Areas Of Two Similar Triangles
Theorem 1: The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides.
Given: Two triangles ABC and DEF such that ∆ABC ~ ∆DEF.
To Prove: (frac{Area (Delta ABC)}{Area (Delta DEF)}=frac{A}{D}=frac{B}{E}=frac{A}{D})
Construction: Draw AL ⊥ BC and DM ⊥ EF.
Proof: Since similar triangles are equiangular and their corresponding sides are proportional. Therefore,
∆ABC ~ ∆DEF
⇒ ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and (frac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF} ) ….(i)
Thus, in ∆ALB and ∆DME, we have
⇒ ∠ALB = ∠DME [Each equal to 90º]
and, ∠B = ∠E [From (i)]
So, by AA-criterion of similarity, we have
∆ALB ~ ∆DME
(Rightarrow frac{AL}{DM}=frac{AB}{DE} ) ….(ii)
From (i) and (ii), we get
(frac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF}=frac{AL}{DM} )
Now,
(Rightarrow frac{Area (Delta ABC)}{Area (Delta DEF)}=frac{frac{1}{2}(BCtimes AL)}{frac{1}{2}(EFtimes DM)})
(Rightarrow frac{Area (Delta ABC)}{Area (Delta DEF)}=frac{BC}{EF}times frac{AL}{DM} )
(Rightarrow frac{Area (Delta ABC)}{Area (Delta DEF)}=frac{BC}{EF}times frac{BC}{EF}text{ }left[ From (iii), frac{BC}{EF}=frac{AL}{DM} right])
(Rightarrow frac{Area (Delta ABC)}{Area (Delta DEF)}=frac{B}{E} )
(Butfrac{BC}{EF}=frac{AB}{DE}=frac{AC}{DF} )
(Rightarrow frac{B}{E}=frac{A}{D}=frac{A}{D} )
(Hence,frac{Area (Delta ABC)}{Area (Delta DEF)}=frac{A}{D}=frac{B}{E}=frac{A}{D} )
Theorem 2: If the areas of two similar triangles are equal, then the triangles are congruent i.e. equal and similar triangles are congruent.
Given: Two triangles ABC and DEF such that
∆ABC ~ ∆DEF and Area (∆ABC) = Area (∆DEF).
To Prove: We have,
∆ABC ≅ ∆DEF
Proof: ∆ABC ~ ∆DEF
∠A = ∠D, ∠B = ∠E, ∠C = ∠F and (frac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF} )
In order to prove that ∆ABC ≅ ∆DEF, it is sufficient to show that AB = DE, BC = EF and AC = DF.
Now, Area (∆ABC) = Area (∆DEF)
(Rightarrow frac{Area (Delta ABC)}{Area (Delta DEF)}=1)
(Rightarrow frac{A}{D}=frac{B}{E}=frac{A}{D}=1text{ }left[ because frac{Area (Delta ABC)}{Area (Delta DEF)}=frac{A}{D}=frac{B}{E}=frac{A}{D} right] )
⇒ AB2 = DE2, BC2 = EF2 and AC2 = DF2
⇒ AB = DE, BC = EF and AC = DF
Hence, ∆ABC ≅ ∆DEF.
Read More:
- Angle Sum Property of a Triangle
- Median and Altitude of a Triangle
- The Angle of An Isosceles Triangle
- Area of A Triangle
- To Prove Triangles Are Congruent
- Criteria For Similarity of Triangles
- Construction of an Equilateral Triangle
- Classification of Triangles
Areas Of Two Similar Triangles With Examples
Example 1: The areas of two similar triangles ∆ABC and ∆PQR are 25 cm2 and 49 cm2 respectively. If QR = 9.8 cm, find BC.
Sol. It is being given that ∆ABC ~ ∆PQR,
ar (∆ABC) = 25 cm2 and ar (∆PQR) = 49 cm2.
