Question

Area of two similar triangles are 225cm2 and 81cm2 if side of smaller triangles is 12 cm then find corresponding side of the triangle

Answer 1

Two Triangles are said to be similar if their i)corresponding angles are equal and ii)corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)

SOLUTION:

GIVEN:

ar(∆ABC) / ar(∆DEF) = 225/81 cm²

Side of a larger ∆ABC = 30 cm

ar(larger ∆ABC ) / ar(Smaller ∆DEF ) = (Side of a larger ∆ / Side of a Smaller ∆)²

[The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides]

225 / 81 = (30 /Side of a Smaller ∆)²

Taking square root on both sides

15 / 9 = 30 /Side of a Smaller ∆

15 × Side of a Smaller ∆ = 30 × 9

Side of a Smaller ∆ = (30 × 9)/15

Side of a Smaller ∆ = 2× 9

Side of a Smaller ∆ = 18 cm

Hence, the the longest side of the smaller ∆ DEF is 18 cm

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Answer 2

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• Two similar ∆
• ∆ABC and ∆PQR
• Areas 225 and 81

${\sf{\green{\underline{\large{To\:find}}}}}$

• Corresponding side of ∆ PQR

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Diagram

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Solution

From Theorm

• if two ∆s are similar then their area is proportional to its square of side

∆PQR~∆ABC

• We conclude that

$\tt\rightarrow\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

Now from theorm

$\tt\rightarrow\frac{225}{81}=\frac{AB^2}{PQ^2}=\frac{BC^2}{QR^2}\frac{AC^2}{PR^2}$

$\tt\rightarrow\frac{225}{81}=\frac{AB^2}{PQ^2}$

$\tt\rightarrow\frac{225}{81}=\frac{AB^2}{12^2}$

$\tt\rightarrow\frac{225}{81}=\frac{AB^2}{144}$

$\tt\rightarrow\frac{225}{81}\times144={AB^2}$

=>15/9 ×12 =AB

=>AB=20

Hence AB =20