- The ratio of corresponding sides of similar triangles is 3 : 5; then find the ratio of…
- If ΔABC ~ ΔPQR and AB: PQ = 2:3, then fill in the blanks. a (deltaabc)/a (deltapqr) =…
- If ΔABC ~ ΔPQR, A (ΔABC) = 80, A(ΔPQR) = 125, then fill in the blanks. a (deltaabc)/a…
- ΔLMN ~ ΔPQR, 9 × A(ΔPQR) = 16 × A (ΔLMN). If QR = 20 then find MN.…
- Areas of two similar triangles are 225 sq.cm 81 sq.cm. If a side of the smaller…
- Δ ABC and ΔDEF are equilateral triangles. If A(ΔABC) : A(ΔDEF) = 1 : 2 and AB = 4, find…
- 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.

Answer:

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.

Answer:

∵ 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.

Answer:

∵ 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.

Answer:

∵ 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.

Answer:

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.

Answer:

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

Answer:

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.