Understanding Quadrilaterals — NCERT Solutions, AI Tutor & Practice

A polygon is a simple closed curve made up entirely of line segments.

• (a) Triangle — made of 3 line segments, closed, and no side crosses another. It is a polygon and a simple polygon.
• (b) Two intersecting circles — made of curves, not line segments. Not a polygon.
• (c) Five-sided closed figure with non-crossing sides — made of line segments, closed, no crossing. It is a polygon and a simple polygon.
• (d) Closed figure of curved lines — contains curves. Not a polygon.

The sum of interior angles of a polygon with $n$ sides is: $S = (n-2)\times 180^\circ$

(a) $n = 8$: $S = (8-2)\times 180^\circ = 6 \times 180^\circ = 1080^\circ$

(b) $n = 12$: $S = (12-2)\times 180^\circ = 10 \times 180^\circ = 1800^\circ$

For a regular polygon, each exterior angle = $180^\circ - \text{interior angle}$.

$\text{Exterior angle} = 180^\circ - 150^\circ = 30^\circ$

The sum of all exterior angles of any polygon is $360^\circ$. So: $n = \frac{360^\circ}{\text{each exterior angle}} = \frac{360^\circ}{30^\circ} = 12$

The polygon has 12 sides.

The sum of exterior angles of any polygon = $360^\circ$.

$n = \frac{360^\circ}{24^\circ} = 15$

So the polygon has 15 sides.

Each interior angle: $180^\circ - 24^\circ = 156^\circ$

The sum of interior angles of a quadrilateral is: $(4-2)\times 180^\circ = 360^\circ$

Let the fourth angle be $x$. $80^\circ + 95^\circ + 112^\circ + x = 360^\circ$ $287^\circ + x = 360^\circ$ $x = 360^\circ - 287^\circ = 73^\circ$

The fourth angle is $73^\circ$.

Let the angles be $x$, $2x$, $3x$ and $4x$.

Sum of angles of a quadrilateral = $360^\circ$: $x + 2x + 3x + 4x = 360^\circ$ $10x = 360^\circ$ $x = 36^\circ$

So the angles are:

• $x = 36^\circ$
• $2x = 72^\circ$
• $3x = 108^\circ$
• $4x = 144^\circ$

Check: $36 + 72 + 108 + 144 = 360^\circ$ ✓

In a parallelogram:

• Opposite angles are equal.
• Adjacent angles are supplementary (add to $180^\circ$).

Given $\angle P = 70^\circ$.

$\angle R$ is opposite to $\angle P$: $\angle R = 70^\circ$

$\angle Q$ is adjacent to $\angle P$: $\angle Q = 180^\circ - 70^\circ = 110^\circ$

$\angle S$ is opposite to $\angle Q$: $\angle S = 110^\circ$

In a parallelogram, opposite sides are equal. So the sides are $5\text{ cm}$, $8\text{ cm}$, $5\text{ cm}$, $8\text{ cm}$.

Perimeter: $P = 2 \times (\text{sum of two adjacent sides})$ $P = 2 \times (5 + 8) = 2 \times 13 = 26\text{ cm}$

The perimeter is $26\text{ cm}$.

In a rhombus, the diagonals bisect each other at right angles.

So $\angle AOB = 90^\circ$.

In triangle $AOB$, the angles sum to $180^\circ$: $\angle OAB + \angle AOB + \angle ABO = 180^\circ$ $35^\circ + 90^\circ + \angle ABO = 180^\circ$ $\angle ABO = 180^\circ - 125^\circ = 55^\circ$

So $\angle AOB = 90^\circ$ and $\angle ABO = 55^\circ$.

(a) True. A square has all four angles equal to $90^\circ$ and opposite sides equal — exactly the defining properties of a rectangle. So every square is a rectangle.

(b) True. A rhombus has both pairs of opposite sides parallel, which is the defining property of a parallelogram.

(c) False. A rectangle need not have all sides equal, but a square does. So not every rectangle is a square.

(d) False. A trapezium has only one pair of parallel sides, whereas a parallelogram needs both pairs of opposite sides parallel.

In a kite, the pair of angles between the unequal sides are equal. Here $\angle B$ and $\angle D$ lie between the unequal sides, so: $\angle B = \angle D$ $\angle D = 100^\circ$

Verification using angle sum: $\angle A + \angle B + \angle C + \angle D = 360^\circ$ $60^\circ + 100^\circ + \angle C + 100^\circ = 360^\circ$ $\angle C = 360^\circ - 260^\circ = 100^\circ$

So $\angle D = 100^\circ$ (and $\angle C = 100^\circ$).

Let the interior angle be $i$ and the exterior angle be $e$.

Given: $e = \dfrac{2}{5}i$.

Since interior and exterior angles are supplementary: $i + e = 180^\circ$ $i + \frac{2}{5}i = 180^\circ$ $\frac{7}{5}i = 180^\circ$ $i = 180^\circ \times \frac{5}{7} = \frac{900^\circ}{7} \approx 128.57^\circ$

Then: $e = 180^\circ - \frac{900^\circ}{7} = \frac{1260^\circ - 900^\circ}{7} = \frac{360^\circ}{7}$

Number of sides: $n = \frac{360^\circ}{e} = \frac{360^\circ}{\frac{360^\circ}{7}} = 7$

The polygon has 7 sides.