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
(therefore text{ }frac{ar (Delta ABC)}{ar (Delta PQR)}=frac{B}{Q} )
(Rightarrow frac{25}{49}=frac}} )
(Rightarrow =left( frac{25}{49}times 9.8times 9.8 right) )
(Rightarrow x=sqrt{frac{25}{49}times 9.8times 9.8}=left( frac{5}{7}times 9.8 right)=left( 5times 1.4 right)=7 )
Hence BC = 7 cm.
Example 2: In two similar triangles ABC and PQR, if their corresponding altitudes AD and PS are in the ratio 4 : 9, find the ratio of the areas of ∆ABC and ∆PQR.
Sol. Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.
(therefore text{ }frac{Area (Delta ABC)}{Area (Delta PQR)}=frac{A}{P})
(Rightarrow frac{Area,(Delta ABC)}{Area (Delta PQR)}==frac{16}{81} ) [∵ AD : PS = 4 : 9]
Hence, Area (∆ABC) : Area (∆PQR) = 16 : 81
Example 3: If ∆ABC is similar to ∆DEF such that ∆DEF = 64 cm2, DE = 5.1 cm and area of ∆ABC = 9 cm2. Determine the area of AB.
Sol. Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
(therefore text{ }frac{Area (Delta ABC)}{Area,,(Delta DEF)}=frac{A}{D} )
(Rightarrow frac{9}{64}=frac{A}} )
(Rightarrow AB=sqrt{3.65} )
⇒ AB = 1.912 cm
Example 4: If ∆ABC ~ ∆DEF such that area of ∆ABC is 16cm2 and the area of ∆DEF is 25cm2 and
BC = 2.3 cm. Find the length of EF.
Sol. We have,
(frac{text{Area} text{(}Delta text{ABC})}{Area (Delta DEF)}=frac{B}{E} )
(Rightarrow frac{16}{25}=frac}{E} )
(Rightarrow EF=sqrt{8.265}~~=text{ }2.875text{ }cm)
Example 5: In a trapezium ABCD, O is the point of intersection of AC and BD, AB || CD and AB = 2 × CD. If the area of ∆AOB = 84 cm2. Find the area of ∆COD.
Sol. In ∆AOB and ∆COD, we have
∠OAB = ∠OCD (alt. int. ∠s)
∠OBA = ∠ODC (alt. int. ∠s)
∴ ∆AOB ~ ∆COD [By AA-similarity]
(Rightarrow frac{ar (Delta AOB)}{ar (Delta COD)}=frac{A}{C}=frac}{C} ) [∵ AB = 2 × CD]
(Rightarrow frac{4times C}{C}=4 )
⇒ ar (∆COD) = 1/4 × ar (∆AOB)
(Rightarrow left( frac{1}{4}times 84 right)c=21c )
Hence, the area of ∆COD is 21 cm2.
Example 6: Prove that the area of the triangle BCE described on one side BC of a square ABCD as base is one half the area of the similar triangle ACF described on the diagonal AC as base.
Sol. ABCD is a square. ∆BCE is described on side BC is similar to ∆ACF described on diagonal AC.
Since ABCD is a square. Therefore,
AB = BC = CD = DA and, AC = √2 BC [∵ Diagonal = √2 (Side)]
Now, ∆BCE ~ ∆ACF
(Rightarrow frac{Area (Delta BCE)}{Area (Delta ACF)}=frac{B}{A} )
(Rightarrow frac{Area (Delta BCE)}{Area (Delta ACF)}=frac{B}}=frac{1}{2} )
⇒ Area (∆BCE) = (frac { 1 }{ 2 }) Area (∆ACF)
Example 7: D, E, F are the mid-point of the sides BC, CA and AB respectively of a ABC. Determine the ratio of the areas of ∆DEF and ∆ABC.
