An isosceles triangle of sides 13cm, 13cm, 10cm is inscribed in a circle. What is the radius of the circle?
r = 169/24
r=169/24
a=13, b=13cm,c=10cs=a+b+c/2 =13+13+10/2 = 36/2= 18cm.
Area of triangle∆= √s(s-a)(s-b)(s-c)
=√18(18-13)(18-13)(18-10)
=√18×5×5×8
=√3600
= 60cm^2
* R= a×b×c/4×Area of∆
= 13×13×10/ 4×60
= 7.02cm 【Ans】
To find the radius of the circle inscribing the isosceles triangle, we can use a geometric property of such triangles. In an isosceles triangle, the circle inscribed is such that the center of the circle lies on the perpendicular bisector of the base of the triangle.
Let's denote the two equal sides of the triangle as AB = AC = 13 cm, and the base as BC = 10 cm. The perpendicular bisector of BC will pass through the midpoint of BC, let's call it point M.
To find the length of the perpendicular bisector, we can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In this case, we have a right-angled triangle BMC, where BM is the perpendicular bisector (or the height of the triangle) and BC is the base. Since BM is the perpendicular bisector, it divides BC into two equal parts, so MC = MB = 5 cm.
Using the Pythagorean theorem, we can calculate the length of BM:
BM^2 = MC^2 + BC^2
BM^2 = 5^2 + 10^2
BM^2 = 25 + 100
BM^2 = 125
BM = √125
BM = 11.18 cm (rounded to two decimal places)
So, the length of the perpendicular bisector BM is approximately 11.18 cm. Since the center of the inscribed circle lies on BM and is equidistant from the vertices of the triangle, the radius of the circle is equal to BM.
Therefore, the radius of the circle inscribing the isosceles triangle is approximately 11.18 cm.
13*13=169
13-1=12*2=24
169/24=ans