QUESTION 2a
We want to find the area of the given right angle triangle.
We use the formula
[tex]Area=\frac{1}{2}\times base\times height[/tex]
The height of the triangle is [tex]=a cm[/tex].
The base is [tex]12cm[/tex].
We substitute the given values to obtain,
[tex]Area=\frac{1}{2}\times 12\times a cm^2[/tex].
This simplifies to get an expression for the area to be
[tex]Area=6a cm^2[/tex].
QUESTION 2b
The given diagram is a rectangle.
The area of a rectangle is given by the formula
[tex]Area=length \times width[/tex]
The length of the rectangle is [tex]l=7cm[/tex] and the width of the rectangle is [tex]w=ycm[/tex].
We substitute the values to obtain the area to be
[tex]Area=7 \times y[/tex]
The expression for the area is
[tex]Area=7y[/tex]
QUESTION 2c.
The given diagram is a rectangle.
The area of a rectangle is given by the formula
[tex]Area=length \times width[/tex]
The length of the rectangle is [tex]l=2x cm[/tex] and the width of the rectangle is [tex]w=4 cm[/tex].
We substitute the values to obtain the area to be
[tex]Area=2x \times 4[/tex]
The expression for the area is
[tex]Area=8x[/tex]
QUESTION 2d
The given diagram is a square.
The area of a square is given by,
[tex]Area=l^2[/tex].
where [tex]l=b m[/tex] is the length of one side.
The expression for the area is
[tex]Area=b^2 m^2[/tex]
QUESTION 2e
The given diagram is an isosceles triangle.
The area of this triangle can be found using the formula,
[tex]Area=\frac{1}{2}\times base\times height[/tex].
The height of the triangle is [tex]4cm[/tex].
The base of the triangle is [tex]6a cm[/tex].
The expression for the area is
[tex]Area=\frac{1}{2}\times 6a \times 4cm^2[/tex]
[tex]Area=12a cm^2[/tex]
QUESTION 3a
Perimeter is the distance around the figure.
Let P be the perimeter, then
[tex]P=x+x+x+x[/tex]
The expression for the perimeter is
[tex]P=4x mm[/tex]
QUESTION 3b
The given figure is a rectangle.
Let P, be the perimeter of the given figure.
[tex]P=L+B+L+B[/tex]
This simplifies to
[tex]P=2L+2B[/tex]
Or
[tex]P=2(L+B)[/tex]
QUESTION 3c
The given figure is a parallelogram.
Perimeter is the distance around the parallelogram
[tex]Perimeter=3q+P+3q+P[/tex]
This simplifies to,
[tex]Perimeter=6q+2P [/tex]
Or
[tex]Perimeter=2(3q+P) [/tex]
QUESTION 3d
The given figure is a rhombus.
The perimeter is the distance around the whole figure.
Let P be the perimeter. Then
[tex]P=5b+5b+5b+5b[/tex]
This simplifies to,
[tex]P=20b mm[/tex]
QUESTION 3e
The given figure is an equilateral triangle.
The perimeter is the distance around this triangle.
Let P be the perimeter, then,
[tex]P=2x+2x+2x[/tex]
We simplify to get,
[tex]P=6x mm[/tex]
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,
[tex]Perimeter=3m+5m+5m[/tex]
[tex]Perimeter=13m mm[/tex]
QUESTION 3g
The given figure is a scalene triangle.
The perimeter is the distance around the given triangle.
Let P be the perimeter. Then
[tex]P=(3x+1)+(2x-1)+(4x+5)[/tex]
This simplifies to give us,
[tex]P=3x+2x+4x+5-1+1[/tex]
[tex]P=9x+5[/tex]
QUESTION 3h
The given figure is a trapezium.
The perimeter is the distance around the whole trapezium.
Let P be the perimeter.
Then,
[tex]P=m+(n-1)+(2m-3)+(n+3)[/tex]
We group like terms to get,
[tex]P=m+2m+n+n-3+3-1[/tex]
We simplify to get,
[tex]P=3m+2n-1[/tex]mm
QUESTION 3i
The figure is an isosceles triangle.
We add all the distance around the figure to obtain the perimeter.
Let [tex]P[/tex] be the perimeter.
Then [tex]P=(2a-b)+(a+2b)+(a+2b)[/tex]
We regroup the terms to get,
[tex]P=2a+a+a-b+2b+2b[/tex]
This will simplify to give us the expression for the perimeter to be
[tex]P=4a+3b[/tex]mm.
QUESTION 4a
The given figure is a square.
The area of a square is given by the formula;
[tex]Area=l^2[/tex]
where [tex]l=2m[/tex] is the length of one side of the square.
We substitute this value to obtain;
[tex]Area=(2m)^2[/tex]
This simplifies to give the expression of the area to be,
[tex]Area=4m^2[/tex]
QUESTION 4b
The given figure is a rectangle.
The formula for finding the area of a rectangle is
[tex]Area=l\times w[/tex].
where [tex]l=5a cm[/tex] is the length of the rectangle and [tex]w=6cm[/tex] is the width of the rectangle.
We substitute the values into the formula to get,
[tex]Area =5a \times 6[/tex]
[tex]Area =30a cm^2[/tex]
QUESTION 4c
The given figure is a rectangle.
