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

Given that \$ABCD\$ is a square.To prove : \$AC = BD\$ and \$AC\$ and \$BD\$ bisect each other at right angles.Proof: (i) In a \$Δ ABC\$ and \$Δ BAD\$,\$AB = AB\$ ( common line)\$BC = AD\$ ( opppsite sides of a square)\$∠ABC = ∠BAD\$ ( = 90° )\$Δ ABC ≅ Δ BAD \$( By SAS property)\$AC = BD\$ ( by CPCT).(ii) In a \$Δ OAD\$ and \$Δ OCB,\$\$AD = CB\$ ( opposite sides of a square)\$∠OAD = ∠OCB\$ ( transversal AC )\$∠ODA = ∠OBC\$ ( transversal BD )\$ΔOAD ≅ ΔOCB\$ (ASA property)\$OA = OC \$ ---------(i)Similarly \$OB = OD\$ ----------(ii)From (i) and (ii) AC and BD bisect each other.Now in a \$ΔOBA\$ and \$ΔODA\$,\$OB = OD\$ ( from (ii) )\$BA = DA\$\$OA = OA\$ ( common line )\$ΔAOB = ΔAOD \$----(iii) ( by CPCT\$∠AOB + ∠AOD = 180° \$ (linear pair)\$2∠AOB = 180°\$\$∠AOB = ∠AOD = 90°\$∴\$AC\$ and \$BD\$ bisect each other at right angles.

Q: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Here, we are joining A and C.In \$\Delta\$ABCP is the mid point of ABQ is the mid point of BCPQ\$||\$AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and also is half of it]PQ\$=\dfrac{1}{2}\$ACIn \$\Delta\$ADCR is mid point of CDS is mid point of ADRS\$||\$AC [Line segments joining the mid points of two sides of a triangle is parallel to third side and also is half of it]RS\$=\dfrac{1}{2}\$ACSo, PQ\$||\$RS and PQ\$=\$RS [one pair of opposite side is parallel and equal]In \$\Delta\$APS & \$\Delta\$BPQAP\$=\$BP [P is the mid point of AB)\$\angle\$PAS\$=\angle\$PBQ(All the angles of rectangle are \$90^o\$)AS\$=\$BQ\$\therefore \Delta\$APS\$\cong\Delta\$BPQ(SAS congruency)\$\therefore\$PS\$=\$PQBS\$=\$PQ & PQ\$=\$RS (opposite sides of parallelogram is equal)\$\therefore\$ PQ\$=\$RS\$=\$PS\$=\$RQ[All sides are equal]\$\therefore\$ PQRS is a parallelogram with all sides equal\$\therefore\$ So PQRS is a rhombus.

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

Given,Diagonals are equal\$AC=BD\$ .......\$(1)\$and the diagonals bisect each other at right angles\$OA=OC; OB=OD\$ ...... \$(2)\$\$\angle{AOB}=\$ \$\angle{BOC}=\$ \$\angle{COD}=\$ \$\angle{AOD}=\$ \$90^{0}\$ ..........\$(3)\$ Proof:Consider \$\triangle {AOB} \$ and \$\triangle {COB}\$\$ OA=OC \$ ....[from \$(2)\$]\$\angle{AOB}=\$ \$\angle{COB}\$\$OB\$ is the common sideTherefore,\$\triangle AOB\cong\$ \$\triangle COB\$From \$SAS\$ criteria, \$AB=CB \$Similarly, we prove\$\triangle AOB\cong\$ \$\triangle DOA\$, so \$AB=AD\$\$\triangle BOC\cong\$ \$\triangle COD\$, so \$CB=DC\$So, \$AB=AD=CB=DC\$ ....\$(4)\$So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence \$ABCD\$ is parallelogramIn \$\triangle {ABC} \$ and \$\triangle {DCB}\$\$AC=BD\$ ...(from \$(1)\$)\$AB=DC\$ ...(from \$(4)\$)\$BC\$ is the common side\$\triangle ABC\cong\$ \$\triangle DCB\$So, from SSS criteria, \$\angle{ABC}=\$ \$\angle{DCB}\$Now,\$AB \parallel CD, BC \$ is the tansversal\$\angle B + \angle C=\$ \$180^{0}\$\$\angle B + \angle B=\$ \$180^{0}\$\$\angle B=\$ \$90^{0}\$Hence, \$ ABCD\$ is a parallelogram with all sides equal and one angle is \$90^{0}\$So, ABCD is a square.Hence proved.

