Class 9 Mathematics Chapter 8 Quadrilaters

Key Concept: Diagonals of Square

c) Square

[Solution Description]

Key Concept: Theorem 8.9 Statement

b) Bisects $AC$ into two equal parts

[Solution Description] By Theorem 8.9, a line drawn through the midpoint of one side ($AB$) parallel to another side ($BC$) will bisect the third side ($AC$). Thus, $AF = FC$.

Key Concept: Theorem Application

c) Parallelogram

[Solution Description] According to Theorem 8.3, if both pairs of opposite sides are equal in a quadrilateral, it is a parallelogram. Here $WX = YZ$ (opposite sides) and $WZ = XY$ (remaining opposite sides). Hence, $WXYZ$ must be a parallelogram.

Key Concept: Opposite sides equality

b) $AB = CD$

[Solution Description]

By Theorem 8.2, opposite sides in a parallelogram are equal. Therefore,

$AB = DC$ and $AD = BC$. This matches option b).

Key Concept: Theorem 8.2 – Opposite Sides Equality

c) Opposite sides are equal

[Solution Description] Theorem 8.2 states that in a parallelogram, opposite sides are equal. This fundamental property distinguishes parallelograms from other quadrilaterals.

Key Concept: Rhombus Diagonal Property

c) 15 cm

[Solution Description]

1. Diagonals of rhombus bisect at right angles: $\triangle POS$ is right-angled at $O$.

2. Apply Pythagoras: $PS = \sqrt{PO^2 + SO^2} = \sqrt{9^2 + 12^2}$.

3. Calculate: $\sqrt{81 + 144} = \sqrt{225} = 15$ cm.

Key Concept: Theorem Statement

c) They bisect each other

[Solution Description]

According to Theorem\,8.6, the diagonals of a parallelogram bisect each other. Diagonals being perpendicular applies to rhombuses, while equal length applies to rectangles. Bisecting angles occurs only in specific cases.

Key Concept: Adjacent Sides

b) 2

[Solution Description]

A kite is defined as a quadrilateral with exactly two distinct pairs of adjacent sides that are equal in length. Adjacent sides mean sides that meet at a vertex. Non-adjacent (opposite) sides are typically unequal in a kite.

Step 1: Recall the definition of a kite – “two pairs of adjacent sides are equal”.

Step 2: Count these pairs – first pair shares a common vertex, second pair shares another vertex.

Final answer: 2 pairs.

Key Concept: Congruence Rule

b) ASA

[Solution Description]

In $\Delta ABC$ and $\Delta CDA$, we have $\angle BCA = \angle DAC$ (alternate angles), $\angle BAC = \angle DCA$ (alternate angles), and $AC = CA$ (common side). These satisfy the ASA congruence rule.

Key Concept: Opposite angles in parallelogram

d) $\angle A$ and $\angle C$

[Solution Description] By Theorem 8.4, opposite angles in a parallelogram are equal. Therefore $\angle A = \angle C$ and $\angle B = \angle D$.

Key Concept: Mid-Point Theorem Application

b) 6 cm

[Solution Description]

By Theorem 8.8, the line segment joining mid-points of two sides ($DE$) is parallel to the third side ($BC$) and equal to half its length. Calculation: $DE = \frac{1}{2} BC = \frac{1}{2} \times 12 = 6$ cm.

Key Concept: Diagonals of Rectangle

c) Bisect each other and are equal

[Solution Description]

Step 1: Recall that in a rectangle, diagonals are equal in length and bisect each other (as per syllabus point 115).

Step 2: Verify logical consistency: “Bisect each other and are equal” matches the property of a rectangle’s diagonals. The “vice-versa” part applies only if a parallelogram satisfies these conditions to become a rectangle.

Key Concept: Theorem 8.7 – Diagonals Bisect Each Other

c) It is a parallelogram

[Solution Description]

Key Concept: Mid-point Theorem Application in Trapezium

a) Parallelogram

[Solution Description] By the Mid-point Theorem, joining midpoints of any quadrilateral always forms a parallelogram. In trapezium $ABCD$, connecting midpoints $P$, $Q$, $R$, $S$ creates opposite sides $PQ \parallel SR$ (both parallel to $AC$) and $PS \parallel QR$ (both parallel to $BD$). Hence, $PQRS$ is a parallelogram.

Key Concept: Theorem 8.2 Application

c) 7 cm

[Solution Description] By Theorem 8.2, opposite sides of parallelogram are equal. So $PQ = RS$. Given $PQ = 7$ cm, therefore $RS = 7$ cm.

Key Concept: Diagonal Properties

b) 8 cm

[Solution Description] Diagonals bisect each other into equal segments. Each segment forms a leg of a right-angled isosceles triangle with the square’s side as hypotenuse. Let $OA = OB = 4\sqrt{2}$ cm. Using Pythagoras:

$\text{Side}^2 = OA^2 + OB^2 = (4\sqrt{2})^2 + (4\sqrt{2})^2 = 32 + 32 = 64 \\ \implies \text{Side} = \sqrt{64} = 8\ \text{cm}$

Key Concept: Mid-Point Theorem Application

c) $DE \parallel BC$ and $DE = \frac{1}{2}BC$

[Solution Description]

By Theorem 8.8: The segment joining midpoints of two sides is parallel to the third side and half its length. Here, $DE$ connects midpoints $D$ and $E$. Thus, $DE \parallel BC$ and $DE = \frac{1}{2}BC$.

Key Concept: Midpoints forming a rectangle

c) Rectangle

[Solution Description]

As per the syllabus example, connecting midpoints of a rhombus’s sides forms a rectangle.

Key Concept: Basic Definition

b) 4

[Solution Description] A quadrilateral is defined as a polygon with exactly four sides. This comes from the prefix “quad-” meaning four and “-lateral” referring to sides.

Key Concept: Properties of Rectangles

b) 7 cm

[Solution Description]

In a rectangle, diagonals bisect each other equally. This means PO = OR and QO = OS. Since PO is given as 7 cm, OS will also be 7 cm because O is the midpoint of both diagonals.

Key Concept: Real-Life Quadrilaterals

c) Rectangular door

[Solution Description]

A quadrilateral has four straight sides. Doors typically have four edges arranged in a rectangular shape with right angles. Other options like circles (\$0\$ sides), triangles (\$3\$ sides), and pentagons (\$5\$ sides) do not match.

Key Concept: Definition of Parallelogram

b) 2

[Solution Description]

By definition, a parallelogram is a quadrilateral where **both pairs** of opposite sides are parallel. This means there are 2 pairs of parallel sides.

Key Concept: Opposite sides in parallelogram

c) AB = DC

[Solution Description]

According to Theorem 8.2, opposite sides of a parallelogram are equal. Therefore AB = DC and AD = BC.

Key Concept: Angles in Isosceles Trapezium

c) $\angle A = \angle B$

[Solution Description]

When $AD = BC$, trapezium becomes isosceles. Draw CE parallel to DA meeting extended AB at E. Triangles AED and BEC become congruent $\implies$ $\angle DAB = \angle ABC$. Thus $\angle A = \angle B$.

Key Concept: Diagonals Bisect Each Other

d) Diagonals bisect each other

[Solution Description]

According to Theorem 8.7, if the diagonals of a quadrilateral bisect each other, then it is a parallelogram. This is the main condition specified in the theorem.

Key Concept: Theorem 8.5 Converse Application

c) Parallelogram

[Solution Description] According to Theorem 8.5, if a quadrilateral has pairs of opposite angles equal, it must be a parallelogram. No additional calculation is needed as this is a direct application of the theorem.