Verified answer

Solution:

Area of a triangle;

[tex] \sf{ \boxed{\sf{}Triangle \: = \dfrac{1}{2} \times \: base \times time}}[/tex]

[tex]\sf{ Area \: of \:( ∆ODB ) = \dfrac{1}{2} \times \: 3.5\times 2cm² = 3.5 cm²}[/tex]

Therefore Area of Triangle (ODB) is equals to 3.5 cm²

[tex]\sf {Sector\: angle({\theta} )\: = 90° }[/tex]

[tex]\sf{Radius \: of \: sector \: (R) = 3.5 cm}[/tex]

[tex]\sf{OD = 2cm }[/tex]

As we know that, Area of sector ;

[tex] \: \: \: \: \rightarrow \boxed{ \dfrac{\pi \: r ^{2}{ \theta} }{360} }[/tex]

By substituting values ;

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \dfrac{3.14 \times 3.5 \times 3.5 \times 90}{360} cm²\\ \\ = 9.625 cm²} [/tex]

Therefore, Area of the quadrant is equals to 9.625 cm².

[tex]\sf{Shaded \:Area\: = Area\: of \:quadrant\: OACB\: - Area\: of \:( ∆ODB )} [/tex]

[tex]\sf{= 9.635 cm² - 3.5 cm²}\\\\= 6.125cm²[/tex]

[tex]\therefore[/tex] Hence, Area of shaded region is equals to 6.125 cm²

Solution:

We use the formula for area of sector of the circle to solve the problem.

Since quadrant means 1/4th of a complete circle, therefore, the angle at the centre of a quadrant of a circle, θ = 360°/4 = 90°

• Area of the quadrant OACB = 1/4πr²

We can get the area of quadrant OACB with radius r = 3.5 cm

Area of shaded region

[tex]\sf \: Area \: of \: quadrant OACB- Area \: of \: ΔBDO[/tex]

[tex]\sf \: Since ∠BOD = 90°, ΔBDO \: is \: a \: right-angled \: triangle[/tex]

[tex]\sf Area \: of \: triangle = 1/2 × base × height [/tex]

we can find area of ΔBDO with base = OB = 3.5 cm (radius of quadrant) and height = OD = 2 cm

(i) Area of quadrant OACB = 1/4πr

1/4 × 22/7 × (3.5 cm)²

1/4 × 22/7 × 7/2 × 7/2 cm²

[tex]\implies\sf 77/8 cm {}^{2} [/tex]

(ii) In ΔBDO, OB = r = 3.5 cm = 7/2 cm and OD = 2 cm

Area of ΔBDO = 1/2 × base × height

[tex]\implies \sf1/2 × OB × OD[/tex]

1/2 × 7/2 cm × 2 cm

[tex]\implies\sf7/2 cm {}^{2} [/tex]

From figure, it is observed that:

Area of shaded region = Area of Quadrant OACB - Area of ΔBDO

[tex]\implies \sf77/8 cm2 - 7/2 cm {}^{2} [/tex]

[tex]\implies\sf(77 - 28)/8 cm {}^{2} [/tex]

[tex]\implies \sf49/8 cm {}^{2} [/tex] = 6.125cm²

So, the area of shaded region is 6.125cm².