Question

Solve For The Area And Perimeter Of The Swan. Show All Your Work

Answer 1

Answer:

pls upload the clear picture

Answer 2

Concept

The area of a form or figure is the area covered by the shape, whereas the perimeter is the distance covered by the shape's outside boundary.

Given

We are given a figure in the question.

Find

We have to find the area and the perimeter of the given figure.

Solution

Let, side of square be $a.$

Area of $C=16 \mathrm{~cm}^{2}$

$\text{(side)^{2}}=16$

$\text{side}(a)=4 \mathrm{~cm}$

In the figure,

Sides of triangle $A=$side of square $C$

Side of triangle $A=4 \mathrm{~cm}$

Area of triangle $A=\frac{\sqrt{3}}{4} a^{2}$

$\begin{aligned}A&=\frac{\sqrt{3}}{4} \times(4)^{2} \\&=4 \sqrt{3} \mathrm{~cm}^{2}\end{aligned}$

Also, Area of triangle $A=\frac{1}{2} \times\text{ base }\times\text{ height }$

$\begin{aligned}4\sqrt{3}&=\frac{1}{2} \times 4 \times x\\ x&=2 \sqrt{3}\end{aligned}$

Now, For parallelogram $B$,

$\begin{aligned}\text { Breadth } &=2 x \\&=2 \times 2 \sqrt{3} \\&=4 \sqrt{3}\end{aligned}$

Now, for the trapezoid $\mathrm{H}$,

$\begin{aligned}\text{Area of }H &=\frac{1}{2} \times(3+4) \times 2 \\&=7 \mathrm{~cm}^{2}\end{aligned}$

$\begin{aligned}\text{area of the swan }=& \text{ar}(A)+\text{ar}(B)+\text{ar}(C)+\text{ar}(D)+\text{ar}(E)+\text{ar}(F) +\text{ar}(G)+\text{ar}(H) \\=& 4 \sqrt{3}+24+16+45+24+14+17.5+7 \\=& 4 \sqrt{3}+147.5 \\=& 6.93+147.5 \\=& 154.5 \mathrm{~cm}^{2} \approx 155 \mathrm{~cm}^{2}\end{aligned}$

$\begin{aligned}\text{Perimeter of swan }&=(4+4)+(4+7)+4+8+(4+6)+(4+2) +5+(8+6)+4+3 \\&=73 \mathrm{~cm}\end{aligned}$

Now for trapezoid $F$,

$\begin{aligned}\text{parallel sides }&= y\text{ and }y+2\\&=(6\text{ and }8) \mathrm{cm}\end{aligned}$

$\begin{aligned} \text { Perpendicular } &=\text {Side of G }-3 \mathrm{~cm} \\ &=5-3 \\ &=2 \mathrm{~cm} \end{aligned}$

$\begin{aligned}\text{Area of }F&=\frac{1}{2}\times \text{(Sum of parallel sides)} \times \text{height}\\ &=\frac{1}{2}\times \left(6+8\right)\times 2\\ &=14\text{cm}^2\end{aligned}$

Now, for a triangle $G$,

$\begin{aligned}\text { Area of } G &=\frac{1}{2} \times 5 \times 7 \\&=\frac{35}{2} \\&=17.5 \mathrm{~cm}^{2}\end{aligned}$

Now, for trapezoid $H$,

$\begin{aligned}\text { Area of }H &=\frac{1}{2} \times(3+4) \times 2 \\&=7 \mathrm{~cm}^{2}\end{aligned}$

$\begin{aligned}\text{area of swan }=& \text{ar}(A)+\text{ar}(B)+\text{ar}(C)+\text{ar}(D)+\text{ar}(E)+\text{ar}(F) +\text{ar}(\mathrm{G})+\text{ar}(H)\\&=4 \sqrt{3}+24+16+45+24+14+17.5 \\&=4 \sqrt{3}+147.5 \\&=6.93+1475 \\&=154.5 \mathrm{~cm}^{2} \approx 155 \mathrm{~cm}^{2}\end{aligned}$

$\begin{aligned}\text{Perimeter of swan }&=\frac{(4+4)+(4+7)+4+8+(4+6)+(4+2)}{+5+(8+6)+4}+3 \\&=73 \mathrm{~cm}\end{aligned}$

Hence, the area of swan is $155cm^2$ and the perimeter of swan is $73cm.$

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