### Practice Set 1.4 Similarity Class 10th Mathematics Part 2 MHB Solution

Practice Set 1.4
1. The ratio of corresponding sides of similar triangles is 3 : 5; then find the ratio of…
2. If ΔABC ~ ΔPQR and AB: PQ = 2:3, then fill in the blanks. a (deltaabc)/a (deltapqr) =…
3. If ΔABC ~ ΔPQR, A (ΔABC) = 80, A(ΔPQR) = 125, then fill in the blanks. a (deltaabc)/a…
4. ΔLMN ~ ΔPQR, 9 × A(ΔPQR) = 16 × A (ΔLMN). If QR = 20 then find MN.…
5. Areas of two similar triangles are 225 sq.cm 81 sq.cm. If a side of the smaller…
6. Δ ABC and ΔDEF are equilateral triangles. If A(ΔABC) : A(ΔDEF) = 1 : 2 and AB = 4, find…
7. In figure 1.66, seg PQ || seg DE, A(Δ PQF) = 20 units, PF = 2 DP, then find A(DPQE) by…
###### Practice Set 1.4
Question 1.

The ratio of corresponding sides of similar triangles is 3 : 5; then find the ratio of their areas.

Theorem: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

⇒ Ratio of areas = 32:52

⇒ Ratio of areas = 9 : 25

Question 2.

If ΔABC ~ ΔPQR and AB: PQ = 2:3, then fill in the blanks. ∵ Δ ABC~Δ PQR and AB:PQ = 2:3

⇒ Question 3.

If ΔABC ~ ΔPQR, A (ΔABC) = 80, A(ΔPQR) = 125, then fill in the blanks. ∵ Δ ABC~Δ PQR

⇒ (∵ A(Δ PQR) = 125 is given)

⇒ ⇒ ⇒ Question 4.

ΔLMN ~ ΔPQR, 9 × A(ΔPQR) = 16 × A (ΔLMN). If QR = 20 then find MN.

∵ Δ ABC~Δ PQR

⇒ Given that, 9× A(Δ ABC) = 16× A(Δ PQR)

⇒ And, ⇒ ⇒ ⇒ MN2 = ⇒ MN = 15

Question 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.

Let area of one(bigger) triangle be ‘A’, other(smaller) triangle be ‘B’,corresponding side of smaller triangle be ‘a’ and bigger triangle be ‘b’.

⇒ (By theorem)

And a = 12cm, A = 225 sq.cm, B = 81 sq.cm ……(Given)

⇒ ⇒ b2 = ⇒ b = √400

⇒ b = 20 cm

Question 6.

Δ ABC and ΔDEF are equilateral triangles. If A(ΔABC) : A(ΔDEF) = 1 : 2 and AB = 4, find DE.

We know that, all the angles of an equilateral triangles are equal, i.e., 60°.

⇒ Δ ABC~Δ DEF ……(By AAA Similarity Test)

⇒ And, (Given)

⇒ ⇒ DE2 = 2× 42 (∵ AB = 4)

⇒ DE = √32

DE = 4√2

Question 7.

In figure 1.66, seg PQ || seg DE, A(Δ PQF) = 20 units, PF = 2 DP, then find A(DPQE) by completing the following activity.

A(Δ PQF) = 20 units, PF = 2 DP, Let us assume DP = x. ∴ PF = 2x In Δ FDE and Δ FPQ,

∠FDE ≅ ∠ .......... corresponding angles

∠FED ≅ ∠ ......... corresponding angles

∴ Δ FDE ~ Δ FPQ .......... AA test  A(Δ PQF) = 20units,PF = 2DP,Let us assume DP = x,∴ PF = 2x

⇒ DF = DP + PF = x + 2x = 3x

In Δ FDE & Δ FPQ

∠ FDE ≅ ∠ FPQ (Corresponding angles)

∠ FEP ≅ ∠ FQP (Corresponding angles)

∴ Δ FDE~ Δ FPQ (AA Test)

∴ A(Δ FDE) = A(ΔFPQ) = × 20 = 45

A(□DPQE) = A (Δ FDE) - A(ΔFPQ)

= 45-20

= 25 sq.unit.

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