Make Pythagoras Proud

Geometry Level 3

Using the image above, find $N$ .

(The image is not given to scale.)

2 3 4 5

1 solution

Jason Chrysoprase
Jan 28, 2016

The drawn image above is given to scale

By Pythagoras' Theorem,

$ED^2 + DB^2 = 5^2$

$AD^2 + CD^2 = 10^2$

$CD^2 + DB^2 = 11^2$

By adding the first 2 equations and taking away the last shows that :

$(ED^2+DB^2) + (AD^2+CD^2) - (CD^2+DB^2)$

$= ED^2 + DB^2+ AD^2+CD^2-CD^2-DB^2$

$= AD^2 +ED^2$

$= N^2$

So,

$N^2 = 5^2 + 10^2 - 11^2 = 4$

$N = 2$

I would use your image but solve it in different way.

$\vartriangle BDE$ have: $BD^{2} + DE^{2} = N^{2}$ (1)

$\vartriangle BDA$ have: $AD^{2} + DE^{2} = AB^{2} = 10^{2}$ (2)

$\vartriangle CDE$ have: $CD^{2} + DE^{2} = 5^{2}$ (3)

$\vartriangle CDA$ have: $BD^{2} + DE^{2} = 11^{2}$ (4)

$I \ am \ going \ to \ write \ it \ later.$

Evan Huynh - 5 years, 4 months ago

Great, try to solve it with different way

Jason Chrysoprase - 5 years, 4 months ago

Thx m8 I'll make my best

Jason Chrysoprase - 5 years, 4 months ago

Sravanth Chebrolu

Tell me the geometry drawing you use to change my Problem picture !!

I'll use it

Jason Chrysoprase - 5 years, 4 months ago

Hmm... It wasn't me who did that, maybe some other moderator did that, Thanks!

Sravanth C. - 5 years, 4 months ago

Ohh, ok

Thx for following me

Jason Chrysoprase - 5 years, 4 months ago

Is there a way to use similar triangles to solve for the sides? Just curious

Syed Hamza Khalid - 2 years, 5 months ago

Well done! +1! Keep posting such solutions.

Sravanth C. - 5 years, 4 months ago