<u>QUESTION 2a</u>
We want to find the area of the given right angle triangle.
We use the formula
![Area=\frac{1}{2}\times base\times height](https://tex.z-dn.net/?f=Area%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20base%5Ctimes%20height)
The height of the triangle is
.
The base is
.
We substitute the given values to obtain,
.
This simplifies to get an expression for the area to be
.
<u>QUESTION 2b</u>
The given diagram is a rectangle.
The area of a rectangle is given by the formula
![Area=length \times width](https://tex.z-dn.net/?f=Area%3Dlength%20%5Ctimes%20width)
The length of the rectangle is
and the width of the rectangle is
.
We substitute the values to obtain the area to be
![Area=7 \times y](https://tex.z-dn.net/?f=Area%3D7%20%5Ctimes%20y)
The expression for the area is
![Area=7y](https://tex.z-dn.net/?f=Area%3D7y)
<u>QUESTION 2c.</u>
The given diagram is a rectangle.
The area of a rectangle is given by the formula
![Area=length \times width](https://tex.z-dn.net/?f=Area%3Dlength%20%5Ctimes%20width)
The length of the rectangle is
and the width of the rectangle is
.
We substitute the values to obtain the area to be
![Area=2x \times 4](https://tex.z-dn.net/?f=Area%3D2x%20%5Ctimes%204)
The expression for the area is
![Area=8x](https://tex.z-dn.net/?f=Area%3D8x)
<u>QUESTION 2d</u>
The given diagram is a square.
The area of a square is given by,
.
where
is the length of one side.
The expression for the area is
![Area=b^2 m^2](https://tex.z-dn.net/?f=Area%3Db%5E2%20m%5E2)
<u>QUESTION 2e</u>
The given diagram is an isosceles triangle.
The area of this triangle can be found using the formula,
.
The height of the triangle is
.
The base of the triangle is
.
The expression for the area is
![Area=\frac{1}{2}\times 6a \times 4cm^2](https://tex.z-dn.net/?f=Area%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%206a%20%5Ctimes%204cm%5E2)
![Area=12a cm^2](https://tex.z-dn.net/?f=Area%3D12a%20cm%5E2)
<u>QUESTION 3a</u>
Perimeter is the distance around the figure.
Let P be the perimeter, then
![P=x+x+x+x](https://tex.z-dn.net/?f=P%3Dx%2Bx%2Bx%2Bx)
The expression for the perimeter is
![P=4x mm](https://tex.z-dn.net/?f=P%3D4x%20mm)
<u>QUESTION 3b</u>
The given figure is a rectangle.
Let P, be the perimeter of the given figure.
![P=L+B+L+B](https://tex.z-dn.net/?f=P%3DL%2BB%2BL%2BB)
This simplifies to
![P=2L+2B](https://tex.z-dn.net/?f=P%3D2L%2B2B)
Or
![P=2(L+B)](https://tex.z-dn.net/?f=P%3D2%28L%2BB%29)
<u>QUESTION 3c</u>
The given figure is a parallelogram.
Perimeter is the distance around the parallelogram
![Perimeter=3q+P+3q+P](https://tex.z-dn.net/?f=Perimeter%3D3q%2BP%2B3q%2BP)
This simplifies to,
![Perimeter=6q+2P](https://tex.z-dn.net/?f=Perimeter%3D6q%2B2P%20)
Or
![Perimeter=2(3q+P)](https://tex.z-dn.net/?f=Perimeter%3D2%283q%2BP%29%20)
<u>QUESTION 3d</u>
The given figure is a rhombus.
The perimeter is the distance around the whole figure.
Let P be the perimeter. Then
![P=5b+5b+5b+5b](https://tex.z-dn.net/?f=P%3D5b%2B5b%2B5b%2B5b)
This simplifies to,
![P=20b mm](https://tex.z-dn.net/?f=P%3D20b%20mm)
<u>QUESTION 3e</u>
The given figure is an equilateral triangle.
The perimeter is the distance around this triangle.
