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DerKrebs [107]
3 years ago
12

I will mark as the brainliest answer

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0
1. each number visual always has a shape in each one. all you do is add one each time.
2. the numbers are shown in shapes inside of shapes.
3. each pattern has one more shape in it than the others.
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Draw two number lines that show 0.200 and 1/5 are equivalent
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Consider the triangles shown.
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DiKsa [7]

\bold{\huge{\blue{\underline{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • <u>The </u><u>right </u><u>angled </u><u>below </u><u>is </u><u>formed </u><u>by </u><u>3</u><u> </u><u>squares </u><u>A</u><u>, </u><u> </u><u>B </u><u>and </u><u>C</u>
  • <u>The </u><u>area </u><u>of </u><u>square </u><u>B</u><u> </u><u>has </u><u>an </u><u>area </u><u>of </u><u>1</u><u>4</u><u>4</u><u> </u><u>inches </u><u>²</u>
  • <u>The </u><u>area </u><u>of </u><u>square </u><u>C </u><u>has </u><u>an </u><u>of </u><u>1</u><u>6</u><u>9</u><u> </u><u>inches </u><u>²</u>

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>area </u><u>of </u><u>square </u><u>A</u><u>? </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u><u> </u></h3>

The right angled triangle is formed by 3 squares

<u>We </u><u>have</u><u>, </u>

  • Area of square B is 144 inches²
  • Area of square C is 169 inches²

<u>We </u><u>know </u><u>that</u><u>, </u>

\bold{ Area \: of \: square =  Side × Side }

Let the side of square B be x

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 144 =  x × x }

\sf{ 144 =  x² }

\sf{ x = √144}

\bold{\red{ x = 12\: inches }}

Thus, The dimension of square B is 12 inches

<h3><u>Now, </u></h3>

Area of square C = 169 inches

Let the side of square C be y

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 169 =  y × y }

\sf{ 169 =  y² }

\sf{ y = √169}

\bold{\green{ y = 13\: inches }}

Thus, The dimension of square C is 13 inches.

<h3><u>Now, </u></h3>

It is mentioned in the question that, the right angled triangle is formed by 3 squares

The dimensions of square be is x and y

Let the dimensions of square A be z

<h3><u>Therefore</u><u>, </u><u>By </u><u>using </u><u>Pythagoras </u><u>theorem</u><u>, </u></h3>

  • <u>The </u><u>sum </u><u>of </u><u>squares </u><u>of </u><u>base </u><u>and </u><u>perpendicular </u><u>height </u><u>equal </u><u>to </u><u>the </u><u>square </u><u>of </u><u>hypotenuse </u>

<u>That </u><u>is</u><u>, </u>

\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}

<u>Here</u><u>, </u>

  • Base = x = 12 inches
  • Perpendicular = z
  • Hypotenuse = y = 13 inches

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ (z)² + (x)² = (y)² }

\sf{ (z)² + (12)² = (169)² }

\sf{ (z)² + 144 = 169}

\sf{ (z)² = 169 - 144 }

\sf{ (z)² = 25}

\bold{\blue{ z = 5 }}

Thus, The dimensions of square A is 5 inches

<h3><u>Therefore</u><u>,</u></h3>

Area of square

\sf{ = Side × Side }

\sf{ = 5 × 5  }

\bold{\orange{ = 25\: inches }}

Hence, The area of square A is 25 inches.

6 0
2 years ago
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