A quadrilateral field PQRS is divided by diagonal PR = 10 m. In right-angled triangle PQR, PQR = 90° and the base PQ = 8 m. Triangle PRS has a perpendicular height to base PR of 6 m. What is the total area of field PQRS?

54 m^2. First, calculate the perpendicular height QR using Pythagoras' theorem: QR = √(10^2 - 8^2) = √(100 - 64) = √(36) = 6 m. The area of PQR = (1)/(2) × 8 × 6 = 24 m^2. The area of PRS = (1)/(2) × 10 × 6 = 30 m^2. Total area = 24 + 30 = 54 m^2.

Grade 10 · Core Mathematics · Lesson 5 of 12

Determining Area of Quadrilaterals in Different Situations: Parallelograms and Trapeziums

A Grade 10 Core Mathematics CBE lesson from the sub-strand Area of Polygons. In it, learners decompose quadrilaterals into right-angled triangles and use Pythagoras' theorem to calculate unknown perpendicular heights.

Strand Measurements and GeometrySub-strand Area of PolygonsLesson code 10CM-25-L540 minutes · 3 slides

KICD learning outcome (d)

Determine the area of quadrilaterals in different situations.

By the end of this lesson, learners should be able to

Decompose quadrilaterals into right-angled triangles and use Pythagoras' theorem to calculate unknown perpendicular heights.
Calculate the area of a parallelogram using adjacent sides and the sine of the included angle.
Determine the area of a trapezium by calculating its perpendicular height using trigonometric ratios.

Key vocabulary

Quadrilateral: A closed plane figure bounded by four straight line segments.; A plot shaped as a quadrilateral is split into two triangles to find area.
Parallelogram: A quadrilateral with two pairs of opposite parallel sides.; A parallelogram with side lengths $10 m$ and $6 m$ has area $a b \sin θ$ .
Trapezium: A quadrilateral with exactly one pair of parallel sides.; A trapezium with parallel sides $15 cm$ and $25 cm$ has area $\frac{1}{2} h (a + b)$ .

Lesson preview

Slide 1 · Area of Decomposed Quadrilaterals and Environmental ShapesOpening · 12 min

Figure 1: A quadrilateral is decomposed into two right-angled triangles by a diagonal to calculate its total area.

Many real-world objects in our immediate environment, such as flower beds, plots of land, and vegetable gardens, are shaped as quadrilaterals. When determining the area of quadrilaterals in different situations, we can decompose the shape into simpler figures like triangles. If a right-angled triangle forms part of the quadrilateral, we can calculate any unknown perpendicular height using Pythagoras' theorem before finding the combined surface area.

Real-life items like agricultural plots and tiles are often quadrilateral in shape.
Decomposing complex quadrilaterals into two triangles simplifies the total area calculation.
Pythagoras' theorem helps find an unknown perpendicular height when right angles exist.
Summing the individual triangular areas gives the total surface area of the quadrilateral.

Worked example. Find height

B C

using Pythagoras' theorem in right triangle

A B C

B C = \sqrt{15^{2} - 12^{2}} = \sqrt{225 - 144} = \sqrt{81} = 9 m

. Calculate area of

△ A B C

{Area}_{1} = \frac{1}{2} \times 12 \times 9 = 54 m^{2}

. Calculate area of right-angled

△ A C D

{Area}_{2} = \frac{1}{2} \times A C \times A D = \frac{1}{2} \times 15 \times 9 = 67.5 m^{2}

. Sum the areas for total plot area:

Area = 54 + 67.5 = 121.5 m^{2}

. Answer: Height $B C = 9 m$ and total area $= 121.5 m^{2}$ ..

Check question: A quadrilateral field $P Q R S$ is divided by diagonal $P R = 10 m$ . In right-angled triangle $P Q R$ , $∠ P Q R = 90^{\circ}$ and the base $P Q = 8 m$ . Triangle $P R S$ has a perpendicular height to base $P R$ of $6 m$ . What is the total area of field $P Q R S$ ?

$54 m^{2}$
$70 m^{2}$
$80 m^{2}$
$108 m^{2}$

Show answer

$54 m^{2}$ . First, calculate the perpendicular height $Q R$ using Pythagoras' theorem: $Q R = \sqrt{10^{2} - 8^{2}} = \sqrt{100 - 64} = \sqrt{36} = 6 m$ . The area of $△ P Q R = \frac{1}{2} \times 8 \times 6 = 24 m^{2}$ . The area of $△ P R S = \frac{1}{2} \times 10 \times 6 = 30 m^{2}$ . Total area $= 24 + 30 = 54 m^{2}$ .

Slide 2 · Area of a Parallelogram Using TrigonometryModel · 14 min

Figure 2: A parallelogram has its adjacent sides and included angle labelled to demonstrate calculating area with trigonometry.

Teach the full lesson on the projector

This lesson, and every other Grade 1–10 CBE lesson, comes ready to project with the EAP ClassKit. It runs without electricity or internet.

All 3 slides with diagrams and class activities
Teacher notes, SNE adaptations and assessment checks
Homework and the other 11 lessons in this sub-strand

See the school projector WhatsApp us

Determining Area of Quadrilaterals in Different Situations: Parallelograms and Trapeziums

By the end of this lesson, learners should be able to

Key vocabulary

Lesson preview

Teach the full lesson on the projector

All 12 lessons in Area of Polygons