{"id":112,"date":"2021-11-24T17:22:25","date_gmt":"2021-11-24T17:22:25","guid":{"rendered":"https:\/\/visualfractions.com\/blog\/?p=112"},"modified":"2023-02-23T17:08:46","modified_gmt":"2023-02-23T17:08:46","slug":"area-of-a-square","status":"publish","type":"post","link":"https:\/\/visualfractions.com\/blog\/area-of-a-square\/","title":{"rendered":"Area of a Square"},"content":{"rendered":"\n

The area of a square is an important concept in geometry. It allows us to understand the properties of a square even further and also opens a wide range of applications and problems involving squares. In the past, we\u2019ve learned that squares are quadrilaterals with equal sides, so it\u2019s time to establish the rules for finding the area of a square.<\/p>\n\n\n\n

There are different ways to calculate the area of a square such as: 1) counting one-unit squares that make up the larger square, 2) using the square\u2019s sides, and 3) utilizing the square\u2019s diagonals. This article covers all these methods and breaks down the formulas for each method.<\/p>\n\n\n\n

This discussion covers the fundamentals of a square and its area, the important dimensions and formulas we can use to find its area, and word problems involving areas of squares. By the end of this discussion, we want you to feel confident when dealing with a square\u2019s area.<\/p>\n\n\n\n

What is the Area of a Square?<\/h2>\n\n\n\n

The area of a square represents the amount of area covered by a given surface enclosed by a square region. As a refresher, a square is a four-sided figure that has four identical sides. Squares also have opposite sides that are parallel and corners that form right triangles. The area of a square has defined as the product of its length and width and is measured in squared units.<\/p>\n\n\n\n

The images below are examples of square-shaped surfaces. The measure of the region that covers these surfaces represents their areas.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

For each surface, we can define its area by counting the smaller squares found within, the square\u2019s dimensions, or even its diagonals \u2013 depending on what\u2019s given. This leads to a wide range of applications and problems that we can finally solve after learning about the area of a square.<\/p>\n\n\n\n

Understanding the Area of a Square<\/h2>\n\n\n\n

Before establishing the formula for the area of square, let\u2019s first understand how we can visualize it. Suppose that we have a [katex] 4\\text{ cm} \\times 4\\text{ cm} [\/katex] square, we can illustrate it using unit squares measuring [katex] 1\\text{ cm} \\times 1\\text{ cm} [\/katex] and has an area of [katex] 1\\text{ cm}^2[\/katex].<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The image above shows that we can represent [katex] 4\\text{ cm} \\times 4\\text{ cm} [\/katex] square as a region filled by sixteen [katex] 1\\text{ cm} \\times 1\\text{ cm} [\/katex] squares. This means that the area of the square is simply equal to [katex]16 \\times 1 \\text{cm}^2 = 16\\text{cm}^2[\/katex].\u00a0 A faster way of counting the squares is to multiply the number of squares in each row and in each column. Extending this to find the area of a square, simply multiply the dimensions of the square. This leads to a faster way to find areas of squares given their sides\u2019 lengths.<\/p>\n\n\n\n

The Formula for the Area of the Square Using Its Side<\/h2>\n\n\n\n

Now that we\u2019ve broken down the idea behind the area of a square, let\u2019s establish a general formula for it. Since squares have four equal sides, a square\u2019s length and width will always be equal. This means that when given the length of the square\u2019s side, its area is equal to the product of the side\u2019s length and itself or the square of the side\u2019s length.<\/p>\n\n\n\n

[katex]\\text{Area of a Square} = \\text{Side} \\times \\text{Side} \\text{ unit}^2[\/katex]<\/p>\n\n\n\n

[katex]\\text{Area of a Square} = \\text{Side} \\times \\text{Side} \\text{ unit}^2[\/katex]<\/p>\n\n\n\n

[katex] \\begin{aligned}\\text{Area of a Square} &= \\text{Side} \\times \\text{Side} \\text{ unit}^2\\\\ &= \\text{Side}^2 \\text{ unit}^2\\\\A &= s^2\\\\\\end{aligned} [\/katex]<\/p>\n\n\n\n

