Practice Set 1.1 Geometry 10th Standard Maths Part 2 Chapter 1 Similarity Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.1 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.1 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Solution:
Let the base, height and area of the first triangle be b1, h1, and A1 respectively.
Let the base, height and area of the second triangle be b2, h2 and A2 respectively.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 1

[Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ The ratio of areas of the triangles is 3:4.

Question 2.
In the adjoining figure, BC ± AB, AD _L AB, BC = 4, AD = 8, then find \(\frac{A(\Delta A B C)}{A(\Delta A D B)}\)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 2
Solution:
∆ABC and ∆ADB have same base AB.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 3
[Since Triangles having equal base]

Question 3.
In the adjoining figure, seg PS ± seg RQ, seg QT ± seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 4
Solution:
In ∆PQR, PR is the base and QT is the corresponding height.
Also, RQ is the base and PS is the corresponding height.
\(\frac{A(\Delta P Q R)}{A(\Delta P Q R)}=\frac{P R \times Q T}{R Q \times P S}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{1}{1}=\frac{P R \times Q T}{R Q \times P S}\)
∴ PR × QT = RQ × PS
∴ 12 × QT = 6 × 6
∴ QT = \(\frac { 36 }{ 12 } \)
∴ QT = 3 units

Question 4.
In the adjoining figure, AP ⊥ BC, AD || BC, then find A(∆ABC) : A(∆BCD).
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 5
Solution:
Draw DQ ⊥ BC, B-C-Q.

Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 6
AD || BC [Given]
∴ AP = DQ   (i)  [Perpendicular distance between two parallel lines is the same]
∆ABC and ∆BCD have same base BC.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 7

Question 5.
In the adjoining figure, PQ ⊥ BC, AD ⊥ BC, then find following ratios.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 8
Solution:
i. ∆PQB and tPBC have same height PQ.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 9
ii. ∆PBC and ∆ABC have same base BC.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 10
iii. ∆ABC and ∆ADC have same height AD.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 11

Question 1.
Find \(\frac{A(\Delta A B C)}{A(\Delta A P Q)}\)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 12
Solution:
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 13

Class 10 Maths Digest

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