Contents

**GSEB Solutions for Class 10 mathematics – Similarity of Triangles (English Medium)**

### Exercise-6.1

**Question 1:**

According to the definition of similarity of triangles, which are the conditions for correspondence DEF ↔ ZXY between ∆DEF and ∆XYZ to be a similarity ?

**Solution :**

**Question 2:**

For ∆PQR and ∆XYZ, the correspondence PQR ↔ YZX is a similarity. m∠p = 2m∠Q and m∠x = 120. Find m∠Y.

**Solution :
**

**Question 3:**

The correspondence ABC ↔ PQR between ∆ABC and ∆PQR is a similarity. AB : PQ = 4 : 5. If AC = 6, then find PR. If QR = 15, then find BC.

**Solution :
**

**Question 4:**

∆PQR ~ ∆DEF for the correspondence PQR ↔ EDF. If PQ + QR = 15, DE + DF = 10 and PR = 6, find EF.

**Solution :**

**Question 5:**

In ∆ABC and ∆PQR, ABC ↔ QPR is a similarity. The perimeter of ∆ABC is 15 and perimeter of ∆PQR is 27. If BC = 7 and QR = 9, find PR and AC.

**Solution :**

**Question 6:**

In ∆XYZ ~ ∆DEF consider the correspondence xyz ↔ EDF.

If \(\frac{perimeter\text{ of }\Delta \text{XYZ}}{perimeter\text{ of }\Delta DEF}=\frac{3}{4}\) , find \(\frac{XY}{ED}\) and \(\frac{XZ+\text{ YZ}}{EF\text{ + DF}}\)

**Solution :**

**Question 7:**

Using the definition of similarity prove that all the isosceles right angled triangles are similar.

**Solution :**

Data: In ∆ ABC, AB = BC and m∠B = 90°.

In ∆ DEF, DE = EF and m∠E = 90°. i.e., ∆ABC and ∆DEF are isosceles right angled triangles.

To prove: ∆ABC and ∆DEF are similar triangles.

Proof: ∆ABC and ∆DEF are both isosceles right angled triangles in which m∠B = m∠Q = 90°.

Also AB = BC and DE = EF, then

In ∆ ABC, m∠B = 90°

m∠A + m∠B + m∠C = 180°

∴m∠A + 90° + m∠C = 180°

∴m∠A + m∠C = 90° … … (1)

Moreover, AB = BC

∴m∠C = m∠A … … (2)

From (1) and (2),

Thus, for the correspondence ABC ↔ DEF between the isosceles right angled triangles ∆ABC and ∆DEF the corresponding angles are congruent and the lengths of the corresponding sides are in proportion. Hence, the correspondence ABC ↔ DEF is a similarity. Therefore, all the isosceles right angled triangles are similar.

**Question 8:**

For ∆ABC and ∆xyz, ABC ↔ xyz is a similarity. If \(\frac{AB}{4}\) = \(\frac{BC}{6}\) = \(\frac{AC}{3}\), AC = 3 and XY = 5, find YZ and XZ.

**Solution :**

**Question 9:**

State whether the following statements are true or false. Give reasons for your answer :

- If ∆PQR and ∆ABC are similar and none of them is equilateral, then all the six correspondences between ∆PQR and ∆ABC are similarities.
- All congruent triangles are similar.
- All similar triangles are congruent.
- If the correspondences ABC ↔ BAC is similarity, then AABC is an isosceles triangle.
- The correspondence PQR ↔ YZX between ∆PQR and ∆YZX is a similarity. If m∠P = 60, m∠R = 40, then m∠Z = 80.

**Solution :**

- Given statement is false.

For equilateral triangles, all the six correspondences are similarity. But in triangles other than equilateral triangles, the measures of all the angles are not same. Hence, any one of the angles of the first triangle cannot be congruent to all the angles of the second triangle. - Given statement is true.

Congruent triangles are equal with respect to size and shape, while for triangles to be similar, it is sufficient that their shapes are same. - Given statement is false .

Similar triangles are equal in shape but not in size. For triangles to be congruent, they must be equal in both, shape as well as size. Hence, all similar triangles are not congruent. - Given statement is true.

