On a piece of paper, draw and then cut out two copies of the same triangle. Keep one copy aside. On the other, mark the midpoint of each side. (You can do this by folding the paper to join two vertices at a time.) Draw the triangle made by the three midpoints and cut the paper along each of these lines. Now you have four smaller triangles. Compare them to each other and to the intact copy of the original triangle. What do you notice?
Step-by-step solution
Idea: Each small triangle has sides equal to half the sides of the big triangle. The reason is the Midpoint Theorem: the segment joining two midpoints is half the third side and parallel to it.
- Place the pieces on one another. Corner triangles 1 (APQ), 2 (PBR) and 3 (QRC) match by sliding. The middle triangle 4 (PRQ) matches after turning it through 180°. So all four are congruent.1 mark
- Compare a small triangle with the intact copy: its angles are the same as the angles A, B, C of the original, and each side is half of the corresponding side (PQ = ½BC, QR = ½AB, PR = ½AC). So each piece is a half-size copy of the original.1 mark
- Four equal pieces make the whole triangle, so each has one quarter of the area. The pattern holds for every shape of triangle (acute, right, obtuse); the Midpoint Theorem explains why.1 mark
Answer to write in the exam
∆APQ, ∆PBR, ∆QRC, ∆PRQ fit exactly on one another ⇒ all four congruent
PQ = ½BC, QR = ½AB, RP = ½CA; angles equal to ∠A, ∠B, ∠C
∴ Each small triangle is a half-size copy of ∆ABC, with ¼ of its area.
Common mistakes that cost marks
- Saying the middle triangle is different because it “points the other way”. Turned through 180° it fits exactly.
- Saying the small triangles have half the area. Their sides are halved, so the area is one quarter.
- Checking only one triangle shape. Try a long thin triangle and a right triangle too.
How this can come in the exam
P, Q, R are the midpoints of the sides of ∆ABC, whose area is 64 cm². The area of ∆PQR is
- 32 cm²
- 16 cm²
- 21⅓ cm²
- 8 cm²
Show answer
(B) 16 cm²
The midpoints cut ∆ABC into four congruent triangles, so ar(PQR) = 64 ÷ 4 = 16 cm².
Try one yourself
The sides of a triangle are 8 cm, 10 cm and 12 cm. Find the perimeter of the triangle formed by joining the midpoints of its sides.
Show answer
Its sides are 4, 5 and 6 cm, so the perimeter is 15 cm (half of 30 cm).
More questions like this
- Try to prove directly that the four smaller triangles are congruent. You will see that no congruence test applies, because many quantities are unknown.
- The segment joining the midpoints of two sides of a triangle is parallel to the third side and has half the length.
- Read the statement of the Midpoint Theorem again. The following questions seem natural to ask. What can we say about (1) a segment PQ (with P on side AB and Q on side AC) that is parallel to the third side and has half the length, and (2) the line that passes through the midpoint of one side and is parallel to another side? For now, experiment and see if you can make a guess about question (2).
- The line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side (and the resulting segment has half the length of the parallel side).
- BCRP is a parallelogram, so CR = BP. BP = PA and ∠AQP = ∠CQR (vertically opposite angles). Combining with CR ‖ BA, ∆APQ ≅ ∆CRQ. (Why?)
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