NCERT Solutions for Class 9 Maths Chapter 12 — Exercise Set 12.2#

True. Adjacent angles of a parallelogram add up to 180°, so the angles next to the 90° angle are $180° - 90° = 90°$, and the opposite angle is also 90°. All four angles are right angles, so it is a rectangle.

False. The diagonals of every rhombus are perpendicular. A rhombus with angles 60° and 120° has perpendicular diagonals but is not a square.

True. Let $ABCD$ be a parallelogram with $AC = BD$. In $\triangle ABC$ and $\triangle DCB$:

Part	Reason
$AB = DC$	opposite sides of a parallelogram
$BC = CB$	common side
$AC = DB$	given

By SSS, $\triangle ABC \cong \triangle DCB$, so $\angle ABC = \angle DCB$. These are adjacent angles of a parallelogram, so they add up to 180°, and each is 90°. So $ABCD$ is a rectangle.

$AC$ bisects $\angle A$:

$$\angle DAC = \angle CAB \quad \cdots (1)$$

$AB \parallel DC$ with transversal $AC$ gives equal alternate angles:

$$\angle CAB = \angle ACD \quad \cdots (2)$$

$AD \parallel BC$ with transversal $AC$:

$$\angle DAC = \angle ACB \quad \cdots (3)$$

From equation (1), equation (2) and equation (3):

$$\angle ACD = \angle CAB = \angle DAC = \angle ACB$$

So $\angle ACD = \angle ACB$, and $AC$ bisects $\angle C$.

Also, in $\triangle ABC$, $\angle CAB = \angle ACB$, so the sides opposite them are equal: $BC = AB$. A parallelogram with two adjacent sides equal has all four sides equal (opposite sides are equal), so $ABCD$ is a rhombus.

Yes. $AC$ bisects $\angle A$ and $\angle C$. In $\triangle ABC$ and $\triangle ADC$: $\angle BAC = \angle DAC$, $AC$ is common, and $\angle BCA = \angle DCA$. By ASA, they are congruent, so $AB = AD$ and $CB = CD$.

In the same way, $BD$ bisects $\angle B$ and $\angle D$, which gives $BA = BC$ and $DA = DC$.

Together: $AB = BC = CD = DA$, so $ABCD$ is a rhombus.

Yes. The diagonals bisect each other, so $ABCD$ is a parallelogram (Theorem 4). Let them meet at $O$. In $\triangle AOB$ and $\triangle AOD$: $OB = OD$, $AO$ is common, and $\angle AOB = \angle AOD = 90°$. By SAS, $AB = AD$. A parallelogram with two adjacent sides equal is a rhombus.

No. An isosceles trapezium (or a kite with equal diagonals) has equal diagonals but is not a rectangle.

Extra condition: if the diagonals are equal and bisect each other, then $ABCD$ is a parallelogram with equal diagonals, which is a rectangle (Q1(iii)).

Adjacent angles of a parallelogram add up to 180°. Take the bisectors of $\angle A$ and $\angle B$ and let them meet at a point $X$. In $\triangle ABX$:

$$\angle XAB + \angle XBA = \dfrac{1}{2}\angle A + \dfrac{1}{2}\angle B$$

$$\angle XAB + \angle XBA = \dfrac{1}{2}(\angle A + \angle B)$$

$$\angle XAB + \angle XBA = 90°$$

So the third angle of the triangle is:

$$\angle AXB = 180° - 90° = 90°$$

The same argument works for each pair of adjacent angles ($B$ and $C$, $C$ and $D$, $D$ and $A$). So each of the four intersection points is a vertex with an angle of 90°, and the quadrilateral $PQRS$ they form has all four angles 90° — it is a rectangle.

Why AB ≠ BC: if $AB = BC$, the parallelogram is a rhombus. Then the angle bisectors are its diagonals, and all four bisectors pass through one point (the centre). The four “intersection points” are the same point, so there is no rectangle — it shrinks to a point.