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Triangle area · finding base and height
P6 · Area & perimeter · PSLE Maths
Question
Ali drew a rectangle ABCD on a school notice board. AB = 30 cm and BC = 18 cm. E is on AB such that AE = 12 cm. F is on CD such that DF = 20 cm. Ali joined ED, EF and EC. The shaded part is triangle EFC. Find the area of the shaded triangle.
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Hint
First find the length of FC, then use FC as the base. The perpendicular distance from E to the line containing FC is the height of the rectangle.
Show the answer and step-by-step explanation
Answer: Area of shaded triangle 90 cm²
- See the shaded part — The shaded part is triangle EFC. We will use: area of triangle = base × height ÷ 2.
- Find the base — Choose FC as the base. Opposite sides of a rectangle are equal, so CD = AB = 30 cm. Since DF = 20 cm, FC = 30 − 20.
- Find the height — The base FC lies on the bottom side CD. The corresponding height is the perpendicular distance from E to the line containing the base. This height may fall outside the triangle, but we can still use it.
- Calculate area — Now the base is 10 cm and the height is 18 cm. Put them into the triangle area formula.
- Write the answer — So the area of the shaded triangle is 90 cm². 📌 Key idea: after choosing a base, find the height perpendicular to that base. In a rectangle, the distance between parallel sides is often the triangle's height.
Method: For a triangle inside a rectangle, choose a known or findable side as the base, then find the perpendicular height to that base. If the base lies on a side of the rectangle, the height is often the other side of the rectangle.
Practice: 3 similar questions
- Mei Ling drew a rectangle ABCD, where AB = 24 cm and BC = 14 cm. E is on AB such that AE = 8 cm. F is on CD such that DF = 15 cm. ED, EF and EC are joined. The shaded part is triangle EFC. Find the area of the shaded triangle.
Answer
63 cm² — CD = AB = 24 cm, so FC = 24 − 15 = 9 cm. The base FC lies on CD, so the corresponding height is BC = 14 cm. Area = 9 × 14 ÷ 2 = 63 cm².
- Siti drew on a rectangular card ABCD. AB = 28 cm and BC = 16 cm. G is on AD such that AG = 6 cm. H is on BC such that BH = 10 cm. GH and GC are joined. The shaded part is triangle GHC. Find the area of the shaded triangle.
Answer
84 cm² — BC = 16 cm and BH = 10 cm, so HC = 16 − 10 = 6 cm. The base HC is vertical. The corresponding height is the horizontal distance from G to BC, which is AB = 28 cm. Area = 6 × 28 ÷ 2 = 84 cm².
- Raju drew a rectangle ABCD, where AB = 36 cm and BC = 22 cm. E is on AB. F and G are on CD such that DF = 11 cm and CG = 7 cm. EF and EG are joined. The shaded part is triangle EFG. Find the area of the shaded triangle.
Answer
198 cm² — CD = AB = 36 cm. FG = 36 − 11 − 7 = 18 cm. The base FG lies on CD, so the corresponding height is the height of the rectangle, BC = 22 cm. Area = 18 × 22 ÷ 2 = 198 cm².
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