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Mathematics

Areas of Two Similar Triangles

Gk Scientist May 23, 2022 No Comments
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Areas of Two Similar Triangles:

Theorem 1: Prove that the ratio of the areas of two similar triangles is equal to the ratio of the square of any two corresponding sides.

Areas of Two Similar Triangles in Diagram - Areas of Two Similar Triangles
Given: △ABC ~ △DEF
So, AB/DE = BC/EF = AC/DF
Also, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

To prove: area △ABC/area △DEF = AB2/DE2 = BC2/EF2 = AC2/DF2

Construction: Through A draw AP ⊥ BC and through D draw DQ ⊥ EF.

Proof: area (△ABC) = (1/2) BC x AP; area (△DEF) = (1/2) EF x DQ.
Thus, area △ABC/area △DEF = [(1/2) BC x AP]/[(1/2) EF x DQ]
⇒ area △ABC/area △DEF = (BC/EF) x (AP/DQ) ……….(i)

In △APB and △DQE,
∠1 = ∠2 = 90° [By construction]
∠3 = ∠4 [given]
∴ △APB ~ △DQE [AA corollary]
∴ AB/DE = AP/DQ = BP/EQ
⇒ AB/DE = AP/DQ ……….(ii)
Also, AB/DE = BC/EF [given] ……….(iii)

From (ii) and (iii), we get
AP/DQ = BC/EF ……….(iv)

Putting the value of AP/DQ from (iv) in (i), we get-
area △ABC/area △DEF = (BC/EF) x (BC/EF) = BC2/EF2 ……….(v)

Similarly. it can also be proved that
area △ABC/area △DEF = AB2/DE2 ……….(vi)
and area △ABC/area △DEF = AC2/DF2 ……….(vii)

From (v), (vi) and (vii), we obtain
area △ABC/area △DEF = AB2/DE2 = BC2/EF2 = AC2/DF2

Some More Theorems On Areas:

Theorem 1: The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.

The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes - Areas of Two Similar Triangles
Given: △ABC ~ △DEF and AL ⊥ BC, DM ⊥ EF

To prove: area (△ABC)/area (△DEF) = AL2/DM2

Proof: Since, the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

∴ area (△ABC)/area (△DEF) = AB2/DE2 ……….(i)

In △ALB and △DME,
∠ALB = ∠DME = 90° and ∠B = ∠E [∵ △ABC ~ △DEF]
∴ △ALB ~ △DME [By AAcorollary]
So, AB/DE = AL/DM and therefore, AB2/DE2 = AL2/DM2 ……….(ii)

Substituting for AB2/DE2 in (i), we get
area (△ABC)/area (△DEF) = AL2/DM2

Theorem 2: The areas of two similar triangles are in the ratio of the squares of the corresponding medians.

The areas of two similar triangles are in the ratio of the squares of the corresponding medians - Areas of Two Similar Triangles
Given: △ABC ~ △DEF and AP, DQ are their medians.

To prove: area (△ABC)/area (△DEF) = AP2/DQ2

Proof: Since, the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

∴ area (△ABC)/area (△DEF) = AB2/DE2 ……….(i)

Now, △ABC ~ △DEF
⇒ AB/DE = BC/EF = 2BP/2EQ = BP/EQ ……….(ii)

Thus, AB/DE = BP/EQ and ∠B = ∠E [∵ △ABC ~ △DEF]
∴ △APB ~ △DQE [By SAS similarity]
So, BP/EQ = AP/DQ ……….(iii)

From (ii) and (iii), we get
AB/DE = AP/DQ and therefore, AB2/DE2 = AP2/DQ2 ……….(iv)
∴ area (△ABC)/area (△DEF) = AP2/DQ2 [from (i) and (iv)]

Theorem 3: The areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments.

The areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments diagram - Areas of Two Similar Triangles
Given: △ABC ~ △DEF and AX, DY are the bisectors of ∠A and ∠D respectively.

To prove: area (△ABC)/area (△DEF) = AX2/DY2

Proof: Since, the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have

area (△ABC)/area (△DEF) = AB2/DE2 ……….(i)

Now, △ABC ~ △DEF ⇒ ∠A = ∠D
⇒ (1/2) ∠A = (1/2) ∠D
⇒ ∠BAX = ∠EDY ……….(ii)

∴ In △ABX and △DEY, from (ii)
⇒ ∠BAX = ∠EDY and ∠B = ∠E [∵ △ABC ~ △DEF]
∴ △ABX ~ △DEF [By AA corollary]
∴ AB/DE = AX/DY and therefore, AB2/DE2 = AX2/DY2

Substituting AX2/DY2 for AB2/DE2 from (iii) in (i), we get
area (△ABC)/area (△DEF) = AX2/DY2

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