Question

If each side of a triangle is doubled, find the % increase in the area of the triangle​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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 %.