Lesson 1: Revision on Right-angled Triangles

Video Lesson

Competence

Dear learner; At the end of the lesson you will be able to:

• Identify the hypotenuse, opposite and adjacent of a right-angled triangle.
• Describe the basic properties of a right-angled triangle.
• Apply the Pythagorean Theorem to determine if a set of three given side lengths can form a right-angled triangle.
• Analyze and compare right-angled triangles to determine their similarity based on their acute angles.

Key terms

• A right-angled triangle
• Hypotenuse
• Legs
• Pythagorean Theorem
• Isosceles Right-Angled Triangle
• Altitude on Hypotenuse
• Similarity and
• Proportionality

Brainstorming Questions

Solution:

From your previous grades if a triangle has side lengths a, b and c with longer side length c, then the following criteria holds:

$$\begin{gathered} a^2+b^2=c^2\mathrm{The}\mathrm{side}\mathrm{length}\mathrm{marked}\mathrm{on}b\mathrm{satisfies}\mathrm{the}\mathrm{property:} \\ {3}^2+4^2=5^2\Rightarrow 9+16=25 \end{gathered}$$

1.1. Right angeled Triangle

Definition

• A right-angled triangle is a triangle in which one of its three interior angles is exactly 90 degrees, known as a right angle.
• The side opposite this right angle (side c) is called the hypotenuse, and it is the longest side of the triangle.
• The two remaining sides (side a and side b), which form the right angle, are referred to as the legs of the triangle.

Fig 1: right angled triangle

1.1.1. Pythagorean Theorem

the pythagorean theorem states that the square of the hypotenuse’s length is equal to the sum of the squares of the lengths of the two legs:

$\mathrm{i.e.}(a^2+b^2=c^2)$

1.1.2. Properties of a Right-Angled Triangle:

• One Right Angle: A right-angled triangle has one angle that is exactly 90 degrees.
• Hypotenuse: The side opposite the right angle is called the hypotenuse and is the longest side of the triangle.
• Legs: The two sides that form the right angle are known as the legs of the triangle.
• The sum of the other two interior angles is equal to 90° (each angle is acute).
• If one of the angles is 90° and the other two angles are equal to 45° each, then the triangle is called an isosceles right-angled triangle, where the adjacent sides to 90° are equal in length.
• Altitude on Hypotenuse: The altitude drawn from the right angle to the hypotenuse divides the triangle into two smaller right-angled triangles that are similar to the original triangle and to each other.
• Similarity: All right-angled triangles that have the same acute angles are similar, meaning their corresponding sides are in proportion.

Example 1:

Identify whether the given sides can form a right-angled triangle or not provided that the units for each length is identical.

$$\begin{gathered} \mathrm{a)}4,6,8 \\ \mathrm{b)}6,8,10 \\ \mathrm{c)}\sqrt{3},3,\sqrt{6} \end{gathered}$$

Solution:

$$\begin{gathered} \mathrm{a)}{4}^2+6^2\ne 8^2 \\ 16+36\ne 64 \\ 52\ne 64.\mathrm{Thus},4,6,8\mathrm{cannot}\mathrm{form}a\mathrm{right-angled}\mathrm{triangle}. \\ \mathrm{b)}{6}^2+8^2={10}^2 \\ 36+64=100 \\ 100=100.\mathrm{Thus},6,8,10\mathrm{can}\mathrm{form}a\mathrm{right-angled}\mathrm{triangle}. \\ \mathrm{c)}(\sqrt{3}{)}^2+(\sqrt{6}{)}^2=(3{)}^2 \\ 3+6=9 \\ 9=9.\mathrm{Thus},\sqrt{3},3,\sqrt{6}\mathrm{can}\mathrm{form}a\mathrm{right-angled}\mathrm{triangle}. \end{gathered}$$

2 determine the value of x if the triangles are right angled triangles?

a.

b.

