- 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.
