{"id":3732,"date":"2026-05-24T06:58:59","date_gmt":"2026-05-24T06:58:59","guid":{"rendered":"https:\/\/www.mathros.net.ua\/en\/?p=3732"},"modified":"2026-05-24T07:10:24","modified_gmt":"2026-05-24T07:10:24","slug":"height-of-an-isosceles-triangle","status":"publish","type":"post","link":"https:\/\/www.mathros.net.ua\/en\/height-of-an-isosceles-triangle.html","title":{"rendered":"Height of an Isosceles Triangle: Theory and Examples"},"content":{"rendered":"

The height of an isosceles triangle is a perpendicular segment drawn from the vertex of the triangle to its base. It helps us better understand the structure of this figure<\/a> and is also an important value for calculating its area.<\/p>\n

In this article, we will look at how to find the height of an isosceles triangle, where the main formula comes from, and how to apply it in problems.<\/p>\n

Height of an Isosceles Triangle: Main Formula<\/h2>\n

The height of an isosceles triangle can be found if the length of its base and the length of its equal side are known. Suppose we have an isosceles triangle \\( ABC \\), where \\( AB=BC \\), and \\( AC \\) is the base. The height \\( BH \\) is drawn from vertex \\( B \\) to side \\( AC \\).<\/p>\n

\"Image:<\/p>\n

For this triangle, the height \\( BH \\) can be calculated using the formula:<\/p>\n

\\[
\nBH=\\sqrt{AB^2-\\frac{AC^2}{4}}.
\n\\]<\/p>\n

Note.<\/strong> This formula can be used only when the given sides can actually form an isosceles triangle. For this, the following conditions must be satisfied:
\n\\[
\nAB>0,\\qquad AC>0,\\qquad AC<2\\cdot AB.
\n\\]
\nThat is, the base must be less than the sum of the two equal sides.<\/p><\/blockquote>\n

Why This Formula Works: Explanation Through Triangle Properties<\/h2>\n

To understand the formula, it is important to remember one useful property of an isosceles triangle. The height drawn from the vertex to the base is also a median and an angle bisector at the same time. This means that it divides the base in half and forms two equal right triangles.<\/p>\n

Therefore, if \\( BH \\) is the height, then point \\( H \\) is the midpoint of the base \\( AC \\). So,<\/p>\n

\\[
\nAH=HC=\\frac{AC}{2}.
\n\\]<\/p>\n

Now let us look at the right triangle \\( ABH \\). In it, \\( AB \\) is the hypotenuse, \\( BH \\) is one leg, and \\( AH \\) is the other leg. By the Pythagorean theorem, we have:<\/p>\n

\\[
\nAB^2=BH^2+AH^2.
\n\\]<\/p>\n

Since \\( AH=\\frac{AC}{2} \\), we can write:<\/p>\n

\\[
\nAB^2=BH^2+\\left(\\frac{AC}{2}\\right)^2.
\n\\]<\/p>\n

Next, let us rearrange the equation:<\/p>\n

\\[
\nAB^2=BH^2+\\frac{AC^2}{4},\\qquad BH^2=AB^2-\\frac{AC^2}{4},\\qquad BH=\\sqrt{AB^2-\\frac{AC^2}{4}}.
\n\\]<\/p>\n

This is exactly how we get the formula for the height of an isosceles triangle.<\/p>\n

Note.<\/strong> If we denote the length of the equal side by \\( a \\), the base by \\( b \\), and the height by \\( h \\), then the formula will have a more familiar form:
\n\\[
\nh=\\sqrt{a^2-\\frac{b^2}{4}}.
\n\\]<\/p><\/blockquote>\n

Height of an Isosceles Triangle: Examples with Answers<\/h2>\n

Let us look at several problems where we need to find the height of an isosceles triangle or, conversely, find the equal side when the height is known. Each example includes a step-by-step solution, so you can easily follow the logic of the calculations.<\/p>\n

Example 1. What is the height of an isosceles triangle if its base is \\( 8 \\) cm and its equal sides are \\( 6 \\) cm each?<\/h3>\n

In this problem, we have \\( a=6 \\) and \\( b=8 \\). Let us substitute the given values into the formula:<\/p>\n

\\[
\nh=\\sqrt{a^2-\\frac{b^2}{4}}.
\n\\]<\/p>\n

We get:<\/p>\n

\\[
\nh=\\sqrt{6^2-\\frac{8^2}{4}}
\n=\\sqrt{36-\\frac{64}{4}}
\n=\\sqrt{36-16}
\n=\\sqrt{20}
\n\\approx 4.47.
\n\\]<\/p>\n

