Quadrilaterals

Concept: In a rectangle, all four corner angles are 90°. When a diagonal is drawn, it divides each 90° corner into two parts. The angles formed by the diagonal are related by the properties of parallel lines (alternate interior angles) and the angle-sum property of triangles.

(i) Given: One of the angles formed by the diagonal inside the rectangle is 35°.

Since the corner angle of a rectangle = 90°, the other part of that corner = 90° − 35° = 55°.

By alternate interior angles (diagonal acts as a transversal between parallel sides):
- The angle alternate to 35° = 35°
- The angle alternate to 55° = 55°

So the four angles inside the rectangle (formed by the diagonal) are: 35°, 55°, 35°, 55°.

(ii) Given: One of the angles formed by the diagonal inside the rectangle is 55°.

Other part of the corner = 90° − 55° = 35°.

By alternate interior angles:
- Angle alternate to 55° = 55°
- Angle alternate to 35° = 35°

So the four angles inside the rectangle are: 55°, 35°, 55°, 35°.

Concept: A quadrilateral whose diagonals are equal in length and bisect each other is a rectangle. The angle at which the diagonals intersect determines the shape of the rectangle (or square when 90°).

Steps of Construction (same for all parts, angle varies):

1. Draw diagonal $AC = 8$ cm.
2. Find the midpoint $O$ of $AC$.
3. At $O$, draw the second diagonal $BD = 8$ cm such that $OB = OD = 4$ cm, making the required angle with $AC$.
4. Join $A$–$B$, $B$–$C$, $C$–$D$, $D$–$A$ to get the quadrilateral $ABCD$.

(i) Angle = 30°: Draw $BD$ at 30° to $AC$ at $O$, with $OB = OD = 4$ cm. The resulting quadrilateral $ABCD$ is a rectangle.

(ii) Angle = 40°: Draw $BD$ at 40° to $AC$ at $O$, with $OB = OD = 4$ cm. The resulting quadrilateral is a rectangle.

(iii) Angle = 90°: Draw $BD$ perpendicular to $AC$ at $O$, with $OB = OD = 4$ cm. Since the diagonals are equal, bisect each other, and are perpendicular, the resulting quadrilateral is a square.

(iv) Angle = 140°: Draw $BD$ at 140° to $AC$ at $O$, with $OB = OD = 4$ cm. The resulting quadrilateral is a rectangle.

Note: In all cases the quadrilateral formed is a rectangle (special case: square when angle = 90°), because equal diagonals that bisect each other characterise a rectangle.

Given: Circle with centre $O$; $PL$ and $AM$ are two perpendicular diameters, so $PL \perp AM$.

Finding the figure $APML$:

Since $PL$ and $AM$ are diameters:
$$OA = OP = OM = OL = r \text{ (radius)}$$

The diagonals of quadrilateral $APML$ are $PL$ and $AM$.
- Both diagonals = $2r$ (equal lengths, as both are diameters).
- They bisect each other at $O$ (centre of the circle).
- They are perpendicular to each other ($PL \perp AM$).

Conclusion: A quadrilateral whose diagonals are equal, bisect each other, and are perpendicular to each other is a square.

Therefore, $APML$ is a square.

Verification of sides: Using the Pythagorean theorem in $\triangle AOP$:
$$AP = \sqrt{OA^2 + OP^2} = \sqrt{r^2 + r^2} = r\sqrt{2}$$
Similarly, $PM = ML = LA = r\sqrt{2}$. All sides are equal, confirming it is a square.

Given: Two sticks of equal length and a thread.

Method:

1. Place the two sticks so that they cross each other at their midpoints. (Since both sticks are of equal length, their midpoints divide each stick into two equal halves.)

2. Use the thread to tie the two sticks together at their crossing point (midpoint of each stick), ensuring the midpoints coincide exactly.

3. Now, hold the four endpoints of the two sticks and adjust the sticks so that all four endpoints are equidistant from the crossing point (which is already ensured since both sticks have equal length and cross at midpoints).

4. Use the thread to connect the four endpoints in order, forming a quadrilateral. Adjust the angle between the sticks until this quadrilateral (formed by the thread) is a rhombus — i.e., all four sides of thread are equal.

5. When all four sides of the thread quadrilateral are equal, the diagonals (the two sticks) bisect each other at 90°.

Reasoning: The two sticks act as diagonals. When they bisect each other (crossing at midpoints) and the figure formed by joining their endpoints is a rhombus (all sides equal), the diagonals of a rhombus are perpendicular. Hence the angle between the sticks is exactly 90°.

Answer: No, this cannot be chosen as the definition of a rectangle.

Reasoning:

A quadrilateral with opposite sides parallel and equal is a parallelogram. However, not every parallelogram is a rectangle.

For example, a rhombus has all four sides equal (so opposite sides are equal) and opposite sides are parallel — but its angles are not necessarily 90°. Similarly, a general parallelogram has opposite sides equal and parallel, but its angles can be any value (not necessarily 90°).