Let P be the perimeter, then,
![P=2x+2x+2x](https://tex.z-dn.net/?f=P%3D2x%2B2x%2B2x)
We simplify to get,
![P=6x mm](https://tex.z-dn.net/?f=P%3D6x%20mm)
QUESTION 3f
The figure is an isosceles triangle so two sides are equal.
We add all the distance around the triangle to find the perimeter.
This implies that,
![Perimeter=3m+5m+5m](https://tex.z-dn.net/?f=Perimeter%3D3m%2B5m%2B5m)
![Perimeter=13m mm](https://tex.z-dn.net/?f=Perimeter%3D13m%20mm)
<u>QUESTION 3g</u>
The given figure is a scalene triangle.
The perimeter is the distance around the given triangle.
Let P be the perimeter. Then
![P=(3x+1)+(2x-1)+(4x+5)](https://tex.z-dn.net/?f=P%3D%283x%2B1%29%2B%282x-1%29%2B%284x%2B5%29)
This simplifies to give us,
![P=3x+2x+4x+5-1+1](https://tex.z-dn.net/?f=P%3D3x%2B2x%2B4x%2B5-1%2B1)
![P=9x+5](https://tex.z-dn.net/?f=P%3D9x%2B5)
<u>QUESTION 3h</u>
The given figure is a trapezium.
The perimeter is the distance around the whole trapezium.
Let P be the perimeter.
Then,
![P=m+(n-1)+(2m-3)+(n+3)](https://tex.z-dn.net/?f=P%3Dm%2B%28n-1%29%2B%282m-3%29%2B%28n%2B3%29)
We group like terms to get,
![P=m+2m+n+n-3+3-1](https://tex.z-dn.net/?f=P%3Dm%2B2m%2Bn%2Bn-3%2B3-1)
We simplify to get,
mm
QUESTION 3i
The figure is an isosceles triangle.
We add all the distance around the figure to obtain the perimeter.
Let
be the perimeter.
Then ![P=(2a-b)+(a+2b)+(a+2b)](https://tex.z-dn.net/?f=P%3D%282a-b%29%2B%28a%2B2b%29%2B%28a%2B2b%29)
We regroup the terms to get,
![P=2a+a+a-b+2b+2b](https://tex.z-dn.net/?f=P%3D2a%2Ba%2Ba-b%2B2b%2B2b)
This will simplify to give us the expression for the perimeter to be
mm.
QUESTION 4a
The given figure is a square.
The area of a square is given by the formula;
![Area=l^2](https://tex.z-dn.net/?f=Area%3Dl%5E2)
where
is the length of one side of the square.
We substitute this value to obtain;
![Area=(2m)^2](https://tex.z-dn.net/?f=Area%3D%282m%29%5E2)
This simplifies to give the expression of the area to be,
![Area=4m^2](https://tex.z-dn.net/?f=Area%3D4m%5E2)
QUESTION 4b
The given figure is a rectangle.
The formula for finding the area of a rectangle is
.
where
is the length of the rectangle and
is the width of the rectangle.
We substitute the values into the formula to get,
![Area =5a \times 6](https://tex.z-dn.net/?f=Area%20%3D5a%20%5Ctimes%206)
![Area =30a cm^2](https://tex.z-dn.net/?f=Area%20%3D30a%20cm%5E2)
QUESTION 4c
The given figure is a rectangle.
The formula for finding the area of a rectangle is
.
where
is the length of the rectangle and
is the width of the rectangle.
We substitute the values into the formula to get,
![Area =7y \times 2x](https://tex.z-dn.net/?f=Area%20%3D7y%20%5Ctimes%202x)
The expression for the area is
![Area =14xy cm^2](https://tex.z-dn.net/?f=Area%20%3D14xy%20cm%5E2)
QUESTION 4d
The given figure is a rectangle.
The formula for finding the area of a rectangle is
.
where
is the length of the rectangle and
is the width of the rectangle.
We substitute the values into the formula to get,
![Area =3p \times p](https://tex.z-dn.net/?f=Area%20%3D3p%20%5Ctimes%20p)
The expression for the area is
![Area =3p^2 cm^2](https://tex.z-dn.net/?f=Area%20%3D3p%5E2%20cm%5E2)
See attachment for the continuation