An isosceles ΔMNP is inscribed in a circle. If MN =MP = $16\sqrt{5}$ cm, and NP = 32 cm, what is the radius (in cm) of the circle ?
Answer & explanation
Correct answer: option 1
We know that,
Area of a triangle = \(\frac{1}{2}\) × Base × Height
Circumradius = \(\frac{ ABC }{4Δ }\)
Δ = area of the triangle
MN = MP = 16\(\sqrt {5}\) cm
NP = 32 cm
MD is perpendicular to NP
ND = DP = \(\frac{32}{2}\) = 16 cm
MD = \(\sqrt {(16\sqrt {5})^2 – (16)^2}\) = \(\sqrt {1280 - 256}\) = 32 cm
Area of ΔMNP = \(\frac{1}{2}\) × 32 × 32 = 512 cm2
Circumradius = (16\(\sqrt {5}\) × 16\(\sqrt {5}\) × 32)/(4 × 512) = 20 cm