Math, asked by adya1896, 5 hours ago

Find the area of the following figure​

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Answered by krishnahandicraft628
0

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.

Answered by ananya4das
0

 \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.... :)

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