Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E.

Other Exercises

Question 1.
In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q1.1
Solution:
In the given figure,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q1.3

Question 2.
In the following figure, ABCD to a trapezium with AB // DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm. Calculate:
(i) EC
(ii) AF
(iii) PE
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q2.1
Solution:
In the figure,
ABCD is a trapezium
AB || DC
AB = 9 cm, DC = 18 cm, CF = 13.5 cm AP = 6 cm and BE = 15 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q2.3

Question 3.
In the following figure, AB, CD and EF are perpendicular to the straight line BDF.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q3.1
Solution:
In the given figure,
AB, CD and EF are perpendicular to the line BDF
AB = x, CD = z, EF = y
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q3.3

Question 4.
Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that:
\(\frac { AB }{ PQ }\) = \(\frac { AD }{ PM }\)
Solution:
∆ABC ~ ∆PQR
AD and PM are the medians of ∆ABC and ∆PQR respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q4.2

Question 5.
Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, prove that: \(\frac { AB }{ PQ }\) = \(\frac { AD }{ PM }\).
Solution:
Given, ∆ABC ~ ∆PQR
AD and PM are altitude of these two triangles
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q5.2

Question 6.
Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: \(\frac { AB }{ PQ }\) = \(\frac { AD }{ PM }\)
Solution:
Given, ∆ABC ~ ∆PQR
AD and PM are the angle bisectors of ∠A and ∠P respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q6.2

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q7.3
But ∠AXY = ∠AYX is given
∠B = ∠C
AC = AB (Side opposite to equal angles)
∆ABC is an isosceles triangle.

Question 8.
In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q8.1
Solution:
In the given figure,
l || m || n
Transversal p and q intersects them at A, B, C and P, Q, R respectively as shown in the given figure.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q8.3

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q9.2

Question 10.
In the figure given below, AB // EF // CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.
Calculate:
(i) EF
(ii) AC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q10.1
Solution:
In the given figure,
AB || EF || CD
AB = 22.5 cm, EP = 7.5 cm
PC = 15 cm and DC = 27 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q10.3

Question 11.
In quadrilateral ABCD, its diagonals AC and BD intersect at point O such that
\(\frac { OC }{ OA }\) = \(\frac { OD }{ OB }\) = \(\frac { 1 }{ 3 }\)
Prove that:
(i) ∆OAB ~ ∆OCD
(ii) ABCD is a trapezium.
Further if CD = 4.5 cm; find the length of AB.
Solution:
In quadrilateral ABCD, diagonals AC and BD intersect each other at O and
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q11.2

Question 12.
In triangle ABC, angle A is obtuse and AB = AC. P is any point in side BC. PM ⊥ AB and PN x AC.
Prove that: PM x PC = PN x PB
Solution:
Given, AB = AC
Since equal sides has equal angle opposite to it
∠B = ∠C …(i)
In ∆PMB and ∆PNC, we have
∠B = ∠C [using (i)]
∠PMB = ∠PNC (each 90°)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q12.1

Question 13.
In triangle ABC, AB = AC = 8 cm, BC = 4 cm and P is a point in side AC such that AP = 6 cm. Prove that ∆BPC is similar to ∆ABC. Also, find the length of BP.
Solution:
In ∆ABC,
AB = AC = 8 cm
BC = 4cm
P is a point on AC such that AP = 6 cm
PB and PC are joined
To prove: ∆BPC ~ ∆ABC
and find length of BP
Proof: AC = 8 cm and AP = 6 cm
PC = AC – AP = 8 – 6 = 2 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q13.1

Question 14.
In ∆ABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA.
(ii) Find BC and CD.
(iii) Find area of ∆ACD : area of ∆ABC. (2014)
Solution:
In ∆ACD and ∆BCA
∠C = ∠C (common)
∠ABC = ∠CAD (Given)
∆ACD = ∆BCA (by AA axiom)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q14.2

