Question

Cw 11.1 11.4 round to the tenths

Answer 1

Answer:

1. A = 21,924.0

2. A = 1,584.0

3. A = 104.0

4. A = 126.0

5. A = 81.0

6. A = 217.0

7. A = 67.5

8. x = 6.0

9. x = 20.0

Step-by-step explanation:

Kindly find the attached for the step by step explanation

Answer 2

QUESTION.1

The area of the shaded region equals area of the bigger rectangle minus area of the smaller rectangle.

By equation, area of a rectangle $=l\times w$

This implies that the area of the bigger rectangle $=236\times105$

$=24780ft^2$

Also,the area of the smaller rectangle$68\times42$

$=2856 ft^2$

Hence,area of the shaded region$=24780-2856=21924ft^2$  

QUESTION.2

The polygon is a trapezium.

The area of a trapezium is given as;

$\frac{1}{2}\times(a+b)\times h$

where a and b denote the the two parallel sides and h denotes the height

From the question $a=40, b=48, h=36$

 By substitution,the area of the trapezium

$=\frac{1}{2}\times(40+48)\times36=\frac{1}{2}\times88\times36=1584 sq.units$

QUESTION 3

The polygon is a trapezium.

The area of a trapezium is given as;

$\frac{1}{2}\times(a+b)\times h$

where a and b denote the the two parallel sides and h denotes the height

From the question $a=6, b=20, h=8$

 By substitution,the area of the trapezium

$=\frac{1}{2}\times(6+20)\times8=\frac{1}{2}\times26\times8=104 sq.units$

QUESTION 4

The polygon is a parallelogram.

The area of a parallelogram is given as;

$A=b\times h$

where b denotes of the length of any base and h denotes the height

From the question b=14 and h=9

By substitution,the area of the  parallelogram

$A=14\times9=126sq.units$

QUESTION 5

The polygon is a rhombus.

The area of a rhombus is given as;

$A=s^2$

where s is the length any side

From the question the value of s is 9

By substitution,the area of the rhombus

$A=9^2=81 sq.units$

QUESTION 6

The polygon is a kite.

The area of a kite is given as;

$A=\frac{1}{2}(a\times b)$

where a and b denote the length of the two diagonals

From the question $a=7+7=14$

$b=24+7=31$

 By substitution,the area of the kite

$A=\frac{1}{2}(14\times31)=217sq.units$

QUESTION.7

The polygon is a triangle.

The area of a triangle is given as;

$A=\frac{1}{2}(b\times h)$

where b is length of the base and h denotes the length of the h of the height.

From the question $b=15$

$h=9$

 By substitution,the area of the triangle

$A=\frac{1}{2}(15\times9)=67.5sq.units$

QUESTION.8

The area of a triangle is given as;

$A=\frac{1}{2}(b\times h)$

where b is length of the base and h denotes the length of the height.

From the question $b=12inches$

$h=x$  and the area, $A=36in^2$

 By substitution,

$36=\frac{1}{2}(12\times x)$

$\implies 36=6x$

Dividing both sides by 6

$\implies x=6inches$

QUESTION. 9

The area of a kite is given as;

$A=\frac{1}{2}(a\times b)$

where a and b denote the length of the two diagonals

From the question $a=10yards$

$b=x=?$  and area,$A=100yd^2$  

 By substitution,

$100=\frac{1}{2}(10\times x)$

this implies that$100=5x$

Multiplying both sides by $\frac{1}{5}$,

We obtain,$x=20yards$