Conclusion: The condition "opposite sides are parallel and equal" defines a parallelogram, not a rectangle. To define a rectangle, we additionally need the condition that all angles are 90° (or equivalently, that the diagonals are equal).

So, every rectangle is a parallelogram, but not every parallelogram is a rectangle.

Answer: No, it is not possible.

Reasoning:

The sum of all four angles of a quadrilateral = 360°.

If three angles are each 90°, their sum = $3 \times 90° = 270°$.

Fourth angle = $360° - 270° = 90°$.

So the fourth angle must also be 90°. It is impossible to have a quadrilateral with exactly three right angles; if three angles are 90°, the fourth must also be 90°.

Why $\triangle GEO \cong \triangle MEO$:

In rhombus $GAME$, let $O$ be the intersection of diagonals $GM$ and $AE$.

- $GE = ME$ (sides of a rhombus are all equal, so $GE = ME$)
- $EO = EO$ (common side)
- $GO = MO$ (diagonals of a rhombus bisect each other)

By SSS congruence, $\triangle GEO \cong \triangle MEO$.

Therefore: $\angle GOE = \angle MOE$ (corresponding parts of congruent triangles).

Since $\angle GOE$ and $\angle MOE$ are supplementary (they form a straight line $GM$):
$$\angle GOE + \angle MOE = 180°$$
$$2\angle GOE = 180°$$
$$\angle GOE = 90°$$

Conclusion: The diagonals of a rhombus intersect at 90° (right angles).

Concept used: In a parallelogram:
- Opposite angles are equal.
- Adjacent angles are supplementary (add up to 180°).

(i) Given one angle = 70°.
- Opposite angle = 70°
- Each adjacent angle = 180° − 70° = 110°
- Remaining angles: 70°, 110°, 110°

(ii) Given one angle = 110°.
- Opposite angle = 110°
- Each adjacent angle = 180° − 110° = 70°
- Remaining angles: 110°, 70°, 70°

(iii) Given one angle = 60°.
- Opposite angle = 60°
- Each adjacent angle = 180° − 60° = 120°
- Remaining angles: 60°, 120°, 120°

(iv) Given one angle = 130°.
- Opposite angle = 130°
- Each adjacent angle = 180° − 130° = 50°
- Remaining angles: 130°, 50°, 50°

*(Note: The exact given angles depend on the figures in the textbook which are not fully visible. The method above applies to each case using the properties of a parallelogram.)*

Concept: The diagonals of a parallelogram bisect each other.

Steps of Construction:

1. Draw diagonal $AC = 7$ cm.
2. Find the midpoint $O$ of $AC$ (i.e., $OA = OC = 3.5$ cm).
3. At point $O$, draw a ray making an angle of 140° with $AC$.
4. On this ray, mark point $B$ such that $OB = 2.5$ cm, and on the opposite ray, mark point $D$ such that $OD = 2.5$ cm. (So $BD = 5$ cm and $O$ is the midpoint of $BD$.)
5. Join $A$–$B$, $B$–$C$, $C$–$D$, $D$–$A$.

The quadrilateral $ABCD$ is the required parallelogram with diagonals 7 cm and 5 cm intersecting at 140°.

Verification: Since the diagonals bisect each other, $ABCD$ is a parallelogram by property.

Concept: The diagonals of a rhombus bisect each other at right angles (90°).

Steps of Construction:

1. Draw diagonal $AC = 5$ cm.
2. Find the midpoint $O$ of $AC$ (i.e., $OA = OC = 2.5$ cm).
3. At $O$, draw a perpendicular to $AC$ (i.e., at 90°).
4. On this perpendicular, mark $B$ such that $OB = 2$ cm and $D$ such that $OD = 2$ cm. (So $BD = 4$ cm.)
5. Join $A$–$B$, $B$–$C$, $C$–$D$, $D$–$A$.

The quadrilateral $ABCD$ is the required rhombus.

Verification of side length:
$$AB = \sqrt{OA^2 + OB^2} = \sqrt{(2.5)^2 + (2)^2} = \sqrt{6.25 + 4} = \sqrt{10.25} \approx 3.2 \text{ cm}$$
All four sides are equal (each $\approx 3.2$ cm), confirming it is a rhombus.

Given: Two equilateral triangles, each with side 4 cm, joined along one common side.

Shape formed: A rhombus (kite-shaped figure — actually a rhombus).

When two equilateral triangles are joined along a common side:
- All four sides of the resulting quadrilateral = 4 cm (sides of the equilateral triangles).
- The common side becomes an internal diagonal.

Angles:
- Each angle of an equilateral triangle = 60°.
- At the two vertices where triangles are joined (top and bottom): angle = 60° + 60° = 120° each.
- At the two side vertices (left and right): angle = 60° each.

Summary:
- All four sides = 4 cm
- Two angles (at joined vertices) = 120° each
- Two angles (at outer vertices) = 60° each
- Check: 120° + 60° + 120° + 60° = 360° ✓

The quadrilateral is a rhombus (all sides equal).

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.

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(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.

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(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.)

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(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.

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(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.

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(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.

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(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.