Please explain the answer to the question you posted earlier.
Well we were told that the perimeter of rectangle is 28cm. So if we denote the length of the rectangle as y, then the perimeter would be the whole distance around the rectangle. Which would be 2x + 2y.
We now that the perimeter is 2x +2y, therefore we can equate this to 28 as we were told that the perimeter of the rectangle is 28cm.
Equation 1:
2x + 2y =28
Dividing through by two. Therefore x + y = 14. Then we can rearrange the equation making y the subject. Which gives us, y = 14 - x
Another fact the we now is that the width is xcm and the length of the diagonal is 12 cm. As we have let the length of the rectangle equal to y, we can now you Pythagoras which gives us x^2 + y^2 = (12)^2. Which simplifies to x^2 +y^2 =144
We now have two equations.
Equation 1: x + y = 14 and Equation 2: x^2 + y^2 =144. I would now like you to solve these simultaneous equations. Hope this helped!
Well we were told that the perimeter of rectangle is 28cm. So if we denote the length of the rectangle as y, then the perimeter would be the whole distance around the rectangle. Which would be 2x + 2y.
We now that the perimeter is 2x +2y, therefore we can equate this to 28 as we were told that the perimeter of the rectangle is 28cm.
Equation 1:
2x + 2y =28
Dividing through by two. Therefore x + y = 14. Then we can rearrange the equation making y the subject. Which gives us, y = 14 - x
Another fact the we now is that the width is 8cm and the length of the diagonal is 12 cm. As we have let the length of the rectangle equal to y, we can now you Pythagoras which gives us x^2 + y^2 = (12)^2. Which simplifies to x^2 +y^2 =144
We now have two equations.
Equation 1: x + y = 14 and Equation 2: x^2 + y^2 =144. I would now like you to solve these simultaneous equations. Hope this helped!
Quick question: how would you know that the width is 8?
Well we were told that the perimeter of rectangle is 28cm. So if we denote the length of the rectangle as y, then the perimeter would be the whole distance around the rectangle. Which would be 2x + 2y.
We now that the perimeter is 2x +2y, therefore we can equate this to 28 as we were told that the perimeter of the rectangle is 28cm.
Equation 1:
2x + 2y =28
Dividing through by two. Therefore x + y = 14. Then we can rearrange the equation making y the subject. Which gives us, y = 14 - x
Another fact the we now is that the width is 8cm and the length of the diagonal is 12 cm. As we have let the length of the rectangle equal to y, we can now you Pythagoras which gives us x^2 + y^2 = (12)^2. Which simplifies to x^2 +y^2 =144
We now have two equations.
Equation 1: x + y = 14 and Equation 2: x^2 + y^2 =144. I would now like you to solve these simultaneous equations. Hope this helped!
It is not necessary to solve simultaneous equations for this question. Simply one of the side lengths is x and the other is 144−x2 (Pythagoras). Then simply solve the equation x+144−x2=14.
Quick question: how would you know that the width is 8?
Sorry my fault the which is xcm as they stated that in the question. I must have put 8cm by accident. I have edited my written explanation and changed it to xcm. Sorry about that!