In the diagram, PQRS is a quadrilateral, < PQR = <... | WAEC SSCE - General Mathematics - 2019 (Essay)

Question 1 Report

In the diagram, PQRS is a quadrilateral, < PQR = < PRS = 90\(^o\), |PQ| =3cm, |QR| = 4cm and |PS| = 13 cm. Find the area of the quadrilateral.

Answer

From the diagram, \(PQRS\) is split by the diagonal \(PR\) into two right-angled triangles: triangle \(PQR\) (right-angled at \(Q\)) and triangle \(PRS\) (right-angled at \(R\)). Given \(|PQ| = 3\) cm, \(|QR| = 4\) cm and \(|PS| = 13\) cm.

Step 1 - find the diagonal \(|PR|\). In right-angled triangle \(PQR\) (\(\angle PQR = 90^{\circ}\)), \(PR\) is the hypotenuse:

\[ |PR| = \sqrt{|PQ|^2 + |QR|^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}. \]

Step 2 - find \(|RS|\). In right-angled triangle \(PRS\) (\(\angle PRS = 90^{\circ}\)), \(PS\) is the hypotenuse:

\[ |RS| = \sqrt{|PS|^2 - |PR|^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}. \]

Step 3 - add the two triangle areas.

\[ \text{Area of } PQR = \tfrac{1}{2}\times |PQ| \times |QR| = \tfrac{1}{2}\times 3 \times 4 = 6\text{ cm}^2. \]

\[ \text{Area of } PRS = \tfrac{1}{2}\times |PR| \times |RS| = \tfrac{1}{2}\times 5 \times 12 = 30\text{ cm}^2. \]

\[ \text{Area of } PQRS = 6 + 30 = 36\text{ cm}^2. \]

Area of the quadrilateral \(= 36\) cm\(^2\).

Read lesson note on Areas (WAEC)

Read lesson note on Triangles And Polygons (WAEC)

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