet cannot make for you.
True or False?
In $a^{2} + b^{2} = c^{2}$ , any of the three sides can be labelled $c$ .
False.
$c$ must always be the hypotenuse, so it cannot be swapped with either of the others.
$a$ and $b$ can be either way round, because they are simply added together.
A ladder 10 m long leans against a wall with its foot 6 m from the base. How far up the wall does it reach?
It reaches 8 m up the wall.
The ladder is the hypotenuse, so the wall is a shorter side and the squares are subtracted: $10^{2} - 6^{2} = 64$ .
Taking the square root gives $\sqrt{64} = 8$ .
One ramp has a slope of 12 m over a horizontal distance of 9 m, and a second ramp of the same height continues to a total horizontal distance of 22 m. How long is the second slope?
The second slope is 15.2 m.
The shared height comes from the first ramp: $12^{2} - 9^{2} = 63$ , so the height is $\sqrt{63}$ , which is left unrounded.
The second ramp covers $22 - 9 = 13$ m horizontally, and its slope is the hypotenuse: $63 + 13^{2} = 232$ , giving $\sqrt{232} = 15.2$ m.
What is the first step in a Pythagoras problem that is not just a bare triangle?
Find a right-angled triangle in which you already know two of the sides.
You may have to draw a line yourself to create one, for instance splitting an isosceles triangle down the middle into two right-angled halves.
What sorts of problem might need Pythagoras before you can finish them?
Any problem where a length is missing before the real calculation can start. Common cases are:
- perimeters and areas
- lengths in compound shapes made of two right-angled triangles
- diameters of circles
- the perpendicular height of an isosceles triangle
In each of these the theorem supplies a length that the question does not give you directly.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Tolerance & Reading ScalesNext:Perimeter & Area

More Revision Notes you might like