Question

ASAP 30 + Brainliest Please only solve 2 - 5

Answer 1

QUESTION 2a

We want to find the area of the given right angle triangle.

We use the formula

$Area=\frac{1}{2}\times base\times height$

The height of the triangle is $=a cm$.

The base is $12cm$.

We substitute the given values to obtain,

$Area=\frac{1}{2}\times 12\times a cm^2$.

This simplifies to get an expression for the area to be

$Area=6a cm^2$.

QUESTION 2b

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=7cm$ and the width of the rectangle is $w=ycm$.

We substitute the values to obtain the area to be

$Area=7 \times y$

The expression for the area is

$Area=7y$

QUESTION 2c.

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=2x cm$ and the width of the rectangle is $w=4 cm$.

We substitute the values to obtain the area to be

$Area=2x \times 4$

The expression for the area is

$Area=8x$

QUESTION 2d

The given diagram is a square.

The area of a square is given by,

$Area=l^2$.

where $l=b m$ is the length of one side.

The expression for the area is

$Area=b^2 m^2$

QUESTION 2e

The given diagram is an isosceles triangle.

The area of this triangle can be found using the formula,

$Area=\frac{1}{2}\times base\times height$.

The height of the triangle is $4cm$.

The base of the triangle is $6a cm$.

The expression for the area is

$Area=\frac{1}{2}\times 6a \times 4cm^2$

$Area=12a cm^2$

QUESTION 3a

Perimeter is the distance around the figure.

Let P be the perimeter, then

$P=x+x+x+x$

The expression for the perimeter is

$P=4x mm$

QUESTION 3b

The given figure is a rectangle.

Let P, be the perimeter of the given figure.

$P=L+B+L+B$

This simplifies to

$P=2L+2B$

Or

$P=2(L+B)$

QUESTION 3c

The given figure is a parallelogram.

Perimeter is the distance around the parallelogram

$Perimeter=3q+P+3q+P$

This simplifies to,

$Perimeter=6q+2P$

Or

$Perimeter=2(3q+P)$

QUESTION 3d

The given figure is a rhombus.

The perimeter is the distance around the whole figure.

Let P be the perimeter. Then

$P=5b+5b+5b+5b$

This simplifies to,

$P=20b mm$

QUESTION 3e

The given figure is an equilateral triangle.

The perimeter is the distance around this triangle.

Let P be the perimeter, then,

$P=2x+2x+2x$

We simplify to get,

$P=6x mm$

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,

$Perimeter=3m+5m+5m$

$Perimeter=13m mm$

QUESTION 3g

The given figure is a scalene triangle.

The  perimeter is the distance around the given triangle.

Let P be the perimeter. Then

$P=(3x+1)+(2x-1)+(4x+5)$

This simplifies to give us,

$P=3x+2x+4x+5-1+1$

$P=9x+5$

QUESTION 3h

The given figure is a trapezium.

The perimeter is the distance around the whole trapezium.

Let P be the perimeter.

Then,

$P=m+(n-1)+(2m-3)+(n+3)$

We group like terms to get,

$P=m+2m+n+n-3+3-1$

We simplify to get,

$P=3m+2n-1$mm

QUESTION 3i

The figure is an isosceles triangle.

We add all the distance around the figure to obtain the perimeter.

Let $P$ be the perimeter.

Then $P=(2a-b)+(a+2b)+(a+2b)$

We regroup the terms to get,

$P=2a+a+a-b+2b+2b$

This will simplify to give us the expression for the perimeter to be

$P=4a+3b$mm.

QUESTION 4a

The given figure is a square.

The area of a square is given by the formula;

$Area=l^2$

where $l=2m$ is the length of one side of the square.

We substitute this value to obtain;

$Area=(2m)^2$

This simplifies to give the expression of the area to be,

$Area=4m^2$

QUESTION 4b

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$.

where $l=5a cm$ is the length of the rectangle and $w=6cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =5a \times 6$

$Area =30a cm^2$

QUESTION 4c

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$.

where $l=7y cm$ is the length of the rectangle and $w=2x cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =7y \times 2x$

The expression for the area is

$Area =14xy cm^2$

QUESTION 4d

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$.

where $l=3p cm$ is the length of the rectangle and $w=p cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =3p \times p$

The expression for the area is

$Area =3p^2 cm^2$

See attachment for the continuation

Answer 2

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