Practice Set 1.4 Algebra 10th Std Maths Part 2 Answers Chapter 1 Similarity
Question 1
Maharashtra Board Solution
The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
If ∆ABC ~ ∆PQR and AB : PQ = 2:3, then fill in the blanks.
Solution & Step-by-Step Answer:
Question 3
Maharashtra Board Solution
If ∆ABC ~ ∆PQR, A(∆ABC) = 80, A(∆PQR) = 125, then fill in the blanks.
Solution & Step-by-Step Answer:
Question 4
Maharashtra Board Solution
∆LMN ~ ∆PQR, 9 × A(∆PQR) = 16 × A(∆LMN). If QR = 20, then find MN.
Solution & Step-by-Step Answer:
9 × A(∆PQR) = 16 × A(∆LMN) [Given] ∴ (i) Now, ∆LMN ~ ∆PQR [Given] ∴ (ii) [Theorem of areas of similar triangles] ∴ [From (i) and (ii)] ∴ [Taking square root of both sides] ∴ ∴ MN = ∴ MN = 15 units
Question 5
Maharashtra Board Solution
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 & Step-by-Step Answer:
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
Maharashtra Board Solution
∆ABC and ∆DEF are equilateral triangles. If A(∆ABC): A(∆DEF) = 1:2 and AB = 4, find DE.
Solution & Step-by-Step Answer:
In ∆ABC and ∆DEF,
Question 7
Maharashtra Board Solution
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 & Step-by-Step Answer:
A(∆PQF) = 20 sq.units, PF = 2 DP, [Given] Let us assume DP = x. ∴ PF = 2x