# Struggling with geometry :(( (1 Viewer)

#### _caecilius_iucundum_

##### New Member
IM STRUGGLING WITH GEOMETRY AAAAHHHH!!!

I've been doing some work on geometry (mostly triangle congruency/similarity proofs, finding lengths, angles etc.) for the past year or so both at school (in my 'competition maths' class) and at the tutoring college I go to.

The trouble is, I seem to never get the 'hang' of understanding what I'm doing.

I have trouble identifying when I should be using certain rules to answer questions with and I guess, in general, finding a direction as to how to solve the question at all. And in the end, geometry questions just end up overwhelming me and I just don't know what to do.

Example of question I have trouble with:

I feel like geometry doesn't come to me naturally like algebra - and I get that algebra is pretty straightforward most of the time so maybe that is why. But lately, geometry has got me so overwhelmed, I start hating maths and not wanting to do any maths at all. I feel like I'm falling behind in both my school maths comp class and my tutoring class and I feel like everyone else in my classes know exactly what they're doing and I'm just dumb and not meant for geometry.

I just want to know if there are any tips to overcoming the problems I'm experiencing with geometry (and hard high school maths in general) and how I can maybe start enjoying geometry.

Any help is desperately needed!

From, a struggling not-so-happy geometrician

#### DarkOperator618

##### Well-Known Member
IM STRUGGLING WITH GEOMETRY AAAAHHHH!!!

I've been doing some work on geometry (mostly triangle congruency/similarity proofs, finding lengths, angles etc.) for the past year or so both at school (in my 'competition maths' class) and at the tutoring college I go to.

The trouble is, I seem to never get the 'hang' of understanding what I'm doing.

I have trouble identifying when I should be using certain rules to answer questions with and I guess, in general, finding a direction as to how to solve the question at all. And in the end, geometry questions just end up overwhelming me and I just don't know what to do.

Example of question I have trouble with:

View attachment 31343

I feel like geometry doesn't come to me naturally like algebra - and I get that algebra is pretty straightforward most of the time so maybe that is why. But lately, geometry has got me so overwhelmed, I start hating maths and not wanting to do any maths at all. I feel like I'm falling behind in both my school maths comp class and my tutoring class and I feel like everyone else in my classes know exactly what they're doing and I'm just dumb and not meant for geometry.

I just want to know if there are any tips to overcoming the problems I'm experiencing with geometry (and hard high school maths in general) and how I can maybe start enjoying geometry.

Any help is desperately needed!

From, a struggling not-so-happy geometrician
first off, wtf is this lol

this stuff isn't available in the maths advanced syllabus btw so you'll be fine (like HSC advanced maths)

#### CM_Tutor

##### Moderator
Moderator
@_caecilius_iucundum_

Firstly, are you a fan of Latin or of Dr Who? Just curious from the user name.

Second, most of the geometry has been removed from the year 11 and 12 syllabi. What is left will require formulae for areas and volumes, and occasionally a similarity / congruence needs to be recognised. It's a shame, really, geometry teaches some useful skills in setting out proofs.

Third, this is not a straight-forward geometry problem. I'll try to do a solution but I wanted to make these comments first. I also want to say that I am not certain that the question is complete. The line $\bg_white IF$, for example, is given no definition other that it forming a side of two of the triangles. In other words, we know that $\bg_white IF$ divides $\bg_white \Delta AIE$ into two triangles of equal area ($\bg_white \Delta AIF$ and $\bg_white \Delta FIE$) and it appears that $\bg_white IF$ might be parallel to $\bg_white AC$ and that $\bg_white F$ might be the midpoint of $\bg_white AE$, but neither of those is certain from the information given. I make these observations to illustrate the kinds of thoughts / impressions that I have before attempting to solve the problem. I also note that the information that $\bg_white \Delta BIE$ is isosceles does not actually specify which pair of sides is equal. Now, since $\bg_white \angle EBI$ is a right angle and so can't be a base angle, we do actually know that $\bg_white BI = BE$.

My final thought is that competition maths and high school maths are quite different. I found competition maths to be useful for developing skills for the more difficult exam questions where insight is required, but many of the problems are unlike any specific syllabus content.

#### Life'sHard

##### Well-Known Member
Yup most geometry is gone from yr 11 and 12. However, some circle geo can be found in 4u and it's still important that you at least know the basics of some of these theorems. Also knowledge of parallelograms, kites, trapeziums etc and there properties (Intersection, bisections, parallel, equal bisection etc) is important.

#### Drongoski

##### Well-Known Member
No, you're not necessarily struggling with Geometry. This question itself is not straightforward. I bet half of 2U Maths Advanced student will find this question hard.

#### Drongoski

##### Well-Known Member
I got 18 cm for BG; not sure if correct.

#### Eagle Mum

##### Well-Known Member
These are problem solving exercises - usually all the given information (words & diagrams) are relevant and the challenge is to work out what order to use the information.

1. The total area of 2016 cm^2 is divided into 7 equal triangles, so each triangle is 288 cm^2.

2. Triangle BIE comprises four of these smaller triangles, so its total area is 1152 cm^2.

3. Triangle BIE is isosceles so BI = BE = X (turning this into algebra); it’s also a right angle triangle with BE as its height (h) and BI as its base (b).

