Question
An archery target has three regions formed by the concentric circles as shown in the figure. If the diameters of the concentric circles are in the ratio 1:2:3, then find the ratio of the areas of three regions.
✓
Answer
Let the diameters of concentric circles be k,
$\therefore$ Radius of concentric circles are $\frac{\text{k}}{2},\text{k}\text{ and }\frac{3\text{k}}{2}.$
$\therefore$ Area of inner circle , $\text{A}_1=\pi\Big(\frac{\text{k}}{2}\Big)^2=\frac{\text{k}^2\pi}{4}$
$\therefore$ Area of middle region,
$\text{A}_2=\pi(\text{k})^2-\frac{\text{k}^2\pi}{4}=\frac{3\text{k}^2\pi}{4}$
$[\therefore\text{area of ring}=\pi(\text{R}^2-\text{r}^2),$ where R is radius of outer and r is radius of inner ring]
and area of outer region, $\text{A}_3=\pi\Big(\frac{3\text{k}}{2}\Big)-\pi\text{k}^2$
$=\frac{9\pi\text{k}^2}{4}-\pi\text{k}^2=\frac{5\pi\text{k}^2}{4}$
$\therefore\text{Required ratio}=\text{A}_1:\text{A}_2:\text{A}_3$
$=\frac{\text{k}^2\pi}{4}:\frac{3\text{k}^2\pi}{4}:\frac{5\pi\text{k}^2}{4}=1:3:5$
$\therefore$ Radius of concentric circles are $\frac{\text{k}}{2},\text{k}\text{ and }\frac{3\text{k}}{2}.$
$\therefore$ Area of inner circle , $\text{A}_1=\pi\Big(\frac{\text{k}}{2}\Big)^2=\frac{\text{k}^2\pi}{4}$
$\therefore$ Area of middle region,
$\text{A}_2=\pi(\text{k})^2-\frac{\text{k}^2\pi}{4}=\frac{3\text{k}^2\pi}{4}$
$[\therefore\text{area of ring}=\pi(\text{R}^2-\text{r}^2),$ where R is radius of outer and r is radius of inner ring]
and area of outer region, $\text{A}_3=\pi\Big(\frac{3\text{k}}{2}\Big)-\pi\text{k}^2$
$=\frac{9\pi\text{k}^2}{4}-\pi\text{k}^2=\frac{5\pi\text{k}^2}{4}$
$\therefore\text{Required ratio}=\text{A}_1:\text{A}_2:\text{A}_3$
$=\frac{\text{k}^2\pi}{4}:\frac{3\text{k}^2\pi}{4}:\frac{5\pi\text{k}^2}{4}=1:3:5$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Write the value of $a_{30} - a_{10}$_ for the A.P. $4, 9, 14, 19, .....$
→Show that the square of any positive integer cannot be of the form $3m + 2$, where m is a natural number.
→In a quadrilateral $ABCD$, $\angle\text{B}=90^\circ,$ $AD^2 = AB^2 + BC^2 + CD^2$, prove that $\angle\text{ACD}=90^\circ.$
→Tiptop Electronics’ supplied an AC of 1.5 ton to a company. Cost of the AC supplied is Rs. 51,200 (with GST). Rate of CGST on AC is 14%. Then find the following amounts as shown in the tax invoice of Tiptop Electronics.
(1) Rate of SGST
(2) Rate of GST on AC
(3) Taxable value of AC
(4) Total amount of GST
(5) Amount of CGST
(6) Amount of SGST
(1) Rate of SGST
(2) Rate of GST on AC
(3) Taxable value of AC
(4) Total amount of GST
(5) Amount of CGST
(6) Amount of SGST
Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions $14\ cm \times 7\ cm$. Find the area of the remaining card board.$\Big(\text{Use }\pi=\frac{22}{7}\Big).$
→If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2+ px + q$, find a quadratic polynomial whose zeroes are:
→- $\alpha+2 ,\beta+2$
- $\frac{\alpha-1}{\alpha+1},\frac{\beta-1}{\beta+1}.$
Show that the following numbers are irrational.
$3-\sqrt{5}$
→$3-\sqrt{5}$
$3 m^2+2 m-7=0$
→Seg AB and seg AD are the chords of the circle. C is a point on tangent of the circle at point A. If m(arc APB) 80° = and $\angle$BAD = 30°. Then find (i) $\angle$BAC (ii) m(arc BQD).
→If $5\text{x}=\sec\theta\text{ and }\frac{5}{\text{x}}=\tan\theta,$ find is the value of $5\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)$.
→