finding sides of a triangle when circumradius and inradius are given

Question

The radius of the circumscribed circle of a right triangle is $15 cm$ and the radius of its inscribed circle is $6 cm$. Find sides of triangle.

From another site I got, $c=30$, $a+b=2(15+6)=42$. $a+b+c=72$. $ab=6\times 72=432$. So, sides are $18$, $24$.

I didn't get how we wrote $a+b$ and $ab$ equations. What's the relation of sum and product of sides with circumscribed and inscribed radii.

i would write it as comment http://en.wikipedia.org/wiki/Triangle — dato datuashvili, Jan 09 '14 at 18:31
this one third http://www.mathopenref.com/trianglecircumcircle.html — dato datuashvili, Jan 09 '14 at 18:33
http://www.ajdesigner.com/phptriangle/right_triangle_inscribed_circle_radius_r.php — dato datuashvili, Jan 09 '14 at 18:35
please see last one,if i put all of this,it may be as answer right :D:) — dato datuashvili, Jan 09 '14 at 18:36
good aevening @Ramit is my answer relevant to your question or you are looking different one? — dato datuashvili, Jan 09 '14 at 19:05

score 1 · Answer 1 · edited Feb 09 '14 at 11:00

1

Because radius of circumcircle is $15$, it means yes that $c$ or hypotenuse is $30$.

Also,

$r=(a+b-30)/2$

Putting values we get,

$a+b=42$

Now we have

$a^2+b^2=900$

$a+b=42$

From which, we have $b=42-a$.

We get,

$a^2+(42-a)^2=900$

$a^2+1764-84\cdot a+a^2=900$

$2\cdot a^2-84\cdot a+864=0$

Or, $a^2-42\cdot a+432-0$

$D=1764-1728=36$

Could you continue please? Also reject negative values.

EDITED:

$a_1=(42+6)/2=24$

and $a_2=(42-6)/2=18$

Therefore,

$b_1=18$ and $b_2=24$

Now you know what is the relationship between small radius and sides, and also hypotenuse and big radius. Sure you can find a general formula for relationship between sides combination and radius, but it would be a little tricky, you should express from Pythagorean theorem sides and put in radius calculation formulas, or at least use angles formula, which you can find easily on the internet.

edited Feb 09 '14 at 11:00

Sawarnik

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answered Jan 09 '14 at 18:46

dato datuashvili

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sorry, how did we write r=(a+b-30)/2? – aarbee Jan 10 '14 at 11:07
because $r=(a+b-c)/2$ it is existed formula – dato datuashvili Jan 10 '14 at 13:44
If I am given a right angle triangle with circumradius= x cm then can we directly conclude that hypotenuse = 2x cm? – Seema Bhukar Mar 08 '16 at 09:44

score -1 · Answer 2 · answered Aug 20 '17 at 15:26

Cracked by Adarsh (8th grade)

Step by step process:

1) For a Right Angled Triangle, if Circumradius (R) = 15 then Hypotenuse (c) = 2*R = 2*15=30 CM

2) For a Right Angled Triangle, Inradius (r) = (a+b-c)/2 ==> 6 = (a+b-30)/2 ==> a+b=42

3) Area = s*r = (a+b+c)*r/2= (a+b+30)*6/2 = (a+b+30)*3 = (42+30)*3 = 216 sq.Cm

4) Area = ab/2 ==> ab = 2* Area = 2 *216 = 432 Sq.Cm

5) a^2 + b^2 = c^2 ==> a^2 + b^2 = 900

6) (a+b)^2 = a^2 + b^2 + 2a*b = 900 + (2*432) = 1764 = 42^2 ==> a+b=42

7) (a-b)^2 = a^2 + b^2 - 2a*b = 900 - (2*432) = 36 = 6^2 ==> a-b=6

8) solving 6 and 7, a = 24 and b = 18 c = 30 (ref step 1)

Please use MathJax. – José Carlos Santos Aug 20 '17 at 15:43 — José Carlos Santos, Aug 20 '17 at 15:43

score -2 · Answer 3 · answered Oct 16 '18 at 03:17

-2

In radius=(a+b-c)/2 a+b=42...........1 a²+b²=30²(in right angle triangle).............2 Square on both side in equation 1so ab=432 a=42-b Put the value of a in equation 1 (42-b)b=432 b²-42b-432=0 (B-18)(b-24)=0 Then Sides are 18 and 24

answered Oct 16 '18 at 03:17

Shahrukh sheikh

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finding sides of a triangle when circumradius and inradius are given

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