In Fig. 11.17, ABCD is a trapezium with AB || DC, AB = \[18\] cm, DC = \[32\] cm and distance between AB and DC = \[14\] cm. If arcs of equal radii \[7\] cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure. - Noon Academy

In Fig. 11.17, ABCD is a trapezium with AB || DC, AB =

$18$

cm, DC =

$32$

cm and distance between AB and DC =

$14$

cm. If arcs of equal radii

$7$

cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.

Areas Related to Circles | Exercise 11.4 | Maths | NCERT Exemplar

In Fig. 11.17, ABCD is a trapezium with AB || DC, AB =

$18$

cm, DC =

$32$

cm and distance between AB and DC =

$14$

cm. If arcs of equal radii

$7$

cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.

Solution

Given AB =

$18$

cm, DC =

$32$

cm

Given, Distance between AB and DC = Height =

$14$

cm

We know that Area of the trapezium = (

$1/2$

) × (Sum of parallel sides) × Height

=

$(1/2)\times (18+32)\times 14$

=

$350$

$c{{m}^{2}}$

Given AB ∥ DC, ∴

$\angle A+\angle D={{180}^{\circ }}$

and

$\angle B+\angle C={{180}^{\circ }}$

Also given, radius of each arc =

$7$

cm

Therefore,

Area of the sector with central angle as A =

$(1/2)\times (\angle A/180)\times \pi \times {{r}^{2}}$

Area of the sector with central angle as D =

$(1/2)\times (\angle D/180)\times \pi \times {{r}^{2}}$

Area of the sector with central angle as B =

$(1/2)\times (\angle B/180)\times \pi \times {{r}^{2}}$

Area of the sector with central angle as C =

$(1/2)\times (\angle C/180)\times \pi \times {{r}^{2}}$

Hence, Total area of the sectors,

=

$\frac{\angle A}{360}\times \pi \times {{r}^{2}}+\frac{\angle D}{360}\times \pi \times {{r}^{2}}+\frac{\angle B}{360}\times \pi \times {{r}^{2}}+\frac{\angle C}{360}\times \pi \times {{r}^{2}}$

=

$\left( \frac{\angle A+\angle D}{360}\times \pi \times {{r}^{2}} \right)+\left( \frac{\angle B+\angle C}{360}\times \pi \times {{r}^{2}} \right)$

=

$\left( \frac{180}{360}\times \frac{22}{7}\times 49 \right)+\left( \frac{180}{360}\times \frac{22}{7}\times 49 \right)$

=

$77+77$

=

$154$

∴ we will get Area of shaded region = Area of trapezium – (Total area of sectors)

=

$350-154$

=

$196$

$c{{m}^{2}}$

therefore, the area of shaded region is

$196$

$c{{m}^{2}}$

.

i

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