Areas of similar triangles

Two similar triangles look the same but can be wildly different sizes — and their areas follow a rule that surprises most students the first time they meet it. If you double every side of a triangle, you do not double the area: you quadruple it. The ratio of the areas of two similar triangles equals the square of the ratio of their corresponding sides. This single theorem sits in the CBSE Class 10 chapter Triangles, right after the similarity criteria (AA, SSS, SAS) and the Basic Proportionality Theorem, and it is a favourite exam question precisely because the squaring step is so easy to forget. Per CBSE, the result is used and applied — no formal proof is demanded — but seeing why it is true makes it impossible to misuse. This article builds that picture, then drills the two skills examiners actually test: reading off the correct side ratio, and squaring it correctly before touching any area.

The core insight

For similar triangles, area ratio equals the square of the corresponding side ratio: ar(△ABC)/ar(△DEF)=(AB/DE)2\text{ar}(\triangle ABC)/\text{ar}(\triangle DEF) = (AB/DE)^2.

The core idea

The theorem. If △ABC∼△DEF\triangle ABC \sim \triangle DEF, then the ratio of their areas equals the square of the ratio of any pair of corresponding sides (equivalently, corresponding altitudes, medians, or perimeters):

ar(△ABC)ar(△DEF)=(ABDE)2=(BCEF)2=(CAFD)2\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{CA}{FD}\right)^2

where: ar(△ABC)\text{ar}(\triangle ABC) and ar(△DEF)\text{ar}(\triangle DEF) are the areas of the two similar triangles, and AB,BC,CAAB, BC, CA correspond respectively to DE,EF,FDDE, EF, FD under the similarity.

Why squared, not linear. Area is a two-dimensional quantity — length times length — while a side is a one-dimensional, linear quantity. Scaling every linear measurement of a shape by a factor kk scales its area by k2k^2. This is true for any similar shapes, not just triangles, but it is examined specifically for triangles here.

The scale-factor reading. Writing k=AB/DEk = AB/DE for the scale factor between the two triangles,

ar(△ABC)ar(△DEF)=k2\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = k^2

where: kk is the ratio of any one pair of corresponding sides (the same value for every pair, since the triangles are similar). Doubling the sides (k=2k = 2) quadruples the area (k2=4k^2 = 4); tripling the sides (k=3k = 3) makes the area nine times as large (k2=9k^2 = 9).

Similarity first, always. The area theorem applies only once similarity has been established. Before writing the area ratio, confirm △ABC∼△DEF\triangle ABC \sim \triangle DEF using AA, SSS, or SAS — the relationship does not hold for triangles that merely look alike or share a few equal parts without full similarity.

Figure
Two similar triangles with sides scaled by factor two and their areas shaded in proportion four to one
Scale sides by k, area scales by k squared DEF, side = s area = 1 unit ABC, side = 2s area = 4 units k = 2
Derivation AProving the area ratio equals the square of the side ratio

The goal is to show, using only the formula for the area of a triangle and the definition of similarity, that when △ABC∼△DEF\triangle ABC \sim \triangle DEF the ratio of their areas is forced to equal the square of the ratio of any pair of corresponding sides — the same squaring that appears whenever a linear scale factor acts on a two-dimensional area, derived here specifically for the triangle case that CBSE examines.

Step 1. Draw altitudes AMAM from AA to BCBC in △ABC\triangle ABC, and DNDN from DD to EFEF in △DEF\triangle DEF. Using the standard area formula, ar(△ABC)=12×BC×AM\text{ar}(\triangle ABC) = \tfrac{1}{2}\times BC \times AM and ar(△DEF)=12×EF×DN\text{ar}(\triangle DEF) = \tfrac{1}{2}\times EF \times DN.

Step 2. Divide the two area formulas: ar(△ABC)ar(△DEF)=12×BC×AM12×EF×DN=BCEF×AMDN\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \dfrac{\tfrac{1}{2}\times BC\times AM}{\tfrac{1}{2}\times EF\times DN} = \dfrac{BC}{EF}\times\dfrac{AM}{DN}. The 12\tfrac{1}{2} factors cancel, leaving the product of a side ratio and an altitude ratio.

Step 3. Show the altitude ratio equals the side ratio. Since △ABC∼△DEF\triangle ABC \sim \triangle DEF, ∠B=∠E\angle B = \angle E. Also ∠AMB=∠DNE=90∘\angle AMB = \angle DNE = 90^\circ by construction. So △ABM∼△DEN\triangle ABM \sim \triangle DEN by AA similarity, which gives AMDN=ABDE\dfrac{AM}{DN} = \dfrac{AB}{DE}. But similarity of the original triangles also gives ABDE=BCEF\dfrac{AB}{DE} = \dfrac{BC}{EF}, so AMDN=BCEF\dfrac{AM}{DN} = \dfrac{BC}{EF} as well.

Step 4. Substitute back into Step 2: ar(△ABC)ar(△DEF)=BCEF×BCEF=(BCEF)2\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \dfrac{BC}{EF}\times\dfrac{BC}{EF} = \left(\dfrac{BC}{EF}\right)^2. Since BC/EFBC/EF equals every other pair of corresponding side ratios under the similarity, the same argument gives (ABDE)2\left(\dfrac{AB}{DE}\right)^2 and (CAFD)2\left(\dfrac{CA}{FD}\right)^2 too, completing the theorem.

