Quadrilaterals — NCERT Solutions

Concept: In a kite, the diagonals are perpendicular to each other. One diagonal (the main diagonal) bisects the other diagonal at right angles, but is not necessarily bisected itself.

Steps of Construction:

1. Draw the main diagonal $AC = 8$ cm.
2. Find the midpoint $O$ of $AC$ (i.e., $OA = OC = 4$ cm). (The shorter diagonal bisects the longer one in a standard kite, or vice versa — here we take the 6 cm diagonal to be bisected.)
3. At $O$, draw a perpendicular to $AC$.
4. On this perpendicular, mark $B$ such that $OB = 3$ cm and $D$ such that $OD = 3$ cm. (So $BD = 6$ cm.)
5. Join $A$–$B$, $B$–$C$, $C$–$D$, $D$–$A$.

The quadrilateral $ABCD$ is the required kite with diagonals 8 cm and 6 cm.

Note: $AB = AD$ and $CB = CD$, which is the defining property of a kite.

Concept: In a trapezium with one pair of parallel sides, co-interior angles (same-side interior angles) between the parallel sides are supplementary (add up to 180°).

In an isosceles trapezium, the base angles are equal.

(i) Given: One base angle = 70° (trapezium with one pair of parallel sides).

• Co-interior angle on the same side = 180° − 70° = 110°
• If isosceles: the other base angle = 70°, and the remaining angle = 110°.

(ii) Given: One angle = 120°.

• Co-interior angle = 180° − 120° = 60°
• If isosceles: base angles are equal, so the angles are 120°, 120°, 60°, 60°.

(Note: The exact figures are not fully visible from the OCR. The method is: for each pair of co-interior angles between parallel sides, their sum = 180°. Apply this to find missing angles.)

Venn Diagram Description:

• Quadrilaterals (outermost set)
• Parallelograms (subset of quadrilaterals)
• Rectangles (subset of parallelograms)
• Squares (subset of both rectangles and rhombuses)
• Rhombuses (subset of parallelograms)
• Squares (intersection of rectangles and rhombuses)
• Kites (partially overlapping with rhombuses — rhombuses that are kites = squares/rhombuses)
• Trapeziums

(i) Quadrilateral that is both a kite and a parallelogram:

A kite has two pairs of adjacent equal sides. A parallelogram has opposite sides equal. For a shape to be both, all four sides must be equal.

Answer: A rhombus (and specifically a square if angles are also 90°). A rhombus is both a kite (two pairs of adjacent equal sides) and a parallelogram.

(ii) Can a quadrilateral be both a kite and a rectangle?

A rectangle has all angles = 90° and opposite sides equal. A kite has two pairs of adjacent equal sides (not opposite). For a shape to be both a kite and a rectangle, all four sides must be equal and all angles 90°.

Answer: Yes — a square is both a kite and a rectangle (it satisfies both definitions).

(iii) Is every kite a rhombus?

No. A general kite has two pairs of adjacent equal sides, but the two pairs need not be equal to each other (e.g., sides of 3 cm and 5 cm). A rhombus requires all four sides to be equal.

Correct relationship: Every rhombus is a kite (since a rhombus has all sides equal, it trivially has two pairs of adjacent equal sides), but not every kite is a rhombus. The set of rhombuses is a subset of kites.

Given: $PAIR$ and $RODS$ are two rectangles sharing a common vertex/side arrangement (based on the figure, $I$ and $O$ are intersection points of the diagonals of the two rectangles).

Concept: In a rectangle, the diagonals bisect each other. The diagonals of a rectangle are equal in length.

Reasoning (based on standard configuration of this problem):

In rectangle $PAIR$: diagonal $PR$ and diagonal $AI$ intersect at their midpoint.
In rectangle $RODS$: diagonal $RS$ and diagonal $OD$ intersect at their midpoint.

Since $PAIR$ is a rectangle, $\angle PAI = 90°$ (corner angle), and the diagonals make specific angles.

In the standard version of this problem:

• $\angle RIO$ is formed at the intersection.
• Using the exterior angle theorem or properties of the rectangles:

$$\angle IOD = \angle IOR + \angle ROD$$

In rectangle $PAIR$: $\angle PIR$ (angle at $I$ in triangle formed by diagonal) can be found.
In rectangle $RODS$: similarly.

Standard Answer: $\angle IOD = 90°$

(This follows because the diagonals of the two rectangles, when the rectangles share vertex $R$, create perpendicular lines at the intersection point $O$, giving $\angle IOD = 90°$.)

Note: The exact figure is not fully visible. The answer $\angle IOD = 90°$ is based on the standard configuration of this NCERT problem where the two rectangles are arranged so that their diagonals meet at right angles.

Concept: The diagonals of a square are equal, bisect each other, and are perpendicular to each other.

Steps of Construction:

1. Draw diagonal $AC = 6$ cm.
2. Find the midpoint $O$ of $AC$ by drawing the perpendicular bisector of $AC$ using a compass (without protractor).
3. The perpendicular bisector passes through $O$ and is perpendicular to $AC$.
4. On the perpendicular bisector, mark point $B$ such that $OB = 3$ cm (above $AC$) and point $D$ such that $OD = 3$ cm (below $AC$).
5. Join $A$–$B$, $B$–$C$, $C$–$D$, $D$–$A$.

The quadrilateral $ABCD$ is the required square with diagonal 6 cm.

Verification:
$$\text{Side} = \sqrt{OA^2 + OB^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \approx 4.24 \text{ cm}$$
All sides are equal and all angles are 90° (since diagonals are perpendicular bisectors of each other).

Given: $CASE$ is a square with side $a$. $U$, $V$, $W$, $X$ are midpoints of sides $CA$, $AS$, $SE$, $EC$ respectively.

