Triangles ABC and DEF are similar.
(i) If area (∆ABC) = 16 cm², area (∆DEF) = 25 cm² and BC = 2.3 cm, find EF.
(ii) If area (∆ABC) = 9 cm², area (∆DEF) = 64 cm² and DE = 5.1 cm , find AB.
Answers
SOLUTION :
(i) Given : ∆ ABC ∼ ∆DEF ,area of (ΔABC) = 16 cm², area (ΔDEF) = 25 cm² and BC = 2.3 cm.
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
ar(ΔABC)/ar(ΔDEF) =(BC/EF)²
16/25 = (2.3/EF)²
√16/25 = (2.3/EF)
⅘ = 2.3/EF
4EF = 2.3 × 5
EF = 11.5/4
EF = 2.875 cm
Hence, the length of EF is 2.875 cm.
(ii) Given : ∆ ABC ∼ ∆DEF , area (ΔABC) = 9 cm² , area (ΔDEF) = 64 cm² and DE = 5.1 cm
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
ar(ΔABC)/ar(ΔDEF) = (AB/DE)²
9/64 = (AB/DE)²
√9/64 = (AB/DE)
⅜ = AB/5.1
3×5.1 = 8AB
AB = (3 × 5.1) / 8
AB = 15.3/8
AB = 1.9125 cm
Hence, the length of AB is 1.9125 cm.
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SOLUTION :
(i) Given : ∆ ABC ∼ ∆DEF ,area of (ΔABC) = 16 cm², area (ΔDEF) = 25 cm² and BC = 2.3 cm.
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
ar(ΔABC)/ar(ΔDEF) =(BC/EF)²
16/25 = (2.3/EF)²
√16/25 = (2.3/EF)
⅘ = 2.3/EF
4EF = 2.3 × 5
EF = 11.5/4
EF = 2.875 cm
Hence, the length of EF is 2.875 cm.
(ii) Given : ∆ ABC ∼ ∆DEF , area (ΔABC) = 9 cm² , area (ΔDEF) = 64 cm² and DE = 5.1 cm
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
ar(ΔABC)/ar(ΔDEF) = (AB/DE)²
9/64 = (AB/DE)²
√9/64 = (AB/DE)
⅜ = AB/5.1
3×5.1 = 8AB
AB = (3 × 5.1) / 8
AB = 15.3/8
AB = 1.9125 cm
Hence, the length of AB is 1.9125 cm.