For ∆ABC, the correspondence ABC ↔ BAC is a similarity.

Hence, two sides of ∆ABC are congruent and therefore ∆ABC is an isosceles triangle. - Given statement is true.

The correspondence PQR ↔ YZX between ∆PQR and ∆YZX is a similarity.

∴m∠Y = m∠P = 60° and m∠X = m∠R = 40°

In ∆XYZ,

m∠X + m∠Y + m∠Z = 180°

∴40° + 60° + m∠Z = 180°

∴m∠Z = 180 – 100

∴m∠Z = 80

**Question 10:**

Select a proper option (a), (b), (c) or (d) From given options :

- ∆ABC ~ ∆PQR for the correspondence ABC ↔ QRP. If m∠A = 50, m∠C = 30, then m∠R = ……..
- ∆LMN ~ ∆XYZ for the correspondence LMN ↔ ZYX. If m∠Z = 50, m∠X = 40, then m∠L + m∠N = …….
- If the correspondence ABC ↔ EFD is a similarity in ∆ABC and ∆DEF, then .. of the following is not true.
- ABC ~ PQR is a similarity in ∆ABC and ∆PQR. If the perimeter of ∆ABC is 12 and perimeter of a ∆PQR is 20, then AB : PQ = …….

**Question 10(1):**

**Solution :
**d. 100°

In ∆ABC ∽ ∆PQR, the correspondence ABC ↔ QRP is a similarity.

∴m∠Q = m∠A = 50° and m∠P = m∠C = 30°

In ∆ PQR,

m∠P + m∠Q + m∠R = 180°

∴30° + 50° + m∠R = 180°

∴m∠R = 180° – 80°

∴m∠R = 100°

**Question 10(2):**

**Solution :
**b. 90°

In ∆LMN and ∆XYZ, the correspondence ABC ↔ QRP is a similarity

∴m∠L = m∠Z =50° and m∠N = m∠X = 40°

Now,

m∠L + m∠N = 50° + 40° = 90°

**Question 10(3):**

**Solution :**

**Question 10(4):**

**Solution :**

### Exercise-6.2

**Question 1:**

In ∆ABC, a line parallel to \(\overline{BC}\) intersects \(\overline{AB}\) and \(\overline{AC}\) in D and E respectively. Fill in the blanks shown in the table :

**Solution :**

No. |
AD |
DB |
AB |
AE |
EC |
AC |

1. | 3.6 | 2.4 | 6 |
2.7 |
1.8 | 4.5 |

2. | 1.55 |
4.65 |
6.2 | 1.05 |
3.15 | 4.2 |

3. | 9.6 |
12 | 21.6 |
6.4 | 8.0 | 14.4 |

4. | 7.2 | 11.2 |
18.4 | 5.4 | 8.4 |
13.8 |

5. | 3.4 |
3.4 | 6.8 |
2.55 |
2.55 | 5.10 |

**Question 2:**

In ∆ABC, the bisector of ∠C intersects \(\overline{AB}\) in F. If 2AF = 3FB and AC = 7.2 find BC.

**Solution :
**

**Question 3:**

In ∆XYZ, the bisector of ∠Y intersects \(\overline{ZX}\) in P.

- if XP : PZ = 4 : 5and YZ = 6.5, find XY.
- if XY : YZ = 2 : 3and XP = 3.8, find PZ and ZX.

**Solution :**

**Question 4:**

In ∆ABC, the bisector of ∠A intersects BC in D. Prove that BD = \(\frac{BC\times \text{AB}}{AB+BC}\) and DC = \(\frac{BC\times \text{AC}}{AC+AB}\)

**Solution :**

**Question 5:**

\(\square \) ABCD is a trapezium such that\(\overline{AB}\parallel \overline{CD}\) . M ∈ \(\overline{AD}\) and N ∈ \(\overline{AD}\) such that \(\overline{MN}\parallel \overline{AB}\). Prove that \(\frac{AM}{MD}=\frac{BN}{NC}\)

**Solution :**

**Question 6:**

In ∆ABC, D and E are the mid-points of \(\overline{BC}\) and \(\overline{AC}\) respectively. \(\overline{AD}\) and \(\overline{BE}\) intersect in G. A line m passing through D and parallel to \(\overleftrightarrow{BE}\) intersects \(\overline{AC}\) in K. Prove that AC = 4CK.

