Answer:
there is no picture
Step-by-step explanation:
No because a 2 digit divisor always has less digits than a 3 digit divisor
Answer:
(a) See attachment
(b) Mara's rectangle
Step-by-step explanation:
Given
Mara
![Area =15cm^2](https://tex.z-dn.net/?f=Area%20%3D15cm%5E2)
Ashton
![Area =9cm^2](https://tex.z-dn.net/?f=Area%20%3D9cm%5E2)
Solving (a): Draw the rectangles
First, we need to calculate the possible dimension of both rectangles.
Area is calculated as:
![Area = Length * Width](https://tex.z-dn.net/?f=Area%20%3D%20Length%20%2A%20Width)
For Mara
![15 = Length * Width](https://tex.z-dn.net/?f=15%20%3D%20Length%20%2A%20Width)
Possible values of length and width are:
![Length = 5cm](https://tex.z-dn.net/?f=Length%20%3D%205cm)
![Width = 3cm](https://tex.z-dn.net/?f=Width%20%3D%203cm)
This is so because:
![5cm * 3cm = 15cm^2](https://tex.z-dn.net/?f=5cm%20%2A%203cm%20%3D%2015cm%5E2)
For Ashton
![9 = Length * Width](https://tex.z-dn.net/?f=9%20%3D%20Length%20%2A%20Width)
Possible values of length and width are:
![Length = 4.5cm](https://tex.z-dn.net/?f=Length%20%3D%204.5cm)
![Width = 2cm](https://tex.z-dn.net/?f=Width%20%3D%202cm)
This is so because:
![4.5cm * 2cm = 9cm^2](https://tex.z-dn.net/?f=4.5cm%20%2A%202cm%20%3D%209cm%5E2)
<em>See attachment for diagram</em>
Solving (b): Rectangle with bigger area
Mara's rectangle has a bigger area because ![15cm^2 > 9cm^2](https://tex.z-dn.net/?f=15cm%5E2%20%3E%209cm%5E2)
Answer:
Step-by-step explanation:
12). By applying Pythagoras theorem in the given right triangle,
Hypotenuse² = (Leg 1)² + (Leg 2)²
x² = 7² + 7²
x² = 2(7)²
x = ![\sqrt{2(7)^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B2%287%29%5E%7B2%7D%20%7D)
x = ![7\sqrt{2}](https://tex.z-dn.net/?f=7%5Csqrt%7B2%7D)
13). By sine rule,
sin(60)° = ![\frac{\text{Opposite side}}{\text{Hypotenuse}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BOpposite%20side%7D%7D%7B%5Ctext%7BHypotenuse%7D%7D)
![\frac{\sqrt{3}}{2}=\frac{9\sqrt{3} }{y}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D%5Cfrac%7B9%5Csqrt%7B3%7D%20%7D%7By%7D)
y = ![\frac{18\sqrt{3} }{\sqrt{3} }](https://tex.z-dn.net/?f=%5Cfrac%7B18%5Csqrt%7B3%7D%20%7D%7B%5Csqrt%7B3%7D%20%7D)
y = 18
By cosine rule,
cos(60)° = ![\frac{\text{Adjacent side}}{\text{Hypotenuse}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BAdjacent%20side%7D%7D%7B%5Ctext%7BHypotenuse%7D%7D)
![\frac{1}{2}=\frac{x}{y}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%3D%5Cfrac%7Bx%7D%7By%7D)
x = ![\frac{y}{2}](https://tex.z-dn.net/?f=%5Cfrac%7By%7D%7B2%7D)
x = ![\frac{18}{2}=9](https://tex.z-dn.net/?f=%5Cfrac%7B18%7D%7B2%7D%3D9)
14). By applying Pythagoras theorem in the given right triangle,
12² = x² + x²
144 = 2x²
72 = x²
x = √72
x = 6√2