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**Hint:**In part (a), find the area of the square by the formula $A={{\left( side \right)}^{2}}$ and the area of the triangle by $A=\dfrac{1}{2}\times b\times h$. Then subtract the area of the triangle from the area of the square to get the area of the shaded region. In part (b), find the area of the rectangle by the formula $A=l\times b$ and the area of the semi-circle by $A=\dfrac{1}{2}\pi {{r}^{2}}$. Then subtract the area of the semicircle from the area of the rectangle to get the area of the shaded region. In part (c), find the area of the rectangle by the formula $A=l\times b$ and the area of the triangle by $A=\dfrac{1}{2}\times b\times h$. Then subtract the area of the triangle from the area of the rectangle to get the area of the shaded region. In part (d), there are two semi-circles with the same diameter. So, find the area of the complete circle by the formula $A=\pi {{r}^{2}}$ to get the area of the shaded region.

**Complete step-by-step answer:**

$(a)$ The area of the square is given by,

${{A}_{S}}={{\left( side \right)}^{2}}$

Here side = 10 cm,

$\Rightarrow$${{A}_{S}}={{\left( 10 \right)}^{2}}$

Square the term on the right side,

$\Rightarrow$${{A}_{S}}=100c{{m}^{2}}$

The area of the triangle is given by,

$\Rightarrow$${{A}_{T}}=\dfrac{1}{2}\times b\times h$

Here, the base of the triangle, \[b=10cm\] and height of the triangle, $h=2cm$.

$\Rightarrow$${{A}_{T}}=\dfrac{1}{2}\times 10\times 2$

Cancel out the common factors,

$\Rightarrow$${{A}_{T}}=10c{{m}^{2}}$

Thus, the area of the shaded region is,

$\Rightarrow$$A={{A}_{S}}-{{A}_{T}}$

Substitute the values,

$\Rightarrow$$A=100-10$

Subtract the terms,

$\Rightarrow$$A=90c{{m}^{2}}$

**Hence, the area of the shaded region is $90c{{m}^{2}}$.**

$(b) $The area of the rectangle is given by,

${{A}_{R}}=l\times b$

Here the length of the rectangle, $l=10m$ and breadth of the rectangle, $b=7m$.

$\Rightarrow$${{A}_{R}}=10\times 7$

Multiply the term on the right side,

$\Rightarrow$${{A}_{R}}=70{{m}^{2}}$

The area of the semi-circle is given by,

${{A}_{C}}=\dfrac{1}{2}\pi {{r}^{2}}$

Here, the radius of the circle, $r=1.5m$.

$\Rightarrow$${{A}_{C}}=\dfrac{1}{2}\times 3.14\times {{\left( 1.5 \right)}^{2}}$

Cancel out the common factors and square the terms,

$\Rightarrow$${{A}_{C}}=1.57\times 2.25$

Multiply the terms,

$\Rightarrow$${{A}_{C}}=3.5325{{m}^{2}}$

Round-off to two decimals,

$\Rightarrow$${{A}_{C}}=3.53{{m}^{2}}$

Thus, the area of the shaded region is,

$A={{A}_{R}}-{{A}_{C}}$

Substitute the values,

$\Rightarrow$$A=70-3.53$

Subtract the terms,

$\Rightarrow$$A=66.47{{m}^{2}}$

**Hence, the area of the shaded region is $66.47{{m}^{2}}$.**

$(c) $ The area of the rectangle is given by,

${{A}_{R}}=l\times b$

Here the length of the rectangle, $l=32m$ and breadth of the rectangle, $b=18m$.

$\Rightarrow$${{A}_{R}}=32\times 18$

Multiply the term on the right side,

$\Rightarrow$${{A}_{R}}=576{{m}^{2}}$

The area of the triangle is given by,

${{A}_{T}}=\dfrac{1}{2}\times b\times h$

Here, the base of the triangle, \[b=18m\] and height of the triangle, $h=14m$.

$\Rightarrow$${{A}_{T}}=\dfrac{1}{2}\times 18\times 14$

Cancel out the common factors,

$\Rightarrow$${{A}_{T}}=9\times 14$

Multiply the terms,

$\Rightarrow$${{A}_{T}}=126{{m}^{2}}$

Thus, the area of the shaded region is,

$A={{A}_{S}}-{{A}_{T}}$

Substitute the values,

$\Rightarrow$$A=576-126$

Subtract the terms,

$\Rightarrow$$A=450{{m}^{2}}$

**Hence, the area of the shaded region is $450{{m}^{2}}$.**

$(d)$ Since the two semi-circles are of the same diameter. So, the sum of these semi-circles will give a full circle.

The area of the circle is given by,

${{A}_{C}}=\pi {{r}^{2}}$

Here, the radius of the circle, $r=\dfrac{7}{2}cm$.

$\Rightarrow$${{A}_{C}}=\dfrac{22}{7}\times {{\left( \dfrac{7}{2} \right)}^{2}}$

Square the term,

$\Rightarrow$${{A}_{C}}=\dfrac{22}{7}\times \dfrac{49}{4}$

Cancel out the common factors and multiply the terms,

$\Rightarrow$${{A}_{C}}=38.5c{{m}^{2}}$

**Hence, the area of the shaded region is $38.5c{{m}^{2}}$.**

**Note:**Mensuration 2D mainly deals with problems on perimeter and area. The shape is two dimensional, such as triangle, square, rectangle, circle, parallelogram, etc.

Perimeter: The length of the boundary of a 2D figure is called the perimeter.

Area: The region enclosed by the 2D figure is called the area.

Pythagoras Theorem: In a right-angled triangle, \[{{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Height \right)}^{2}}\].

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