**Solution :**

**Question 7:**

In ∆PQR, X ∈ \(\overleftrightarrow{QR}\), such that Q – X – R. A line parallel to \(\overline{PR}\) and passing through X intersects \(\overline{PQ}\) in Y. A line parallel to \(\overline{PX}\) and passing through Y intersects \(\overline{QR}\) in Z. Prove that \(\frac{QZ}{ZX}=\frac{QX}{XR}\)

**Solution :**

**Question 8:**

In ∆ABC, X ∈ \(\overleftrightarrow{BC}\) and B – X – C. A line passing through X and parallel to \(\overline{AB}\) intersects \(\overline{AC}\) in Y. A line passing through X and parallel to \(\overline{BY}\) intersects \(\overline{AC}\) in Z. Prove that CY2 = AC ∙ CZ.

**Solution :**

**Question 9:**

In ∆ABC, the bisector of ∠A intersects \(\overline{BC}\) in D and the bisector of ∠ADC intersects \(\overline{AC}\) in E. Prove that AB × AD × EC = AC × BD × AE.

**Solution :**

**Question 10:**

In ∆ABC, D is the mid-point of \(\overline{BC}\) and P is the mid-point of \(\overline{AD}\). \(\overrightarrow{BP}\) intersects \(\overline{AC}\) in Q. Prove that (i) CQ = 2AQ (ii) BP = 3PQ

**Solution :**

### Exercise-6.3

**Question 1:**

In figure 6.27, \(\overline{PQ}\parallel \overline{BC}\), \(\overline{PR}\parallel \overline{AC}\), \(\overline{QS}\parallel \overline{AB}\). Prove that BR = CS.

**Solution :**

**Question 2:**

\(\square \) ABCD is a parallelogram. Prove that ∆ABD and ∆BDC are similar.

**Solution :**

**Question 3:**

In \(\square \)ABCD, M and N are the mid-points of \(\overline{AD}\) and \(\overline{BC}\). If \(\overline{AB}\parallel \overline{CD}\), prove that \(\overline{MN}\parallel \overline{AB}\).

**Solution :**

**Question 4:**

In \(\square \)ABCD, A – P – D, B – Q – C. If \(\overline{AB}\parallel \overline{PQ}\) and \(\overline{PQ}\parallel \overline{DC}\), prove that AP × QC = PD × BQ.

**Solution :**

**Question 5:**

In ∆ABC, the bisector of ∠A intersects \(\overline{BC}\) in D. The bisector of ∠ADB intersects \(\overline{AB}\) in F and the bisector of ∠ADC intersects \(\overline{AC}\) in E. Prove that AF × AB × CE = AE × AC × BF.

**Solution :**

**Question 6:**

In ∆ABC, ∆PQR and ∆XYZ correspondences ABC ↔ PQR, PQR ↔ xyz are similarity. Prove that ABC ↔ XYZ is similarity.

**Solution :**

**Question 7:**

State giving reasons, whether the following statements are True or false :

In all the following questions the line does not contain a side of the triangle.

- A line can be drawn in the plane of a triangle not intersecting any of the sides of a triangle.
- A line can be drawn in the plane of a triangle which is not passing through any of the three vertices and intersecting all the three sides of the triangle.
- If a line drawn in the plane of a triangle intersects the triangle at only one point, the line passes through a vertex of the triangle.
- If a line intersects two of the three sides of a triangle in two distinct points and does not intersect the third side, then the line is parallel to the third side.
- In the plane of ∆ABC, a line l can be drawn such that l ∩ \(\overline{BC}\) = {P}, l ∩ \(\overline{AC}\) = {Q} and l ∩ \(\overline{AB}\) = ∅

**Solution :
**1. The given statement is true.

Reason:

For a line in the plane of a triangle, there are three possibilities:

i. The line does not intersect the triangle.

ii. The line intersects the triangle at one point

iii. The line intersects the triangle at two points.

