This question was uploaded on 02/09/21 on social media accounts.
Three lines are drawn inside a circle such that one line with length 5 is a chord to the circle. The second line is drawn perpendicular from one end of the chord and having length 6. The third line is drawn perpendicular to the end of the second line and having a length of 4. The end of the third line is the touches circle. Find the radius of the circle.
Solution-1
Step 1:
Extend line having length 4 to the opposite side of the circle.
Extend lane having length 6 to the opposite side of the circle.
Draw a line parallel to line with length 6 from a point on the chord.
BC = DE = 5;
AD = EF = 4;
Step 2:
Step 3:
Puzzle related to Geometry, Circle, Length
Now apply the chord theorem at point D:
(AD)(DF) = (BD)(DG)
⇒ (4)(5 + 4) = (6)(x)
⇒ 36 = 6x
⇒ x = 6
Step 3:
Observe BD = DG and AF is perpendicular on BG,
⇒ AF is perpendicular bisector of BG;
⇒ AF is diameter of circle.
⇒ 2R = 4 + 5 + 4
⇒ 2R = 13
⇒ R = 13/2
Solution-2
Step 1:
Join free end (Intersection with circle) of the line having length 4 to both ends of the chord having length 5.
AB² = AD² + BD²
⇒ AB² = 6² + 4² = 52
⇒ AB = 2√13
Also,
AC² = (AD + BC)² + BD² -----(By forming rectangle as in above solution)
⇒ AC² = 9² + 6² = 117
⇒ AC = 3√13
Consider triangle ABC,
Given circle is circumcircle of triangle ABC
Apply cosine rule on Angle BAC:
\[cosA = \frac{AB^2 + AC^2 + BC^2}{2 \times AB\times BC}\]
\[\Rightarrow cosA = \frac{(2\sqrt{13})^2 + (3\sqrt{13})^2 + 5^2}{2 \times 2\sqrt{13}\times 3\sqrt{13}}\]
\[\Rightarrow cosA = \frac{144}{12\times13} = \frac{12}{13}\]
Now,
\[sinA = \sqrt{1 - cos^2A}\]
\[\Rightarrow sinA = \sqrt{1 - \left(\frac{12}{13}\right)^2}\]
\[\Rightarrow sinA = \frac{5}{13}\]
We know that,
By sine rule:
\[\frac{BC}{sinA} = 2R\]
\[\Rightarrow \frac{5}{\frac{5}{13}} = 2R\]
\[\Rightarrow R = \frac{13}{2}\]
Image Solutions:
Solution-1
Solution-2
