## Maharashtra State Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.4

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 S_{1} and S_{2}.

Let A_{1} and A_{2} 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 A_{1} and A_{2}.

A_{1} = 225 sq. cm. A_{2} = 81 sq. cm.

Let the corresponding sides of triangles be S_{1} and S_{2} respectively.

S_{1} = 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