PROPERTIES OF SIMILAR TRIANGLES

Property 1 :

Two triangles are said to be similar if their corresponding sides are proportional.

Property 2 :

A perpendicular line drawn from the vertex of a right angled triangle divides the triangle into two triangles similar to each other and also to original triangle.

Property 3 :

If two triangles are similar, then the ratio of the  corresponding sides are equal to the ratio of their corresponding altitudes.

Property 4 :

If two triangles are similar, then the ratio of the corresponding sides are equal to the ratio of the corresponding perimeters.

Problem 5 :

The ratio of the area of two similar triangles are equal to the ratio of the squares of their corresponding sides.

Problem 6 :

If two triangles have common vertex and their bases are on the same straight line, the ratio between their areas is equal to the ratio between the length of their bases.

Solved Problems

Problem 1 :

Find BH.

Solution :

The sides HI and CD are parallel.

BH/HC = BI/ID

BH/9 = 9/15

BH = 81/15

BH = 5.4

Problem 2 :

Verify WX and DE are parallel.

Solution :

Here

FW/WD = FX/XE

1.5/2.5 = 2.1/3.5

0.6 = 0.6

By Converse of Triangle Proportionality Theorem, WX and DE are parallel.

Problem 3 :

Find x.

Solution :

EDG is a right triangle, EF is the perpendicular drawn from the right angle D.

ΔFDG, ΔEDF and ΔEDG are similar triangles to each other.

In triangles FDG and EDG.

∠DFG = ∠GDE   (A)

∠FGD = ∠EGD  (A)

ΔFDG ~ ΔEDG

In ΔFDG and ΔEDG :

DG/EG = DF/DE

8/10 = x/6

0.8 = x/6

Multiply both sides by 6.

4.8 = x 

Problem 4 :

Find x and y.

Solution :

RST is a right triangle, SV is the perpendicular drawn from the right angle S.

ΔRSV, ΔSVT and ΔRST are similar triangles to each other.

∠RST = ∠SVT (Right Angles)

∠STR = ∠STV (Reflexive Property)

By Angle-Angle Similarity Theorem,

ΔRST ~ ΔSVT

Since ΔRST ~ ΔSVT, the corresponding sides are proportional.

RT/ST = RS/SV = ST/VT

20/x = y/SV = x/4

20/x = x/4

x² = 20(4)

x = √80

x = 4√5

In triangle RST,

RS² + ST² = RT²

y² + (4√5)² = 20²

y² + 80 = 400

Subtract 80 from both sides.

y² = 320

Take square root on both sides.

y = √320

y = 8√5

Problem 5 :

The perimeters of two similar triangles is in the ratio 3 : 4. The sum of their areas is 75 cm². Find the area of each triangle.

Solution :

Given :  Perimeters of two similar triangles is in the ratio

3 : 4

Then,

Perimeter of the 1^st Δ = 3x

Perimeter of the 2^nd Δ = 4x

And also, 

Area of 1^st Δ : Area 2^nd Δ = (3x)² : (4x)²

Area of 1^st Δ : Area 2^nd Δ = 9x² : 16x²

Given : Sum of the areas is 75 cm².

9x² + 16x² = 75

25x² = 75

Divide both sides by 25. 

x² = 3

Area of 1^st Δ = 9(3) = 27 cm²

Area of 2^nd Δ = 16(3) = 48 cm²

Problem 6 :

ΔHIJ ~ ΔXYZ, the ratio of their areas is 25/16, if XY has a length of 40, what is the length of HI?

Solution :

Area of ΔHIJ/Area of ΔXYZ :

= (ratio of corresponding sides)²

25/16 = (HI/XY)²

HI/XY = 5/4

Given XY = 40.

HI/40 = 5/4

HI = (5/4) ⋅ 40

HI = 50

Problem 7 :

Figure A and figure B are similar with a ratio of similarity of 5/4. If the perimeter of figure A is 18 units, what is the perimeter of figure B?

Solution :

Ratio of corresponding sides :

= Perimeter of figure A/ Perimeter of figure A

5/4 = 18/Perimeter of figure B

(Perimeter of figure B)(5/4) = 18

Perimeter of figure B = 18(4/5)

= 72/5

= 14.4

Problem 8 :

If figure A and figure B are similar and the ratio of their perimeters is 17/6, what is their ratio of similarity?

Solution :

Ratio of corresponding sides :

= Perimeter of figure A/Perimeter of figure A

Ratio of corresponding sides :

= 17/6

So, the required ratio is 17:6.

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