Area of a triangle · 2 marks

Find the area of triangle ADE in the figure.

Answer: Take AD = 8 cm as the base; the height from E to AD is the length of the rectangle, 10 cm. Area = 12 × 8 × 10 = 40 cm² (half the rectangle), wherever E is on BC.

Step-by-step solution

Given: ABCD is a rectangle, DC = 10 cm, BC = 8 cm; E lies on BC
To find: Area of △ADE

Idea: Use AD as the base. BC is parallel to AD, so every point E on BC is the same distance (10 cm) from AD.

Base AD = BC = 8 cm (opposite sides of a rectangle). Height = distance from E to AD = AB = DC = 10 cm, because E lies on BC and BC ∥ AD.1 mark
Area = 12 × base × height = 12 × 8 × 10 = 40 cm².1 mark

Area of triangle ADE = 40 cm² (half of the 80 cm² rectangle).

Check: The rectangle has area 80 cm², and a triangle on one side of a rectangle with its third vertex on the opposite side is always half of it ✓.

Answer to write in the exam

Base AD = BC = 8 cm (opposite sides of rectangle)

Height = distance between AD and BC = DC = 10 cm

ar(△ADE) = 12 × 8 × 10

∴ ar(△ADE) = 40 cm²

Common mistakes that cost marks

Thinking the position of E must be known. Any point on BC gives the same height.
Taking AD as 10 cm. AD is the short side, 8 cm.
Forgetting the 12 and answering 80 cm².

How this can come in the exam

MCQ (1 mark)

PQRS is a rectangle with PQ = 12 cm and QR = 5 cm. T is any point on QR. The area of △PST is

15 cm²
30 cm²
60 cm²
Cannot be found

Show answer

(B) 30 cm²
Base PS = 5, height = 12: 12 × 5 × 12 = 30 cm².

Try one yourself

A rectangle is 15 cm by 6 cm. A triangle has one long side of the rectangle as its base and its third vertex on the opposite side. Find its area.