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Home » Area of Geometric Shapes » Area of Similar Triangles

Area of Similar Triangles

Last updated: Mar 27, 2021 by Ido Sarig · This website generates income via ads and uses cookies · Terms of use · Privacy policy

How do we find the area of similar triangles?

The area of a triangle is given by the formula (base x height)/2. If we have similar triangles, their sides are proportional with a ratio given by a number called the scale factor.

The same scale factor also applies to other lines in the similar triangles - like their height, or to a combination of those lines like the perimeter.

The area of similar triangles is also proportional, but here the scale factor will be squared - because we are multiplying two line segments (base and height) which are both proportional, and have the same scale factor.

Problem

In the following drawing, AD is parallel to CB, and CD is perpendicular to AB. Find the area of triangle ΔAED.

Area of similar triangles

Strategy

The area of ΔBEC is (base x height)/2, and both the base and the height are given (CE is the height because CE is perpendicular to EB).

The hint in the problem is that AB||CD, so we can use our knowledge of angles created between parallel lines and an intersecting transversal line to show that the two triangles are similar.

∠DAE ≅ ∠CBE, as alternating interior angles between two parallel lines, and as m∠AED=m∠BEC=90 °, ΔAED∼ ΔBEC. The scale factor is BE/AE=6/3=2

We can now either use the scale factor to find the length of ED (ED/AE=CE/EB or ED=8/6 x 3= 4) and then use the formula for an area of a triangle (base x height)/2 to find the area, or find the area of triangle ΔBEC , whose base and height are given, and then use the scale factor squared to find the area of ΔAED.

Solution

(1) CD⊥AB //Given
(2) m∠AED=m∠BEC=90° //(1), defintion of perpendicular lines
(3) AB||CD //Given
(4) ∠DAE ≅ ∠CBE //Alternating interior angles between two parallel lines
(5) ΔAED∼ ΔBEC //Angle-Angle
(6) ED/AE=CE/EB //corresponding sides in similar triangles
(7) ED=8/6 x 3= 4 //CE=8, EB=6, AE=3, given
(8) AreaAED=ED x AE /2 //Formula for area of a triangle
(9) AreaAED=4 x 3 /2 = 6

or

(8) AreaBEC = CE x EB / 2 //Formula for area of a triangle
(9) AreaBEC= 8x 6 /2 = 24
(10) AreaAED = AreaBEC/4 //Area of similar triangle is proportional to scale factor squared
(11) AreaAED=24/4=6

« Concentric Circles Intersected by a Secant
Area of a Rhombus with a 60° angle »

About the Author

Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. His goal is to help you develop a better way to approach and solve geometry problems. You can contact him at [email protected]

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About

Welcome to Geometry Help! I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree in Management of Technology. I'm here to tell you that geometry doesn't have to be so hard! My goal with this website is to help you develop a better way to approach and solve geometry problems, even if spatial awareness is not your strongest quality. Read More…

Geometry Topics

  • Area of Geometric Shapes
  • Circles
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    • Tangent Lines
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    • Congruent Triangles
    • Equilateral Triangles
    • Isosceles Triangles
    • Pythagorean Theorem
    • Right Triangles
    • Similar Triangles
    • Triangle Inequalities

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