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Triangles are classified as scalene, equilateral, or isosceles based on the sides.

In this section, we will learn about the isosceles triangle definition and their properties.

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Before we learn the definition of isosceles triangles, let us do a small activity. 

Take a rectangular sheet of paper and fold it into half.

Draw a line from the top folded corner to the bottom edge.

Cut it along the line as shown.

 

You can see a triangle when you open the sheet.

Mark the vertices of the triangles as \(\text{A}\), \(\text{B}\), and \(\text{C}\).

Now measure \(\text{AB}\) and \(\text{AC}\).

Repeat this activity with different measures and observe the pattern.

 

 

We can observe that \(\text{AB}\) and \(\text{AC}\) are always equal.

This type of triangle where two sides are equal is called an isosceles triangle.

Here are a few isosceles triangle real-life examples.

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• The two equal sides of an isosceles triangle are called the legs and the angle between them is called the vertex angle or apex angle.
• The side opposite the vertex angle is called the base and base angles are equal.
• The perpendicular from the vertex angle bisects the base and it also bisects the vertex angle.

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In an isosceles triangle, if the vertex angle is \(90^\circ\), the triangle is a right triangle.

A right isosceles triangle is a special triangle where the base angles are \(45 ^\circ\) and the base is also the hypotenuse.

The hypotenuse of an isosceles right triangle with side \({a}\) is

\(  \sqrt{2}a\)

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The area of an isosceles triangle can be calculated in many ways based on the known elements of the isosceles triangle.

Known parameters	Isosceles triangle formula
When the base \(b\) and height \(h\) are known	\[\frac{1}{2} \times b \times h\]
When all the sides \( a\) and the base \(b\) are known	\[\frac{b}{2}\sqrt{\text{a}^2 - \frac{b^2}{4}}\]
When the length of the two sides \(a\) and \(b\) and the angle between them \(\angle \text{α}\) is known	\[\frac{1}{2} ab\:sin(\text{α}) \]

 Use the calculator below to find the area of an isosceles triangle when the base and height are given.

Use the calculator below to find the area of an isosceles triangle when the base and the equal side are given.

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Example 1

 

 

In the given isosceles triangle \(\text{ABC}\), find the measure of the vertex angle and base angles.

 

Solution:

\[\begin{align}
3x &= x +42 (\because\angle \text{ABC} \! =\! \angle \text{BCA} )\\
\therefore  2x &= 42\\
x &=21\\
\angle \text{ABC} &= x+42\\
&= 63^\circ\\
\Rightarrow \angle \text{BCA} &=63^\circ(\!\because\!3x \!=\!3 \!\times\! 21\! =\!63\! )\\
\therefore \angle\text{BAC} &= (180-(63+63)\\
&=180-126\\
&=54^\circ
\end{align}\]

\(\begin{align}\angle \text{ABC}\!=\!\angle \text{BCA}\!=\!63^\circ \text{and} \:\angle\text{BAC}\!=\!54^\circ\end{align}\)

Example 2

 

 

In the given figure, \(\text{AC = BC}\) and \( \angle A = 30^\circ\). What is the measure of \(\angle\text{ECD}\)?

Solution:

\(\Delta\text{ACB}\) is isosceles as \(\text{AC = BC}\)

Therefore,

\(\angle A= \angle B\)

In \(\Delta\text{ACB}\)

\[\begin{align}
\angle\text{CAB} +\angle\text{ABC}+\angle\text{BCA} &= 180^\circ\\
(\text{Sum of the angles of a triangle})\\
\Rightarrow \angle\text{BCA}\!&\!=\!180^\circ-(\!30^\circ\!+\!30^\circ) \\
\therefore \angle\text{BCA} &=120^\circ \\
\angle\text{BCA} &= \angle\text{DCE}\\
[\because \text{Vertically opposite angles are equal}]\\
\therefore x&=120^\circ
\end{align}\]

\( \therefore \angle \text{ECD} =120^\circ \)

Example 3

 

 

In the given triangle, find the measure of BD and area of triangle ADB.