4. Since the formula for the area of a triangle is A = (b x h)/2, we substitute 1152 = (X^2)/2

5. Rearranging our equation: BE = BI = X = √2304 = 48

6. Area of HIE = (HI x BE)/2 = 288 Therefore, HI = 576 / 48 = 12

7. BH = BI - HI = 48 - 12 = 36

8. Area of BHD = (BH x BD)/2 = 576 Therefore, BD = 1152 / 36 = 32

9. Area of BGD = (BG x BD)/2 = 288 Therefore, BG = 576 / 32 = 18

Note that one of the crucial points to recognise in this question is that the common angle at B is a right angle, so equations can incorporate the areas of the different triangles (which are multiples of 288) to find the length of either the base or perpendicular height of individual triangles once the other has been determined and sequentially determine various lengths to arrive at the length of BG.

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#### CM_Tutor

##### Moderator
Moderator
• I've commented above on some first impressions.
• I didn't comment that the only number we have here is an area, so we'll obviously need to be using the areas of the triangles
• All seven small triangles have areas where using $\bg_white A = \cfrac{1}{2} \times \text{base} \times \text{perpendicular height}$ is straight-forward as, in each case, the base is on one side of the given right angled $\bg_white \Delta ABC$ and the height is measured along the other.
• I will start by giving the distance I seek a name, and introducing a pronumeral for the other side of $\bg_white \Delta DBG$, since I can then use the area formula for that triangle easily.

We seek the length $\bg_white BG$, so let $\bg_white BG = b\ \text{cm}$ and let $\bg_white BD = h\ \text{cm}$.

We have seven triangles, equal in area, with a total area of $\bg_white 2016\ \text{cm}^2$, and so the area of each is $\bg_white \cfrac{1}{7} \times 2016 = 288\ \text{cm}^2$. Using this fact with $\bg_white \Delta DBG$ gives

\bg_white \begin{align*} \text{In}\ \Delta DBG,\ \text{Area}\ &= \cfrac{1}{2} \times BG \times BD \\ 288 &= \cfrac{1}{2} \times b \times h \\ 576 &= bh \qquad \text{. . . . . . . . . . (1)} \end{align*}

and, in $\bg_white \Delta DGH$ gives

\bg_white \begin{align*} \text{In}\ \Delta DGH,\ \text{Area}\ &= \cfrac{1}{2} \times GH \times BD \\ 288 &= \cfrac{1}{2} \times GH \times h \\ 576 &= GH \times h \\ bh &= GH \times h \qquad \text{using (1)} \\ GH &= b \qquad \text{. . . . . . . . . . (2)} \end{align*}

Now, we know that $\bg_white \Delta BIE$ is isosceles and $\bg_white \angle BIE = 90^\circ$ must be the apex angle, so $\bg_white BI = BE$. $\bg_white \Delta BIE$ is made up of four of triangles of area 288 cm2 and so

\bg_white \begin{align*} \text{In}\ \Delta BIE,\ \text{Area}\ &= \cfrac{1}{2} \times BI \times BE \\ 4 \times 288 &= \cfrac{1}{2} \times x \times x \qquad \text{letting the sides BI and BE be x cm} \\ 1152 &= \cfrac{x^2}{2} \\ x &= \sqrt{2 \times 1152} \qquad \text{as x > 0} \\ &= 48\ \text{cm} \end{align*}

So, we have shown that $\bg_white BE = 48\ \text{cm}$ and, as $\bg_white BD = h\ \text{cm}$, we can deduce that $\bg_white DE = 48 - h\ \text{cm}$, and so we can examine the area of $\bg_white \Delta EDH$:

\bg_white \begin{align*} \text{In}\ \Delta EDH,\ \text{Area}\ &= \cfrac{1}{2} \times ED \times BH \\ 288 &= \cfrac{1}{2} \times (48 - h) \times 2b \qquad \text{as BH = BG + GH = b + b = 2b, using (2)} \\ 288 &= (48 - h)b \\ &= 48b - bh \\ &= 48b - 576 \qquad \text{using (1)} \\ 288 + 576 &= 48b \\ b &= 18\ \text{cm} \end{align*}

#### CM_Tutor

##### Moderator
Moderator
And overall, we have distances:

$\bg_white BC = 56\ \text{cm}$, made up of $\bg_white BG = GH = 18\ \text{cm}$, $\bg_white HI = 12\ \text{cm}$, and $\bg_white IC = 8\ \text{cm}$

and

$\bg_white BA = 72\ \text{cm}$, made up of $\bg_white BD = 16\ \text{cm}$, $\bg_white DE = 32\ \text{cm}$, and $\bg_white EF = FA = 12\ \text{cm}$

Note that the diagram is quite misleading in some ways. $\bg_white DE$ does not appear to be double $\bg_white BD$, nor to be longer than $\bg_white GI$. My initial impression that $\bg_white F$ is the midpoint of $\bg_white AE$ is correct, but trigonometry tells us that

$\bg_white \tan{\angle ACB} = \cfrac{AB}{BC} = \cfrac{72}{56} = \cfrac{9}{7}$

$\bg_white \tan{\angle FIB} = \cfrac{FB}{BI} = \cfrac{60}{48} = \cfrac{5}{4} \neq \cfrac{9}{7}$

and thus $\bg_white AC$ and $\bg_white IF$ are not parallel as their corresponding angles are not equal.