The formula for finding the area of a rectangle is
[tex]Area=l\times w[/tex].
where [tex]l=7y cm[/tex] is the length of the rectangle and [tex]w=2x cm[/tex] is the width of the rectangle.
We substitute the values into the formula to get,
[tex]Area =7y \times 2x[/tex]
The expression for the area is
[tex]Area =14xy cm^2[/tex]
QUESTION 4d
The given figure is a rectangle.
The formula for finding the area of a rectangle is
[tex]Area=l\times w[/tex].
where [tex]l=3p cm[/tex] is the length of the rectangle and [tex]w=p cm[/tex] is the width of the rectangle.
We substitute the values into the formula to get,
[tex]Area =3p \times p[/tex]
The expression for the area is
[tex]Area =3p^2 cm^2[/tex]
See attachment for the continuation
Answer:
1a) P=(a+27) cm
1b) P=2(y+7) cm
1c) P=4(x+2) cm
1d) P=4b m
1e) P=16a cm
2a) A=6a cm^2
2b) A=7y cm^2
2c) A=8x cm^2
2d) A=b^2 m^2
2e) A=12a cm^2
3a) P=4x mm
3b) P=2(B+L) mm
3c) P=2(3q+p) mm
3d) P=20b mm
3e) P=6x mm
3f) P=(3m+10n) mm
3g) P=(9x+5) mm
3h) P=(3m+2n-1) mm
3i) P=(4a+3b) mm
4a) A=4m^2 cm^2
4b) A=30a cm^2
4c) A=14xy cm^2
4d) A=3p^2 cm^2
4e) A=20b cm^2
4f) A=30h^2 cm^2
5a) We bought altogether 3p pens and 4q pencils
5b) I have now 14a pears
5c) The total cost will be 40x cents
5d) You will have altogether 9m chocolates and 8n lollies
Step-by-step explanation:
Perimeter: P=?
1a) P=a cm+12 cm+15 cm=(a+12+15) cm→P=(a+27) cm
1b) P=2(y cm+7 cm)→P=2(y+7) cm
1c) P=2(2x cm+4cm)=2(2x+4) cm=2(2)(2x/2+4/2) cm→P=4(x+2) cm
1d) P=4b m
1e) P=5a cm+5a cm+6a cm=(5a+5a+6a) cm→P=16a cm
Area: A=?
2a) A=bh/2=(12 cm)(a cm)/2=12a cm^2/2→A=6a cm^2
2b) A=bh=(7 cm)(y cm)→A=7y cm^2
2c) A=bh=(2x cm)(4 cm)→A=8x cm^2
2d) A=s^2=(b m)^2=(b)^2 (m)^2→A=b^2 m^2
2e) A=bh/2=(6a cm)(4 cm)/2=(24a cm^2)/2→A=12a cm^2
3a) P=4(x mm)→P=4x mm
3b) P=2(B mm+L mm)→P=2(B+L) mm
3c) P=2(3q mm+p mm)→P=2(3q+p) mm
3d) P=4(5b mm)→P=20b mm
3e) P=3(2x mm)→P=6x mm
3f) P=3m mm+5n mm+5n mm=(3m+5n+5n) mm→P=(3m+10n) mm
3g) P=(4x+5) mm+(2x-1) mm+(3x+1) mm=(4x+5+2x-1+3x+1) mm→P=(9x+5) mm
3h) P=(2m-3) mm+(n-1) mm+(m) mm+(n+3) mm=(2m-3+n-1+m+n+3) mm→
P=(3m+2n-1) mm
3i) P=(2a-b) mm+(a+2b) mm+(a+2b) mm=(2a-b+a+2b+a+2b) mm→
P=(4a+3b) mm
4a) A=s^2=(2m cm)^2=(2)^2 (m)^2 (cm)^2→A=4m^2 cm^2
4b) A=bh=(5a cm)(6 cm)→A=30a cm^2
4c) A=bh=(7y cm)(2x cm)→A=14xy cm^2
4d) A=bh=(3p cm)(p cm)→A=3p^2 cm^2
4e) A=bh/2=(4b cm)(10 cm)/2=(40b cm^2)/2→A=20b cm^2
4f) A=bh/2=(10h cm)(6h cm)/2=(60h^2 cm^2)/2→A=30h^2 cm^2
5a) (p pens+3q pencils)+(2p pens+q pencils)=
p pens+3q pencils+2p pens+q pencils=(p+2p) pens+(3q+q) pencils=
3p pens and 4q pencils
5b) (10a pears)-(3a pears)+(7a pears)=10a pears-3a pears+7a pears=
(10a-3a+7a) pears=14a pears
5c) Total cost: T=4(5x cents)+4(3x cents)+4(2x cents)→
T=4(5x+3x+2x) cents=4(10x) cents→T=40x cents
5d) Total sweets: T=3(m chocolates+2n lollies)+2(3m chocolates+n lollies)→
T=3(m chocolates)+3(2n lollies)+2(3m chocolates)+2(n lollies)→
T=3m chocolates+6n lollies+6m chocolates+2n lollies→
T=(3m+6m) chocolates+(6n+2n) lollies→T=9m chocolates+8n lollies