Solution:

$$\begin{gathered} \text{a)}\text{BD is the altitude of an isosceles triangle ACD. It passes through the mid point of the base of the} \\ \text{triangle. The triangle formed by this line segment is a right-angled triangle. i.e.,}\overline{AB}=\overline{BC}. \\ \text{implies}\Delta \text{ABD is a right-angled triangle with hypotenuse}\overline{AD}\text{.}\text{and}\overline{AB}=\frac{4}{2}=2,\overline{AD}=3\text{and}\overline{BD}=x. \end{gathered}$$

$$\begin{gathered} \text{then}{x}^2+(\mathrm{AB}{)}^2=(\mathrm{BD}{)}^2\mathrm{implies}{x}^2+2^2=3^2 \\ \mathrm{implies}{x}^2=9–4=5\mathrm{implies}x=\sqrt{5} \\ \mathrm{b.}\mathrm{The}\mathrm{triangle}\mathrm{is}a\mathrm{right-angled}\mathrm{triangle}\mathrm{and}{x}^2+x^2=8^2 \\ \mathrm{implies}2x^2=64 \\ \mathrm{implies}{x}^2=\frac{64}{2}=32 \\ \mathrm{implies}x=\sqrt{32}=4\sqrt{2} \end{gathered}$$

1.2. Converse of the phytagorean theorem

$$\begin{gathered} \text{Converse of the Pythagoras theorem states that if a triangle has side lengths}a,b,\text{and}c,\text{where}c \\ \text{is the longest side, and if the equation}{a}^2+b^2=c^2\text{holds true, then the triangle is a right angled triangle.} \\ (\angle BAC=90°). \end{gathered}$$

Example 2:

1. Show that the following triangles are right -angled triangles or not?

Solution:

$$\begin{gathered} \mathrm{a).}\mathrm{The}\mathrm{greatest}\mathrm{length}\mathrm{is}28,\mathrm{and}\mathrm{check}{20}^2+{19}^2={28}^2 \\ \mathrm{implies}400+361=761,{28}^2=784. \\ \mathrm{implies}761\ne 784,\mathrm{therefore}\mathrm{the}\mathrm{triangle}\mathrm{is}\mathrm{not}a\mathrm{right-angled}\mathrm{triangle}. \end{gathered}$$
$$\begin{gathered} \mathrm{b).}\mathrm{the}\mathrm{longest}\mathrm{side}\mathrm{is}25\mathrm{and}\mathrm{check}{8}^2+{24}^2={25}^2 \\ \mathrm{implies}64+576=640\mathrm{and}{25}^2=625. \\ \mathrm{and}640\ne 625\mathrm{therefore}\mathrm{the}\mathrm{triangle}\mathrm{is}\mathrm{not}a\mathrm{right-angled}\mathrm{triangle}. \end{gathered}$$
$$\begin{gathered} \mathrm{c).}\mathrm{the}\mathrm{longest}\mathrm{side}\mathrm{is}65\mathrm{and}\mathrm{check}{33}^2+{56}^2={65}^2 \\ \mathrm{implies}1089+3136=4225\mathrm{and}{65}^2=4225. \\ \mathrm{They}\mathrm{are}\mathrm{equal},\mathrm{therefore}\mathrm{the}\mathrm{triangle}\mathrm{satisfies}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{Pythagorean}\mathrm{theorem},\mathrm{and}\mathrm{the}\mathrm{triangle}\mathrm{is}a\mathrm{right-angled}\mathrm{triangle}. \end{gathered}$$
$$\begin{gathered} 2.\mathrm{Point}B\mathrm{is}5m\mathrm{south}\mathrm{of}\mathrm{point}A,\mathrm{point}C\mathrm{is}8m\mathrm{east}\mathrm{of}\mathrm{point}B,\mathrm{and}\mathrm{point}D\mathrm{is}hm\mathrm{east}\mathrm{of}\mathrm{point}C. \\ \mathrm{If}\mathrm{the}\mathrm{distance}\mathrm{from}A\mathrm{to}D\mathrm{is}13m,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}h\mathrm{is}: \end{gathered}$$

Solution:

$$\begin{gathered} \text{angle B is a right angle, applying pythagorean relations} \\ {5}^2+(8+h){)}^2={13}^2 \\ 25+(8+h){)}^2=169 \\ (8+h){)}^2=169–25 \\ (8+h){)}^2=144 \\ 8+h=12 \\ h=12–8=4. \end{gathered}$$