Similarity

We know that area of triangle = × Base× Height

⇒ Area (triangle 1) = ×9× 5

=

⇒ Area (triangle 2) = ×10× 6

= 30

∴ the ratio of areas of these triangles will be =

=

=

=

Here,ΔABC and ΔADB has common Base.

∴

(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)

⇒

=

=

Considering, Area of (ΔPQR) with base QR

⇒ PS will be the Height

Now, consider the Area of (ΔPQR) with base PR

⇒ QT will be the Height

∵ , the triangle is the same


⇒ the area will be the same irrespective of the base taken.

And we know that area of triangle = × Base× Height

⇒ ×QR×PS

= ×PR×QT

⇒ ×6×6


= ×12×QT

⇒ QT = 3

We can re-draw the fig.1.15(as shown above) where we add DO

which will be height of ΔBCD.

Now,

(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)

⇒

⇒

(∵ the distance between the two parallel lines is always equal ⇒ AP = DO)

⇒ = 1:1

We know that area of triangle = × Base× Height

(i)

(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)

(ii)

(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)

(iii)

(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)

(iv)

=

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

Therefore, we’ll find the ratio for all the triangle.Hence, for

(1)

= 2.33

And

= 2.33

⇒

⇒ In (1), ray PM is a bisector.

(2)

= 0.75

And

= 0.7

⇒

⇒ In (2), ray PM is not a bisector.

(3)

= 1.1

And

= 1.11

⇒

⇒ In (3), ray PM is a bisector.

By Converse of basic Proportionality Theorem

(Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.)

⇒ If, then line NM is parallel to side RQ.

∴ we’ll check if.

⇒

=

=

=

And,

=

=

=

⇒ , therefore line NM || side RQ

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

⇒

⇒

⇒ QP× 5 = 2.5× 7

⇒ QP =

⇒ QP = 3.5

Here, PQ||BC (∵ ∠ APQ≅ ∠ ABC)

(PROPERTY: If a transversal intersects two lines so that corresponding angles are congruent, then the lines are parallel)

∴ By Basic Proportionality Theorem

(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)

⇒

By Basic Proportionality Theorem

(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)

⇒

⇒

⇒ BQ =

⇒ BQ = 17.5

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

And ∵ NQ is angle bisector of ∠N

⇒

⇒

⇒ QP =

⇒ QP = 22.4

Theorem: The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.

⇒

⇒

⇒ x =

⇒ x = 6

Now,AE = AC + CE

= 12 + x

= 12 + 6

⇒ AE = 18

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

⇒

⇒ LT =

⇒ LT =

⇒ LT = 4.8

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

⇒

⇒

⇒ x(x + 2) = (x-2)(x + 5)

⇒ x² + 2x = x²-2x + 5x-10

⇒ x² + 2x-x² + 2x-5x + 10 = 0

⇒ x = 10

Proof: In ΔXDE, PQ||DE….. (Given)

∴ …..(I)

(Basic proportionality theorem)

In ΔXDE, QR||EF …….(Given)

∴ ………(II) (Basic Proportionality Theorem)

∴ ……… from (I) and (II)

∴ seg PR||Seg DE ………..

(converse of basic proportionality theorem)

PROOF:

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

⇒ ……(1)

and …..(2) (∵ , BD and CE are angle bisectors of ∠B and ∠C respectively.)

Now, ∵ seg AB ≅ seg AC

⇒ AB = AC

⇒

⇒ R.H.S of (1) & (2) are equal.

⇒ L.H.S of (1) & (2) will be equal.

∴ Equating L.H.S of (1) &(2), we get-

⇒

⇒ ED||BC (By converse basic proportionality theorem)

With one- to-one correspondence ABC ↔ EDC

∵ ∠ABC ≅ ∠EDC = 75°

∠ACB ≅ ∠ECD (Is common in both the triangles ABC and EDC)

⇒ Δ ABC~Δ EDC ………(By AA Test)

In Δ PQR and Δ LMN

And

And

⇒

⇒ ΔPQR~ΔLMN …………(By SSS Similarity Test)

∵ the shadows are measured at the same time

⇒ angle of of elevation will be equal for both the pole

⇒ Δ PQR~Δ ABC ………(By AA Test)

⇒

⇒ BC

⇒ x =

⇒ x = 12 m

From fig.

