High School Mathematics - 2 8.15 Areas of Similar Triangles

Theorem: The ratio of the areas of similar triangles is equal to the ratio of the squares of their corresponding sides.

Given: Triangle ABC and Triangle PQR are similar

To prove: (Area of triangle ABC)/(Area of triangle PQR) = AB^{2}/PQ^{2} = BC^{2}/QR^{2} = CA^{2}/RP^{2} Construction: Draw AD perpendicular to BC and PS perpendicular to QR Proof: (Area of triangle ABC)/(Area of triangle PQR) = (1/2 x BC x AD)/(1/2 x QR x PS)
or (area of ABC)/(area of PQR) = BC/QR x AD/PS ----------equation 1
Now in triangles ADB and PSQ
angle B = angle Q (they are similar triangles)
angle ADB = angle PSQ (90^{o})
Hence the triangles ADB and PSQ are similar
Consequently AD/PS = AB/PQ
But AB/BC = BC/QR ( As ABC and PQR are similar)
Hence AD/PS = BC/QR ---------equation 2
Substituting 2 in 1 we get
(Area of ABC)/(Area of triangle PQR) = BC/QR x BC/QR
(Area of ABC)/(Area of triangle PQR) = BC^{2}/QR

Example: Prove that PM^{2} = QM x MR

Solution: As PM is perpendicular to QR we find that triangles PMQ and PMR are similar
Hence PM/MR = QM/PM (In similar triangles corresponding sides are proportional)
Hence PM^{2} = MR x QM

Directions: Solve the following problems. Also write at least 10 examples of your own.