Angles in Quadrilaterals Textbook Exercise - Corbettmaths - Free Printable

Let's solve each part of Question 1 step by step. The task is to find the missing angle $ x $ in each quadrilateral.

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🔷 Key Concept:

The sum of interior angles in any quadrilateral is always:

$$
360^\circ
$$

We’ll use this fact for all parts.

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(a)

Given angles:
- $ 155^\circ $
- $ 55^\circ $
- $ 50^\circ $
- $ x $

Add known angles:
$$
155 + 55 + 50 = 260^\circ
$$

So,
$$
x = 360 - 260 = \boxed{100^\circ}
$$

✔ Answer: $ x = 100^\circ $

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(b)

Given angles:
- $ 70^\circ $
- $ 50^\circ $
- Right angle $ = 90^\circ $
- $ x $

Sum of known angles:
$$
70 + 50 + 90 = 210^\circ
$$

So,
$$
x = 360 - 210 = \boxed{150^\circ}
$$

✔ Answer: $ x = 150^\circ $

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(c)

Given angles:
- $ 85^\circ $
- $ 70^\circ $
- $ 45^\circ $
- $ x $

Sum:
$$
85 + 70 + 45 = 200^\circ
$$

$$
x = 360 - 200 = \boxed{160^\circ}
$$

✔ Answer: $ x = 160^\circ $

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(d)

Given angles:
- $ 97^\circ $
- $ 86^\circ $
- $ 114^\circ $
- $ x $

Sum:
$$
97 + 86 + 114 = 297^\circ
$$

$$
x = 360 - 297 = \boxed{63^\circ}
$$

✔ Answer: $ x = 63^\circ $

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(e)

This one looks tricky — it’s a concave quadrilateral with an angle labeled $ 212^\circ $. That’s greater than 180°, so it's a reflex angle.

But we still use the rule: sum of interior angles is $ 360^\circ $, even for concave shapes.

Given angles:
- $ 77^\circ $
- $ 212^\circ $
- $ 40^\circ $
- $ x $

Sum:
$$
77 + 212 + 40 = 329^\circ
$$

$$
x = 360 - 329 = \boxed{31^\circ}
$$

✔ Answer: $ x = 31^\circ $

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(f)

This is a rectangle with arrows indicating opposite sides are parallel. Two right angles ($ 90^\circ $) are shown.

One angle is $ 52^\circ $, and $ x $ is opposite or adjacent?

Wait — it shows two right angles (square corners), and one angle is $ 52^\circ $. But in a rectangle, all angles are $ 90^\circ $. So likely this is a trapezium (trapezoid) with two right angles.

From the diagram:
- Two right angles: $ 90^\circ $ each
- One angle: $ 52^\circ $
- $ x $ is the fourth angle

Sum of known angles:
$$
90 + 90 + 52 = 232^\circ
$$

$$
x = 360 - 232 = \boxed{128^\circ}
$$

✔ Answer: $ x = 128^\circ $

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(g)

Angles given:
- $ 88^\circ $
- $ 142^\circ $
- $ 106^\circ $
- $ x $

Sum:
$$
88 + 142 + 106 = 336^\circ
$$

$$
x = 360 - 336 = \boxed{24^\circ}
$$

✔ Answer: $ x = 24^\circ $

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(h)

This is a pentagon? Wait — no! It has five sides? Let's count.

Wait — actually, this shape has five vertices, but the label says "Quadrilaterals" — so probably a mistake?

Wait — looking closely:
It has four sides — it's a quadrilateral with one side extended or drawn oddly. But the angles are labeled at four corners.

Given angles:
- $ 115^\circ $
- $ 139^\circ $
- $ 41^\circ $
- $ x $

Wait — that's only three angles? No — there are four angles labeled: $ 115^\circ $, $ 139^\circ $, $ 41^\circ $, and $ x $. Yes.

Sum:
$$
115 + 139 + 41 = 295^\circ
$$

$$
x = 360 - 295 = \boxed{65^\circ}
$$

✔ Answer: $ x = 65^\circ $

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(i)

Angles:
- $ 76^\circ $
- $ 135^\circ $
- $ 81^\circ $
- $ x $

Sum:
$$
76 + 135 + 81 = 292^\circ
$$

$$
x = 360 - 292 = \boxed{68^\circ}
$$

✔ Answer: $ x = 68^\circ $

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✔ Final Answers Summary:

| Part | Missing Angle $ x $ |
|------|------------------------|
| (a) | $ 100^\circ $ |
| (b) | $ 150^\circ $ |
| (c) | $ 160^\circ $ |
| (d) | $ 63^\circ $ |
| (e) | $ 31^\circ $ |
| (f) | $ 128^\circ $ |
| (g) | $ 24^\circ $ |
| (h) | $ 65^\circ $ |
| (i) | $ 68^\circ $ |

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