Solution:

Given:

\[\begin{align}
\text{AB} &= 5 \: \text{cm}\\
\text{AC} &= 5 \: \text{cm}\\
\text{AD} &= 4 \:\text{cm}\\
\text{DC} &= 3 \: \text{cm}\\
\text{AD}&\perp \text{BC}
\end{align}\]

In an isosceles triangle, the perpendicular from the vertex angle bisects the base.

Therefore,

\( \text{BD} = \text{DC} = 3 \: \text{cm} \)

In \(\Delta \text{ADB}\),

\[\begin{align}
 \text{Base}&=3\:\text{cm} \\
\text{Height}&=4\:\text{cm} (\text{given)}\\
\text{Area of} \Delta\text{ADC}&=\frac{1}{2}\times 3 \times 4 \\
&= 6\: \text{cm}^2
\end{align}\]

\(\therefore \text{Area of } \Delta\text{ADB} = 6\: \text{cm}^2\)

Example 4

 

 

In the given triangle \(\Delta \text{PQR}\), find the measure of the perpendicular \(\text{QS}\) (approx. to \(2\) decimal places).

Solution:

In the isosceles right triangle \(\Delta{PQR}\), we have:

\[\begin{align}
 \text{PQ} &=6\: \text{cm} \\
 \text{QR} &=6\: \text{cm} \\
\angle \text{PQR} &= 90^\circ \\
 \text{QS} &\perp  \text{PR}
\end{align}\]

Area of \(\Delta{PQR}\)

\[\begin{align}
&=\frac{1}{2} \times  \text{Base}  \times  \text{Height} \\
&=\frac{1}{2} \times\text{PQ} \times \text{QR}\\
&=\frac{1}{2} \times 6 \times 6 \\
&=18 \:\text{cm}^2
\end{align}\]

Therefore, \(\text{PR}\)

\[\begin{align}
&=6\sqrt{2} \: (\because \text{hypotenuse} = side\! \times\!\sqrt{2}) \\
&≈ 8.485\: \text{cm}
\end{align}\]

Considering \(\text{PR}\) as the base and \(QS\) as the altitude, we have

\[\begin{align}
\text{Area of }\Delta \text{PQR} &=\frac{1}{2} \times\text{Base} \times \text{Height}  \\
\Rightarrow18 &=\frac{1}{2} \times\text{PR} \times \text{QS}\\
18 &=\frac{1}{2} \times 8.485 \times\text{QS} \\
\therefore \text{QS} &= 4.24\: \text{cm}
\end{align}\]

\( \therefore \text{QS} = 4.24\: \text{cm} \)

Example 5

 

 

Find the perimeter of an isoselese triangle, if the base is \(24\: \text{cm}\) and the area is \(60 \:\text{cm}^2\).

Solution:

We know that,

\[\begin{align}
 \text{base} &=  24\: \text{cm}\\
\text{area} &=60 \:\text{cm}^2
\end{align}\]

Area of the triangle =

\(\frac{\text{b}}{2}\sqrt{\text{a}^2 - \frac{\text{b}^2}{4}}\)

Therefore,

\[\begin{align}
\Rightarrow 60 &= \frac{24}{2}\sqrt{\text{a}^2 - \frac{24^2}{4}} \\
60 &= 12\sqrt{\text{a}^2 - 144} \\
5 &=\!\sqrt{\text{a}^2 \!-\!144} \: (\text{Squaring both sides}) \\
25 &= \text{a}^2 -144 \\
\therefore \text{a}^2 &= 169 \\
\Rightarrow \text{a}&=13\: \text{cm}
\end{align}\]

Perimeter of the isoselese triangle

\[\begin{align}
&2\text{a}+\text{b} \\
&(2 \times 13) +24 \\
&=26+24 \\
&=50\: \text{cm}
\end{align}\]

\( \therefore\) Perimeter of given triangle = \(50\: \text{cm}\)

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