Question

Can u help please ...............

Answer 1

Step-by-step explanation:

Given -

• A, B, C and D lies on a circle, Centre O, radius 8 cm.
• AB and CD are tangents to a circle, centre O, radius 4 cm
• ABCD is a rectangle

To Find -

• Distance AE
• Area of the shaded region

As we know that :-

(Hypotenuse)² = (Perpendicular)² + (Base)²

from Pythagoras Theorem

Now,

In ΔOAE,

» (OA)² = (OE)² + (AE)²

» (AE)² = (8)² - (4)²

» (AE)² = 64 - 16

» (AE)² = 48

» AE = √48

• » AE = 4√3 cm

Now,

AD = 2OE

AD = 2×4

• » AD = 8 cm

And

AB = 2AE

AB = 2×4√3

• » AB = 8√3 cm

Now,

Area of rectangle ABCD = l × b

here,

l = length

b = breadth

» 8√3 × 8

» 64√3

substituting √3 = 1.73

» 64 × 1.73

• » 110.72 cm²

And

Area of Circle = πr²

here,

r = radius of the circle

» π(4)²

» 16π

Substituting π = 3.14

» 16 × 3.14

• » 50.24 cm²

Now,

• Area of shaded region = Area of rectangle ABCD - Area of circle

» 110.72 - 50.24

• » 60.48 cm²

Hence,

The Distance AE is 8√3 cm

And

Area of shaded region is 60.48 cm²

Note :-

Figure in the attachment

(NOT TO SCALE)

Answer 2

$\huge \mathfrak{Answer:-}$

$\large\underline\mathrm{The \: distance \: AE \: is \: 8√3cm.}$

$\large\underline\mathrm{Area \: of \: shaded \: region \: is \: 60.48cm².}$

$\large\underline\mathrm{Given:-}$

• A, B, C and D lies on a circle, centre O, radius 8cm.
• AB and CD are tangent to a circle O, radius 4cm.
• ABCD is a rectangle.

$\large\underline\mathrm{To \: find}$

• Distance AE
• Area of the shaded region.

$\large\underline\mathrm{Using}: formula:-}$

• H² = P² + B²

$\large\underline\mathrm{Solution}$

• ∆OAE

$\implies$ OA² = OE² + AE²

$\implies$ AE² = 8² - 4²

$\implies$ AE² = 64 - 16

$\implies$ AE² = 48

$\implies$ AE = √48

$\implies$ AE = 4√3cm

$\large\underline\mathrm{Now,}$

$\implies$ AD = 2OE

$\implies$ AD = 2 × 4

$\implies$ AD = 8cm

$\large\underline\mathrm{And}$

$\implies$ AB = 2AE

$\implies$ AB = 2 × 4√3

$\implies$ AB = 8√3cm

$\large\underline\mathrm{Now,}$

$\large\underline\mathrm{Area \: of \: rectangle \: ABCD \: = \: l \: × \: b}$

$\implies$ 8√3 × 8

$\implies$ 64√3

$\large\underline\mathrm{So, \: √3 \: = \: 1.73cm²}$

$\implies$ 64 × 1.73

$\implies$ 110.72cm²

$\large\underline\mathrm{And,}$

$\large\underline\mathrm{Area \: of \: circle \: = \: πr²}$

$\implies$ π4²

$\implies$ 16π

$\large\underline\mathrm{So, \: π \: = \: 3.14}$

$\implies$ 16 × 3.14

$\implies$ 50.24cm²

$\large\underline\mathrm{Thus,}$

$\large\underline\mathrm{Area \: of \: shaded \: region \: = \: Area \: of \: rectangle \: ABCD \: - \: Area \: of \: circle.}$

$\implies$ 110.72 - 50.24

$\implies$ 60.48cm²

$\large\underline\mathrm{Hence,}$

$\large\underline\mathrm{The \: distance \: AE \: is \: 8√3cm.}$

$\large\underline\mathrm{And,}$

$\large\underline\mathrm{Area \: of \: shaded \: region \: is \: 60.48cm².}$

$\large\underline\mathrm{Hope \: it \: helps \: you \: plz \: mark \: me \: brainlist}$