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Rus_ich [418]
3 years ago
11

Solve for the area using heron´s formula

Mathematics
2 answers:
Sveta_85 [38]3 years ago
6 0

Given =>

<em><u>A </u></em><em><u>figure</u></em><em><u> </u></em><em><u>ABCD </u></em><em><u>with</u></em><em><u> </u></em><em><u>two </u></em><em><u>adjacent</u></em><em><u> </u></em><em><u>triangles</u></em><em><u> </u></em><em><u>i.e </u></em><em><u>ABC </u></em><em><u>&</u></em><em><u>ACD</u></em>

To Acquire =>

<em><u>The </u></em><em><u>Area </u></em><em><u>of </u></em><em><u>the </u></em><em><u>figure</u></em>

Points to know while solving this problem=>

  • <em><u>Area of a ∆ According to heron = </u></em>
  • <em><u>[</u></em><em><u>tex] \sqrt{s(s - a)(s - b)(s - c)} [/tex]</u></em>
  • <em><u>Where, s= semi-perimeter </u></em>
  • <em><u> a= length of side a </u></em>
  • <em><u> b = length of side b </u></em>
  • <em><u> с = length of side c</u></em>
  • <em><u>Perimeter of a ∆ = sum of it's all 3 sides</u></em>
  • <em><u>The area of the Given Figure would be Area of ∆ABC Area of ∆ACD</u></em>

Step-by-step explanation:

In ∆ABC,

Perimeter of the ∆

= mAB +mBC+mAC

Where,

mAB =20cm

mAC = 16cm

mBC= ?

<em>Also</em>

mBC

= <u>Base</u> of the ∆ ABC where AB = <u>perpendicular</u> & AC = <u>Hypotenuse</u>

So

mBC = √AC²-AB².......(Using Pythagoras theorem)

mBC = √20²-16²

mBC =√400-256

mBC =√144

mBC = 12.....( The square root of 144 = 12)

Hence,

The perimeter of ∆

= mAB +mBC+mAC

= 16+12+20

= 48cm

Semiperimeter = 48/2

= > 24 cm

<em><u>.°.</u></em>

<em><u>The </u></em><em><u>area </u></em><em><u>of </u></em><em><u>∆</u></em><em><u> </u></em><em><u>ABC </u></em><em><u>according</u></em><em><u> </u></em><em><u>to </u></em><em><u>heron's</u></em><em><u> </u></em><em><u>formula</u></em>

<em><u>=</u></em><em><u>></u></em>

<em><u>\sqrt{24(24 - 16)(24 - 12)(24 - 20)}  \\  =  >  \sqrt{24 \times 8 \times 12 \times 4 }  \\  =  >  \sqrt{12 \times 2 \times 8 \times 12 \times 4}  \\  =  >  \sqrt{12 \times 12 \times 2 \times 4 \times 8}  \\  =  >  \sqrt{12 \times 12 \times 8 \times 8}  \\  =  > 12 \times 8 \\  =  > 96cm {}^{2}</u></em>

<em><u>Now,</u></em>

<em><u>In </u></em><em><u>∆</u></em><em><u>A</u></em><em><u>C</u></em><em><u>D</u></em><em><u>,</u></em>

<em><u>The </u></em><em><u>perimeter</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>mAC+</u></em><em><u>mCD+</u></em><em><u>mAD</u></em>

<em><u>where</u></em><em><u>,</u></em>

<em><u>mAC</u></em><em><u>=</u></em><em><u>1</u></em><em><u>6</u></em><em><u>c</u></em><em><u>m</u></em>

<em><u>mCD</u></em><em><u>=</u></em><em><u>2</u></em><em><u>3</u></em><em><u>c</u></em><em><u>m</u></em>

<em><u>mAD=</u></em><em><u>2</u></em><em><u>0</u></em><em><u>cm</u></em>

<em><u>=</u></em><em><u>></u></em><em><u> </u></em>

<em><u>16cm + 23cm + 20cm \\  =  > 59cm</u></em>

<em><u>So.</u></em><em><u>.</u></em><em><u>.</u></em>

<em><u>Semiperimeter</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>5</u></em><em><u>9</u></em><em><u>/</u></em><em><u>2</u></em><em><u>c</u></em><em><u>m</u></em>

<em><u>=</u></em><em><u>></u></em><em><u> </u></em><em><u>2</u></em><em><u>9</u></em><em><u>.</u></em><em><u>5</u></em>

<em><u>Area </u></em><em><u>of </u></em><em><u>the </u></em><em><u>Same</u></em><em><u>∆</u></em><em><u> </u></em><em><u>using</u></em><em><u> </u></em><em><u>Heron's</u></em><em><u> formula</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>5</u></em><em><u>6</u></em><em><u> </u></em><em><u>.</u></em><em><u>9</u></em><em><u>7</u></em><em><u>c</u></em><em><u>m</u></em><em><u>²</u></em>

<em><u>Hence,</u></em>

<em><u>The </u></em><em><u>area </u></em><em><u>of </u></em><em><u>the </u></em><em><u>figure</u></em>

<em><u>=</u></em><em><u>></u></em><em><u>9</u></em><em><u>6</u></em><em><u>c</u></em><em><u>m</u></em><em><u>²</u></em><em><u>+</u></em><em><u>1</u></em><em><u>5</u></em><em><u>6</u></em><em><u>.</u></em><em><u>9</u></em><em><u>7</u></em><em><u>c</u></em><em><u>m</u></em><em><u>²</u></em>

<h2><em><u>=</u></em><em><u>></u></em><em><u>2</u></em><em><u>5</u></em><em><u>2</u></em><em><u>.</u></em><em><u>9</u></em><em><u>7</u></em><em><u> </u></em><em><u>cm²</u></em></h2>

Vadim26 [7]3 years ago
4 0

9514 1404 393

Answer:

  252.8 cm²

Step-by-step explanation:

The missing side of the right triangle can be found from the Pythagorean theorem:

  s² = 20² -16² = 400 -256 = 144

  s = 12 . . . . cm

The area of a right triangle is more easily found using the traditional area formula:

  A = 1/2bh

  A = 1/2(12 cm)(16 cm) = 96 cm² (left-side triangle)

The area of the triangle on the right can be found from Heron's formula. The semiperimeter is ...

  s = (16 +20 +23)/2 = 29.5

The area is ...

  A = √(29.5(29.5 -16)(29.5 -20)(29.5 -23)) = √(29.5·13.5·9.5·6.5)

  A = √24591.9375 ≈ 156.818 . . . . . cm² (right-side triangle)

Then the total area of the figure is ...

  A = 96 cm² +156.818 cm² = 252.818 cm² . . . . total area

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