Question 15.
In the given triangle P, Q and R are the mid-points of sides AB, BC and AC respectively. Prove that triangle PQR is similar to triangle ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q15.1
Solution:
Given : P and R the mid points of AB and AC respectively.
PR || BC and PR = \(\frac { 1 }{ 2 }\) BC = BQ.
PRQB is a || gm.
∠B = ∠PRQ ….(i)
Similarly, Q and R are the mid points of sides. BC and AC respectively
RQ || AB and QR = \(\frac { 1 }{ 2 }\) AB = AP
APQR is a ||gm.
∠A = ∠PQR ….(ii)
Similarly, we can prove that ∠C = ∠RPQ.
Now in ∆PQR and ∆ABC,
∠PQR = ∠A , ∠PRQ = ∠B and ∠RPQ = ∠C
(i) In ∆BCE
D is mid-point of BC and DF || CE
∆PQR ~ ∆ABC (AAA criterion of similarity)

Question 16.
In the following figure, AD and CE are medians of A ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB
(ii) AG : GD = 2 : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q16.1
Solution:
Proof:
(i) In ∆BCE
D is the mid point of BC and DF || CE
E is mid-point of BE and EF = FB
(ii) AE = EB (E is mid point of AB)
and EF = FB (Proved)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q16.2

Question 17.
In the given figure, triangle ABC is similar to triangle PQR. AM and PN are altitudes whereas AX and PY are medians.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q17.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q17.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q17.3

Question 18.
The two similar triangles are equal in area. Prove that the triangles are congruent.
Solution:
Given : ∆ABC ~ ∆PQR and are equal in area
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q18.2

Question 19.
The ratio between the altitudes of two similar triangles is 3 : 5; write the ratio between their:
(i) medians
(ii) perimeters
(iii) areas
Solution:
∆ABC ~ ∆PQR
AL ⊥ BC and PM ⊥ QR
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q19.2

Question 20.
The ratio between the areas of two similar triangles is 16 : 25. Find the ratio between their:
(i) perimeters
(ii) altitudes
(iii) medians.
Solution:
∆ABC ~ ∆DEF,
AL ⊥ BC and DM ⊥ EF
and AP and DQ are the medians and also
area ∆ABC : area ∆DEF = 16 : 25
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q20.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q20.2

Question 21.
The following figure shows a triangle PQR in which XY is parallel to QR. If PX : XQ = 1 : 3 and QR = 9 cm, find the length of XY. Further, if the area of ∆PXY = x cm²; find in terms of x, the area of :
(i) triangle PQR.
(ii) trapezium XQRY.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q21.1
Solution:
In ∆PQR, XY || QR and PX : XQ = 1 : 3, QR = 9 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q21.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q21.3

Question 22.
On a map, drawn to a scale of 1 : 20000, a rectangular plot of land ABCD has AB = 24 cm, and BC = 32 cm. Calculate :
(i) The diagonal distance of the plot in kilometre
(ii) The area of the plot in sq. km.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q22.1

Question 23.
The dimensions of the model of a multistoreyed building are 1 m by 60 cm by 1.20 m. If the scale factor is 1 : 50,. find the actual dimensions of the building. Also, find :
(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq cm.
(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90 m3.
Solution:
The scale factor is 1 : 50 or k = \(\frac { 1 }{ 50 }\)
Dimension of the building = 100 cm x 60 cm x 120 cm.
k x actual dimensions of the building = Dimension of the model.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q23.1

Question 24.
In a triangle PQR, L and M are two points on the base QR, such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP. Prove that:
(i) ∆PQL ~ ∆RPM
(ii) QL x RM = PL x PM
(iii) PQ² = QR x QL [2003]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q24.1
Solution:
(i) In ∆PQL and ∆RPM
∠PQL = ∠RPM (Given)
∠LPQ = ∠MRP (Given)
∆PQL ~ ∆RPM (AA criterion of similarity)
(ii) ∆PQL ~ ∆RPM (Proved)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q24.2

Question 25.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q25.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q25.2