Q: A diagonal of a rectangle is inclined to one side of the rectangle at \$25 ^{ o }\$. The acute angle between the diagonals is:

In \$ \triangle BOC\$, \$\angle OBC=\angle OCB\$ ----- (Opposite angle of isosceles triangle)\$\angle OBC+\angle OCB+\angle BOC = 180^\circ\$\$25^\circ+25^\circ+\angle BOC = 180^\circ\$Therefore, \$\angle BOC = 130^\circ\$ So, \$\angle AOB+ \angle BOC = 180^\circ\$--------(Linear pair)\$\angle AOB = 180^\circ- 130^\circ\$\$\angle AOB = 50^\circ\$So, the acute angle between the diagonal is \$50^\circ\$.

Q: Sum of the area of two squares is \$640\$ m\$^2\$. If the difference of their perimeters is \$64\$ m, find the sides of the two squares.

Let the sides of the two squares be of \$a\$ and \$b\$.The according to the problem\$a^2+b^2=640\$.........(1).Again \$4(a-b)=64\$or, \$a-b=16\$.......(2).Using (2) in (1) we get,\$a^2+(-16+a)^2=640\$or, \$2a^2-32a+256=640\$or, \$2a^2-32a-384=0\$or, \$a^2-16a-192=0\$or, \$(a-24)(a+8)=0 \$Since \$a\$ is the side of a square then \$a=24\Rightarrow b=8\$.So the sides of the squares be \$24\$ m and \$8\$ m.

Q: Show that the quadrilateral, formed by joining the mid-points of the sides of a square is also a square.

In a square \$ABCD,\$ \$P,Q,R\$ and \$S\$ are the mid-points of \$AB,BC,CD\$ and \$DA\$ respectively.\$\Rightarrow\$ \$AB=BC=CD=AD\$ [ Sides of square are equal ]In \$\triangle ADC,\$\$SR\parallel AC\$ and \$SR=\dfrac{1}{2}AC\$ [ By mid-point theorem ] ---- ( 1 )In \$\triangle ABC,\$\$PQ\parallel AC\$ and \$PQ=\dfrac{1}{2}AC\$ [ By mid-point theorem ] ---- ( 2 )From equation ( 1 ) and ( 2 ),\$SR\parallel PQ\$ and \$SR=PQ=\dfrac{1}{2}AC\$ ---- ( 3 )Similarly, \$SP\parallel BD\$ and \$BD\parallel RQ\$\$\therefore\$ \$SP\parallel RQ\$ and \$SP=\dfrac{1}{2}BD\$and \$RQ=\dfrac{1}{2}BD\$\$\therefore\$ \$SP=RQ=\dfrac{1}{2}BD\$Since, diagonals of a square bisect each other at right angle.\$\therefore\$ \$AC=BD\$\$\Rightarrow\$ \$SP=RQ=\dfrac{1}{2}AC\$ ----- ( 4 )From ( 3 ) and ( 4 )\$SR=PQ=SP=RQ\$We know that the diagonals of a square bisect each other at right angles.\$\angle EOF=90^o.\$Now, \$RQ\parallel DB\$\$RE\parallel FO\$Also, \$SR \parallel AC\$\$\Rightarrow\$ \$FR \parallel OE\$\$\therefore\$ \$OERF\$ is a parallelogram.So, \$\angle FRE =\angle EOF=90^o\$ (Opposite angles are equal)Thus, \$PQRS\$ is a parallelogram with \$\angle R=90^o\$ and \$SR=PQ=SP=RQ\$\$\therefore\$ \$PQRS\$ is a square.

Q: The measures of two adjacent angles of a parallelogram are in the ratio \$3 : 2\$. Find the measure of each of the angles of the parallelogram

\$\textbf{Step - 1: Finding the adjacent angles of the parallelogram}\$ \$\text{Since the adjacent angles are in the ratio }3:2\text{, let the angles be }3x\text{ and }2x\$ \$\text{Adjacent angles of a parallelogram are supplementary}\$ \$\therefore 3x+2x=180^\circ\$ \$\Rightarrow5x=180^\circ\$ \$\Rightarrow x=36^{\circ}\$ \$\therefore 3x=108^{\circ} \text{ and }2x=72^{\circ}\$\$\textbf{Step - 2: Finding the opposite angles of the parallelogram}\$ \$\text{Opposite angles of a parallelogram are equal.}\$ \$\text{Therefore the angle opposite to }108^{\circ}\text{ is }108^{\circ}\$ \$\text{And , the angle opposite to }72^{\circ}\text{ is }72^{\circ}\$ \$\textbf{Hence, the angles of the parallelogram are }\mathbf{108^{\circ},\;72^{\circ},\;108^{\circ}\text{ and }\;72^{\circ}}.\$

