10th Standard Maths 2 Practice Set 1.4 Chapter 1 Similarity Textbook Answers Maharashtra Board
Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.4 Algebra 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.
Class 10 Maths Part 2 Practice Set 1.4 Chapter 1 Similarity Questions With Answers Maharashtra Board
Question 1.
 The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.
 Solution:
 Let the corresponding sides of similar triangles be S1 and S2.
 Let A1 and A2 be their corresponding areas.
 
 ∴ Ratio of areas of similar triangles = 9 : 25
Question 2.
 If ∆ABC ~ ∆PQR and AB : PQ = 2:3, then fill in the blanks.
 Solution:
 
Question 3.
 If ∆ABC ~ ∆PQR, A(∆ABC) = 80, A(∆PQR) = 125, then fill in the blanks.
 Solution:
 
Question 4.
 ∆LMN ~ ∆PQR, 9 × A(∆PQR) = 16 × A(∆LMN). If QR = 20, then find MN.
 Solution:
 9 × A(∆PQR) = 16 × A(∆LMN) [Given]
 ∴ \(\frac{\mathrm{A}(\Delta \mathrm{LMN})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{9}{16}\) (i)
 Now, ∆LMN ~ ∆PQR [Given]
 ∴ \(\frac{\mathrm{A}(\Delta \mathrm{LMN})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{\mathrm{MN}^{2}}{\mathrm{QR}^{2}}\) (ii) [Theorem of areas of similar triangles]
 ∴ \(\frac{\mathrm{MN}^{2}}{\mathrm{QR}^{2}}=\frac{9}{16}\) [From (i) and (ii)]
 ∴ \(\frac{M N}{Q R}=\frac{3}{4}\) [Taking square root of both sides]
 ∴ \(\frac{\mathrm{MN}}{20}=\frac{3}{4}\)
 ∴ MN = \(\frac{20 \times 3}{4}\)
 ∴ MN = 15 units
Question 5.
 Areas of two similar triangles are 225 sq. cm. and 81 sq. cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle.
 Solution:
 Let the areas of two similar triangles be A1 and A2.
 A1 = 225 sq. cm. A2 = 81 sq. cm.
 Let the corresponding sides of triangles be S1 and S2 respectively.
 S1 = 12 cm
 
 ∴ The length of the corresponding side of the bigger triangle is 20 cm.
Question 6.
 ∆ABC and ∆DEF are equilateral triangles. If A(∆ABC): A(∆DEF) = 1:2 and AB = 4, find DE.
 Solution:
 In ∆ABC and ∆DEF,
 
Question 7.
 In the adjoining figure, seg PQ || seg DE, A(∆PQF) = 20 sq. units, PF = 2 DP, then find A (꠸ DPQE) by completing the following activity.
 Solution:
 A(∆PQF) = 20 sq.units, PF = 2 DP, [Given]
 Let us assume DP = x.
 ∴ PF = 2x
 
Maharashtra State Board Class 10 Maths Solutions Part 2
- Similarity Practice Set 1.1 Class 10 Maths Solutions
 - Similarity Practice Set 1.2 Class 10 Maths Solutions
 - Similarity Practice Set 1.3 Class 10 Maths Solutions
 - Similarity Practice Set 1.4 Class 10 Maths Solutions
 - Similarity Problem Set 1 Class 10 Maths Solutions
 - Pythagoras Theorem Practice Set 2.1 Class 10 Maths Solutions
 - Pythagoras Theorem Practice Set 2.2 Class 10 Maths Solutions
 - Pythagoras Theorem Problem Set 2 Class 10 Maths Solutions