Question

A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, then find the cost of making the design at the rate of Rs. 0.35 per sq cm.

Answer 1

Hint: We can observe from the figure that ABCDEF is a hexagon. In the question, it is given that all the designs are equal and we can conclude that the lengths of the sides, AB, BC, CD, DE, EF, FA are equal. So, the hexagon is a regular hexagon. The angles $\angle BOC=\angle COD=\angle DOE=\angle EOF=\angle FOA=\angle AOB$ and the sum of all of them is ${{360}^{\circ }}$, so each of the angle is equal to ${{60}^{\circ }}$. The triangles formed by the radii and each of the above mentioned sides are congruent. Area of the triangle whose lengths of the sides are a, b and the included angle is ${{C}^{\circ }}$ is given by $A=\dfrac{1}{2}ab\sin C$. Let us consider $\Delta OAB$, using this property taking $OA=OB=28$ and $\angle BOA={{60}^{\circ }}$, we get the area of $\Delta OAB$ and multiplying this area by 6, we get the area of the hexagon. We can write that the area of the circle with radius 28 cm is $Area=\pi {{\left( 28 \right)}^{2}}$. By subtracting the area of the hexagon from the area of the circle, we get the area of the shaded region. By multiplying the area of the shaded region with 0.35, we get the answer in rupees.

Complete step-by-step answer:
Let us consider the triangles $\Delta OAB\text{ and }\Delta OAF$, we can write
$OF=OA=OB=28\text{ }cm$(Radius)
$FA=AB$ (Equal design)
We can conclude that the triangles $\Delta OAB\text{ and }\Delta OAF$ are congruent. Likewise all the triangles in the figure are congruent to each other. This means that the hexagon in the figure is a regular hexagon.
We know that the sum of all angles at O is equal to ${{360}^{\circ }}$, we can write that
$\angle BOC=\angle COD=\angle DOE=\angle EOF=\angle FOA=\angle AOB={{60}^{\circ }}\to \left( 1 \right)$
We know the formula of area of the triangle whose lengths of the sides are a, b and the included angle is ${{C}^{\circ }}$ is given by $A=\dfrac{1}{2}ab\sin C\to \left( 2 \right)$.
We can infer from the point that that A, B are points on the circle with centre O and radius 28 cm. We can write that, OA = OB = 28 cm
From equation -1, we can write that $\angle BOA={{60}^{\circ }}$,
From equation-2, we can write the area of the triangle $\Delta OAB$ as
$Area=\dfrac{1}{2}\times 28\times 28\times \sin {{60}^{\circ }}=\dfrac{28\times 28\times \sqrt{3}}{2\times 2}=196\sqrt{3}=196\times 1.73=339.08\text{ c}{{\text{m}}^{2}}\text{ }\left( \sqrt{3}=1.73 \right)$
We know that the hexagon is a regular hexagon and total area will be 6 times the area of$\Delta OAB$
Total area of the hexagon = $6\times Area\left( \Delta OAB \right)=6\times 339.08=2034.48\text{ c}{{\text{m}}^{2}}$
We know that the area of the circle of radius r cm is $Area=\pi {{\left( r \right)}^{2}}$,
In our question, r = 28 cm.
$Area=\pi {{\left( 28 \right)}^{2}}=\dfrac{22}{7}\times 28\times 28=22\times 4\times 28=88\times 28=2464\text{ c}{{\text{m}}^{2}}$
From the figure, we can infer that
Area of the design = Area of the circle – Area of the hexagon
By substituting the calculated values of area of circle and hexagon in the above equation, we get
Area of the design = $2464-2034.48=429.52\text{ c}{{\text{m}}^{2}}$
We can infer from the question that the cost of per sq cm is Rs. 0.35.
We can write the total cost as,
Total cost = Cost per sq cm $\times $ Area in sq cm
Total cost = $0.35\times 429.52=Rs.150.332$
$\therefore $ The cost of the total design is Rs. 150.332.


Note: The alternate way to find the area is by knowing the property of a regular hexagon which is that each of the triangles in the regular hexagon is an equilateral triangle. In our question, each of the triangles is an equilateral triangle with side = 28 cm. Area of an equilateral triangle with side = a units is given by $Area=\dfrac{\sqrt{3}}{4}{{a}^{2}}$ sq units. Using this formula, we get the area of $\Delta OAB$ is $\dfrac{\sqrt{3}}{4}{{\left( 28 \right)}^{2}}=339.08\text{ c}{{\text{m}}^{2}}$.