In ∆ABC, the midpoint of BC is D (see the figure). Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).
Answer: ar(ABD) = ar(ACD) (AD is a median of △ABC) and ar(PBD) = ar(PCD) (PD is a median of △PBC). Subtracting: ar(ABP) = ar(ACP).
Step-by-step solution
Idea: A median divides a triangle into two triangles of equal area. Use it in the big triangle ABC and in the small triangle PBC, then subtract.
- AD is a median of △ABC, so ar(△ABD) = ar(△ACD).1 mark
- PD is a median of △PBC (D is the midpoint of BC), so ar(△PBD) = ar(△PCD).1 mark
- Subtract: ar(△ABD) − ar(△PBD) = ar(△ACD) − ar(△PCD), i.e. ar(△ABP) = ar(△ACP). Shown.1 mark
area(ΔABP) = area(ΔACP), by subtracting the equal areas of ΔPBD and ΔPCD from the equal areas of ΔABD and ΔACD.
Answer to write in the exam
ar(△ABD) = ar(△ACD) (AD is a median of △ABC)
ar(△PBD) = ar(△PCD) (PD is a median of △PBC)
ar(△ABD) − ar(△PBD) = ar(△ACD) − ar(△PCD)
∴ ar(△ABP) = ar(△ACP)
Common mistakes that cost marks
- Claiming △ABP ≅ △ACP. They are not congruent in general; only their areas are equal.
- Adding instead of subtracting the small triangles.
- Forgetting to give the reason ‘median’ for each equality.
How this can come in the exam
MCQ (1 mark)
AD is a median of △ABC and P is the midpoint of AD. If ar(△ABC) = 40 cm2, then ar(△ABP) is
- 5 cm2
- 10 cm2
- 20 cm2
- 15 cm2
Show answer
(B) 10 cm2
ar(△ABD) = 20; BP is a median of △ABD, so ar(△ABP) = 10 cm2.
Try one yourself
In △XYZ, M is the midpoint of YZ and N is any point on XM. If ar(△XYN) = 7 cm2, what is ar(△XZN)?
Show answer
By the same argument, 7 cm2.
More questions like this
- Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (see the figure). What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)?
- In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined (see the figure). Prove that Area (∆BPQ) = 12 Area (∆ABC).
- Why were human beings so fond of using circular shapes? Was this only for practical reasons, or could there have been other reasons too? What kinds of uses have human beings found for the circular shape?
- Such reasoning suggests that for a circle too, if C = circumference and A = area, the ratio C2 : A must be some fixed constant. But what is this constant?
- As the slices become smaller and smaller, the arcs in the figure (B) become more and more closer to a line. This makes the figure more and more closer to a parallelogram with base = half the circumference (why?) = πr, height = radius r.