Find the Area of the Shaded Region

Here we will learn how to find the area of the shaded region.

To find the area of the shaded region of a combined geometrical shape, subtract the area of the smaller geometrical shape from the area of the larger geometrical shape.

1. A regular hexagon is inscribed in a circle of radius 14 cm. Find the area of the circle falling outside the hexagon.

Solution:

The given combined shape is combination of a circle and a regular hexagon.

Required area = Area of the circle – Area of the regular hexagon.

To find the area of the shaded region of the given combined geometrical shape, subtract the area of the regular hexagon (smaller geometrical shape) from the area of the circle (larger geometrical shape).

Area of the circle = πr²

                         = $\frac{22}{7}$ × 14² cm².

                         = 616 cm².

Area of the regular hexagon = 6 × area of the equilateral ∆OPQ

                                         = 6 × $\frac{√3}{4}$  × OP²

                                         = $\frac{3√3}{2}$ × 14² cm².

                                         = 294√3 cm².

                                         = 509.21 cm².

Alternate method

Required area = 6 × area of the segment PQM

                     = 6{Area of the sector OPMQ – Area of the equilateral ∆OPQ

                     = 6{$\frac{60°}{360°}$ × πr² - $\frac{√3}{4}$r²}

                     = 6{$\frac{1}{6}$ ∙ $\frac{22}{7}$ ∙ 14² - $\frac{√3}{4}$ × 14²} cm².

                     = (22 × 2 × 14 - 3√3 × 14 × 7) cm².

                     = (616 - 294 × 1.732) cm².

                     = (616 - 509.21) cm².

                     = 106.79 cm².

2. Three equal circles, each of radius 7 cm, touch each other, as shown. Find the shaded area between the three circles. Also, find the perimeter of the shaded region.

Solution:

The triangle PQR is equilateral, each of whose sides is of length = 7 cm + 7 cm, i.e., 14 cm. So, each of the angles SPU, TRU, SQT has the measure 60°.

Area of the ∆PQR = $\frac{√3}{4}$ × (Side)²

                          = $\frac{√3}{4}$ × 14² cm².

Area of each of the three sectors = $\frac{60°}{360°}$ × πr²

                                                = $\frac{1}{6}$ ∙ $\frac{22}{7}$ ∙ 7² cm².

Now, the shaded area = Area of the triangle ∆PQR - Area of the sector ∆SPU - Area of the sector ∆TRU - Area of the sector ∆SQT

                                 = $\frac{√3}{4}$ × 14² cm² – 3 × ($\frac{1}{6}$ × $\frac{22}{7}$ × 7²) cm².

                                 = (49√3 – 77) cm².

                                 = (49 × 1.732 – 77) cm².

                                 = 7.87 cm².

Next, perimeter of the shaded region

                                = Sum of arcs SU, TU and TS, which are equal.

                                 = 3 × arc SU

                                 = 3 × $\frac{60°}{360°}$ × 2πr

                                 = 3 × $\frac{1}{6}$ × 2 × $\frac{22}{7}$ × 7 cm

                                 = 22 cm.

10th Grade Math

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