Can you provide solutions for geometry problems involving parallelograms, triangles, and angle/side measurements, as well as instructions for constructing a triangle?

Here are the solutions to the problems.

Question 3:

• Assumption: The problem statement for Question 3 is incomplete as it does not provide the cost per kilometer or any other pricing information for hiring the car. Therefore, it cannot be solved.

Question (e):

• Assumption: As an AI, I cannot perform physical constructions or measurements. However, I can provide the steps for constructing the triangle.

• Step 1: Draw a line segment $AB$ of length $6.5 \text{ cm}$.

• Step 2: With $A$ as the center, draw an arc with a radius of $8.5 \text{ cm}$ (length of $AC$).

• Step 3: With $B$ as the center, draw an arc with a radius of $9 \text{ cm}$ (length of $BC$).

• Step 4: The intersection point of the two arcs is point $C$. Join $AC$ and $BC$ to form triangle $ABC$.

• Step 5: To measure angle $CAB$, place the protractor's center at point $A$ and align its baseline with $AB$. Read the angle where $AC$ crosses the protractor scale.

Question (a) - Geometry:
Given: $AB \parallel DC$, $ABDE$ is a parallelogram, $BCE$ is an isosceles triangle with $|BC|=|BE|$.
$|AD|=5 \text{ cm}$, $|AB|=10 \text{ cm}$, $|CD|=16 \text{ cm}$.

• Step 1: Determine unknown side lengths. Since $ABDE$ is a parallelogram, opposite sides are equal: $|DE| = |AB| = 10 \text{ cm}$ $|BE| = |AD| = 5 \text{ cm}$ Since $\triangle BCE$ is isosceles with $|BC|=|BE|$, then: $|BC| = 5 \text{ cm}$ The total length of $CD$ is $16 \text{ cm}$. We know $CD = DE + EC$. $16 \text{ cm} = 10 \text{ cm} + |EC|$ $|EC| = 16 \text{ cm} - 10 \text{ cm} = 6 \text{ cm}$.

• Step 2: Calculate the height of the trapezium ABCD. Draw an altitude from $B$ to $DC$, let the foot of the altitude be $F$. This altitude $BF$ is the height of the trapezium. In isosceles $\triangle BCE$, the altitude from $B$ to $EC$ bisects $EC$. So, $FC = \frac{EC}{2} = \frac{6}{2} = 3 \text{ cm}$. Now consider the right-angled $\triangle BFC$: $BF^2 + FC^2 = BC^2$ $BF^2 + 3^2 = 5^2$ $BF^2 + 9 = 25$ $BF^2 = 25 - 9$ $BF^2 = 16$ $BF = \sqrt{16} = 4 \text{ cm}$. The height of the trapezium is $h = 4 \text{ cm}$.

• Step 3: Calculate the area of ABCD. The area of a trapezium is given by the formula: Area $= \frac{1}{2}(a+b)h$, where $a$ and $b$ are the lengths of the parallel sides. Area of ABCD $= \frac{1}{2}(|AB| + |CD|) \times h$ Area of ABCD $= \frac{1}{2}(10 \text{ cm} + 16 \text{ cm}) \times 4 \text{ cm}$ Area of ABCD $= \frac{1}{2}(26 \text{ cm}) \times 4 \text{ cm}$ Area of ABCD $= 13 \text{ cm} \times 4 \text{ cm}$ Area of ABCD $= \boxed{\text{52 cm}^2}$

• Step 4: Calculate the perimeter of ABCD. The perimeter of ABCD is the sum of the lengths of its sides: $AB + BC + CD + DA$. Perimeter of ABCD $= 10 \text{ cm} + 5 \text{ cm} + 16 \text{ cm} + 5 \text{ cm}$ Perimeter of ABCD $= \boxed{\text{36 cm}}$

Question (b):
Mr. Abban withdrew some money. He used $\frac{3}{4}$ for rent and gave $\frac{1}{5}$ to his children. He had GH¢ 150.00 left.

• Step 1: Calculate the total fraction of money spent. Fraction for rent $= \frac{3}{4}$ Fraction for children $= \frac{1}{5}$ Total fraction spent $= \frac{3}{4} + \frac{1}{5}$ To add these fractions, find a common denominator, which is 20. Total fraction spent $= \frac{3 \times 5}{4 \times 5} + \frac{1 \times 4}{5 \times 4} = \frac{15}{20} + \frac{4}{20} = \frac{19}{20}$

• Step 2: Calculate the fraction of money left. Fraction left $= 1 - \text{Total fraction spent}$ Fraction left $= 1 - \frac{19}{20} = \frac{20}{20} - \frac{19}{20} = \frac{1}{20}$

• Step 3: Determine the total amount of money withdrawn. We know that $\frac{1}{20}$ of the total money is GH¢ 150.00. Let $M$ be the total money withdrawn. $\frac{1}{20} M = 150$ $M = 150 \times 20$ $M = \boxed{\text{GH¢ 3000.00}}$

Question (a) - Pie chart:
The pie chart shows the distribution of students. The sum of angles in a circle is $360^\circ$.

• Step 1: Sum the given angles. Angles given: $140^\circ$ (Red House), $110^\circ$ (Yellow House), $70^\circ$ (Blue House). Sum of known angles $= 140^\circ + 110^\circ + 70^\circ = 320^\circ$.

• Step 2: Calculate the value of x. The angle for Green House is $x$. $x + \text{Sum of known angles} = 360^\circ$ $x + 320^\circ = 360^\circ$ $x = 360^\circ - 320^\circ$ $x = \boxed{\text{40}^\circ}$