Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1 Practice Set 1.4 – Similarity
Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1: Similarity. Marathi or English Medium Students of Class 10 get here Similarity full Exercise Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 2 Solution |
Chapter |
Similarity |
Practice Set |
1.4 |
Practice Set – 1.4
(1) The ratio of corresponding sides of similar triangles is 3:5 ;then find the ratio of their areas
Solution:
Let, the smaller ∆ be ∆ABC
The bigger ∆ be ∆PQR
Given, ∆ABC ~ ∆PQR, ratio of corresponding sides AB:PQ = 3:5
∴A(∆ABC)/A(∆PQR) = AB2/PQ2 – theorem area of similar triangle
= 32/52
= 9/25
∴Ratio of their areas is 9:25
(2) If ∆ABC ~∆PQR and AB:PQ = 2:3, then fill in the blanks.
A(∆ABC)/A(∆PQR) = AB2/– = 22/32 = –/–
Solution:
Given, ∆ABC ~ ∆PQR, AB:PQ = 2:3
∴A(∆ABC)/A(∆PQR) = AB2/PQ2 [ratio of similar triangles]
= 22/32 = 4/9
(3) If ∆ABC ~∆PQR, A (∆ABC) = 80 (∆PQR) = 125, then fill in the blanks.
A(∆ABC)/A(∆….) = 80/125
∴AB/PQ = —-/—-
Solution:
Given, ∆ABC ~ ∆PQR, A(∆ABC) – 30, A (∆PQR) = 125
∴A(∆ABC)/A(∆(PQR) = AB2/PQ2 [Ratio of areas of similar triangle]
Or, 80/125 = AB2/PQ2
Or, AB/PQ = √16/25
Or, AB/PQ = 4/5
(4) ∆LMN ~∆PQR, 9×A (∆PQR) = 16×A (∆LMN). If QR = 20 then find MN.
Solution:
Given, ∆LMN ~ ∆PQR, QR = 20
9×A (∆PQR) = 16×A (∆LNN)
Or, 9/16 = MN2/202 [From equation (i)]
Or, MN2 = 9×20×20/16
Or, MN = √9×25
= 3×5
= 15
(5) Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle.
Solution:
Let, the smaller ∆ be ∆ABC
The bigger ∆ be ∆PQR
Given, A (∆ABC) = 81 cm2, A (∆PQR) = 225 cm2
Let, one of the side of ∆ABC be AB Now,
∴ Corresponding side of ∆PQR be PQ AB = 12cm [Given]
∴ A (∆ABC)A(∆PQR) = AB2/PQ2 [Ratio of areas of similar triangles]
Or, 81/225 = 122/PQ2
Or, PQ = √122×225/81
Or, PQ = 12×15/9
Or, PQ = 20 cm
(6) ∆ABC and ∆DEF are equilateral triangles. If A (∆ABC) : A (∆DEF) = 1:2 and AB = 4, find DE.
Solution:
Given, ∆ABC and ∆DEF are equilateral triangle.
∴Each angle of ∆ABC and ∆DEF is 60.
∴All angles of ∆ABC & ∆DEF are congruent.
∴∆ABC ~ ∆DEF by AA test
Now, A (∆ABC)/A(∆DEF) = 1/2 [Given]
AB = 4 [Given]
∴A(∆ABC)/A(∆DEF) = AB2/DE2 [Theorem of areas of similar triangles]
Or, 1/2 = 42/DE2
Or, DE = √42×2
Or, DE = 4√2
(7) In figure 1.66, seg PQ ||seg DE, A (∆PQF) = 20 units, PF = 2 DP, then find A (DPQE) by completing the following activity.
Solution:
Given, in the adjacent figure,
PQ||DE, A (∆PQF) = 20 unit2
PF = 2DP
Let, assume DP = x
∴ PF = 2x
DF = DP + PF = x + 2x = 3x
In ∆FDE & ∆FPQ,
∠FDE ≅∠FPQ [Corresponding angles]
∠FED ≅∠FQP [Corresponding angles]
∴∆FDE ~ ∆FPQ by AA test
∴ A (∆FDE)/A(∆FPQ) = DF2/PF2 = (3x)/(2x) = 9/4
A (∆FDE) = 9/4 A (∆FPQ)
= 9/4 × 20
= 45 unit2
A (OPQE) = A (∆FDE) – A (∆FPQ)
= 45-20
= 25 unit2
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 1 Similarity Practice Set 1.4
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