[Closed] Help with maths homework

Do you know what the arrows mean on some of the lines?

parallel lines

arrows = lines are parallel
double cross hatch = lines are same length (left hand diagram)

My gut feeling is we are missing something. Like the triangle being isocelles

Left hand side
d, e, f, g = 47.5
a = 85
b, c = 95

Left hand side is easy - I think there's something missing from the right side.

ampthill - Member
My gut feeling is we are missing something. Like the triangle being isocelles

POSTED 2 MINUTES AGO # REPORT-POST

Yip

The two areas to look are angles of triangles isosceles flavour (d&e on the left are 47.5)
On the right you need angles of [url= http://www.coolmath.com/reference/trapezoids ]trapeziods.[/url]
iirc.
hth.

On the right I think the captain's method tells us c and d (which is the other bit of c (complement?).

Assuming that the short sides of the trapezoid are the same (35 deg angle to d = c to f) then we can work out the rest.

If that's not the case then the question seems somewhat harder than the left hand side! Either way, I think we need to know more (i.e. confirm if it is an isosceles trapezoid)

I can confirm what has already been suspected the right side is missing an angle to be possible to solve. I'm a Maths teacher.

Is the solution to go old skool and get out a ruler and find if..
Opposite sides of an isosceles trapezoid are the same length (congruent).
By my monitor they are, hence it is a isosceles trapezoid and possible to solve.
I'm a science teacher. a=30, b,c,f=35 d,e=45 ???

Ditto to chilled76, I'm also a Maths teacher and we need one more piece of info to solve.

Call me old fashioned but could you not just get a protractor on it and show some initiative? 🙂

No need for additional info. On the left you have an isosceles triangle and some Z things which indicate the same angle. On the right you have some F things where the internal angles add to 180 as well as some Z things with equal angles.

I've had beer so I can't type much more but it's all pretty basic geometry.

Edit: a ruler and protractor won't work unless you know it's a scale drawing. It could just be a sketch.

[s]C=35, d=145
A+f+35=180 :.a+f=145
F+e=180
E=a+35
Solve the above a=110, e=145 f=35

Edit and b=f=35[/s]

Ahhh rubbish

I can confirm what has already been suspected the right side is missing an angle to be possible to solve.

and

I'm a Maths teacher.Ditto to chilled76, I'm also a Maths teacher and we need one more piece of info to solve.

Thanks all - a strongly worded note in his homework diary is required. 8)

Gone fishing, everything you've said is right but it doesn't change the fact you need one more angle to solve the right hand side presuming IT ISN'T DRAWN TO SCALE.

I'd be careful with the "I'm a maths teacher " approach as it is a bit like the "I'm an engineer " approach to bike shops. You'll be pretty embarrassed if there is a solution! I'd try extending the parallel lines and n the sketch in the right.

Although I accept that there may not be a solution.

On the right, there's nothing that constrains the angle of the line across the bottom. So it's undefined and thus insoluble without that.
(IANAMT)
(but I am an engineer)

I'm an engineer and it definitely needs another bit of information 😉

You can get c and d easily with angles on parallel lines but that's all I can see.

Except gone fishing it's so simple I'm not going to get burnt. If it was a tricky further pure question I'd be cautious.. but this is as plain as adding up.

Seen literally 100s of similar questions aimed at gcse students to test if they understand alternate and corresponding angle rules on parallel lines. Draw it out, you can draw infinite versions of the same problem with the information presented.

In fact just to illustrate, put a ruler along the Base of the right problem. You can change the angle of the ruler and that would change where the line is with 4 unknowns still being labelled the same.

You've effectively got two similar (mathematical definition not coloqualistic) one inside the other with only one angle known. The other two must complement to 180 but that's all you can say unless there is scale in the drawing. But the point of these problems generally is that they aren't drawn to scale to avoid students measuring the angles.

Original OP, trust me there's info missing.

OP I agree, right hand diagram is missing something. You can write many equations but you cannot solve them, eg

a+f = 180-35
b=f
b+e=180
c+d=180
a+b+c=180

Is there not something about equivalent/similar triangles (or something) that will help with this? It's been a little big time since I did a lot of geometry.

I reckon it'll come out as an isosceles trapezoid, obvs.
Some of my best friends are engineers. 😐

Draw a line from angle d up to the longest line at right angles to the parallel lines, this will form a right angled triangle with 35, (d-90) and 90 as the included angles. Angle d is 145 degrees (I think!)

edit- I think we knew this anyway, sorry!

Draw a line from angle d up to the longest line at right angles to the parallel lines, this will form a right angled triangle with 35, (d-90) and 90 as the included angles. Angle d is 145 degrees (I think!)

This is an interesting idea, was looking at a but I don't think it works as you cannot know a is cut in half (or any other known fraction)

geoff .... the best lesson from this question is NEVER struggle with such a question in an exam, move on and do the questions you can and come back to this if you have time

I assume you could put in an answer, but I can't see how it would be a simple x degrees - you'd have to do some substituting and use sine rule (or wot-evva; it's been along time) to express it in terms of 35 deg and the length of a side or two

Actually scratch what I last said. The fact that there are two triangles means that I don't think there is a solution. Ignoring the smaller triangle for a minute you can't solve a triangle with only two be angle and no other information. Although that could be the point.

You can't use a sine rule you don't have any lengths. And it's not to scale otherwise you'd just use a protractor.

Actually scratch what I last said. The fact that there are two triangles means that I don't think there is a solution. Ignoring the smaller triangle for a minute you can't solve a triangle with only two be angle and no other information. Although that could be the point.

There is no info stating the non parallel sides of the trapezium are congruent. Which means they can be different lengths.

Agreed if they are assumed the same length you can solve... but that's an assumption and the diagram would have info indicating it if the author wanted that to be the case.

Who pencilled in a=85 on the second diagram? That's not one of these things the pupil was told in class and has subsequently forgot?

😀

This is aimed at about grade B GCSE. Modern grade 5 (don't confuse that with levels). You don't aim work at ages, for example a top set year 7 student is already brighter than some students in year 11. You set work according to ability.

Have fun debating this chaps.

Oh and you can't bring 85 for a from the previous diagram. If you look on the question on the left a and c complement to 180 but on the right question a and c are 2 out of 3 angles in a triangle so they have to be two separate questions.

Im finding mistakes in questions not quite daily but certainly on a weekly basis at the minute as all the new GCSE material has been rushed in so fast that text books etc have hit press without proper checking.

Loads of mistakes on the pre-release exam material too and on the testable and exam.pro software for building tests. Wouldn't be too harsh in your note to the teacher, it'll be the first time they've used this question and just glanced at it without noticing the missing angle. Sometimes mistakes happen.

This is S1 (year 8 ) homework for a 12 year old in Scotland.

You can't use a sine rule you don't have any lengths. And it's not to scale otherwise you'd just use a protractor.

Agreed, sorry skim read your post.

chilled76 - Member
I can confirm what has already been suspected the right side is missing an angle to be possible to solve. I'm a Maths teacher.

schmiken - Member
Ditto to chilled76, I'm also a Maths teacher and we need one more piece of info to solve.

pfft.

Those who can't, teach 😉

What can I say. 13 weeks a year to ride my bike 😉