A geometry problem by Jádson Bráz

Geometry Level 2

In the following figure, $ABCD$ is a trapezoid inscribed in a circle. The longer base $AB$ has length 16 cm, the shorter base $CD$ has length 10 cm, and the altitude of the trapezoid has length 9 cm. What is the radius of the circle, in centimeters?

\frac{40}3

\frac73

\frac{15}3

\frac{50}3

\frac{35}3

\frac{25}3

2 solutions

Swastik Dwibedy
Jul 16, 2015

Drop perpendiculars from the center of the circle O on AB and CD. Let the perpendiculars meet AB and DC at E and F respectively. We know that Perpendiculars from center bisects a chord. So EB=8 and FC=5. Let FO= x , So EO=(9-x) Also, OC=OB=r (radius of circle)

Using Pythagoras Theorem, (FC)^2+(OF)^2=r^2 => (5)^2+(x)^2=r^2 .......eqn (1) (OE)^2+(EB)^2=r^2 => (9-x)^2+(8)^2=r^2 .......eqn(2)

Comparing eqn (1) and eqn (2) we get x= 20/3

Putting value of x in any of the eqations will give value of r which is equal to 25/3.

Its a 2 level Question... not 3!!

Shreyansh Singh Solanki - 5 years, 9 months ago

Radhika Saithree
Jul 18, 2015

Lol, I just estimated and I got it right ;)

What I did was guess that (assuming the diagram was accurate) the diameter was larger than 16cm (approximately 17cm). So radius would be 17/2. Options all had a denominator of 3, so 17/2 converted to x/3 = 25.5.../3. Closest option given was 25/3. Clicked that: "Good work! You are in the 16% of people who solved this problem!"

I answered the diameter instead of radius. There's really no need to have the diameter as one of the choices, except to make people mess up even when their math is sound. Sure tested my geometry skills!

Drew Martin - 5 years, 10 months ago

And I thought that I would be the only one to solve it that way, lol

Glen Mast - 5 years, 10 months ago

A geometry problem by Jádson Bráz

2 solutions

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