Question

pls solve the 3rd part of this qestion ​

Answer 1

Step-by-step explanation:

you can use formula of segment and find areas of 3 segment.

area of segment=theta/360×πr^2 minus 1/2×r^2×sin×theta

then multiple the ans with cost

Answer 2

Step-by-step explanation:

Given:

The figure attached.

To find:

a) The sum of area of two designed segments made by the chord AB and BC.

b) The area of design segment made by chord PQ.

c) Total cost of making designs at the rate of Rs. 15.50 per sq.cm (Take π=3.14)

Solution:

a) The sum of area of two designed segments made by the chord AB and BC.

Ans: ABC is a right triangle, right angle at B

Area of Shaded part: Area of semicircle- Area of ∆ABC

Area of Shaded part:

$=\bold{ \frac{\pi \: {r}^{2} }{2} - \frac{1}{2} \times base \times height }\\ \\ = \frac{1}{2} \pi \: {r}^{2} - \frac{1}{2} \times AB \times CB$

Diameter of Circle= Hypotenuse of triangle ∆ABC

$Diameter = \sqrt{( {4)}^{2} + ( {4)}^{2} } \\ \\ = \sqrt{16 + 16} \\ \\ = \sqrt{32} \\ \\ diameter = 4 \sqrt{2}\: cm$

$= \frac{1}{2} \times 3.14 \times ( {2 \sqrt{2} })^{2} \: - \frac{1}{2} \times 4 \times 4\\ \\ = 3.14 \times 4 - 8 \\ \\ = 4.56$

The area of design segment made by the chord AB and BC is 4.56 cm².

b)The area of design segment made by chord PQ.

Sol: Since radius of circle is 2√2 cm

Area of segment made by right ∆ POQ

= Area of sector- area of ∆POQ

$= \bold{ \frac{90}{360} \pi \times {r}^{2} - \frac{1}{2} \times base \times height} \\ \\ = \frac{1}{4} \times 3.14\times {(2 \sqrt{2} })^{2} - \frac{1}{2} \times 2 \sqrt{2} \times 2 \sqrt{2} \\ \\ = 3.14 \times 2 - 4 \\ \\ = 6.28 - 4 \\ \\ = 2.28 \: {cm}^{2}$

The area of design segment made by chord PQ is 2.28 cm².

c)Total cost :

Total area= 4.56+2.28

=6.84 cm²

Total cost=15.50×6.84=106.02 Rs

Hope it helps you.