  • A = area of the square<\/li>
  • s = length of the side<\/li><\/ul>\n\n\n\n
    Square Dimensions<\/th>Area of Square ([katex]A =s^2 [\/katex])<\/th><\/tr><\/thead>
    [katex] 2\\text{ in}\\times 2\\text{ in} [\/katex]<\/td>[katex] \\begin{aligned}A &= 2^2 \\text{ in}^2\\&= 4 \\text{ in}^2\\end{aligned} [\/katex]<\/td><\/tr>
    [katex] 5\\text{ cm}\\times 5\\text{ cm} [\/katex]<\/td>[katex] \\begin{aligned}A &= 5^2 \\text{ in}^2\\&= 25 \\text{ cm}^2\\end{aligned} [\/katex]<\/td><\/tr>
    [katex] 8\\text{ km}\\times 8\\text{ km} [\/katex]<\/td>[katex] \\begin{aligned}A &= 8^2 \\text{ km}^2\\&= 64 \\text{ km}^2\\end{aligned} [\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    The formula is easy to apply, so give it a try and work on the problems we\u2019ve prepared for you to get used to the process. Let\u2019s now see what happens if we\u2019re given the square\u2019s diagonal instead. This requires an understanding of radical numbers and the Pythagorean theorem, so if this hasn\u2019t been discussed yet, head over to the next section to try out different examples!<\/p>\n\n\n\n

    The Formula for the Area of the Square Using Its Diagonal<\/h2>\n\n\n\n

    When given the diagonal of a square, [katex]d[\/katex], use the Pythagorean theorem to find the formula for the square\u2019s area. Through special triangle, we can establish that [katex]d = \\sqrt{2}s[\/katex]. Squaring this equation will lead to the formula for the square\u2019s area in terms of [katex]d[\/katex].<\/p>\n\n\n\n

    [katex]\\begin{aligned}d^2 &= (\\sqrt{2}s)^2\\\\d^2 &= 2s^2\\\\\\dfrac{1}{2}d^2 &= s^2\\\\A&= \\dfrac{d}{s^2}\\end{aligned}[\/katex]<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    This simply means that we can find the area of the square by squaring its diagonal\u2019s length and dividing the result by [katex]2[\/katex]. In the next section, we\u2019ll learn how to decide which approach would be best depending on the given dimensions and figures.<\/p>\n\n\n\n

    How To Find the Area of a Square?<\/h2>\n\n\n\n

    Summarizing what we\u2019ve learned so far, we can find the area of a square in three ways:<\/p>\n\n\n\n

    • By counting the unit squares that make up the larger square then multiplying the number of squares by 1 squared unit.<\/li><\/ul>\n\n\n\n

      [katex]\\begin{aligned}\\text{Area} &= \\text{Number of Squares} \\times 1 \\text{ unit}^2\\end{aligned} [\/katex]<\/p>\n\n\n\n

      • When given the side of the square, square its length to find the area of the square.<\/li><\/ul>\n\n\n\n

        [katex]\\begin{aligned}\\text{Area}&= \\text{Side}^2 \\text{ unit}^2\\end{aligned} [\/katex]<\/p>\n\n\n\n

        • If we have the length of the square\u2019s diagonal, we can simply square it then take the half to find the square\u2019s area.<\/li><\/ul>\n\n\n\n

          [katex]\\begin{aligned}\\text{Area}&= \\dfrac{1}{2}(\\text{Diagonal})^2 \\text{ unit}^2\\end{aligned} [\/katex]<\/p>\n\n\n\n

          Let\u2019s try out different examples to know the three methods by heart and work on different problems that involve the square\u2019s area.<\/p>\n\n\n\n

          Problem 1<\/h3>\n\n\n\n

          We begin with finding the area of the square shown below given that one-unit square has an area of [katex]1\\text{ inch}^2[\/katex].<\/p>\n\n\n\n

          \"\"<\/figure>\n\n\n\n

          Count the total number of unit squares by multiplying the number of unit squares found in one row and one column. The square is a [katex]6 \\times 6[\/katex] square, so it has six unit squares in its row and column. Its area is equal to [katex]6 \\times 6 = 36[\/katex] times the area of one unit square.<\/p>\n\n\n\n

          [katex] \\begin{aligned}A &= 36 \\times 1\\text{ in}^2\\&= 36 \\text{ in}^2\\end{aligned} [\/katex]<\/p>\n\n\n\n

          This means that the square has an area of [katex]36[\/katex] squared inches or [katex]36\\text{ in}^2 [\/katex].<\/p>\n\n\n\n

          Problem 2<\/h3>\n\n\n\n
          \"\"<\/figure>\n\n\n\n

          Now, let\u2019s try to find the area of the square floor that has a side of 40m. Since we\u2019re given the measure of its side, we simply square its value to find the square\u2019s area in squared meters.<\/p>\n\n\n\n

          [katex]\\begin{aligned}A &= s^2\\&= (40)^2 \\text{ m}^2\\&= 1600 \\text{ m}^2\\end{aligned}[\/katex]<\/p>\n\n\n\n

          This shows that the area of the square floor is equal to 1600 squared meters or 1600 m2<\/sup> [\/katex].<\/p>\n\n\n\n

