Area of the Shaded Region

We will learn how to find the Area of the shaded region of combined figures.

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.

Solved Examples on Area of the Shaded Region:

1. In the adjoining figure, PQR is a right-angled triangle in which ∠PQR = 90°, PQ = 6 cm and QR = 8 cm. O is the centre of the incircle.

Find the area of the shaded regions. (Use π = \(\frac{22}{7}\))

Solution:

The given combined shape is combination of a triangle and incircle.

To find the area of the shaded region of the given combined geometrical shape, subtract the area of the incircle (smaller geometrical shape) from the area of the ∆PQR (larger geometrical shape).

Required area = area of the ∆PQR – Area of the incircle.

Now, area of the ∆PQR = \(\frac{1}{2}\) × 6 cm × 8 cm = 24 cm^2.

Let the radius of the incircle be r cm.

Clearly, QR = \(\sqrt{PQ^{2} + QR^{2}}\)

                 = \(\sqrt{6^{2} + 8^{2}}\) cm

                 = \(\sqrt{36 + 64}\) cm

                 = \(\sqrt{100}\) cm

                 = 10 cm

Therefore,

Area of ∆OPR = \(\frac{1}{2}\) × r × PR

                    = \(\frac{1}{2}\) × r × 10 cm^2.

Area of ∆ORQ = \(\frac{1}{2}\) × r × QR

                    = \(\frac{1}{2}\) × r × 8 cm^2.

Area of ∆OPQ = \(\frac{1}{2}\) × r × PQ

                     = \(\frac{1}{2}\) × r × 6 cm^2.

Adding these, area of the ∆PQR = \(\frac{1}{2}\) × r × (10 + 8 + 6) cm^2.

                                              = 12r cm^2.

Therefore, 24 cm² = 12r cm^2.

⟹ r = \(\frac{24}{12}\)

⟹ r = 2

Therefore, the radius of the incircle = 2 cm.

So, the area of the incircle = πr²

                                       = \(\frac{22}{7}\) × 2² cm^2.

                                       = \(\frac{22}{7}\) × 4 cm^2.

                                       = \(\frac{88}{7}\) cm^2.

Therefore, the required area = Area of the ∆PQR – Area of the incircle.

                                          = 24 cm² - \(\frac{88}{7}\) cm^2.

                                          = \(\frac{80}{7}\) cm^2.

                                          = 11\(\frac{3}{7}\) cm^2.

2. In the adjoining figure, PQR is an equailateral triangle of side 14 cm. T is the centre of the circumcircle.

Find the area of the shaded regions. (Use π = \(\frac{22}{7}\))

Solution:

The given combined shape is combination of a circle and an equilateral triangle.

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

The required area = Area of the circle – The area of the equilateral triangle PQR.

Let PS ⊥ QR.

In the equilateral triangle SR = \(\frac{1}{2}\) QR

                                          = \(\frac{1}{2}\) × 14 cm

                                          = 7 cm

Therefore, PS = \(\sqrt{14^{2} – 7^{2}}\) cm

                    = \(\sqrt{147}\) cm

Also, in an equilateral triangle, the circumcentre T coincides with the centroid.

So, PT = \(\frac{2}{3}\)PS

          = \(\frac{2}{3}\)\(\sqrt{147}\) cm

Therefore, the circumradius = PT = \(\frac{2}{3}\)\(\sqrt{147}\) cm

Therefore, area of the circle = πr²

                                         = \(\frac{22}{7}\) × \((\frac{2}{3}\sqrt{147})^{2}\) cm^2.

                                         = \(\frac{22}{7}\) × \(\frac{4}{9}\) × 147 cm^2.

                                         = \(\frac{616}{3}\) cm^2.

And area of the equilateral triangle PQR = \(\frac{√3}{4}\) PR²

                                                          = \(\frac{√3}{4}\) × 14² cm^2.

                                                          = \(\frac{√3}{4}\) × 196 cm^2.

                                                          = 49√3 cm^2.

Therefore, the required area = Area of the circle – The area of the equilateral triangle PQR.

                                          = \(\frac{616}{3}\) cm² -  49√3 cm^2.

                                          = 205.33 – 49 × 1.723 cm^2.

                                          = 205.33 – 84.868 cm^2.

                                          = 120.462 cm^2.

                                          = 120.46 cm^2. (Approx).

10th Grade Math

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