The area of an equilateral triangle ABC is 17320.5 cm2 . With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205) ?

byAmlesh Mahato •October 31, 2021

The Biotic World is the best free Educational Website for Students. The Biotic World Provides NCERT Solution PDF, CBSE Notes PDF, NCERT Books PDF for class 8, 9, 10, 11, 12. The Biotic World also provides GU TDC Syllabus, Previous Years Question Papers, TDC notes on different topics of Physics, Chemistry, Mathematics, Zoology and Botany any many more.

So, Follow us to get tuned with latest updates.

The area of an equilateral triangle ABC is 17320.5 cm2 . With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205) ?

Solution:

⇒Given

Area of Equilateral Triangle = 17320.5 $\large cm^{2}$

$Area of Equilateral Triangle = 17320.5 \large cm^{2}$

$The area of an equilateral triangle ABC is 17320.5 cm2 . With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205) ?$

$The area of an equilateral triangle ABC is 17320.5 cm2 . With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205) ?$

$The area of an equilateral triangle ABC is 17320.5 cm2 . With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205) ?$

$\large \Rightarrow a^{2}= 10000\times 4$

$\large \Rightarrow a^{2}= 40000$

$\large \Rightarrow a^{2}= 200^{2}$

$\large \Rightarrow a= 200 cm$

∴ Radius of each circle, r = 200/2 = 100 cm

⇒In Equilateral triangle each angle, 𝛉 = 60°

∴ Angle of Sector of circle, 𝛉 = 60°

⇒Area of shaded region

= Area of Equilateral Triangle - 3 x Area of Sector

$= Area of Equilateral Triangle - 3 x Area of Sector$

$The area of an equilateral triangle ABC is 17320.5 cm2 . With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205) ?$

$\large =17320.5-\frac{3\times 3.14\times 10000}{6}$

$\large =17320.5-\frac{3.14\times 10000}{2}$

$\large =17320.5-\frac{31400}{2}$

$\large =17320.5-15700$

$\large =1620.5 cm^2$

Exercise:12.3

1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the Centre of the circle.

2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with Centre O are 7 cm and 14 cm respectively and ∠ AOC = 40°.

3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

4. Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as Centre.

5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.

6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.

7. In Fig. 12.25, ABCD is a square of side 14 cm. With canters A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

8. Fig. 12.26 depicts a racing track whose left and right ends are semi circular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :

(i) the distance around the track along its inner edge

(ii) the area of the track.

9. In Fig. 12.27, AB and CD are two diameters of a circle (with center O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

10. The area of an equilateral triangle ABC is 17320.5 cm2 . With each vertex of the triangle as Centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.

12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region.

13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and Centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.

15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

16. Calculate the area of the designed region inFig. 12.34 common between the two quadrantsof circles of radius 8 cm each.

Tags: Chapter-12 Exercise-12.3 Math- Class 10