According to the first possibility, the given statement is true.

2. The given statement is false.

Reason: According to theorem 6.2, if a line lying in the plane of a triangle and not passing through any vertex intersects one side, then it does intersect one more side but does not intersect the third side. Thus, a line not passing through a vertex of a triangle cannot intersect all the three sides.

3. The given statement is true.

Reason: A line intersecting a triangle and not passing through any vertex will intersect the triangle at two points.

4. The given statement is false

Reason: If a line intersects two sides of a triangle and does not intersect the third side, it can intersect the line containing the third side.

5. Correct Answer: The given statement is true.

### Exercise-6.4

**Question 1:**

∠B is a right angle in ∆ABC and \(\overline{BD}\) is an altitude to hypotenuse. AB = 8, BC = 6. Find the area of ∆BDC.

**Solution :**

**Question 2:**

In \(\square \)^{m} ABCD, T ∈ \(\overline{BC}\) and \(\overrightarrow{AT}\) intersects \(\overline{BD}\) in M and \(\overrightarrow{DC}\) in O. Prove that AM^{2} = MT ∙ MO.

**Solution :**

**Question 3:**

In \(\square \)^{m} ABCD, M is the mid-point of \(\overline{BC}\). \(\overrightarrow{DM}\) and \(\overrightarrow{AB}\) intersect in N. Prove that DN = 2MN.

**Solution :**

**Question 4:**

P and Q are the mid-points of \(\overline{AB}\) and \(\overline{AC}\) in AABC. If the area of ∆APQ = 12\(\sqrt{3}\) , find the area of ∆ABC.

**Solution :**

**Question 5:**

\(\square \) ABCD is a rhombus. \(\overline{AC}\) ∩ \(\overline{BD}\) = {O}. Prove that the area of ∆OAB = \(\frac{1}{4}\) (area of \(\square \)ABCD).

**Solution :**

**Question 6:**

In \(\square \) PQRS, \(\overline{PR}\) ∩ \(\overline{QS}\) = {T}, PS = QR, \(\overline{PQ}\parallel \overline{RS}\). Prove that ∆TPS is similar to ∆QTR.

**Solution :**

**Question 7:**

In ∆ABC, a line parallel to \(\overline{BC}\), passes through the mid-point of \(\overline{AB}\). Prove that the line bisects \(\overline{AC}\).

**Solution :**

**Question 8:**

P, Q, R are the mid-points of the sides of ∆ABC. X, Y, Z are the mid-points of the sides of ∆PQR. If the area of ∆XYZ is 10, find the area of ∆PQR and the area of ∆ABC.

**Solution :**

**Question 9:**

In ∆ABC, x ∈ \(\overline{BC}\), Y ∈ \(\overline{CA}\), z ∈ \(\overline{AB}\) such that \(\overline{XY}\parallel \overline{AB}\), \(\overline{YZ}\parallel \overline{BC}\), \(\overline{ZX}\parallel \overline{AC}\). Prove that X, Y, Z are the mid-points of \(\overline{BC}\), \(\overline{CA}\), \(\overline{AB}\) respectively.

**Solution :**

**Question 10:**

Two triangles are similar. Prove that if sides in one pair of corresponding sides are congruent, then the triangles are congruent.

**Solution :**

### Exercise-6

**Question 1:**

In ∆ABC, P ∈ \(\overline{AB}\) such that \(\frac{AP}{PB}=\frac{m}{n}\), m, n are positive real numbers. A line passing through P and parallel to \(\overline{BC}\) intersects \(\overline{AC}\) in Q. Prove that (m + n)^{2} (area of ∆APB) = m^{2}(area of ∆ABC)

**Solution :**

**Question 2:**

D, E and F are the mid-points of \(\overline{BC}\), \(\overline{CA}\) and \(\overline{AB}\) respectively in ∆ABC. Prove that the area of \(\square \) BDEF = \(\frac{1}{2}\) area of ∆ABC.

**Solution :**

**Question 3:**

Can two similar triangles have same area ? If yes, in which case they have the same area ?

**Solution :**

Yes. Two similar triangles can have the same area. If triangles are congruent, then they are similar and have the same area.

