Areas of Parallelograms, Triangles, and Trapezoids

The area of a triangle is half the product of its base and height. A = 1/2bh The base can be any side of the triangle and the height is the segment that is perpendicular to the base and connects the base or its extension with its opposite vertex. In the given diagram, we can see that the three heights of the triangle are given.

When finding the area, it is important to use the correct data pair since the height is different for each base. Let's pair each base to its corresponding height.

Base	Height
50	34.4
80	21.5
40	43

We can find the area of △ ABC by using any of these three pairs of values. For simplicity, let's use the data from the third row.

The area of the triangle is 860 square units.

Let's start by noticing that RS is perpendicular to PQ. If we take PQ as the base of the triangle, then RS is the corresponding height.

The area of a triangle is half the product of its base and height, so let's write an equation in terms of RS. A = 1/2bh ⇒ A = 1/2* 64* RS We know that the area of △ PQR is 1408. Let's substitute 1408 for A and solve the equation for RS.

We found that RS is 44 units long.

The perimeter of a polygon is the sum of all its side lengths. Let's write an equation for the perimeter of △ JKL. P = JK + KL + LJ From the diagram, we know that JK=30 and KL=44. However, we do not know the length of LJ. The good news is that we can find it by using the fact that the area of the triangle is 650 square units and that the height corresponding to LJ measures 26 units.

Let's substitute 650 for A, 26 for h, and LJ for b into the formula for the area of a triangle. Then we can solve for LJ.

Now we are ready to find the perimeter of △ JKL.

The perimeter of the triangle is 124 units.

Property	Justification
The opposite sides are congruent	Parallelogram Opposite Sides Theorem
The opposite angles are congruent	Parallelogram Opposite Angles Theorem
The diagonals bisect each other	Parallelogram Diagonals Theorem

Property	Justification
The diagonals are congruent.	Isosceles Trapezoid Diagonals Theorem
Each pair of base angles is congruent.	Isosceles Trapezoid Base Angles Theorem

Length on the Whiteboard	Length on the Court
$d_{1} = 73$	$d_{1} = 273$
$d_{2} = 29$	$d_{2} = 229$

Length on the Whiteboard	Length on the Court
$b_{1} = 10.2$	$b_{1} = 20.4$
$b_{2} = 5.1$	$b_{2} = 10.2$

Tans Forming the Trapezoid	Dimensions	$A = \frac{1}{2} h (b_{1} + b_{2})$	Area $(cm^{2})$
$7$ and $3$	$b_{1} = 8$ $b_{2} = 4$ $h = 2$	$A_{7, 3} = \frac{1}{2} \cdot 2 (8 + 4)$	$12$
$1,$ $2,$ and $4$	$b_{1} = 12$ $b_{2} = 8$ $h = 4$	$A_{1, 2, 4} = \frac{1}{2} \cdot 4 (12 + 8)$	$40$
$5$ and $6$	$b_{1} = 28$ $b_{2} = 8$ $h = 8$	$A_{5, 6} = \frac{1}{2} \cdot 8 (28 + 8)$	$12$

Areas of Parallelograms, Triangles, and Trapezoids

Catch-Up and Review

Dividing a Rectangle in Half

Deriving the Formula for the Area of a Triangle

Area of a Triangle

Proof

Proof for Right Triangles

Proof for Non-Right Triangles

Tangram Puzzle

Hint

Solution

Finding the Missing Dimension

Parallelograms

Area of a Parallelogram

Tangram Animals

Hint

Solution

Parallelograms With Congruent Sides

Rhombus

Area of a Rhombus

Rhombuses Everywhere

Hint

Solution

Solving Parallelograms and Rhombuses

Trapezoids

Isosceles Trapezoid

Area of a Trapezoid

Extra

A Head Full of Polygons

Hint

Solution

Solving Trapezoids

Futsal Formations

Hint

Solution

Composite Figures

Areas of Parallelograms, Triangles, and Trapezoids

Recommended exercises

	18 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Base	Height
AB=4	EC=1.5
BC=2	DF=3