A focused Grade 6 learning session for Area of Parallelograms and Trapezoids using visual models, worked examples, strategy guidance, and a short mastery check.
Student goal: I can solve problems involving area of parallelograms and trapezoids.
Area of Parallelograms and Trapezoids means measuring how much flat space a figure covers by using equal-sized square units.
Question: What is the area of the parallelogram?
Why it works: The labeled height is \(3\) cm (the perpendicular distance) and the base is \(6\) cm. Area \(=6 \times 3 = 18\) cm\(^2\).
Answer: \(18\) cm\(^2\)
Question: What is the area of the parallelogram? (The height is shown by dashed lines.)
Answer: \(9\) m\(^2\)
Question: What is the area of the parallelogram?
Answer: \(14\) cm\(^2\)
Question: A garden bed is shaped like a trapezoid with parallel sides of \(8\) ft and \(12\) ft and a height of \(6\) ft. How many square feet is the garden bed?
Why it works: Area of trapezoid \(= \frac{1}{2}(8 + 12) \times 6 = \frac{1}{2}(20)(6) = 60\) ft\(^2\).
Answer: \(60\) ft\(^2\)
Area of Parallelograms and Trapezoids becomes easier when students connect the question to a model, use clear steps, and explain why the answer fits.
Area means every square unit inside the figure.
Use scratch paper. The explanation matters more than simply seeing whether the choice is correct.
A parallelogram has a base of \(8\) m and a height of \(5\) m. What is its area?
Answer C: \(40\) m\(^2\)
Area of a parallelogram \(=\) base \(\times\) height \(=8\times5=40\) m\(^2\).
What is the area of the parallelogram?
Answer B: \(18\) cm\(^2\)
The labeled height is \(3\) cm (the perpendicular distance) and the base is \(6\) cm. Area \(=6 \times 3 = 18\) cm\(^2\).
A trapezoid has bases of \(7\) m and \(9\) m with a height of \(4\) m. What is its area?
Answer B: \(32\) m\(^2\)
Area of a trapezoid \(= \frac{1}{2}(b_1 + b_2) \times h = \frac{1}{2}(7 + 9) \times 4 = \frac{1}{2} \times 16 \times 4 = 32\) m\(^2\).
Which measurement is the height of the trapezoid?
Answer C: \(2.5\) in
The height is the perpendicular distance between the two parallel bases, shown by the dashed line, which is \(2.5\) in.
A parallelogram has a base of \(12\) cm and an area of \(60\) cm\(^2\). What is its height?
Answer D: \(5\) cm
Area \(= b \times h\), so \(60 = 12 \times h\). Dividing both sides by \(12\): \(h = 5\) cm.
A trapezoid has an area of \(48\) m\(^2\) and a height of \(6\) m. One base is \(5\) m long. What is the length of the other base?
Answer C: \(11\) m
Using \(A = \frac{1}{2}(b_1 + b_2) \times h\): \(48 = \frac{1}{2}(5 + b_2) \times 6 = 3(5 + b_2) = 15 + 3b_2\), so \(3b_2 = 33\) and \(b_2 = 11\) m.
A parking lot is shaped like a parallelogram with a base of \(40\) m and a height of \(15\) m. What is the area of the parking lot?
Answer C: \(600\) m\(^2\)
Area of the parallelogram \(= 40 \times 15 = 600\) m\(^2\).
What is the area of the parallelogram? (The height is shown by dashed lines.)
Answer C: \(9\) m\(^2\)
Area \(= 4.5 \times 2 = 9\) m\(^2\). The slanted side is not the height.
A garden bed is shaped like a trapezoid with parallel sides of \(8\) ft and \(12\) ft and a height of \(6\) ft. How many square feet is the garden bed?
Answer A: \(60\) ft\(^2\)
Area of trapezoid \(= \frac{1}{2}(8 + 12) \times 6 = \frac{1}{2}(20)(6) = 60\) ft\(^2\).
Which statement correctly identifies the height of a parallelogram?
Answer B: The height is the perpendicular distance between two parallel sides.
The height must be perpendicular to the base, not just any side length.
What is the area of the parallelogram?
Answer D: \(14\) cm\(^2\)
Area \(= \text{base} \times \text{height} = 7 \times 2 = 14\) cm\(^2\).