Question 1

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

Accepted Answer

Given that $ABCD$ is a square.To prove : $AC = BD$ and $AC$ and $BD$ bisect each other at right angles.Proof: (i)  In a $&#916; ABC$ and $&#916; BAD$,$AB =  AB$ ( common line)$BC = AD$ ( opppsite sides of a square)$&#8736;ABC = &#8736;BAD$ ( = 90&#176; )$&#916; ABC &#8773; &#916; BAD $( By SAS property)$AC = BD$ ( by CPCT).(ii) In a $&#916; OAD$ and $&#916; OCB,$$AD = CB$ ( opposite sides of a square)$&#8736;OAD = &#8736;OCB$ ( transversal AC )$&#8736;ODA = &#8736;OBC$ ( transversal BD )$&#916;OAD &#8773; &#916;OCB$ (ASA property)$OA = OC $ ---------(i)Similarly $OB = OD$ ----------(ii)From (i) and (ii)  AC and BD bisect each other.Now in a $&#916;OBA$ and $&#916;ODA$,$OB = OD$ ( from (ii) )$BA = DA$$OA = OA$  ( common line )$&#916;AOB = &#916;AOD $----(iii) ( by CPCT$&#8736;AOB + &#8736;AOD = 180&#176; $  (linear pair)$2&#8736;AOB  = 180&#176;$$&#8736;AOB = &#8736;AOD = 90&#176;$&#8756;$AC$ and $BD$ bisect each other at right angles.

Question 2

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Accepted Answer

Here, we are joining A and C.In $\Delta$ABCP is the mid point of ABQ is the mid point of BCPQ$||$AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and also is half of it]PQ$=\dfrac{1}{2}$ACIn $\Delta$ADCR is mid point of CDS is mid point of ADRS$||$AC [Line segments joining the mid points of two sides of a triangle is parallel to third side and also is half of it]RS$=\dfrac{1}{2}$ACSo, PQ$||$RS and PQ$=$RS [one pair of opposite side is parallel and equal]In $\Delta$APS & $\Delta$BPQAP$=$BP [P is the mid point of AB)$\angle$PAS$=\angle$PBQ(All the angles of rectangle are $90^o$)AS$=$BQ$\therefore \Delta$APS$\cong\Delta$BPQ(SAS congruency)$\therefore$PS$=$PQBS$=$PQ & PQ$=$RS (opposite sides of parallelogram is equal)$\therefore$ PQ$=$RS$=$PS$=$RQ[All sides are equal]$\therefore$ PQRS is a parallelogram with all sides equal$\therefore$ So PQRS is a rhombus.

Question 3

$ABCD$ is a rhombus and $P,Q,R$ and $S$ are the mid point of the sides $AB,BC,CD$ and $DA$ respectively.show that the quadrilateral $PQRS$ is a rectangle

Accepted Answer

Given-  ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.To Prove-PQRS is a rectangleConstruction,AC and BD are joined.Proof,In &#916;DRS and &#916;BPQ,DS = BQ (Halves of the opposite sides of the rhombus)&#8736;SDR = &#8736;QBP (Opposite angles of the rhombus)DR = BP (Halves of the opposite sides of the rhombus)Thus, &#916;DRS &#8773; &#916;BPQ by SAS congruence condition.RS = PQ by CPCT --- (i)In &#916;QCR and &#916;SAP,RC = PA (Halves of the opposite sides of the rhombus)&#8736;RCQ = &#8736;PAS (Opposite angles of the rhombus)CQ = AS (Halves of the opposite sides of the rhombus)Thus, &#916;QCR &#8773; &#916;SAP by SAS congruence condition.RQ = SP by CPCT --- (ii)Now,In &#916;CDB,R and Q are the mid points of CD and BC respectively.&#8658; QR || BD  also,P and S are the mid points of AD and AB respectively.&#8658; PS || BD&#8658; QR || PSThus, PQRS is a parallelogram.also, &#8736;PQR = 90&#176; Now,In PQRS,RS = PQ and RQ = SP from (i) and (ii)&#8736;Q = 90&#176;Thus, PQRS is a rectangle.

