| 1. |
If each side of a triangle is doubled, find the % increase in the area of the triangle |
|
Answer» Answer: Area is INCREASED by 300%. Step-by-step explanation: If that triangle is a right angled triangle. = > Original area = 1 / 2 x base x height { From properties } When sides are DOUBLED : = > NEW area = 1 / 2 x 2base x 2height = > New area = 2 x base x height Increase % : ( 2 x height x base ) / ( 1 / 2 x base x height ) x 100% = > 2 / ( 1 / 2 ) x 100% = > 4 x 100% = > 400% Hence the area is increased by 400%. If that triangle is not right angled triangle : Let the sides are 2a , 2b and 2c. = > Semi-Perimeter = ( 2a + 2b + 2c ) / 2 = > Semi-Perimeter = a + b + c By Using Heron's Formula : = > Area = √[ ( 2a + 2b + 2c )( 2a + 2b + 2c - 2a )( 2a + 2b + 2c - 2b )( 2a + 2b + 2c - 2c ) ] = > Area = √{ ( 2a + 2b + 2c )( 2b + 2c )( 2a + 2c )( 2a + 2b ) } = > Area = √{ 2( a + b + c )2( b + c )2( a + c )2( a + b ) } = > Area = 4√{ ( a + b + c )( b + c )( a + c )( a + b ) } When sides are doubled : Sides are 4a , 4b and 4c. = > Semi-Perimeter = ( 4a + 4b + 4c ) / 2 = > Semi-Perimeter = 2a + 2b + 2c By Using Heron's Formula : = > Area = √[ ( 4a + 4b + 4c )( 4a + 4b + 4c - 4a )( 4a + 4b + 4c - 4b )( 4a + 4b + 4c - 4c ) ] = > Area = √{ ( 4a + 4b + 4c )( 4b + 4c )( 4a + 4c )( 4a + 4b ) } = > Area = √{ 4( a + b + c )4( b + c )4( a + c )4( a + b ) } = > Area = 16√{ ( a + b + c )( b + c )( a + c )( a + b ) } Thus, Increase : 16√{ ( a + b + c )( b + c )( a + c )( a + b ) } - 4√{ ( a + b + c )( b + c )( a + c )( a + b ) } Increase : 12√{ ( a + b + c )( b + c )( a + c )( a + b ) } Increase % : [ 12√{ ( a + b + c )( b + c )( a + c )( a + b ) } ] / [ 4√{ ( a + b + c )( b + c )( a + c )( a + b ) } ] x 100% Increase % : 3 x 100% Increase % : 300% Hence the area is increased by 300 %. |
|