Question

unit 5: quadrilaterals and the value of the pronumerals
question 1 find the value of the pronumerals
a
b
c
d
e
f
question 2 find the value of the pronumerals
a
b
c
d
e
f
question 3 find the value of the pronumerals
a
b
c

Explanation:

Step1: Recall sum - of - interior - angles formula

The sum of the interior angles of an $n$-sided polygon is given by $(n - 2)\times180^{\circ}$. For a quadrilateral ($n = 4$), the sum of interior angles is $(4 - 2)\times180^{\circ}=360^{\circ}$.

Step2: Solve for Question 1a

We know that for the given quadrilateral, $70^{\circ}+50^{\circ}+130^{\circ}+x = 360^{\circ}$.
Combining like - terms: $250^{\circ}+x = 360^{\circ}$.
Subtracting $250^{\circ}$ from both sides: $x=360^{\circ}-250^{\circ}=110^{\circ}$.

Step3: Solve for Question 1b

For the quadrilateral, $110^{\circ}+110^{\circ}+95^{\circ}+m = 360^{\circ}$.
Combining like - terms: $315^{\circ}+m = 360^{\circ}$.
Subtracting $315^{\circ}$ from both sides: $m = 360^{\circ}-315^{\circ}=45^{\circ}$.

Step4: Solve for Question 1c

For the trapezoid, $90^{\circ}+90^{\circ}+110^{\circ}+a = 360^{\circ}$.
Combining like - terms: $290^{\circ}+a = 360^{\circ}$.
Subtracting $290^{\circ}$ from both sides: $a = 360^{\circ}-290^{\circ}=70^{\circ}$.

Step5: Solve for Question 1d

The sum of angles around a point is $360^{\circ}$. The non - reflex angle at the center is $360^{\circ}-220^{\circ}=140^{\circ}$.
For the triangle - like shape, $70^{\circ}+40^{\circ}+x = 180^{\circ}$ (sum of angles in a triangle is $180^{\circ}$).
Combining like - terms: $110^{\circ}+x = 180^{\circ}$.
Subtracting $110^{\circ}$ from both sides: $x = 70^{\circ}$.

Step6: Solve for Question 1e

In a parallelogram, opposite angles are equal. So $x = 110^{\circ}$ and $y = 70^{\circ}$ (since adjacent angles in a parallelogram are supplementary, $180^{\circ}-110^{\circ}=70^{\circ}$).

Step7: Solve for Question 1f

In a rectangle (implied by the right - angles and equal sides markings), opposite angles are equal. So $x = 70^{\circ}$ and $y = 20^{\circ}$ (since $90^{\circ}-70^{\circ}=20^{\circ}$).

Step8: Solve for Question 2a

For the quadrilateral, $90^{\circ}+80^{\circ}+50^{\circ}+a = 360^{\circ}$.
Combining like - terms: $220^{\circ}+a = 360^{\circ}$.
Subtracting $220^{\circ}$ from both sides: $a = 360^{\circ}-220^{\circ}=140^{\circ}$.

Step9: Solve for Question 2b

For the quadrilateral, $150^{\circ}+65^{\circ}+90^{\circ}+m = 360^{\circ}$.
Combining like - terms: $305^{\circ}+m = 360^{\circ}$.
Subtracting $305^{\circ}$ from both sides: $m = 360^{\circ}-305^{\circ}=55^{\circ}$.

Step10: Solve for Question 2c

The non - reflex angle at the center is $360^{\circ}-262^{\circ}=98^{\circ}$.
For the quadrilateral, $98^{\circ}+125^{\circ}+90^{\circ}+x = 360^{\circ}$.
Combining like - terms: $313^{\circ}+x = 360^{\circ}$.
Subtracting $313^{\circ}$ from both sides: $x = 360^{\circ}-313^{\circ}=47^{\circ}$.

Step11: Solve for Question 2d

For the quadrilateral, $105^{\circ}+85^{\circ}+90^{\circ}+y = 360^{\circ}$.
Combining like - terms: $280^{\circ}+y = 360^{\circ}$.
Subtracting $280^{\circ}$ from both sides: $y = 360^{\circ}-280^{\circ}=80^{\circ}$. Also, $x = 95^{\circ}$ (since $180^{\circ}-85^{\circ}=95^{\circ}$).

Step12: Solve for Question 2e

For the quadrilateral, $x + 2x+120^{\circ}+66^{\circ}=360^{\circ}$.
Combining like - terms: $3x+186^{\circ}=360^{\circ}$.
Subtracting $186^{\circ}$ from both sides: $3x = 174^{\circ}$.
Dividing both sides by 3: $x = 58^{\circ}$.

Step13: Solve for Question 2f

For the quadrilateral, $95^{\circ}+140^{\circ}+70^{\circ}+y = 360^{\circ}$.
Combining like - terms: $305^{\circ}+y = 360^{\circ}$.
Subtracting $305^{\circ}$ from both sides: $y = 55^{\circ}$. Also, $a = 45^{\circ}$ (since $90^{\circ}-45^{\circ}=45^{\circ}$).

Step14: Solve for Question 3a

In the triangle on the left, $a=\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ}$ (isosceles triangle).
For the quadrilateral, $100^{\circ}+90^{\circ}+70^{\circ}+x = 360^{\circ}$.
Combining like - terms: $260^{\circ}+x = 360^{\circ}$.
Subtracting $260^{\circ}$ from both sides: $x = 100^{\circ}$.
For the right - hand triangle, $y = 180^{\circ}-(70^{\circ}+40^{\circ})=70^{\circ}$.

Step15: Solve for Question 3b

In the isosceles triangle, $a=\frac{180^{\circ}-(180^{\circ}-110^{\circ})}{2}=55^{\circ}$.
For the quadrilateral, $110^{\circ}+70^{\circ}+105^{\circ}+x = 360^{\circ}$.
Combining like - terms: $285^{\circ}+x = 360^{\circ}$.
Subtracting $285^{\circ}$ from both sides: $x = 75^{\circ}$.

Step16: Solve for Question 3c

The non - reflex angle at the center is $360^{\circ}-190^{\circ}=170^{\circ}$.
In the left - hand triangle, the third angle is $180^{\circ}-(120^{\circ}+30^{\circ})=30^{\circ}$.
For the quadrilateral, $170^{\circ}+30^{\circ}+y + x=360^{\circ}$. Also, since the two small triangles are congruent (by markings), we know that $x = 40^{\circ}$ and $y = 120^{\circ}$.

Answer:

Question 1:
a. $x = 110^{\circ}$
b. $m = 45^{\circ}$
c. $a = 70^{\circ}$
d. $x = 70^{\circ}$
e. $x = 110^{\circ},y = 70^{\circ}$
f. $x = 70^{\circ},y = 20^{\circ}$
Question 2:
a. $a = 140^{\circ}$
b. $m = 55^{\circ}$
c. $x = 47^{\circ}$
d. $x = 95^{\circ},y = 80^{\circ}$
e. $x = 58^{\circ}$
f. $a = 45^{\circ},y = 55^{\circ}$
Question 3:
a. $a = 70^{\circ},x = 100^{\circ},y = 70^{\circ}$
b. $a = 55^{\circ},x = 75^{\circ}$
c. $x = 40^{\circ},y = 120^{\circ}$