⇒ ∠ APC≅ ∠ BQC (∵ AP⊥ BC and BQ⊥ AC)

⇒ Also, ∠ ACP≅ ∠ BCQ (Common)

⇒ Δ CPA~Δ CQB (By AA Test)

⇒

⇒ AC =

⇒AC =

⇒AC = 10.5

Given that, AR = 5AP and AS = 5AQ

⇒ = 5 ………(1)

And = 5……….(2)

⇒

And, ∠ SAR≅ ∠ QAP …… (opposite angles)

⇒ Δ SAR ~ Δ QAP …………(SAS Test of similarity)

⇒ (corresponding sides are proportional)

But, = 5

⇒ = 5

⇒ SR = 5PQ

In Δ AOB and ΔCOD

⇒ ∠ AOB≅ ∠ COD (opposite angles)

⇒ ∠ CDO≅ ∠ ABO (Alternate angles ∵ AB||DC)

⇒ Δ AOB ~ Δ COD (By AA Test)

⇒ (corresponding sides are proportional)

⇒ OD =

⇒ OD =

⇒ OD = 4.5

In Δ CED and ΔBET

⇒ ∠ CED≅ ∠ BET (opposite angles)

⇒ ∠ CDE≅ ∠ BTE (Alternate angles)

(∵ AB||DC ⇒ BT||DC, as BT is extension to AB)

⇒ Δ CED ~ Δ BET (By AA Test)

⇒ (corresponding sides are proportional)

⇒ DE× BE = CE×TE

In Δ APB & Δ CPD

⇒ ……(Given)

And, ∠APB = ∠DPC (vertically opposite angles)

⇒ Δ APB ~ Δ CPD (By SAS Test)

In Δ BAC & Δ ADC

⇒ ∠ BAC ≅ ∠ ADC ……(Given)

And, ∠ACB ≅ ∠DCA ……(common)

⇒ Δ BAC ~ Δ ADC (By AA Test)

⇒ (corresponding sides are proportional)

⇒ CA² = CB×CD

Theorem: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

⇒ Ratio of areas = 3²:5²

⇒ Ratio of areas = 9 : 25

∵ Δ ABC~Δ PQR and AB:PQ = 2:3

⇒

∵ Δ ABC~Δ PQR

⇒ (∵ A(Δ PQR) = 125 is given)

⇒

⇒

⇒

∵ Δ ABC~Δ PQR

⇒ Given that, 9× A(Δ ABC) = 16× A(Δ PQR)

⇒

And,

⇒

⇒

⇒ MN² =

⇒ MN = 15

Let area of one(bigger) triangle be ‘A’, other(smaller) triangle be ‘B’,corresponding side of smaller triangle be ‘a’ and bigger triangle be ‘b’.

⇒ (By theorem)

And a = 12cm, A = 225 sq.cm, B = 81 sq.cm ……(Given)

⇒

⇒ b² =

⇒ b = √400

⇒ b = 20 cm

We know that, all the angles of an equilateral triangles are equal, i.e., 60°.

⇒ Δ ABC~Δ DEF ……(By AAA Similarity Test)

⇒

And, (Given)

⇒

⇒ DE² = 2× 4² (∵ AB = 4)

⇒ DE = _√32

_{⇒DE = 4√2}

A(Δ PQF) = 20units,PF = 2DP,Let us assume DP = x,∴ PF = 2x

⇒ DF = DP + PF = x + 2x = 3x

In Δ FDE & Δ FPQ

∠ FDE ≅ ∠ FPQ (Corresponding angles)

∠ FEP ≅ ∠ FQP (Corresponding angles)

∴ Δ FDE~ Δ FPQ (AA Test)

∴

A(Δ FDE) = A(ΔFPQ) = × 20 = 45

A(□DPQE) = A (Δ FDE) - A(ΔFPQ)

= 45-20

= 25 sq.unit.

