AREAS OF SIMILAR TRIANGLES

We know what similar triangles are. Now, let us learn about an interesting theorem related to areas of similar triangles.

The theorem states that:

The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Let us prove this theorem.

Given: ΔABC ∼ ΔXYZ

To prove: https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_16_12_42_41/mathmlequation2863688963922532593.png

Construction: Draw segment AD perpendicular to BC and segment XP perpendicular to YZ.

Proof:

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_16_14_37_26/10.png

From the figure, we have

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_16_13_43_29/mathmlequation8221083839126981574.png

Thus, the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding sides.

Let A1 and A2 be the areas of similar triangles and s1 and s2 be their corresponding sides. Then

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_16_13_43_29/mathmlequation1863652677814819677.png

Now, let us learn to apply this formula with the help of an example.

Consider a trapezium PQRS in which SR = 3 PQ. The diagonals PR and QS intersect each other at O.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_37e95570.png

If the area of ΔPOQ is 9 square cm, then what will be the area of ΔSOR? Relation between areas, heights, medians and perimeters of similar triangles:

Let A1 and A2 be the areas of two similar triangles such that s1 and s2 are their corresponding sides, h1 and h2 are their corresponding heights, m1 and m2 are their corresponding medians and P1 and P2 are their respective perimeters.

Then,

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_16_13_43_29/mathmlequation719540405593227169.png

Let us go through some examples based on the areas of similar triangles.

Example 1: The ratio of areas of two similar triangles is 16:25. Find the ratio of their corresponding sides.

Solution:

We know that,

Ratio of areas of similar triangles = (Ratio of corresponding sides)2

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_89e0217.gif (Ratio of corresponding sides)2

Ratio of corresponding sideshttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_m4691faaa.gif

= 4:5

Example 2: The areas of two similar triangles are 25 cm2 and 100 cm2. If one side of the first triangle is 4 cm, then find the corresponding side of the other triangle.

Solution:

Let ABC and DEF be two triangles whose areas are 25 cm2 and 100 cm2 respectively. Let AB = 4 cm

Then, we have to find DE.

Since the two triangles ABC and DEF are similar,

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_6ff2179d.gif

(DE)2 = 64

DE = 8 cm

Thus, the corresponding side of the other triangle is 8 cm.

Example 3: In a triangle ABC, X, Y, and Z are the mid-points of the sides BC, AC, and AB respectively. Find the ratios of the areas of ΔABC and ΔXYZ.

Solution:

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_m7ea08c7b.png

Here, X, Y, and Z are the mid-points of sides BC, AC, and AB respectively.

We know that the line joining the mid-points of two sides is parallel to the third side and its length is half of the third side.

∴ XY || AB and XY =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_m1fd4a48e.gif

YZ || BC and YZ =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_644d0f24.gif

Again, XZ || AC andhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_m198132e5.gif

As, XY || AB, YZ || BC, and XZ || AC,

∴ Quadrilaterals AYXZ, BXYZ, and CXZY are parallelograms.

∴ ∠BAC = ∠ZXY, ∠ABC = ∠ZYX, and ∠ACB = ∠XZY

Using AAA similarity criterion, we obtain

ΔABC ∼ ΔXYZ

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_m3d8d0451.gif

Area of ΔABC: Area of ΔXYZ = 4:1

Example 4: In the given figure, AB and CD are perpendiculars to the line segment BC. Also, AB = 5 cm, CD = 8 cm, and area of ΔAOB is 175 cm2. Find the area of ΔCOD.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_m511f739d.png

Solution:

Here, ΔAOB and ΔDOC are similar triangles because

∠ABO = ∠DCO (Each 90°)

∠AOB = ∠COD (Vertically opposite angles)

Therefore, by AAA similarity criterion, ΔAOB ∼ ΔDOC

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/173/447/947/10.1.6.4.1_ok_SS_html_74152b41.gif

Thus, the area of ΔCOD is 448 cm2.