Question 26.
A triangle ABC with AB = 3 cm, BC = 6 cm and AC = 4 cm is enlarged to ∆DEF such that the longest side of ∆DEF = 9 cm. Find the scale factor and hence, the lengths of the other sides of ∆DEF.
Solution:
In ∆ABC.
AB = 3 cm. BC = 6 cm and AC = 4 cm
In ∆DEF,
Longest side EF = 9 cm
and longest side in ∆ABC is BC = 6 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q26.1

Question 27.
Two isosceles triangles have equal vertical angles. Show that the triangles are similar. If the ratio between the areas of these two triangles is 16 : 25, find the ratio between their corresponding altitudes.
Solution:
Let in two ∆ABC and ∆DEF
The vertical angles of two isosceles triangles are equal i.e. ∠A = ∠D
But AB = DE and AC = DF (isosceles ∆s)
Then base angles are also equal (Angles opposite to equal sides)
The two triangles are similar.
∆ABC ~ ∆DEF
Let AL ⊥ BC and DM ⊥ EF
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q27.1

Question 28.
In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q28.1
Find: (i) area ∆APO : area ∆ABC.
(ii) area ∆APO : area ∆CQO.
Solution:
In ∆ABC,
AP : PB = 2 : 3
PQ || BC and CQ || BA
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q28.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q28.3

Question 29.
The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q29.1
Show that:
(i) ∆ADC ~ ∆BEG
(ii) CA x CE = CB x CD
(iii) ∆ABC ~ ∆DEC
(iv) CD x AB = CA x DE
Solution:
In ∆ABC, AD ⊥ BC and BE ⊥ AC, DE is joined
To prove:
(i) ∆ADC ~ ∆BEG
(ii) CA x CE = CB x CD
(iii) ∆ABC ~ ∆DEC
(iv) CD x AB = CA x DE
Proof:
(i) In ∆ADC and ∆BEC,
∠C = ∠C (common)
∠ABE = ∠BEC (each 90°)
∆ADC ~ ∆BEC (AA axiom)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q29.2

Question 30.
In the given figure, ABC is a triangle-with ∠EDB = ∠ACB. Prove that ∆ABC ~ ∆EBD.
If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of ∆BED = 9 cm². Calculate the
(i) length of AB
(ii) area of ∆ABC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q30.1
Solution:
In ∆ABC and ∆EBD
∠1 = ∠2 (given)
∠B = ∠B (common)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q30.2

Question 31.
In the given figure, ABC is a right-angled triangle with ∠BAC = 90°.
(i) Prove ∆ADB ~ ∆CDA.
(ii) If BD = 18 cm, CD = 8 cm, find AD.
(iii) Find the ratio of the area of ∆ADB is to area of ∆CDA.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q31.1
Solution:
(i) In ∆ADB and ∆CDA :
∠ADB = ∠ADC [each = 90°]
∠ABD = ∠CAD [each = 90° – ∠BAD]
∆ADB ~ ∆CDA [by AA similarity axiom]
(ii) Since, ∆ADB ~ ∆CDA
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q31.2

Question 32.
In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm, DE = 4 cm and AC = 15 cm. Calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q32.1
Solution:
(i) To prove : ∆ABC ~ ∆DEC
In ∆ABC and ∆DEC
∠ABC = ∠DEC = 90°
∠C = ∠C (common)
∆ABC ~ ∆DEC (by AA axiom)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q32.2

Question 33.
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q33.1
(i) ∆ADE ~ ∆ACB.
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of ∆ADE : area of quadrilateral BCED. (2015)
Solution:
In the given figure,
∆ABC is right angled triangle right angle at B.
D is any point on AB and DE ⊥ AC
To prove:
(i) ∆ADE ~ ∆ACB.
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm.Find DE and AD.
(iii) Find, area of ∆ADE : area of quadrilateral BCED
Proof:
(i) In ∆ADE and ∆ACB
∠A = ∠A (common)
∠E = ∠B (each = 90°)
∆ADE ~ ∆ACB. (AAaxiom)
(ii) AC = 13 cm, BC = 5 cm, AE = 4 cm
∆ADE ~ ∆ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q33.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q33.3

Question 34.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q34.1
Solution:
In the given figure, AB || DE, BC || EF
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E Q34.2

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