Area of Triangles, Parallelograms, and Trapezoids

Warm-up

Hand out three paper shapes — a right triangle, a slanted (non-right) triangle, and a parallelogram — plus scissors. Challenge: "Turn each into a rectangle. No area may be created or destroyed."

The scissors are the theorem-provers today: every area formula in this lesson is a rectangle story in disguise, and students are about to discover all three by cutting.

Explore

Cut-and-rearrange lab: (1) Parallelogram → slice a right triangle off one end, slide it to the other end: a rectangle with the same base and height appears. (2) Triangle → two copies of ANY triangle (tape pairs' cutouts together) form a parallelogram: so one triangle is half of it. (3) Trapezoid → two copies form a parallelogram whose base is (top + bottom).

Each station records its discovery as a formula-with-picture. The height ruler matters everywhere: height is the PERPENDICULAR distance to the base line, not the slanted side — measure with a set square against the base.

Formalize

Formalize the three formulas, each carrying its rectangle story:

$A_{\text{par}} = bh \qquad A_{\text{tri}} = \tfrac{1}{2}bh \qquad A_{\text{trap}} = \tfrac{(a+b)}{2}\,h$

The one word doing all the work is HEIGHT: perpendicular to the base, possibly landing outside the shape (obtuse triangles — draw one and watch the height fall outside; the formula still holds). Any side can be the base; the height must match the chosen base.

Practice

Practice: six area computations across the three shapes, including one obtuse triangle (height outside), one trapezoid given upside-down, and one figure requiring decomposition into a rectangle plus a triangle.

Exit ticket: two triangles with the same base and equal heights but wildly different slants — same area or different? Prove with the formula and one sentence. (Same: $\frac{1}{2}bh$ never consulted the slant.)

Exit ticket

Practice: six area computations across the three shapes, including one obtuse triangle (height outside), one trapezoid given upside-down, and one figure requiring decomposition into a rectangle plus a triangle.

Exit ticket: two triangles with the same base and equal heights but wildly different slants — same area or different? Prove with the formula and one sentence. (Same: $\frac{1}{2}bh$ never consulted the slant.)