Numbers & Place Value

Depth · one concept from first principles · step 2 of 4

Moving-point triangle area by subtraction

In a right trapezoid ABCD with right angles at A and B, AD = 10 cm, AB = 12 cm and BC = 15 cm. Point P slides along AB. Find all lengths BP for which △ DPC has at most 8/15 of the trapezoid's area.
10 cm 15 cm 12 cm A D B C P
ExpressionsGeometry
Your answer
How to solve
Strategy Change Focus / Count the Complement — The slanted triangle DPC is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in BP, then solve the area inequality.
1STEP 1

Area of the trapezoid

The parallel sides are AD and BC, and the height is AB.

1/2 × (10 + 15) × 12 = 150
2STEP 2

The two right triangles left over

Let BP = x cm, so AP = 12 - x cm. Both leftover triangles are right triangles, so their legs are base and height.

△ APD = 1/2 × 10 × (12 - x) = 60 - 5x, △ PBC = 1/2 × 15 × x = (15/2)x
3STEP 3

Area of the shaded triangle

Subtract the two right triangles from the trapezoid.

△ DPC = 150 - (60 - 5x + (15/2)x) = 90 - (5/2)x
4STEP 4

Solve the area condition

The shaded area must be at most 8/15 of 150, which is 80.

90 - (5/2)x ≤ 150 × 8/15 = 80 → 10 ≤ (5/2)x → x ≥ 4
5STEP 5

Combine with where P can be

P lies on side AB, which is 12 cm long, so x cannot exceed 12.

4 ≤ x ≤ 12
Answer
at least 4 cm and at most 12 cm (4 ≤ BP ≤ 12)
At x = 4 the shaded area is 90 - 10 = 80 = 8/15 × 150, exactly the limit. At x = 12 (P at A) it is 90 - 30 = 60, which is 1/2 × 10 × 12, the area of triangle DAC. At x = 0 (P at B) it is 90 = 1/2 × 15 × 12, too large.
Takeaway

If a shape is hard to measure, measure what is left around it and subtract.

  • Area of the trapezoid
  • The two right triangles left over
  • Area of the shaded triangle
  • Solve the area condition
  • Combine with where P can be
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