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8.10 Area of non-right triangles

Lesson

Recall that you can find the area of a triangle by using the formula $A=\frac{1}{2}bh$A=12bh, where $b$brepresents a side of the triangle, called the base, and $h$h represents the height of the triangle perpendicular to that side. Note that the base can be any side of the triangle as long as the height is drawn perpendicular to it.

Let's suppose we know the lengths of the sides of the triangle, but not the measurement of the height.  There is another way to find the area of the triangle.  

Exploration

Use the applet below and the questions that follow to derive a new formula for the area of a triangle.

  1. Drag the vertices $A$A$B$B, and $C$C of the triangle to make the triangle larger or smaller.
  2. Drag the "construct altitude" slider all the way to the right.
  3. Answer the questions below to derive the new formula for the area of a triangle.

Discussion questions

1. Is the height of $\Delta ABC$ΔABC the same as the height of the yellow triangle?  Is it the same as the height of the pink triangle?  Explain why or why not.

2. What equation relates the height of the yellow triangle to $\angle B$B and side $c$c?  Check that you are correct by clicking the box next to "Height of Yellow Triangle".

3. Explain why the area of $\Delta ABC$ΔABC can be represented by the equation $A=\frac{1}{2}ac\sin B$A=12acsinB.

4. Find the height of the pink triangle in terms of $\angle C$C and side $b$b.  Check that you are correct by clicking the box next to "Height of Pink Triangle".

5. Explain why the area of $\Delta ABC$ΔABC can also be represented by the equation $A=\frac{1}{2}ab\sin C$A=12absinC

6.  It is also true that the area of $\Delta ABC$ΔABC can be represented by the equation $A=\frac{1}{2}bc\sin A$A=12bcsinA.  Explain why this is also true.  (Hint: Would it matter if the triangle were rotated?)

(Adapted from https://www.geogebra.org/m/teZeAxWN)

 

 

Area of a triangle

$Area=\frac{1}{2}ab\sin C$Area=12absinC

Where $a$a and $b$b are the known side lengths, and $C$C is the given angle between them, as per the diagram above.

Practice questions

Question 1

Calculate the area of the following triangle.

Round your answer to two decimal places.

A triangle is depicted with the measurements of its two sides and their included angle. The included angle, highlighted by a blue-shaded arc, measures $44^\circ$44° and is adjacent to sides measuring $3$3 m and $5.7$5.7 m.

Question 2

Calculate the area of the triangle.

Round your answer to two decimal places.

A triangle is illustrated given the measure of one of its interior angle. The angle on the upperpart of the triangle measures $123^\circ$123°, indicated by an arc shaded with blue. The shorter adjacent side of the $123^\circ$123°-degree angle measures $4$4 m while the longer adjacent side of the $123^\circ$123°-degree angle measures $7.4$7.4 m.

Question 3

Calculate the area of the following triangle.

Round your answer to the nearest square centimeter.

A non-right triangle with vertices labeled $P$P, $Q$Q, and $R$R. Angle $QPR$QPR measures 39 degrees, while angle $PQR$PQR measures 25 degrees. Side $QR$QR, which is opposite angle $QPR$QPR measures 33 $cm$cm. Side $PR$PR, which is opposite angle $PQR$PQR measures 22 $cm$cm.

 

 

Outcomes

III.G.SRT.9

Derive the formula a = 1/2 ab sin(c) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side.

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