          Problem 3<\/h3>\n\n\n\n

          Let\u2019s now try calculating the area of the square shown below. As we can see, what\u2019s given now is the diagonal of the square, so we\u2019ll use the third formula we\u2019ve discussed to find its area.<\/p>\n\n\n\n

          \"\"<\/figure>\n\n\n\n

          [katex]\\begin{aligned}A &= \\dfrac{1}{2}d^2\\&= \\dfrac{1}{2}(12)^2 \\text{ cm}^2\\&= \\dfrac{144}{2} \\text{ cm}^2\\&= 72 \\text{ cm}^2\\end{aligned}[\/katex]<\/p>\n\n\n\n

          Hence, the square\u2019s area is equal to [katex]72\\text{ cm}^2 [\/katex].<\/p>\n\n\n\n

          Problem 4<\/h3>\n\n\n\n

          Now, let\u2019s see what happens when we\u2019re given the square\u2019s perimeter instead. Suppose that a square field has a perimeter of 60 yards. What is the field\u2019s area?<\/p>\n\n\n\n

          \"\"<\/figure>\n\n\n\n

          Recall that the perimeter of a square is equal to four times its side\u2019s length, so the square\u2019s side will be equal to the perimeter divided by four.<\/p>\n\n\n\n

          [katex]\\begin{aligned}P &= 4s\\\\s&= \\dfrac{P}{4} = \\dfrac{60}{4} = 15\\end{aligned} [\/katex]<\/p>\n\n\n\n

          The field has a side length of 15 yards, so we can now find the field\u2019s area by squaring the length of its side.<\/p>\n\n\n\n

          [katex] \\begin{aligned}A &=s^2\\&= (15)^2 \\text{ yards}^2\\&= 225 \\text{ yards}^2\\end{aligned} [\/katex]<\/p>\n\n\n\n

          Hence, the square field has an area of [katex]225 \\text{ yards}^2[\/katex].<\/p>\n\n\n\n

          The fourth problem is an example of how areas of squares and the corresponding formulas are used to solve word problems. When a surface is involved in a problem, there\u2019s a high chance that areas are involved. In the next example, we\u2019ll show you a problem involving square walls!<\/p>\n\n\n\n

          Problem 5<\/h3>\n\n\n\n

          Allison plans to redesign her room and she begins by redoing her square walls\u2019 wallpaper. Each wall has a side length of [katex]8 \\text{ feet}[\/katex] and it costs [katex]\\$1.20[\/katex] per squared feet for the new wallpaper. How much will it cost Allison to redo all four walls?<\/p>\n\n\n\n

          \"\"<\/figure>\n\n\n\n

          To find the cost for one wall, calculate the square wall\u2019s area then multiply it with the cost it takes per squared foot. Since the wall has a side length of [katex]8\\text{ feet}[\/katex], so take the square of the length to find its area.<\/p>\n\n\n\n

          [katex] \\begin{aligned}A &=s^2\\&= (8)^2 \\text{ feet}^2\\&= 64\\text{ feet}^2\\end{aligned} [\/katex]<\/p>\n\n\n\n

          Since it costs [katex]\\$1.20[\/katex] per squared foot for the new wallpaper, multiply the wall\u2019s area by [katex]\\$1.20[\/katex].<\/p>\n\n\n\n

          [katex]\\begin{aligned}\\text{Cost} &= (\\$ 1.20 \/\\text{ft}^2)(64\\text{ ft}^2)\\&= \\$76.80\\end{aligned}[\/katex]<\/p>\n\n\n\n

          It will cost Allison [katex]\\$76.80[\/katex] to redo the wallpaper for each wall, so multiply the cost by [katex]4[\/katex] to find the total cost for redoing the wallpaper for all four walls.<\/p>\n\n\n\n

          [katex]\\begin{aligned}\\text{Total Cost} &= 4(\\$76.80)\\&= \\$307.20\\end{aligned} [\/katex]<\/p>\n\n\n\n

          This means that the total cost to redo the wallpaper is [katex]\\$307.20[\/katex].<\/p>\n","protected":false},"excerpt":{"rendered":"

          The fundamentals of a square and its area, the important dimensions and formulas we can use to find its area, and word problems involving areas of squares.<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"class_list":["post-112","post","type-post","status-publish","format-standard","hentry","category-geometry"],"_links":{"self":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/112","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/comments?post=112"}],"version-history":[{"count":20,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/112\/revisions"}],"predecessor-version":[{"id":141,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/112\/revisions\/141"}],"wp:attachment":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/media?parent=112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/categories?post=112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/tags?post=112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}