∵

⇒ Δ CAB~Δ PQR

(A) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

In Δ DEF & Δ PQR

∠ D≅ ∠ Q and ∠ R≅ ∠ E (Given)

⇒ Δ DEF ~ Δ PQR

⇒ (corresponding sides are proportional)

(A) is matching the solution, hence can’t be false.

(C) is matching the solution, hence can’t be false.

(D) is matching the solution, hence can’t be false.

In Δ ABC & Δ DEF

∠ B≅ ∠ E and ∠ C≅ ∠ F (Given)

⇒ Δ ABC ~ Δ DEF (By AA Test)

⇒ The triangles are similar.

And, Δ ABC ≅ Δ DEF, if AB = DE.

But, given that - AB = 3DE.

⇒ The triangles are not congruent.

(A) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

Solution: We know that, all the angles of an equilateral triangles are equal, i.e., 60°.

⇒ Δ ABC~Δ DEF ……(By AAA Similarity Test)

⇒

And, (Given)

⇒

⇒ DE² = 2× 4² (∵ AB = 4)

⇒ DE = _√32

_{⇒DE = 4√2}

(A) doesn’t match the solution.

(B) doesn’t match the solution.

(C) doesn’t match the solution.

∵ segXY||segBC

⇒∠ AXY≅ ∠ ABC

And, ∠ XAY≅ ∠ BAC (Common)

⇒ Δ AXY~ Δ ABC (By AA Test)

⇒ (corresponding sides are proportional)

⇒

(B) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

Theorem: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

(1)

=

=

=

(2)

=

=

(3)

=

=

=

(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)

⇒

⇒

⇒ base(bigger triangle) = 9 cm

We know that, Area of triangle = × base× height

⇒

=

=

=

We know that, Area of triangle = × base× height

⇒

⇒

⇒

⇒ NR = 11 cm

=

=

By Basic Proportionality Theorem-

⇒

⇒

⇒ 3x = 32-5x

⇒ 8x = 32

⇒ x = 4 = BE

(PROPERTY:If line AX || line BY || line CZ and line l and line m are their transversals then )

⇒

⇒

⇒

[1]

And

⇒

⇒

[2]

And,PS = 280

⇒ PQ + QR + RS = 280 ……(3)

From [1] and [2], we have


From [1],

From [2]

∴.......... (I) theorem of angle bisector.

AND

∴ .......... (II) theorem of angle bisector.

By Bisector Theorem-

⇒ ……(1)

⇒ And, ……(2)

Equating (1) & (2), we get-

⇒

⇒

⇒

=

⇒

⇒ CY =

⇒ CY =

Now,

⇒

⇒

In Δ APD and ΔCPB

⇒ ∠ APD≅ ∠ CPB (opposite angles)

⇒ ∠ ADP≅ ∠ PBC (Alternate angles ∵ AD||BC)

⇒ Δ APD ~ Δ CPB (By AA Test)

⇒ (corresponding sides are proportional)

⇒

ACTIVITY: 2AX = 3BX ∴

……(By Componendo)

…….(I)

Δ BCA~ ΔBYX…… (AA test of similarity).

∴ ………. (corresponding sides of similar triangles).

∴ AC = 15……from (I)

Proof:In □ DEFG is a square

⇒ GF||DE

⇒ GF||BC

Now,In Δ AGF and Δ DBG

⇒ ∠ AGF≅ ∠ DBG (corresponding angles)

⇒ ∠ GDB≅ ∠ FAG (Both are 90° )

⇒ Δ AGF~Δ DBG ……(1) (AA similarity)

Now,In Δ AGF and Δ EFC

⇒ ∠ AFG≅ ∠ ECF (corresponding angles)

⇒ ∠ GAF≅ ∠ FEC (Both are 90° )

⇒ Δ AGF~Δ EFC ……(2) (AA similarity)

From (1) & (2), we have-

⇒ Δ EFC~Δ DBG

⇒

⇒ EF× DG = BD× EC

Now,∵ DEFG is a square

⇒ DE = EF = DG

⇒ DE× DE = BD× EC

⇒ DE² = BD× EC