Therefore, the height of an isosceles triangle is \\( 4.47 \\) cm.<\/p>\n

Example 2. An isosceles triangle has a base of \\( 10 \\) cm, and its equal sides are \\( 12 \\) cm each. Find its height<\/h3>\n

In this case, \\( a=12 \\) and \\( b=10 \\). We will use the same formula:<\/p>\n

\\[
\nh=\\sqrt{a^2-\\frac{b^2}{4}}.
\n\\]<\/p>\n

Let us substitute the values:<\/p>\n

\\[
\nh=\\sqrt{12^2-\\frac{10^2}{4}}
\n=\\sqrt{144-\\frac{100}{4}}
\n=\\sqrt{144-25}
\n=\\sqrt{119}
\n\\approx 10.91.
\n\\]<\/p>\n

Therefore, the height of an isosceles triangle is \\( 10.91 \\) cm.<\/p>\n

Example 3. The height of an isosceles triangle is \\( 12 \\) cm, and the length of its base is \\( 20 \\) cm. Find the equal side of the triangle<\/h3>\n

Now the equal side is not known, but the height is given. We have \\( h=12 \\) and \\( b=20 \\). Let us use the formula:<\/p>\n

\\[
\nh=\\sqrt{a^2-\\frac{b^2}{4}}.
\n\\]<\/p>\n

Substitute the known values:<\/p>\n

\\[
\n12=\\sqrt{a^2-\\frac{20^2}{4}}.
\n\\]<\/p>\n

Now simplify the expression under the square root:<\/p>\n

\\[
\n12=\\sqrt{a^2-\\frac{400}{4}},\\qquad 12=\\sqrt{a^2-100}.
\n\\]<\/p>\n

Square both sides of the equation:<\/p>\n

\\[
\n144=a^2-100.
\n\\]<\/p>\n

From here, we get:<\/p>\n

\\[
\na^2=244,\\qquad a=\\sqrt{244}=2\\sqrt{61}\\approx 15.62.
\n\\]<\/p>\n

Therefore, the length of the equal side of the isosceles triangle is \\( 15.62 \\) cm.<\/p>\n

What to Study Next: Useful Topics to Explore<\/h2>\n

Once the height formula is clear, you can move on to other topics related to the isosceles triangle. They will help you see this figure more broadly and better understand how its properties are used in different problems.<\/p>\n

    \n
  1. Isosceles Triangle: Definition and Properties<\/a> \u2014 Learn what features an isosceles triangle has and how it differs from other triangles.<\/li>\n
  2. Perimeter of an Isosceles Triangle: Formulas and Examples<\/a> \u2014 This topic explains how to find the perimeter of an isosceles triangle using its sides or other given data.<\/li>\n
  3. Area of an Isosceles Triangle: Formulas and Examples<\/a> \u2014 Here you will see how to calculate the area of an isosceles triangle and how to apply the formulas correctly in practice.<\/li>\n<\/ol>\n

    Height of an Isosceles Triangle: Geometry in Code<\/h2>\n

    If you are interested in programming, try turning the formula for the height of an isosceles triangle into a small program in your favorite programming language. The flowchart below already shows the logic of such an algorithm: the user enters the length of the equal side and the base, the program checks whether these values can form an isosceles triangle, and then calculates the height or displays an error message.<\/p>\n

    This simple task combines geometry and programming very well, because one mathematical formula turns into a clear algorithm that can be implemented in Pascal<\/em>, Python<\/em><\/a>, C++<\/em>, JavaScript<\/em>, or any other programming language.<\/p>\n

    \"Image:<\/p>\n","protected":false},"excerpt":{"rendered":"

    The height of an isosceles triangle is a perpendicular segment drawn from the vertex of the triangle to its base.<\/p>\n","protected":false},"author":1,"featured_media":3753,"comment_status":"open","ping_status":"closed","sticky":false,"template":"template-centered.php","format":"standard","meta":{"footnotes":""},"categories":[502],"tags":[80,526,511,504,512],"class_list":["post-3732","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-triangles","tag-geometry-basics","tag-isosceles-triangle","tag-triangle-formulas","tag-triangle-geometry","tag-triangle-height"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/3732","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/comments?post=3732"}],"version-history":[{"count":18,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/3732\/revisions"}],"predecessor-version":[{"id":3752,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/3732\/revisions\/3752"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media\/3753"}],"wp:attachment":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media?parent=3732"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/categories?post=3732"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/tags?post=3732"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}