CDS Mathematics · Triangles

Ratio of Areas of Triangles

Similar triangles have areas in the ratio of the squares of their matching lengths, and triangles with the same height have areas in the ratio of their bases.

All Triangles notes CDS Mathematics strategy

Why this matters

Fifteen PYQs, eleven of them MODERATE. Two ideas cover the page: square the length ratio for similar triangles, and compare bases when the height is shared. The repeated-midpoint questions are the same idea used several times over.

Concept 1 of 3: Areas of similar triangles

Scale a triangle by

k

and both its base and its height grow by

k

, so its area grows by

k^2

. Any matching length will do for

k

: a side, an altitude, a median or the perimeter.

Definition

If two triangles are similar with length ratio $k$ , their areas are in the ratio $k^2$ .

Going back: an area ratio $r$ means a length ratio $\sqrt r$ .
A line $DE \parallel BC$ cuts off triangle $ADE$ similar to $ABC$ with $k = \dfrac{AD}{AB}$ . The remaining trapezium has area $(1 - k^2)$ of the whole.

Area ratio of similar triangles

\dfrac{\text{Area}_1}{\text{Area}_2} = \left(\dfrac{\text{side}_1}{\text{side}_2}\right)^2

Worked example

In triangle

ABC

DE \parallel BC

with

AD : DB = 2 : 3

. Find the ratio of the area of triangle

ADE

to the area of the trapezium

DBCE

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (II) 2017 — Elementary Mathematics · Q82Moderate

Example 1 · Triangles · Ratio of Areas of Triangles

One-fifth of the area of a triangle ABC is cut off by a line DE drawn parallel to BC such that D is on AB and E is on AC. If BC = 10 cm, then what is DE equal to ?

The length ratio is the square root

Cutting off one-third of the area makes

DE = \dfrac{BC}{\sqrt3}

, not

\dfrac{BC}{3}

. And cutting a triangle into two EQUAL parts leaves the small triangle as half the whole, so

k = \dfrac{1}{\sqrt2}

Drill 5 more on areas of similar triangles

Concept 2 of 3: Joining the midpoints, again and again

Joining the midpoints halves every side, so the new triangle has a quarter of the area. Do it again and the area quarters again. Only the number of steps between two triangles matters.

Definition

The triangle joining the midpoints (the medial triangle) has $\dfrac14$ the area and $\dfrac12$ the perimeter.
After $n$ such steps the area is $\left(\dfrac14\right)^n$ of the start.
Two triangles $n$ steps apart have areas in the ratio $4^n : 1$ .

Repeated midpoint triangles

\text{Area after } n \text{ steps} = \dfrac{\text{Area}}{4^{\,n}}

Worked example

A triangle has area

320

. The midpoints of its sides are joined, and the same is done to each new triangle, three times in all. Find the area of the last triangle.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — Elementary Mathematics · Q44Easy

Example 2 · Triangles · Ratio of Areas of Triangles

Suppose

P

Q

and

R

are the mid-points of sides of a triangle of area 128 cm

^2

. If a triangle

ABC

is drawn by joining the mid-points of sides of triangle

PQR

, then what is the area of triangle

ABC

Count the steps, not the positions

The 4th and the 7th triangles are

3

steps apart, so the ratio is

4^3 = 64

, not

4^4

4^7

. Whether the original counts as the first does not matter.

Drill 3 more on joining the midpoints, again and again

Concept 3 of 3: Same height, so compare the bases

Area is half base times height. When two triangles share a vertex and their bases lie on one line, they have the same height, so the ratio of their areas is just the ratio of their bases.

Definition

Triangles with the same height have areas in the ratio of their bases.
Triangles on the same base between the same parallel lines have equal areas.
A median splits a triangle into two equal areas; so for any point $P$ on the median $AD$ , $[PAB] = [PAC]$ .
The centroid $G$ splits the triangle into three equal areas: $[GBC] = [GCA] = [GAB]$ .

Common height

\dfrac{[ABD]}{[ADC]} = \dfrac{BD}{DC}

Worked example

D

is on

BC

with

BD : DC = 3 : 5

. The area of triangle

ABC

64

. Find the area of triangle

ABD

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (II) 2020 — Elementary Mathematics · Q85Moderate

Example 3 · Triangles · Ratio of Areas of Triangles

AD is the median of the triangle ABC. If P is any point on AD, then which one of the following is correct ?

The bases must lie on one line

The shared-height rule needs both bases on the same straight line (or on two parallel lines). Two triangles that share a vertex but have bases pointing different ways do not share a height.

Drill 4 more on same height, so compare the bases

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Areas of similar triangles
Area ratio of similar triangles
$\dfrac{\text{Area}_1}{\text{Area}_2} = \left(\dfrac{\text{side}_1}{\text{side}_2}\right)^2$
Joining the midpoints, again and again
Repeated midpoint triangles
$\text{Area after } n \text{ steps} = \dfrac{\text{Area}}{4^{\,n}}$
Same height, so compare the bases
Common height
$\dfrac{[ABD]}{[ADC]} = \dfrac{BD}{DC}$

Watch out for (3)

The length ratio is the square root
→ Areas of similar triangles
Count the steps, not the positions
→ Joining the midpoints, again and again
The bases must lie on one line
→ Same height, so compare the bases

Test yourself on Triangles

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Triangles chapter test Drill all 15 questions