Maharashtra State BoardSSC (English Medium) 9th Standard
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Construct δ Abc, in Which Bc =5.2 Cm, ∠ Acb = 45 ∘ and Perimeter of δ Abc is 10 Cm. - Geometry

Diagram

Construct  Δ ABC, in which BC =5.2 cm,  ∠ ACB = 45 ° and perimeter  of  Δ ABC  is 10 cm.

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Solution

Perimeter of △ABC = 10 cm
⇒ AB + BC + AC = 10 cm
⇒ AB + 5.2 + AC = 10 cm
⇒ AB + AC = 4.8 cm
A triangle can be only possible if the sum of the sides of a traingle is greater than the third side.
But here, AB + AC  < BC
Hence, the construction of △ABC is not possible.

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APPEARS IN

Balbharati Mathematics 2 Geometry 9th Standard Maharashtra State Board
Chapter 4 Constructions of Triangles
Practice Set 4.1 | Q 4 | Page 53
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