Geometric Reasoning:

Let the square $CASE$ have side $a$.

• $CU = UA = \frac{a}{2}$ (U is midpoint of CA)
• Similarly for V, W, X.

Finding side $UV$:
In $\triangle UAV$ (right angle at $A$, since $CASE$ is a square):
$$UV = \sqrt{UA^2 + AV^2} = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} = \frac{a}{\ \sqrt{2}}$$

Similarly, $VW = WX = XU = \dfrac{a}{\sqrt{2}}$. All sides of $UVWX$ are equal.

Finding angle at $U$:
In $\triangle UAV$: $\angle UAV = 90°$, $UA = AV$, so $\angle AUV = 45°$.
Similarly, $\angle XUC$: In $\triangle XCU$, $\angle XCU = 90°$, $XC = CU$, so $\angle XUC = 45°$.
$$\angle XUV = 180° - 45° - 45° = 90°$$

Since all sides are equal and all angles are 90°, $UVWX$ is a square.

By Construction: Draw square $CASE$, mark midpoints, join them — measurement confirms all sides equal and all angles 90°.

Other ways: Instead of midpoints, take points that divide each side in ratio $1:3$ or $1:2$ (not midpoints) — the inner quadrilateral will still be a square (rotated), as long as the division ratio is the same on each side.

Answer: Yes, a quadrilateral with four equal sides and one angle of 90° must be a square.

Geometric Reasoning:

Let quadrilateral $ABCD$ have $AB = BC = CD = DA$ (all sides equal) and $\angle A = 90°$.

Since all four sides are equal, $ABCD$ is a rhombus.

In a rhombus, opposite angles are equal:
$$\angle A = \angle C = 90°$$

Adjacent angles in a rhombus are supplementary:
$$\angle A + \angle B = 180° \Rightarrow \angle B = 180° - 90° = 90°$$
$$\angle D = \angle B = 90°$$

So all four angles = 90°, and all four sides are equal.

Conclusion: The quadrilateral is a square.

By Construction: Draw a rhombus with one angle = 90° and verify by measurement that all angles are 90°.

Answer: A quadrilateral with opposite sides equal is a parallelogram.

Justification:

Let $ABCD$ be a quadrilateral with $AB = CD$ and $BC = DA$.

Draw diagonal $AC$.

In $\triangle ABC$ and $\triangle CDA$:

• $AB = CD$ (given)
• $BC = DA$ (given)
• $AC = CA$ (common side)

By SSS congruence: $\triangle ABC \cong \triangle CDA$.

Therefore, $\angle BAC = \angle DCA$ (corresponding parts).

These are alternate interior angles for lines $AB$ and $CD$ with transversal $AC$.
$$\therefore AB \parallel CD$$

Also, $\angle BCA = \angle DAC$ (corresponding parts of congruent triangles).

These are alternate interior angles for lines $BC$ and $DA$ with transversal $AC$.
$$\therefore BC \parallel DA$$

Since both pairs of opposite sides are parallel, $ABCD$ is a parallelogram.

The figure shown appears to be a non-convex (concave/arrow-shaped) quadrilateral.

Geometric Reasoning:

For any simple quadrilateral (convex or concave), we can draw a diagonal to divide it into two triangles.

The sum of angles of each triangle = 180°.

For a concave quadrilateral $ABCD$ (with one reflex-like interior angle), draw diagonal $AC$:

• $\triangle ABC$: sum of angles = 180°
• $\triangle ACD$: sum of angles = 180°

The four interior angles of the quadrilateral are made up of the angles of these two triangles.

$$\text{Sum of angles of } ABCD = 180° + 180° = 360°$$

Conclusion: Yes, the sum of angles of any simple quadrilateral (including non-convex ones) is 360°.

By Construction and Measurement: Draw the figure, measure all four interior angles with a protractor, and verify their sum = 360°.

(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.

False.

Justification: A rectangle also has diagonals that are equal and bisect each other, but a rectangle is not necessarily a square (unless all sides are also equal). For example, a rectangle with sides 3 cm and 5 cm has equal diagonals that bisect each other, but it is not a square.

(ii) A quadrilateral having three right angles must be a rectangle.

True.

Justification: Sum of all four angles of a quadrilateral = 360°. If three angles are 90° each, their sum = 270°. The fourth angle = 360° − 270° = 90°. So all four angles are 90°, making it a rectangle.

(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.

True.

Justification: If the diagonals of a quadrilateral bisect each other, then by the converse of the parallelogram diagonal property, the quadrilateral is a parallelogram. (The triangles formed are congruent by SAS, giving opposite sides equal and parallel.)

(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.

False.

Justification: A kite also has perpendicular diagonals, but a kite is not necessarily a rhombus (its sides are not all equal). For example, a kite with sides 3 cm and 5 cm has perpendicular diagonals but is not a rhombus.

(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.

True.

Justification: Let the quadrilateral $ABCD$ have $\angle A = \angle C$ and $\angle B = \angle D$. Sum of angles = 360°:
$$2\angle A + 2\angle B = 360° \Rightarrow \angle A + \angle B = 180°$$
This means $AB \parallel CD$ (co-interior angles supplementary). Similarly $BC \parallel AD$. So $ABCD$ is a parallelogram.

(vi) A quadrilateral in which all the angles are equal is a rectangle.

True.

Justification: If all four angles are equal and their sum = 360°, then each angle = 360° ÷ 4 = 90°. A quadrilateral with all angles equal to 90° is a rectangle by definition.

(vii) Isosceles trapeziums are parallelograms.

False.

Justification: An isosceles trapezium has exactly one pair of parallel sides (with the non-parallel sides equal). A parallelogram requires both pairs of opposite sides to be parallel. Since an isosceles trapezium has only one pair of parallel sides, it is not a parallelogram.