Question

In figure is a chord AB of a circle, with centre O and radius 10 cm, that subtends a right angle at the centre of the circle . Find the area of the minor segment AQBP. Hence ,find the area of major segment ALBQA

Answer 1

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Answer 2

Area of the minor segment AQBPA is $28.54cm^2$.

Area of major segment ALBQA is $50(1+ \sqrt{2} )cm^2$.

Step-by-step explanation:

Given:

AB is a chord of a circle having centre O and radius 10 cm.

Also a chord subtends a right angle at the centre of the circle.

To find:

(1) Area of the minor segment AQBP.

(2) Area of major segment ALBQA

Now,

∠ALB=  $\frac{1}{2}$ ∠AOB=  $\frac{90}{2}$ = 45°

LQ is perpendicular bisection on AB. Hence by isoceles triangles property.

LA = LB

OB = 10cm and OA = 10cm      [ radius ]

In Δ AOB,

By pythogpras theorem,

$AB^2=OA^2+OB^2$

$AB=\sqrt{10^{2} +10^{2} }$

$AB=\sqrt{100 +100}$

$AB=\sqrt{200}$

$AB=\sqrt{2 \times 100}$

$AB = 10\sqrt{2} cm$

Also, $QB=\frac{AB}{2}$

$QB=\frac{10\sqrt{2}}{2}$

$QB= 5\sqrt{2} cm$

In Δ OQB,

$OB^2=OQ^2+BQ^2$

$OQ^2=OB^2-BQ^2$

$OQ=\sqrt{OB^2-BQ^2}$

$OQ=\sqrt{100-50}$

$OQ=\sqrt{50}$

$OQ=\sqrt{25\times 2}$

$OQ=5\sqrt{2} cm$

So, LQ = LO + OQ = $(10+5 \sqrt{2} ) cm$         [given LO = radius = 10 cm]

(1) Area of AQBPA = $\frac { { \pi r }^{ 2 } }{ 4 } -Ar.ofAOB$

= $\frac{\pi \times 10^2}{4} - \frac{1}{2} \times 10^2$

= $\frac{\frac{22}{7} \times 100}{4} - \frac{1}{2} \times 100$

=  ${\frac{22}{7} \times 25 - 50$

= $28.54cm^2$

Therefore area of the minor segment AQBPA is $28.54cm^2$

Area of ALBQA = $\frac { 1 }{ 2 } \times AB\times LQ$

= $\frac { 1 }{ 2 } \times 10\sqrt { 2 } \times ( 10+5\sqrt { 2 } )$

= $50(1+ \sqrt{2} )cm^2$

Therefore area of major segment ALBQA is $50(1+ \sqrt{2} )cm^2$