Inscribed Triangle - Easy Geometry

It is obvious that the radius of the circle is five. Because the triangle has all points on the outside of the circle, with two sharing a diameter, the angle formed $ABC$ must have degree measure 90. Thus, letting $a$ and $b$ be equal to $a^{2}+b^{2}=100$ and $ab=22$ . We want to find $a+b$ . Expanding $(a+b)^{2}$ and grouping, $(a+b)^{2}=a^{2}+b^{2}+2ab$ . Thus, by substitution, $(a+b)^{2}=100+44 \Longrightarrow a+b=12$ . From here, because the hypotenuse of the triangle (the diameter) has length 10, we can conclude that the perimeter is $\boxed{22}$ .

Let the unknown sides be x and y. Angle extended in a semi circle with end points of diameter is 90 degrees. Thus are of right angled triangle is half the product of the sides adjacent to right angle i.e. xy/2=11 cm ^2 . Thus xy =22 cm^2. Also given πr^2 = 25 π cm^2. Thus, radius is 5cm. Thus diameter is 10cm. Then by Pythagoras's theorem, x^2 + y^2 =100............(1) and xy =22...................(2) But, (x+y)^2 = x^2 +y^2 +2xy........(3) Then, from (1), (2) and (3) (x+y)^2 = 100+44=144. Thus x+y =12. Thus, Perimeter = x+y+10=22 cm.

Areas of Circle is 3.14 x R^2 = 3.14 x 5^2 -> R = 5. So, The circle have diameter 10 cm. So a^2 + b^2 = 100 and ab/2 = 11 -> ab = 22. if (a + b)^2 = a^2 + b^2 + 2ab = 100 + 2.22 = 144, then a + b = 12 and we have c = diameter = 10, so, the perimeter of this triagle is a + b + c = (12 + 10) cm = 22 cm. Answer : 22

Area of the circle=25pi => radius=5=> AB=10. Angle ACB is a right angle as it is an angle on semi circle. Therefore, AC^2+BC^2=AB^2 = > AC^2+BC^2 = 100. Area of the triangle is AC BC /2 = 11 => 2 AC*BC=44. Therefore, (AC+BC)^2=100+44=144 => AC+BC=12. Therefore, the perimeter of the triangle is 12+10=22 cm.

Hmmm..

I did it in a different method,

since we can figure out the orthogonal height from vertex C to the segment AB which is =11/5

imagine another segment of length = r = 5 from vertex C to center of circle

Now this segment is either equal to orthogonal height or has a higher magnitude,

since we didn't make any assumptions based on other criteria

Now taking the cosine = Orthogonal height/r = 11/5*5

$cos^{-1}(11/25)= 64.$

5*sin(64)= distance between point where orthogonal segment touches diameter from center of circle..

$r+r*sin(64)$ = length of line segment from vertex A to orthogonal point.. Let this point be Y

then AY=5+4.5

AY=9.5

AC^2=AY^2 +CY^2

AC^2=9.5^2 + (11/5)^2

AC=9.75

similarly,

BC^2=BY^2+CY^2

Since BY+AY=10

BY=0.5

BC=2.25

Perimeter =BC + AC + AB

=2.25+9.75+10

=22

diameter is equal to 10 cm. and 1/2 h b=11 >> 2 h b=44------(1) and a^2+b^2=100 ------(2) adding 1 and 2 we get (h+b)^2=144 >> h+b+diameter=10+12=22.