Thus, ¯ AB must be a diameter of the circle, and so the center O of the circle is the midpoint of ¯ AB. Now the area of triangle AOB OA OB AB r (both using formula bh). Here, r is the radius that is to be found using a and, the diagonals whose values are given. The side of rhombus is a tangent to the circle. Note that since R 2.5, the diameter of the circle is 5, which is the same as AB. Circle inscribed in a rhombus touches its four side a four ends. The only advantage of the above approach is its generality: any similar problem on rhombi can be undertaken with this parameter $\theta$. So as we see from Figure 2.5.3, sinA 3 / 5. For a general radius $R$, the final answer isįinal remark: I recognize that my presentation can surely be ranked as "overkill" (the particularity of the 60° angle can be used for a splitting of the rhombus into 2 equilateral triangles, yielding a direct solution, as remarked in comments). (the circumference of circle should touch the edges of the rhombus) AC and BD are. $$L^2:=length(AB)^2 =\frac$$įor a unit radius. Question: The circle with centre E is inscribed inside rhombus ABCD. Indeed, taking into account the different symmetries present in the figure (with respect to horizontal, vertical and diagonal axes), we can restrict our attention to tangent lines with slopes $\ge 1$ "crossing" the first quadrant $x>0, \ y>0$.Īs a consequence of (1), such tangent lines intersect coordinate axes in $A(1/\cos \theta,0)$ and $B(0,1/\sin \theta)$ giving (Pythagoras): We can assume without loss of generality that Let us call $B$ the upper vertex of the rhombus, and still consider the general case but with the unit circle $(R=1)$ the equation of the tangent line touching this circle in point $(\cos \theta, \sin \theta)$ is $$x \cos \theta y \sin \theta = 1$$ Let us first consider the general case : for a given circle with radius $R$, there exists rhombi having it as an inscribed circle with side lengths $L$ varying from $2R$ (the case of a square: black lines) to $\infty$: The distance from the centre of the circle to the. If P is any point on the circle, then value of P A 2 P B 2 P C 2 P D 2 will be Can something hint the starting approach for this question. Click hereto get an answer to your question A circle is inscribed into a rhombus ABCD with one angle 60 o. The distance from centre of circle to the nearest vertex is 1. I am indebted to who has pointed a misunderstanding of the question. touches all four sides) into rhombus ABCD with one angle 60 degree.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |