A triangular board has sides 9 cm, 10 cm, and 17 cm. A circle passes through all three vertices. Find the circumradius.
10.625 cm
Using Heron's formula, the semi-perimeter is \(s = \frac{9 + 10 + 17}{2} = 18\).
The area is \(\text{Area} = \sqrt{18 \times 9 \times 8 \times 1} = \sqrt{1296} = 36\) square cm.
The circumradius formula is \(R = \frac{a \times b \times c}{4 \times \text{Area}}\).
Substituting the values gives \(R = \frac{9 \times 10 \times 17}{4 \times 36} = \frac{1530}{144} = 10.625\) cm.
Hence, the circumradius of the triangle is 10.625 cm.
Important Questions from Geometry
ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?
If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:
If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.
If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?
D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.
