Question

Find the area of the following figure​

Answer 1

Answer:

The area of a rectangle is given by the formula l×b, where l is the length and b is the breadth. From the figure, we can observe that the length of rectangle B is 7+3=10 cm. Finally, we will add the areas of the three rectangles to get the area of the given figure. Thus, we get the area of the figure as 60 cm2.

Answer 2

$\huge \underline\mathfrak \color{aqua}❥ Answer ࿐$

$\longrightarrow\large{\boxed{\mathfrak {\color{deeppink}{120 \: {m}^{2} }}}}$

$\huge \underline \mathfrak \color{lime}❥ Explanation ࿐$

☆To find :-

The area of the given figure.

☆Given :-

The length of AB=5m ,BC=12m,CD=10m,DE=5m ,AE=12m

The angle of ABC=90° and DEA=90°

☆Construction:-

Draw the figure and join AC and AD

☆Formula use:-

$\leadsto {\boxed{\sf{ \color{yellow} {Area \: of \triangle =\frac{1}{2} \times base \times height}}}}$

$\leadsto{\boxed{\sf{\color{lightgreen}{Hypotenuse ²=Base²+Height ²}}}}$

$\leadsto\small{\boxed{\sf{\color{orange}{s\sqrt{(s-a)(s-b)(s-c)}}}}}$

$\sf \: where \: \: s \: = \frac{1}{2} (a + b+ c)$

☆Solution:-

In the fig ABC is a right angle triangle ,where AB is the height, BC is the base and AC hypotenuse.

Now,we will calculate the length of AC by Pythagoras Formula.

$\therefore\sf \: AC ^{2} \: = {5}^{2} + {12}^{2} \\ \rightarrow\sf \: AC^{2} = 25 + 144 \: \: \\ \rightarrow\sf \: AC ^{2} = 169 \qquad \quad \\ \rightarrow\sf \: AC = \sqrt{169} \qquad \: \: \\ \rightarrow\sf \: AC = \mathfrak \color{red} 13m \qquad \quad$

Now the length of AD will be equal to AC as the length of height and base of both right angle triangles are same

$\therefore\sf \: AD ^{2} \: = {5}^{2} + {12}^{2} \\ \rightarrow\sf \: AD^{2} = 25 + 144 \: \: \\ \rightarrow\sf \: AD ^{2} = 169 \qquad \quad \\ \rightarrow\sf \: AD = \sqrt{169} \qquad \: \: \\ \rightarrow\sf \: AD = \mathfrak \color{pink}13m \qquad \: \: \: \:$

Now we will find the area of Triangles

$\sf \therefore \:Area \: of \triangle \: ABC = \frac{1}{2} \times 12 \times 5 \\ \qquad \qquad \quad \: \: = 6 \times 5 \\ \qquad \qquad \quad \quad = \: \mathfrak\color{lightgreen}\: 30 {m}^{2}$

Now we will calculate the area of Triangle ACD

$\therefore \sf \: s = \frac{10 + 13 + 13}{2} = \frac{36}{2} = \mathfrak \color{yellow}18 \: m$

$\therefore \sf Area \: of \triangle \:ACD \qquad\qquad \qquad\qquad \\ \rightarrow \sqrt{18(18 - 10)(18 - 13)(18 - 13)} \\ \rightarrow \sqrt{18 \times 8 \times 5 \times 5} \qquad \qquad\qquad \quad \: \: \\ \rightarrow \sqrt{9 \times 2 \times 4 \times 2 \times 5\times 5} \qquad\qquad \\ \rightarrow 3 \times 2 \times 2 \times 5\qquad\qquad\qquad\qquad \: \: \: \\ \rightarrow \color{aqua}60 {m}^{2} \qquad\qquad\qquad\qquad\qquad\qquad$

$\sf \therefore \:Area \: of \triangle \: ADE = \frac{1}{2} \times 5\times 12\\ \qquad \qquad \quad \: \: = 5 \times 6 \\ \qquad \qquad \quad \quad = \: \mathfrak\color{lightgreen}\: 30 {m}^{2}$

$\therefore \:Area \: of \: figure = (30 + 60 + 30) \\ \qquad \qquad \quad \color{deeppink} \rightarrow120 {m}^{2}$

☆Hence:-

Area of the figure is 120 m ²

hope it helps.... :)