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===================================================
Question 4 ⇒⇒ two partswe 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
[tex]d = \sqrt{ (x2-x1)^{2} + (y2-y1)^{2} } [/tex]
2. calculating the area using Heron's Formula
[tex]area = \sqrt{s(s-a)(s-b)(s-c)} [/tex]
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)
[tex]RS = \sqrt{
(-3+5)^{2} + (-1+5)^{2} } [/tex] = √20
[tex]ST = \sqrt{
(-1+3)^{2} + (-2+1)^{2} } [/tex] = √5
[tex]RT = \sqrt{
(-1+5)^{2} + (-2+5)^{2} } [/tex] = 5
s = (√20 + √5 + 5)/2 ≈ 5.85
∴ Area = [tex] \sqrt{5.85(5.85-√20)(5.85-√5)(5.85-5)} [/tex] = 5
∴ Area of ΔRST = 5
Part (2): The area of MNLM(1,0) , N(3,2) , L(4,-2)
[tex]MN = \sqrt{
(3-1)^{2} + (2-0)^{2} } [/tex] = 2√2
[tex]NL = \sqrt{
(4-3)^{2} + (-2-2)^{2} } [/tex] = √17
[tex]ML= \sqrt{
(4-1)^{2} + (-2-0)^{2} } [/tex] = √13
s = (2√2 + √17+ √13)/2 ≈ 5.28
∴ Area = [tex] \sqrt{5.28(5.28-2√2)(5.28-√17)(5.28-√13)} [/tex] = 5
∴ Area of ΔMNL = 5
====================================================
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
