## Class 9 NCERT Maths

### NCERT

1   The angles of quadrilateral are in the ratio $3:5:9:13$. Find all the angles of the quadrilateral.

Let the common ratio between the angles be $x.$$\\ Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.\\ As the sum of all interior angles of a quadrilateral is 360^o,$$\\$ $\therefore 3x+5x+9x+13x=360^o\\ 30x =360^o\\ x=12^o$$\\ Hence, the angles are\\ 3x=3*12=36^o\\ 5x=5*12=60^o\\ 9x=9*12=108^o\\ 13x=13*12=156^o 2 If the diagonals of a parallelogram are equal, then show that it is a rectangle. ##### Solution : Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90^o.\\ In \Delta ABC and \Delta DCB,$$\\$ $AB = DC$ (Opposite sides of a parallelogram are equal)$\\$ $BC = BC$ (Common)$\\$ $AC = DB$ (Given)$\\$ $\therefore \Delta ABC \cong \Delta DCB$ (By $SSS$ Congruence rule)$\\$ $\Rightarrow \angle ABC = \angle DCB$$\\ It is known that the sum of the measures of angles on the same side of transversal is 180^o.\\ \angle ABC + \angle DCB = 180^o (AB \parallel CD)$$\\$ $\Rightarrow ABC + \angle ABC = 180^o$$\\ \Rightarrow 2\angle ABC = 180^o$$\\$ $\Rightarrow ABC = 90^o$$\\ Since ABCD is a parallelogram and one of its interior angles is 90^o, ABCD is a rectangle. 3 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. ##### Solution : Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e. OA = OC, OB = OD, and \angle AOB = \angle BOC = \angle COD = \angle AOD = 90^o.$$\\$ To prove $ABCD$ a rhombus,$\\$ We have to prove $ABCD$ is a parallelogram and all the sides of $ABCD$ are equal.$\\$ In $\Delta AOD$ and $\Delta COD,$$\\ OA = OC (Diagonals bisect each other)\\ \angle AOD = \angle COD (Given)\\ OD = OD (Common)\\ \therefore \Delta AOD \cong \Delta COD (By SAS congruence rule)\\ \therefore AD = CD ... (1)\\ Similarly, it can be proved that\\ AD = AB and CD = BC ... (2)\\ From Equations (1) and (2),\\ AB = BC = CD = AD$$\\$ Since opposite sides of quadrilateral $ABCD$ are equal, it can be said that $ABCD$ is a parallelogram. Since all sides of a parallelogram $ABCD$ are equal, it can be said that $ABCD$ is a rhombus.

4   Show that the diagonals of a square are equal and bisect each other at right angles.

##### Solution :

Let $ABCD$ be a square.$\\$ Let the diagonals $AC$ and $BD$ intersect each other at a point $O.$$\\ To prove that the diagonals of a square are equal and bisect each other at right angles,\\ we have to prove,\\ AC = BD, OA = OC, OB = OD, and \angle AOB = 90^o.$$\\$ In $\Delta ABC$ and $\Delta DCB$,$\\$ $AB = DC$(Sides of a square are equal to each other)$\\$ $\angle ABC = \angle DCB$ (All interior angles are of $90^o$ )$\\$ $BC = CB$ (Common side)$\\$ $\therefore \Delta ABC \cong \Delta DCB$(By $SAS$ congruency)$\\$ $\therefore AC = DB$ (By $CPCT$)$\\$ Hence, the diagonals of a square are equal in length.$\\$ In $\Delta AOB$ and $\Delta COD,$$\\ \angle AOB = \angle COD (Vertically opposite angles)\\ \angle ABO = \angle CDO (Alternate interior angles)\\ AB = CD (Sides of a square are always equal)\\ \therefore \Delta AOB \cong \Delta COD (By AAS congruence rule)\\ \therefore AO = CO and OB = OD (By CPCT)\\ Hence, the diagonals of a square bisect each other.\\ In \Delta AOB and \Delta COB,\\ As we had proved that diagonals bisect each other, therefore,\\ AO = CO$$\\$ $AB = CB$ (Sides of a square are equal)$\\$ $BO = BO$ (Common)$\\$ $\therefore \Delta AOB \cong \Delta COB$ (By $SSS$ congruency)$\\$ $\therefore \angle AOB = \angle COB$ (By $CPCT$)$\\$ However, $\angle AOB + \angle COB = 180^o$ (Linear pair)$\\$ $2\angle AOB = 180^o$$\\ \angle AOB = 90^o$$\\$ Hence, the diagonals of a square bisect each other at right angles.

5   Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

##### Solution :

Let us consider a quadrilateral $ABCD$ in which the diagonals $AC$ and $BD$ intersect each other at $O$.$\\$ It is given that the diagonals of $ABCD$ are equal and bisect each other at right angles.$\\$ Therefore, $AC = BD, OA = OC, OB = OD,$ and $\angle AOB = \angle BOC = \angle COD = \angle AOD = 90^o$.$\\$ To prove $ABCD$ is a square$\\$ We have to prove that $ABCD$ is a parallelogram, $AB = BC = CD = AD,$ and one of its interior angles is$90^o$.$\\$ In $\Delta AOB$ and $\Delta COD,$$\\ AO = CO (Diagonals bisect each other)\\ OB = OD (Diagonals bisect each other)\\ \angle AOB = \angle COD (Vertically opposite angles)\\ \therefore \Delta AOB \cong \Delta COD (SAS congruence rule)\\ \therefore AB = CD (By CPCT)$$\\$ ... (1) And, $\angle OAB = \angle OCD$(By $CPCT)$$\\ However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.\\ AB \parallel CD ... (2)$$\\$

From Equations (1) and (2), we obtain$\\$ $ABCD$ is a parallelogram.$\\$ In $\Delta AOD$ and $\Delta COD,$$\\ AO = CO (Diagonals bisect each other)\\ \angle AOD = \angle COD (Given that each is 90^o)\\ OD = OD (Common)\\ \therefore \Delta AOD \cong \Delta COD (SAS congruence rule)\\ \therefore AD = DC ... (3)$$\\$ However,$AD = BC$ and $AB = CD$ (Opposite sides of parallelogram $ABCD$)$\\$ $AB = BC = CD = DA$$\\ Therefore, all the sides of quadrilateral ABCD are equal to each other.\\ In \Delta ADC and \Delta BCD,\\ AD = BC (Already proved)\\ AC = BD (Given)\\ DC = CD (Common)\\ \therefore \Delta ADC \cong \Delta BCD (SSS Congruence rule)\\ \therefore \angle ADC = \angle BCD (By CPCT)$$\\$ However, $\angle ADC + \angle BCD = 180^o$ (Co-interior angles)$\\$ $\angle ADC + \angle ADC = 180^o$$\\ 2\angle ADC = 180^o$$\\$ $\therefore \angle ADC = 90^o$$\\ One of the interior angles of quadrilateral ABCD is a right angle.\\ Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = ADand one of its interior angles is 90^o.$$\\$ Therefore, $ABCD$ is a square.