KnowledgeBoat Logo
|

Mathematics

Each side of a triangle is doubled, the ratio of areas of the original triangle to the new (resulting) triangle is :

  1. 1 : 1

  2. 4 : 1

  3. 1 : 4

  4. 1 : 2

Area Trapezium Polygon

12 Likes

Answer

Let a, b and c be sides of the triangle.

Each side of a triangle is doubled, the ratio of areas of the original triangle to the new (resulting) triangle is : Area of a Trapezium and a Polygon, Concise Mathematics Solutions ICSE Class 8.

s=a+b+c2s = \dfrac{a + b + c}{2}

Area of triangle = s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}

When each side of a triangle is doubled, then 2a, 2b and 2c are sides of new triangle.

Each side of a triangle is doubled, the ratio of areas of the original triangle to the new (resulting) triangle is : Area of a Trapezium and a Polygon, Concise Mathematics Solutions ICSE Class 8.

s′=a+b+c2s′=2a+2b+2c2s′=2(a+b+c)2s′=2(a+b+c)2s′=2ss' = \dfrac{a + b + c}{2}\\[1em] s' = \dfrac{2a + 2b + 2c}{2}\\[1em] s' = \dfrac{2(a + b + c)}{2}\\[1em] s' = 2\dfrac{(a + b + c)}{2}\\[1em] s' = 2s

Area of triangle = s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}

= 2s(2s−2a)(2s−2b)(2s−2c)\sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)}

= 2s2(s−a)2(s−b)2(s−c)\sqrt{2s2(s - a)2(s - b)2(s - c)}

= 16s(s−a)(s−b)(s−c)\sqrt{16s(s - a)(s - b)(s - c)}

= 4 s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}

Ratio of original triangle to new triangle = s(s−a)(s−b)(s−c)4s(s−a)(s−b)(s−c)\dfrac{\sqrt{s(s - a)(s - b)(s - c)}}{4\sqrt{s(s - a)(s - b)(s - c)}}

= 14\dfrac{1}{4}

Hence, option 3 is the correct option.

Answered By

4 Likes

Related Questions