Question

A campsite in the shape of a rectangle FGHI has sides (3x+6)m and (x+1) m , and the length of the diagonal FH is (4x+1) m . Find the are of the campsite

Answer 1

Area = product of sides

= (3x+6)* (x+1)

= 3x² + 9x + 6

Answer 2

Step-by-step explanation:

Using Pythagoras theorem,

Diagonal ²= Length ²+ Breadth²

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+37

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0x(x-6)+1(x-6)=0

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0x(x-6)+1(x-6)=0(x-6)(x+1)=0

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0x(x-6)+1(x-6)=0(x-6)(x+1)=0x= 6 or -1

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0x(x-6)+1(x-6)=0(x-6)(x+1)=0x= 6 or -1 We choose the positive value 6 because length and breadth cannot be negative.

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0x(x-6)+1(x-6)=0(x-6)(x+1)=0x= 6 or -1 We choose the positive value 6 because length and breadth cannot be negative.Length Of the Campsite: (3×6)+6=24 m

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0x(x-6)+1(x-6)=0(x-6)(x+1)=0x= 6 or -1 We choose the positive value 6 because length and breadth cannot be negative.Length Of the Campsite: (3×6)+6=24 mBreadth of the campsite: (6+1)=7 m

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0x(x-6)+1(x-6)=0(x-6)(x+1)=0x= 6 or -1 We choose the positive value 6 because length and breadth cannot be negative.Length Of the Campsite: (3×6)+6=24 mBreadth of the campsite: (6+1)=7 mSo,

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0x(x-6)+1(x-6)=0(x-6)(x+1)=0x= 6 or -1 We choose the positive value 6 because length and breadth cannot be negative.Length Of the Campsite: (3×6)+6=24 mBreadth of the campsite: (6+1)=7 mSo,Area= 24×7

Diagonal ²= Length ²+ Breadth²(4x+1)²= (3x+6)²+ (x+1)²16x²+8x+1= 10x²+ 38x+376x²-30x-36=0x²-5x-6=0x²-6x+1x-6=0x(x-6)+1(x-6)=0(x-6)(x+1)=0x= 6 or -1 We choose the positive value 6 because length and breadth cannot be negative.Length Of the Campsite: (3×6)+6=24 mBreadth of the campsite: (6+1)=7 mSo,Area= 24×7 = 168 m²