The area of a circle is \(5\frac{2}{15}\) times the sum of the areas of two triangles, one with sides 20 cm, 21 cm and 29 cm, and the other with sides 11 cm, 60 cm and 61 cm. What is the radius of the circle? (Take \(\pi=\frac{22}{7}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(21\sqrt{2}\) cm
A triangle whose sides satisfy \(a^2+b^2=c^2\) is right-angled, and its area is half the product of the two shorter sides.
\(20^2+21^2=400+441=841=29^2\), so the first triangle is right-angled
Its area \(\displaystyle =\frac{1}{2}\times20\times21=210\) cm2
\(11^2+60^2=121+3600=3721=61^2\), so the second triangle is right-angled
Its area \(\displaystyle =\frac{1}{2}\times11\times60=330\) cm2
Sum of the areas \(=210+330=540\) cm2
Area of the circle \(\displaystyle =5\frac{2}{15}\times540=\frac{77}{15}\times540=2772\) cm2
\(\displaystyle \frac{22}{7}r^2=2772\), so \(\displaystyle r^2=2772\times\frac{7}{22}=882\)
\(r=\sqrt{882}=\sqrt{441\times2}=21\sqrt{2}\) cm
Why the other options are incorrect
Option (1) 21 cm — Incorrect. It gives \(r^2=441\), an area of 1386 cm2, which is half the required 2772 cm2.
Option (3) \(21\sqrt{3}\) cm — Incorrect. It gives \(r^2=1323\), an area of 4158 cm2, which is too large.
Option (4) 42 cm — Incorrect. It gives \(r^2=1764\), an area of 5544 cm2, twice the required area.