∴ Two similar triangles can have the same area.

**Question 4:**

The correspondence ABC ↔ DEF is similarity in ∆ABC and ∆DEF. \(\overline{AM}\) is an altitude of ∆ABC and \(\overline{DN}\) is an altitude of ∆DEF. Prove that AB × DN = AM × DE.

**Solution :**

**Question 5:**

Explain with reasons, whether the following statements are true or false :

- AAA criterion of similarity of triangles can not be the criterion for congruence of triangles.
- SAS criterion for congruence of triangles can not be a criterion for similarity of triangles.
- Two congruent triangles have the same area.
- Two similar triangles always have the same area.
- Area of similar triangles are proportional to the squares of measures of their corresponding angles.

**Solution :**

- The given statement is true.

Reason: By the AAA criterion, we get the shape of the triangles equal but, for congruence of triangles their size also must be equal. - The given statement is false.

Reason: SAS criterion is also a criterion for similarity. But, for congruence, the corresponding sides should be congruent while for similarity, the corresponding sides must be proportionate. - The given statement is true.

Reason: The shape and size (area) of two congruent triangles is equal. For a closed geometric figure, its size represents its area. - The given statement is false.

Reason: The shape of two similar triangles is equal but their sizes (areas) are not equal. - The given statement is false.

Reason: The areas of similar triangles are proportional to the squares of their corresponding sides. The measures of corresponding angles of two similar triangles are always equal.

**Question 6:**

in the blanks so that the following statements are true :

**Question 6(1):**

\(\overline{AD}\) and \(\overline{BE}\) are the altitudes of ∆ ABC. If AB = 12, AC = 9.9, AD = 8.1 and BE = 7.2 perimeter of ∆ ABC = …….

**Solution :**

**Question 6(2):**

D, E, F are respectively the mid-points of \(\overline{PQ}\), \(\overline{QR}\), \(\overline{PR}\) of ∆ PQR. The correspondence DEF ↔ …….. is similarity.

**Solution :
**

**Question 6(3):**

In ∆ABC, \(\overline{AM}\) and \(\overline{CN}\) are altitudes. If AB = 12, BC = 15 and AM = 9.6, then CN = ……….

**Solution :**

**Question 6(4):**

Areas of two similar triangles are 25 and 16. The ratio of the perimeters of the triangles Is ………..

**Solution :**

**Question 6(5):**

Area of ∆ABC = 36 and area of ∆PQR = 64. The correspondence ABC ↔ PQR is a similarity. If AB = 12, then PQ = ……..

**Solution :**

**Question 6(6):**

In ∆ABC, A – M – B, A – N – C and \(\overline{MN}\parallel \overline{BC}\). If AM = 8.4, AN = 6.4, CN = 19.2, then AB = ………

**Solution :**

**Question 6(7):**

The correspondence ABC ↔ PQR is a similarity in ∆ABC and ∆PQR. AB = 16, AC = 8, PQ = 24, BC = 12, then QR + PR = ……..

**Solution :**

**Question 6(8):**

The correspondence ABC → XYZ is a similarity in ∆ ABC and ∆ XYZ. ABC = 72, BC = 6, YZ = 10. Then XYZ = …… .

**Solution :**

**Question 6(9):**

\(\square \) ABCD is trapezium in which \(\overline{AD}\parallel \overline{BC}\). The diagonals intersect in P. If PD = 9, AP = 5, PB = 7.2, then AC = ……

**Solution :**

**Question 6(10):**

In ∆ABC, m∠B = 90 and \(\overline{BD}\) is an altitude. The correspondence BDA ↔ …….. between ∆ BDA and ∆ BDC is a similarity.

**Solution :**

Answer : CDB

In ∆BDA and ∆BDC,

∠BDA ≅ ∠CDB (right angles)

∠ABD ≅ ∠BCD (measure of both being 90 – m∠A)

Hence, by AA corollary the correspondence BDA ↔ CBD is a similarity

**Question 6(11):**

The correspondence ABC ↔ ZXY is a similarity in ∆ ABC and ∆ XYZ. If AB = 12, BC = 9, CA = 7.5 and ZX = 10, then YZ + XY = ………

**Solution :**

**Question 7:**

Select a proper option (a), (b), (c) or (d) from given options :

**Question 7(1):**

ABC ↔ DEF is a similarity in ∆ABC and ∆DEF, m∠A = 40, m∠E + m∠F = …….

