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**denis_gylaev****Member**- Registered: 2015-03-19
- Posts: 66

1) The angles of a quadrilateral are

. Find the measure of the largest angle of the quadrilateral.2) In the diagram below, quadrilateral ABDE is a parallelogram, and BC = BD. If

, then find , in degrees.3) Let ABCD be a parallelogram. Extend

past B to F, and let E be the intersection of and . If the areas of triangles BEF and ADE are 1 and 9, respectively, find the area of parallelogram ABCD.Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

For 1) all you have to do is add up all those angles and set that equal to 360.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,155

hi denis_gylaev

Q2.

Triangle BCD is isosceles and you know CBD. So you can work out BDC (=BCD)

Then EDB and then BAE.

Q3.

The triangles BEF and ADE are similar with an area ratio of 1:9

So the length ratio AE:EB = 3:1

Triangles BEF and CDF are similar with EB:DC = 4:1

So you can work these areas in order ... FCD, BEDC, ABCD.

Bob

ps. I put

`[img]...[/img]`

tags around your link.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**denis_gylaev****Member**- Registered: 2015-03-19
- Posts: 66

I'm still having trouble, can you post a solution for the problems so that I can understand them better? Thanks for the help

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,155

Q3.

As AD is parallel to BC, angle ADE = angle EFB and FEB = AED (vertically opposite).

So triangles BEF and AED are similar. This means there will be a scale factor that transforms the smaller into the larger.

We know that the ratio of their areas is 1:9, so the ratio of their lengths will be 1:3

Choose a scale so that AE = 3 and EB = 1 (this is not a cheat as the diagram can be measured using any suitable units inches, cm, etc. So I'm just choosing a unit to keep the working simple.)

So, on this scale, CD = 4.

Now consider the triangles BEF and CDF.

They have one angle in common and sides that are parallel, so they are similar as well, with a ratio of sides of 1:4

So triangle CDF has area 16 (square the length scale factor)

So quadrilateral BEDC has area 15, so ABCD has area 24.

Bob

ps. I have been asked to help with a lot of geometry recently, including the same question more than once. I'm not complaining; I like to be helpful; but I was wondering whether it would be helpful to members if I created a post consisting of a geometry contents list with links to the relevant posts. What do you think of this idea?

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**denis_gylaev****Member**- Registered: 2015-03-19
- Posts: 66

I think like a subforum inside this forum with the different topics like, algebra, geometry, trigonometry, etc. Also, thanks for your help on problem 3, can I have some help with 1 and 2?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,155

Q1.

The angles of a quadrilateral always add up to 360 so

x + 5x + 15 + 3x - 25 + 4x - 20 = 360

Solve for x, compute the four angles and identify the largest.

Q2

BC = BD so BCD is isosceles with BCD = BDC

As CBD = 28, => BCD = BDC = (180 - 28)/2

BDE = 180 - CBD = BAE as the opposite angles of a parallelogram are equal.

I haven't got the 'powers' to create new subsections. I thought I could just make a thread that had as the first post, the contents page with links. Over time, I would add new links. Members could ask for a new link (by replying in the thread) and, if I'm able, I'd add it to the contents.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**denis_gylaev****Member**- Registered: 2015-03-19
- Posts: 66

Ok I got number 1 which was 165 degrees, and I tried number 2. Like you said BDE = 180 - CBD = BAE and we are trying to find BAE, since the problem states that CBD is 28 degrees, 180-28 is 152 so now we have BDE=152=BAE so from there we can see that BAE is 152, this feels to easy to be true and I think im doing something wrong and/or missing something

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,155

in BCD, two angles are equal and the third is 28. So subtract 28 from 180 ... 152 and then find half of this for each base angle ...76.

So its angles are 28, 76 and 76.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**denis_gylaev****Member**- Registered: 2015-03-19
- Posts: 66

Whoops, I see what i did wrong, thanks for the help Bob!

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