Worked example

△ABC∼△PQR\triangle ABC \sim \triangle PQR. If AB=6 cmAB = 6\text{ cm}, PQ=9 cmPQ = 9\text{ cm}, and ar(△ABC)=32 cm2\text{ar}(\triangle ABC) = 32\text{ cm}^2, find ar(△PQR)\text{ar}(\triangle PQR).

Step 1. Confirm similarity is already given (△ABC∼△PQR\triangle ABC \sim \triangle PQR) and set up the side ratio between the matching pair ABAB and PQPQ: ABPQ=69=23\dfrac{AB}{PQ} = \dfrac{6}{9} = \dfrac{2}{3}.

Step 2. Square the side ratio to get the area ratio: ar(△ABC)ar(△PQR)=(23)2=49\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)} = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}.

Step 3. Substitute the known area and solve: 32ar(△PQR)=49\dfrac{32}{\text{ar}(\triangle PQR)} = \dfrac{4}{9}, so ar(△PQR)=32×94=2884=72 cm2\text{ar}(\triangle PQR) = \dfrac{32\times 9}{4} = \dfrac{288}{4} = 72\text{ cm}^2.

Step 4. Sanity-check with the scale factor: from ABCABC to PQRPQR, k=9/6=1.5k = 9/6 = 1.5, so area should scale by k2=2.25k^2 = 2.25, giving 32×2.25=72 cm232\times 2.25 = 72\text{ cm}^2. The area of △PQR\triangle PQR is 72 cm272\text{ cm}^2.

Competitive stretch

In △ABC\triangle ABC, DD and EE lie on ABAB and ACAC with DE∥BCDE \parallel BC, AD=4 cmAD = 4\text{ cm}, DB=5 cmDB = 5\text{ cm}. Since DE∥BCDE \parallel BC, ∠A\angle A is common and ∠ADE=∠ABC\angle ADE = \angle ABC (corresponding angles), so △ADE∼△ABC\triangle ADE \sim \triangle ABC by AA, with AB=AD+DB=9 cmAB = AD + DB = 9\text{ cm}. The correct ratio uses the full side: AD/AB=4/9AD/AB = 4/9, not AD/DBAD/DB. Squaring gives ar(△ADE):ar(△ABC)=(4/9)2=16:81\text{ar}(\triangle ADE):\text{ar}(\triangle ABC) = (4/9)^2 = 16:81.

Common traps

  • Forgetting to square the side ratio before applying it to areas — using AB/DEAB/DE directly instead of (AB/DE)2(AB/DE)^2 is the single most common error on this topic.
  • Using the whole side instead of the correct part when only part of a side is given — the ratio must use ADAD over the full AB=AD+DBAB = AD + DB, not ADAD over DBDB.
  • Applying the theorem before proving similarity — equal angles or one proportional side pair is not enough; a full similarity criterion (AA, SSS, or SAS) must be verified first.
  • Mixing up the direction of the ratio — writing ar(△DEF)/ar(△ABC)\text{ar}(\triangle DEF)/\text{ar}(\triangle ABC) when the side ratio used was AB/DEAB/DE, so numerator and denominator no longer match consistently on both sides of the equation.

Ask the tutor: If you keep mixing up when to square the ratio or which side to use when only part of it is given, ask the AI tutor to walk through a couple of area-ratio problems with you step by step.

Why is it the square of the ratio and not just the ratio itself?

Because area is a two-dimensional quantity (length times length) while a side is only one-dimensional. When every linear measurement of a shape is scaled by a factor kk, every area on that shape scales by k×k=k2k \times k = k^2, since area comes from multiplying two lengths together. This is not special to triangles — it happens for any pair of similar shapes — but CBSE examines it specifically for triangles using the theorem above. The derivation confirms it precisely, by splitting the area formula into a base ratio and an altitude ratio that both turn out to equal the same side ratio.

Can I use the ratio of altitudes or perimeters instead of sides to find the area ratio?

Yes. For similar triangles, the ratios of corresponding sides, corresponding altitudes, corresponding medians, and perimeters are all equal to the same scale factor kk. So squaring any one of them gives the same area ratio: ar(△ABC)/ar(△DEF)=k2\text{ar}(\triangle ABC)/\text{ar}(\triangle DEF) = k^2 regardless of whether kk came from a side pair, an altitude pair, or a perimeter pair. Sides are used most often in problems simply because they are usually the measurements given, but there is nothing special about sides mathematically here.

Do I need to prove the triangles are similar every time before using the area formula?

Yes, unless the question already states it. The area theorem is a consequence of similarity, not a standalone fact, so it only applies once △ABC∼△DEF\triangle ABC \sim \triangle DEF (or whichever triangles are involved) has been established via AA, SSS, or SAS. A common mistake is spotting one equal angle or one proportional side and jumping straight to the area ratio — always write the similarity statement first, with the correct vertex correspondence, before squaring any ratio.

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