**Solution :
**c. 140°

For ∆ABC and ∆DEF, the correspondence ABC ↔ DEF is a similarity.

∴ m∠A = m∠D

∴ m∠D = 40°

In ∆DEF,

m∠D + m∠E + m∠F = 180°

∴ 40° + m∠E + m∠F = 180°

∴m∠E + m∠F = 140°

**Question 7(2):**

In ∆ABC, M ∈ \(\overline{AB}\), N ∈ \(\overline{AC}\) such that \(\overline{MN}\parallel \overline{BC}\) ……… is not true.

**Solution :**

**Question 7(3):**

In ∆ABC, B – M – C and A – N – C, \(\overline{MN}\parallel \overline{AB}\) If NC : NA = 1 : 3 and CM = 4, then BC = …….

**Solution :**

**Question 7(4):**

In ∆XYZ and ∆PQR, XYZ ↔ PQR is similarity. XY = 12, YZ = 8, ZX = 16, PR = 8. So, PQ + QR = ………

**Solution :**

**Question 7(5):**

The bisector of ∠P in ∆PQR intersects \(\overline{QR}\) in D. If QD : RD = 4 : 7 and PR = 14, then PQ = …….

**Solution :**

**Question 7(6):**

The lengths of the sides \(\overline{BC}\), \(\overline{CA}\), \(\overline{AB}\) of ∆ABC are in the ratio 3 : 4 : 5. Correspondence ABC ↔ PQR is similarity. If PR = 12, the perimeter of ∆PQR is ……

**Solution :**

**Question 7(7):**

The correspondence ABC ↔ YZX in ∆ABC and ∆XYZ is similarity. m∠B + m∠C = 120. So, m∠Y = …….

**Solution :**

d. 60

In ∆ABC,

m∠A + m∠B + m∠C = 180°

∴m∠A + 120° = 180°

∴m∠A = 60°

In ∆ABC and ∆XYZ, the correspondence ABC ↔ YZX is a similarity.

∴ ∠A ≅ ∠Y

∴ m∠Y = m∠A

∴ m∠Y = 60°

**Question 7(8):**

Correspondence ABC ↔ DEF of ∆ABC and ∆DEF is similarity. If AB + BC = 10 and DE + EF = 12 and AC = 6, then DF = ……

**Solution :**

**Question 7(9):**

The lengths of the sides of ∆DEF are 4, 6, 8. ∆DEF ~ ∆PQR for correspondence DEF ↔ QPR. If the perimeter of ∆PQR = 36, then the length of the smallest side of ∆PQR is ……

**Solution :**

**Question 7(10):**

The bisector of ∠B intersects \(\overline{AC}\) in D. If BA = 12 and BC = 16, AD = 9, then AC = ………

**Solution :**

**Question 7(11):**

In ∆ABC, \(\overline{PQ}\parallel \overline{BC}\), P ∈ \(\overline{AB}\), Q ∈ \(\overline{AC}\). If 4AP = AB and AQ = 4, then AC = ……

**Solution :**

**Question 7(12):**

In ∆ABC, the correspondence ABC ↔ BAC is similarity…… of the following is true.

**Solution :
**c. ∠A ≅ ∠B

In ∆ABC, the correspondence ABC ↔ BAC is a similarity.

∴ ∠A ≅ ∠B, ∠B ≅ ∠A and ∠C ≅ ∠C

**Question 7(13):**

In ∆ABC, A – P – B, A – Q – C and \(\overline{PQ}\parallel \overline{BC}\). If PQ = 5, AP = 4, AB = 12, then BC = ……

**Solution :**

**Question 7(14):**

In ∆PQR, P – M – Q and P – N – R. If PQ = 18, PM = 12, PR = 9 and NR = …….., then \(\overline{MN}\parallel \overline{QR}\).

**Solution :**