How to find the area of the triangle which intersects a line in a rectangle?

Question

The problem is as follows:

$ABCD$ is a rectangle, $EA=5\,cm$ , $BE=3EA=\frac{AD}{2}$. Find the area of the triangle $BNC$.

What I tried so far is pictured below, as from Pitagorean theorem I reached to $EC=\, 15\sqrt{5}$:

but my problem lies on how to find the sides of $EN$ and $NC$ Is there anything that I'm missing?

In this case help which would include a reworked diagram with letters and labels and not segment line notation be much appreciated.

Please, if you are ok, you can accept the answer and set it as solved. Thanks! — user, Jan 28 '18 at 23:48

Narasimham · Answer 1 · 2017-12-19T22:58:17.863

2

Triangles $BEN,BEC$ are similar,

$$ EN\cdot EC = EB^2 $$ $$ EN \cdot 15 \sqrt 5 = 15^2,\quad EN= 3 \sqrt5$$ $$ NC= EC- EN = (15-3)\sqrt5 = 12 \sqrt 5$$ Using Pythagoras $ BC,NC$ $$ BN = 6 \sqrt5$$ Area BNC $$ \frac12 6 \sqrt5 \cdot 12\, \sqrt 5 =180 $$

edited Dec 19 '17 at 22:58

answered Dec 19 '17 at 22:36

Narasimham

40,495

GNUSupporter 8964民主女神地下教會 · Answer 2 · 2017-12-19T23:04:13.437

1

Observe that $\triangle BNC \sim \triangle EBC$. Calculate ratio of the hypotenus. $$\frac{BC}{EC}=\frac{30}{15\sqrt5}=\frac{2}{\sqrt{5}}.$$ $$\frac{S_{\triangle BNC}}{S_{\triangle EBC}}=\left(\frac{2}{\sqrt{5}}\right)^2=\frac45.$$ $$S_{\triangle BNC}=\frac45 S_{\triangle EBC}=\frac45\cdot\frac12 (15\cdot30)=\frac45\cdot225=180$$

edited Dec 19 '17 at 23:04

answered Dec 19 '17 at 22:35

GNUSupporter 8964民主女神地下教會

17,627

Last step to be 180! – Narasimham Dec 19 '17 at 23:00
@Narasimham Thanks. Corrected this mistake. – GNUSupporter 8964民主女神地下教會 Dec 19 '17 at 23:04
Can you add a step to explain why $\frac{S_{BNC}}{S_{EBC}}=(\frac{2}{\sqrt{5}})^2$? This identity or property I'm not aware. Maybe you can hint some source or theoretical base? – Chris Steinbeck Bell Dec 20 '17 at 00:57
@ChrisSteinbeckBell Sorry for the delayed response. You may want to see the reason on Math Open Reference. – GNUSupporter 8964民主女神地下教會 Dec 20 '17 at 01:22
@GNU Supporter Can you help me to understand how do I find triangle similarity in this problem?. In my drawing I did not labeled any angles thus I'm confused at where to look at to establish the similarity. Regarding the identity you mentioned. Does it hold for any side of the triangle?. I mean the ratio of their areas is the square to any of their sides?. – Chris Steinbeck Bell Dec 20 '17 at 02:08
@ChrisSteinbeckBell (1) Triangle similarity: Observe that they have a common angle $EBC$ and a right angle. In this type of question (find area/length) in high school math, we're not asked to give a prove of $\triangle BNC \sim \triangle EBC$, so you can simply state this in tests. (2) The identity "area ratio = side ratio squared" holds for any similar geometric figures. You can apply it to ellipses, parabolae, etc, if possible. – GNUSupporter 8964民主女神地下教會 Dec 20 '17 at 14:50

Seyed · Answer 3 · 2017-12-19T23:06:42.190

1

If you are just looking for the length of $EN$ and $NC$:

edited Dec 19 '17 at 23:06

answered Dec 19 '17 at 22:54

Seyed

8,933

By writing $\cos (BEC)$ is not very clear what you meant. Unless is $\cos (BEN)$ Mind if you check up the steps? – Chris Steinbeck Bell Dec 20 '17 at 00:50
@ChrisSteinbeckBell, I think it is very clear that when you use the sides of triangle $BEC$, you have to use angle $BEC$. – Seyed Dec 20 '17 at 01:33
It was hidden to me but after twisting the triangle I figured out. How about explaining why $EN=BE\times\cos(BEC)$? This part is confusing. – Chris Steinbeck Bell Dec 20 '17 at 01:54
@ChrisSteinbeckBell, this is trigonometry. – Seyed Dec 20 '17 at 02:25

score 1 · Answer 4 · answered Dec 19 '17 at 22:58

1

Area BNC should be equal to 180:

answered Dec 19 '17 at 22:58

user

154,566

Can you include in your answer triangle similarity proofs?. It is very straightforward the way it is.I am stuck and trying to know why is it similar. – Chris Steinbeck Bell Dec 20 '17 at 00:42
triangles EBC, BCN and EBN are similar because the have the same angles; indeed right angles: EBC=BNC=ENB, acute angles ECB=BCN=EBN and BEC=NBC=BEN – user Dec 20 '17 at 00:47
Which case of similarity is it? angle-angle-angle?. Or...? – Chris Steinbeck Bell Dec 20 '17 at 00:58
angle-angle-angle, they are three similar right triangles – user Dec 20 '17 at 01:00
Understood. I would have preferred you mentioned this in your drawing because that's very important unless you gained experience to notice before hand. I'm not very fond in geometry and had to check it twice. Forgot to mention that in the above question I said that variables such as single letters are easier to understand rather than line segment notation as to get them can be confusing. – Chris Steinbeck Bell Dec 20 '17 at 01:02

How to find the area of the triangle which intersects a line in a rectangle?

4 Answers4