Find the missing angles in various polygons using the worksheet provided.

Let's solve each problem step by step using geometric principles, such as the sum of interior angles in polygons and properties of parallel lines, triangles, and quadrilaterals.

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🔷 Problem 1: Quadrilateral

Given angles:
- 135°
- 60°
- 50°
- x°

Sum of interior angles in a quadrilateral = $ (4 - 2) \times 180^\circ = 360^\circ $

$$
135 + 60 + 50 + x = 360 \\
245 + x = 360 \\
x = 360 - 245 = 115^\circ
$$

✔ Answer: $ x = 115^\circ $

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🔷 Problem 2: Quadrilateral with two right angles

Angles:
- 90° (right angle)
- 90° (right angle)
- 50°
- x°

Sum = 360°

$$
90 + 90 + 50 + x = 360 \\
230 + x = 360 \\
x = 130^\circ
$$

✔ Answer: $ x = 130^\circ $

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🔷 Problem 3: Triangle with two equal sides (isosceles)

Marked with tick marks → two sides are equal → base angles are equal.

One angle is 60°, and one angle is $ x^\circ $. The triangle has two equal sides, so the angles opposite them are equal.

But only one angle is labeled: 60°. Let’s see which side it's on.

From the diagram:
- One angle is 60°
- Two sides are marked equal → the angles opposite those sides are equal.
- So if the 60° is at the bottom-left, and the two equal sides are from the top to bottom corners, then the two base angles are equal?

Wait — let's interpret:

The triangle has:
- A 60° angle at the bottom-left
- Two sides marked with ticks: one from the top vertex to bottom-left, and one from top to bottom-right → so the two sides from the top are equal → triangle is isosceles with apex at the top
→ So the two base angles (bottom-left and bottom-right) are equal.

But we are given:
- Bottom-left = 60°
- So bottom-right must also be 60°
- Then the top angle $ x $ is:

$$
x = 180 - 60 - 60 = 60^\circ
$$

So this is an equilateral triangle!

✔ Answer: $ x = 60^\circ $

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🔷 Problem 4: Heptagon (7 sides)

Sum of interior angles in a polygon: $ (n - 2) \times 180^\circ $

For $ n = 7 $:
$ (7 - 2) \times 180 = 5 \times 180 = 900^\circ $

Given angles:
- 130°
- 122°
- 130°
- 130°
- 138°
- 128°
- $ x^\circ $

Add known angles:

$$
130 + 122 = 252 \\
252 + 130 = 382 \\
382 + 130 = 512 \\
512 + 138 = 650 \\
650 + 128 = 778
$$

So:

$$
778 + x = 900 \\
x = 900 - 778 = 122^\circ
$$

✔ Answer: $ x = 122^\circ $

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🔷 Problem 5: Pentagon (5 sides)

Sum of interior angles: $ (5 - 2) \times 180 = 3 \times 180 = 540^\circ $

Angles given:
- 120°
- 107°
- 90° (right angle)
- 90° (right angle)
- $ x^\circ $

Sum of known angles:

$$
120 + 107 = 227 \\
227 + 90 = 317 \\
317 + 90 = 407
$$

Then:

$$
407 + x = 540 \\
x = 540 - 407 = 133^\circ
$$

✔ Answer: $ x = 133^\circ $

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🔷 Problem 6: Quadrilateral

Angles:
- 50°
- 140°
- 70°
- $ x^\circ $

Sum = 360°

$$
50 + 140 + 70 + x = 360 \\
260 + x = 360 \\
x = 100^\circ
$$

✔ Answer: $ x = 100^\circ $

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🔷 Problem 7: Parallelogram-like shape (Quadrilateral)

Looks like a parallelogram or trapezoid. Angles given:
- Top-left: $ x^\circ $
- Top-right: 60°
- Bottom-left: 60°
- Bottom-right: 120°

Note: Opposite angles in a parallelogram are equal.

Check:
- Top-right = 60°, bottom-left = 60° → same → possible
- Bottom-right = 120° → so top-left should be 120°?

But wait: In a parallelogram, adjacent angles are supplementary (add to 180°)

Check:
- If top-right = 60°, then adjacent angle (bottom-right) should be 120° → matches
- Similarly, top-left = $ x $, adjacent bottom-left = 60° → so $ x + 60 = 180 $? No, unless it's a parallelogram.

Wait: Is this a parallelogram?

Look at angles:
- Top-left: $ x $
- Top-right: 60°
- Bottom-left: 60°
- Bottom-right: 120°

If it's a parallelogram:
- Opposite angles equal → top-left = bottom-right = 120°
- Top-right = bottom-left = 60° → matches

So yes! This is a parallelogram.

Thus:
- $ x = 120^\circ $

Alternatively, check sum:

$$
x + 60 + 60 + 120 = 360 \\
x + 240 = 360 \\
x = 120^\circ
$$

✔ Answer: $ x = 120^\circ $

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🔷 Problem 8: Rhombus (or kite-shaped quadrilateral)

Looks like a rhombus (all sides equal), but not necessarily.

Angles:
- Top: 140°
- Left: 40°
- Right: $ x^\circ $
- Bottom: 140°

Wait — notice symmetry.

Opposite angles in a rhombus are equal.

Top = 140°, bottom = 140° → OK

Left = 40°, so right should be 40°? But labeled as $ x $

Wait: But sum of angles:

$$
140 + 40 + x + 140 = 360 \\
320 + x = 360 \\
x = 40^\circ
$$

So $ x = 40^\circ $

But wait — is that consistent?

In a rhombus, opposite angles are equal → top and bottom both 140° → OK

Left and right: both should be equal → so $ x = 40^\circ $

Yes.

Also, adjacent angles are supplementary:

- 140° + 40° = 180° → yes

So all good.

✔ Answer: $ x = 40^\circ $

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $ 115^\circ $ |
| 2 | $ 130^\circ $ |
| 3 | $ 60^\circ $ |
| 4 | $ 122^\circ $ |
| 5 | $ 133^\circ $ |
| 6 | $ 100^\circ $ |
| 7 | $ 120^\circ $ |
| 8 | $ 40^\circ $ |

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