Simultaneous Equation Problems?

Question

Problem 1

Given that the triangle alongside is equilateral, find a and b. (b+2) cm, (a+4) cm, (4a-b) cm.

Problem 2

A rectangle has perimeter 32 cm. If 3 cm is taken from the length and added to the width, the rectangle becomes a square. Find the dimensions of the original rectangle.

Thank You! :)

Answer 1

Problem 1.

As the triangle is equilateral,

b + 2 = a + 4

bringing 2 over,

b = a + 2

similarly, a + 4 = 4a - b

sub b = a + 2 into the eqn above

4a - ( a + 2 ) = a + 4

3a - 2 = a + 4

2a = 6

a = 3

similarly b + 2 = 4a - b

sub a = 3 into the eqn above

b + 2 = 4(3) - b

2b = 10

b = 5

Problem 2

Let length and width of rectangle be L and W respectively.

from statement "A rectangle has perimeter 32 cm. If 3 cm is taken from the length and added to the width, the rectangle becomes a square", you can deduce

2L + 2W = 32 ( its sides make up its perimeter )

L - 3 = W + 3 ( l equals w when u take 3cm from the length and add it to the width )

L = W + 6

sub L = W + 6 into 2L + 2W = 32

2 ( W + 6 ) + 2W = 32

2W + 12 + 2W = 32

4W = 20

W = 5

sub W = 5 into L = W + 6

L = 5 + 6 = 11

original dimensions of rectangle, length = 11cm and width = 5 cm

hope u have a clear picture now ;)