A geometry problem by Adrian Gonzalez

Geometry Level 1

Given the above, what is the length of BC?

The answer is 8.

2 solutions

Mathh Mathh
Aug 13, 2014

Let $OA=OD=c, OB=a, OC=b$ . Our answer is $\sqrt{b^2-a^2}$ .

$\color{royalblue}{\textbf{Pythagorean Theorem:}} \begin{cases}c^2+100=b^2\\ c^2+36=a^2\end{cases}$

$\implies \color{royalblue}{\textbf{answer}}=\sqrt{b^2-a^2}=\sqrt{(c^2+100)-(c^2+36)}=\sqrt{64}=\boxed{8}.$

To visualize the above solution, draw right triangle OBC and apply the Pythagorean Theorm.

Ceazar Sunga - 6 years, 10 months ago

Doesnt the restriction the geometry mean triangles ODC and OBC are identical (symetrically about OC)? In that is the case then that would mean OD=OB, hence OB !=OA or OA !=OD

Sisa James - 6 years, 9 months ago

Triangles ODC and OBC cant be symmetrical because then OB would be equal to OA which is not possible as OB is the hypotenuse in the right angled triangle OAB (OB has to be greater than OA)...and the 2 triangles hav just a right angle and a side in common..they need not be symmetrical..(atleast one more condition is needed for them to be symmetrical)

charvi vitthal - 6 years, 9 months ago

First tyou match OB,OC then you can use pitagorean theorem

Đỗ Lê Phong - 6 years, 9 months ago

Vinit Béléy
Sep 29, 2014

Join O,B and O,C using line segments, and make a ∆OBC

Let OA = OD = x

We find that ∆OAB, ∆ODC and ∆OBC are right angled triangles.

In ∆OAB, OB² = OA² + AB² = 6² + x²

In ∆ODC, OC² = OD² + CD² = 10² + x²

In∆OBC, OC² = OB² + BC² ; 10² + x² = 6² + x² + BC² ; BC² = 10² - 6² = 8² (Pythagorean triplet) ; Therefore, BC = 8

A geometry problem by Adrian Gonzalez

The answer is 8.

2 solutions

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