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Answer 1

Question 3 ⇒⇒ one part

The upper shaded figure represents rhombus
So, the angles 60° and (5x + 15)° are supplementary
∴ 60° + (5x + 15)° = 180°
Solve for x
∴ 5x + 75 = 180
∴ 5x = 180 - 75 = 105
∴ x = 105/5 = 21

So, the value of x = 21
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Question 4 ⇒⇒ two parts

we will use the following sequence to find the area of each triangle
1. calculating the length of each side using the distance between two points (x₁,y₁),(x₂,y₂) = d

`d = \sqrt{ (x2-x1)^(2) + (y2-y1)^(2) }`

2. calculating the area using Heron's Formula

`area = √(s(s-a)(s-b)(s-c))`

where a, b and c are the lengths of sides of the triangle

and s is the half of the triangles perimeter = (a+b+c)/2

Part (1): The area of RST

R(-5,-5) , S(-3,-1) , T(-1,-2)

`RS = \sqrt{ (-3+5)^(2) + (-1+5)^(2) }` = √20

`ST = \sqrt{ (-1+3)^(2) + (-2+1)^(2) }` = √5

`RT = \sqrt{ (-1+5)^(2) + (-2+5)^(2) }` = 5

s = (√20 + √5 + 5)/2 ≈ 5.85

∴ Area =
`√(5.85(5.85-√20)(5.85-√5)(5.85-5))` = 5

∴ Area of ΔRST = 5

Part (2): The area of MNL

M(1,0) , N(3,2) , L(4,-2)

`MN = \sqrt{ (3-1)^(2) + (2-0)^(2) }` = 2√2

`NL = \sqrt{ (4-3)^(2) + (-2-2)^(2) }` = √17

`ML= \sqrt{ (4-1)^(2) + (-2-0)^(2) }` = √13
s = (2√2 + √17+ √13)/2 ≈ 5.28
∴ Area =
`√(5.28(5.28-2√2)(5.28-√17)(5.28-√13))` = 5

∴ Area of ΔMNL = 5

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Question 8 ⇒⇒ four parts

Part (1)⇒⇒⇒ A. hexagonal prism

Part (2)⇒⇒⇒ B. rectangular pyramid

Part (3)⇒⇒⇒ C. triangular prism

Part (4)⇒⇒⇒ D. triangular pyramid

The complete answer is as shown in the attached figure