Maharashtra State Board · Class 10 · Mathematics · Geometry Chapter 1

Similarity — Formula Sheet

Board Formulas

14 formulas

1.Ratio of Areas of Two Triangles★
$\frac{A ( △ A B C )}{A ( △ P QR )} = \frac{B C \times A D}{QR \times P S}$
$A (△ A B C)$ : Area of triangle ABC (cm²) · $B C, QR$ : Bases of the two triangles (cm) · $A D, P S$ : Heights drawn to those bases (cm)
Area of a triangle = ½ × base × height, so the ½ cancels and the ratio of areas is the ratio of the products of base and corresponding height. AD and PS are heights on bases BC and QR.
2.Triangles with Equal Heights
$\frac{A _{1}}{A _{2}} = \frac{b _{1}}{b _{2}}$
$A_{1}, A_{2}$ : Areas of the two triangles · $b_{1}, b_{2}$ : Their corresponding bases
Areas of triangles with equal heights are proportional to their bases. Typical case: two triangles with a common vertex and bases on the same line.
3.Triangles with Equal Bases
$\frac{A _{1}}{A _{2}} = \frac{h _{1}}{h _{2}}$
$A_{1}, A_{2}$ : Areas of the two triangles · $h_{1}, h_{2}$ : Their corresponding heights
Areas of triangles with equal (or common) bases are proportional to their heights.
4.Basic Proportionality Theorem (BPT)★
$D E ∥ B C \Rightarrow \frac{A D}{D B} = \frac{A E}{E C}$
$D, E$ : Points on sides AB and AC where the parallel line cuts them
A line parallel to one side of a triangle, cutting the other two sides in distinct points D and E, divides those sides in the same ratio.
5.Other Forms of BPT
$\frac{A D}{A B} = \frac{A E}{A C}, \frac{D B}{A B} = \frac{E C}{A C}$
Follow from AD/DB = AE/EC by invertendo and componendo (e.g. DB/AD = EC/AE ⇒ AB/AD = AC/AE). Use them when the whole side is given instead of the second part.
6.Converse of BPT
$\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow D E ∥ B C$
If a line divides two sides of a triangle in the same ratio, it is parallel to the third side. Use it to PROVE that two segments are parallel.
7.Property of an Angle Bisector of a Triangle★
$∠ B A D = ∠ D A C \Rightarrow \frac{A B}{A C} = \frac{B D}{D C}$
$A D$ : Bisector of ∠BAC meeting BC at D · $B D, D C$ : Parts of side BC
The bisector of an angle of a triangle divides the opposite side in the ratio of the remaining two sides. The side next to BD is AB — keep them together.
8.Converse of the Angle Bisector Property
$\frac{A B}{A C} = \frac{B D}{D C} \Rightarrow ray A D bisects ∠ B A C$
If D on BC divides it in the ratio of the other two sides, then AD is the bisector of ∠A.
9.Property of Three Parallel Lines and their Transversals
$l ∥ m ∥ n \Rightarrow \frac{A B}{B C} = \frac{P Q}{QR}$
$A B, B C$ : Intercepts on the first transversal · $P Q, QR$ : Corresponding intercepts on the second transversal
Three parallel lines cut any two transversals in the same ratio. A, B, C lie on one transversal and P, Q, R on the other.
10.Similar Triangles
$△ A B C \sim △ P QR \Leftrightarrow ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R and \frac{A B}{P Q} = \frac{B C}{QR} = \frac{A C}{P R}$
Corresponding angles are equal and corresponding sides are in proportion. The order of letters fixes which vertex matches which.
11.AAA / AA Test of Similarity
$∠ A = ∠ P, ∠ B = ∠ Q \Rightarrow △ A B C \sim △ P QR$
Two pairs of equal angles are enough, because the third pair is then equal automatically (angle sum 180°).
12.SAS Test of Similarity
$\frac{A B}{P Q} = \frac{A C}{P R}, ∠ A = ∠ P \Rightarrow △ A B C \sim △ P QR$
Two sides in proportion AND the angle INCLUDED between them equal. An angle that is not between the two sides does not work.
13.SSS Test of Similarity
$\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A C}{P R} \Rightarrow △ A B C \sim △ P QR$
All three pairs of sides in the same ratio. Arrange both triangles' sides in increasing order before comparing.
14.Theorem of Areas of Similar Triangles★
$△ A B C \sim △ P QR \Rightarrow \frac{A ( △ A B C )}{A ( △ P QR )} = \frac{A B ^{2}}{P Q ^{2}} = \frac{B C ^{2}}{Q R ^{2}} = \frac{A C ^{2}}{P R ^{2}}$
$A B, P Q$ : A pair of corresponding sides
Ratio of areas = square of the ratio of corresponding sides. If sides are in the ratio 2 : 3, areas are in the ratio 4 : 9 — not 2 : 3.