Sol. Since D and E are the mid-points of the sides BC and AB respectively of ∆ABC. Therefore,
DE || BA
DE || FA ….(i)
Since D and F are mid-points of the sides BC and AB respectively of ∆ABC. Therefore,
DF || CA ⇒ DF || AE
From (i), and (ii), we conclude that AFDE is a parallelogram.
Similarly, BDEF is a parallelogram.
Now, in ∆DEF and ∆ABC, we have
∠FDE = ∠A [Opposite angles of parallelogram AFDE]
and, ∠DEF = ∠B [Opposite angles of parallelogram BDEF]
So, by AA-similarity criterion, we have
∆DEF ~ ∆ABC
(Rightarrow frac{Area (Delta DEF)}{Area (Delta ABC)}=frac{D}{A}=frac}{A}=frac{1}{4}text{ }left[ because DE =frac{1}{2}AB right] )
Hence, Area (DDEF) : Area (DABC) = 1 : 4.
Example 8: D and E are points on the sides AB and AC respectively of a ∆ABC such that DE || BC and divides ∆ABC into two parts, equal in area. Find .
Sol. We have,
Area (∆ADE) = Area (trapezium BCED)
⇒ Area (∆ADE) + Area (∆ADE)
= Area (trapezium BCED) + Area (∆ADE)
⇒ 2 Area (∆ADE) = Area (∆ABC)
In ∆ADE and ∆ABC, we have
∠ADE = ∠B [∵ DE || BC ∴ ∠ADE = ∠B (Corresponding angles)]
and, ∠A = ∠A [Common]
∴ ∆ADE ~ ∆ABC
( Rightarrow frac{Area (Delta ADE)}{Area (Delta ABC)}=frac{A}{A} )
(Rightarrow frac{Area (Delta ADE)}{2,Area,(Delta ADE)}=frac{A}{A} )
(Rightarrow frac{1}{2}=Rightarrow frac{AD}{AB}=frac{1}{sqrt{2}} )
⇒ AB = √2 AD AB = √2 (AB – BD)
⇒ (√2 – 1) AB = √2 BD
(Rightarrow frac{BD}{AB}=frac{sqrt{2}-1}{sqrt{2}}=frac{2-sqrt{2}}{2} )
Example 9: Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. Find the ratio of their corresponding heights.
Sol. Let ∆ABC and ∆DEF be the given triangles such that AB = AC and DE = DF, ∠A = ∠D.
and ( frac{Area (Delta ABC)}{Area (Delta DEF)}=frac{16}{25} ) …….(i)
Draw AL ⊥ BC and DM ⊥ EF.
Now, AB = AC, DE = DF
( Rightarrow frac{AB}{AC}=1text{ and }frac{DE}{DF}=1 )
( Rightarrow frac{AB}{AC}=frac{DE}{DF}text{ }Rightarrow text{ }frac{AB}{DE}=frac{AC}{DF} )
Thus, in triangles ABC and DEF, we have
( frac{AB}{DE}=frac{AC}{DF}text{ and } ) and ∠A = ∠D [Given]
So, by SAS-similarity criterion, we have
∆ABC ~ ∆DEF
( Rightarrow frac{Area (Delta ABC)}{Area (Delta DEF)}=frac{A}{D} )
( Rightarrow frac{16}{25}=frac{A}{D} ) [Using (i)]
( frac{AL}{DM}=frac{4}{5} )
AL : DM = 4 : 5
Example 10: In the given figure, DE || BC and DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED.
Sol. ∆ADE ~ ∆ABC.
( therefore frac{ar(Delta ADE)}{ar(Delta ABC)}=frac{D}{B}===frac{9}{25} )
Let ar (∆ADE) = 9x sq units
Then, ar (∆ABC) = 25x sq units
ar (trap. BCED) = ar (∆ABC) – ar (∆ADE)
= (25x – 9x) = (16x) sq units
( therefore frac{ar(Delta ADE)}{ar(trap.BCED)}=frac{9x}{16x}=frac{9}{16} )
