QUESTION 2a
We want to find the area of the given right angle triangle.
We use the formula
![Area=\frac{1}{2}\times base\times height](/tpl/images/0251/7090/e86a8.png)
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
.
QUESTION 2b
The given diagram is a rectangle.
The area of a rectangle is given by the formula
![Area=length \times width](/tpl/images/0251/7090/7c438.png)
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](/tpl/images/0251/7090/73994.png)
The expression for the area is
![Area=7y](/tpl/images/0251/7090/45836.png)
QUESTION 2c.
The given diagram is a rectangle.
The area of a rectangle is given by the formula
![Area=length \times width](/tpl/images/0251/7090/7c438.png)
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](/tpl/images/0251/7090/f10ab.png)
The expression for the area is
![Area=8x](/tpl/images/0251/7090/358dd.png)
QUESTION 2d
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](/tpl/images/0251/7090/bfb42.png)
QUESTION 2e
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](/tpl/images/0251/7090/198bc.png)
![Area=12a cm^2](/tpl/images/0251/7090/aef58.png)
QUESTION 3a
Perimeter is the distance around the figure.
Let P be the perimeter, then
![P=x+x+x+x](/tpl/images/0251/7090/c9501.png)
The expression for the perimeter is
![P=4x mm](/tpl/images/0251/7090/8d233.png)
QUESTION 3b
The given figure is a rectangle.
Let P, be the perimeter of the given figure.
![P=L+B+L+B](/tpl/images/0251/7090/2371a.png)
This simplifies to
![P=2L+2B](/tpl/images/0251/7090/9ce6d.png)
Or
![P=2(L+B)](/tpl/images/0251/7090/001e2.png)
QUESTION 3c
The given figure is a parallelogram.
Perimeter is the distance around the parallelogram
![Perimeter=3q+P+3q+P](/tpl/images/0251/7090/d39d9.png)
This simplifies to,
![Perimeter=6q+2P](/tpl/images/0251/7090/773f1.png)
Or
![Perimeter=2(3q+P)](/tpl/images/0251/7090/a5271.png)
QUESTION 3d
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](/tpl/images/0251/7090/318fb.png)
This simplifies to,
![P=20b mm](/tpl/images/0251/7090/f474a.png)
QUESTION 3e
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](/tpl/images/0251/7090/9c223.png)
We simplify to get,
![P=6x mm](/tpl/images/0251/7090/5c9c3.png)
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](/tpl/images/0251/7090/a4a1e.png)
![Perimeter=13m mm](/tpl/images/0251/7090/874fb.png)
QUESTION 3g
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)](/tpl/images/0251/7090/8ce9c.png)
This simplifies to give us,
![P=3x+2x+4x+5-1+1](/tpl/images/0251/7090/d4787.png)
![P=9x+5](/tpl/images/0251/7090/870e4.png)
QUESTION 3h
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)](/tpl/images/0251/7090/a8699.png)
We group like terms to get,
![P=m+2m+n+n-3+3-1](/tpl/images/0251/7090/ef920.png)
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)](/tpl/images/0251/7090/55bc4.png)
We regroup the terms to get,
![P=2a+a+a-b+2b+2b](/tpl/images/0251/7090/a9c02.png)
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](/tpl/images/0251/7090/2620c.png)
where
is the length of one side of the square.
We substitute this value to obtain;
![Area=(2m)^2](/tpl/images/0251/7090/f5ac8.png)
This simplifies to give the expression of the area to be,
![Area=4m^2](/tpl/images/0251/7090/ee3b1.png)
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](/tpl/images/0251/7090/02c8c.png)
![Area =30a cm^2](/tpl/images/0251/7090/c3bcc.png)
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](/tpl/images/0251/7090/95e37.png)
The expression for the area is
![Area =14xy cm^2](/tpl/images/0251/7090/f1b96.png)
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](/tpl/images/0251/7090/b4dfb.png)
The expression for the area is
![Area =3p^2 cm^2](/tpl/images/0251/7090/8e48b.png)
See attachment for the continuation
![Asap 30 + brainliest only solve 2 - 5](/tpl/images/0251/7090/5c3e8.jpg)
![Asap 30 + brainliest only solve 2 - 5](/tpl/images/0251/7090/1aad0.jpg)
![Asap 30 + brainliest only solve 2 - 5](/tpl/images/0251/7090/c8db0.jpg)
![Asap 30 + brainliest only solve 2 - 5](/tpl/images/0251/7090/a25b0.jpg)
![Asap 30 + brainliest only solve 2 - 5](/tpl/images/0251/7090/2e41d.jpg)