Question 4

ABCD is a rhombus. If $\angle ACB = 40^o$, then $\angle ADB$ is

Accepted Answer

In given figure ABCD is a rhombus. We know that diagonals of rhombus bisect each other perpendicularly.Hence, $\angle$BOC= 90$^\circ$$\angle$OCB = 40$^\circ$  (Given)AD$\parallel$BC and BD is the transversal   --- (Opposite sides of rhombus are parallel to each other)$\therefore$ $\angle$ADB = $\angle$DBC   ---- (Alternate angles)In $\triangle$OBC,$\angle$BOC + $\angle$OCB + $\angle$OBC = 180$^\circ$$\Rightarrow$  90$^\circ$+ 40$^\circ$ + $\angle$OBC = 180$^\circ$$\Rightarrow$  $\angle$OBC =180$^\circ$ - 130$^\circ$$\therefore$  $\angle$OBC = 50$^\circ$But $\angle$OBC =$\angle$DBC$\therefore$  $\angle$ADB = 50$^\circ$   ---( Alternate angle)

Question 5

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

Accepted Answer

Given,Diagonals are equal$AC=BD$                  .......$(1)$and the diagonals bisect each other at right angles$OA=OC; OB=OD$           ...... $(2)$$\angle{AOB}=$ $\angle{BOC}=$ $\angle{COD}=$  $\angle{AOD}=$ $90^{0}$    ..........$(3)$ Proof:Consider $\triangle {AOB} $ and $\triangle {COB}$$ OA=OC $  ....[from $(2)$]$\angle{AOB}=$ $\angle{COB}$$OB$ is the common sideTherefore,$\triangle AOB\cong$ $\triangle COB$From $SAS$ criteria, $AB=CB $Similarly, we prove$\triangle AOB\cong$ $\triangle DOA$, so $AB=AD$$\triangle BOC\cong$ $\triangle COD$, so $CB=DC$So, $AB=AD=CB=DC$               ....$(4)$So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence $ABCD$ is  parallelogramIn $\triangle {ABC} $ and $\triangle {DCB}$$AC=BD$            ...(from $(1)$)$AB=DC$            ...(from $(4)$)$BC$ is the common side$\triangle ABC\cong$ $\triangle DCB$So, from SSS criteria, $\angle{ABC}=$ $\angle{DCB}$Now,$AB \parallel CD, BC $ is the tansversal$\angle B + \angle C=$ $180^{0}$$\angle B + \angle B=$ $180^{0}$$\angle B=$ $90^{0}$Hence, $ ABCD$ is a parallelogram with all sides equal and one angle is $90^{0}$So, ABCD is a square.Hence proved.

Question 6

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Accepted Answer

Take quadrilateral ABCD , AC and BD are diagonals which intersect at O.In $\triangle AOB$ and $\triangle AOD$$DO=OB$ $\mid$ $O$ is the midpoint$AO=AO$ $\mid$ Common side$\angle AOB= \angle AOD$ $\mid$ Right angleSo, $\triangle AOB \cong \triangle AOD$So, $AB=AD$Similarly, $AB=BC=CD=AD$ can be proved which means that $ABCD$ is a rhombus.

Question 7

A diagonal of a rectangle is inclined to one side of the rectangle at $25 ^{ o }$. The acute angle between the diagonals is:

Accepted Answer

In $ \triangle BOC$, $\angle OBC=\angle OCB$  ----- (Opposite angle of isosceles triangle)$\angle OBC+\angle OCB+\angle BOC = 180^\circ$$25^\circ+25^\circ+\angle BOC = 180^\circ$Therefore, $\angle BOC = 130^\circ$    So, $\angle AOB+ \angle BOC = 180^\circ$--------(Linear pair)$\angle AOB = 180^\circ- 130^\circ$$\angle AOB = 50^\circ$So, the acute angle between the diagonal is $50^\circ$.

Question 8

Sum of the area of two squares is $640$ m$^2$. If the difference of their perimeters is $64$ m, find the sides of the two squares.

Accepted Answer

Let the sides of the two squares be of $a$ and $b$.The according to the problem$a^2+b^2=640$.........(1).Again $4(a-b)=64$or, $a-b=16$.......(2).Using (2) in (1) we get,$a^2+(-16+a)^2=640$or, $2a^2-32a+256=640$or, $2a^2-32a-384=0$or, $a^2-16a-192=0$or, $(a-24)(a+8)=0 $Since $a$ is the side of a square then $a=24\Rightarrow b=8$.So the sides of the squares be $24$ m and $8$ m.

Question 9

Show that the quadrilateral, formed by joining the mid-points of the sides of a square is also a square.

Accepted Answer

In a square $ABCD,$ $P,Q,R$ and $S$ are the mid-points of $AB,BC,CD$ and $DA$ respectively.$\Rightarrow$  $AB=BC=CD=AD$           [ Sides of square are equal ]In $\triangle ADC,$$SR\parallel AC$ and $SR=\dfrac{1}{2}AC$       [ By mid-point theorem ]  ---- ( 1 )In $\triangle ABC,$$PQ\parallel AC$ and $PQ=\dfrac{1}{2}AC$     [ By mid-point theorem ]  ---- ( 2 )From equation ( 1 ) and ( 2 ),$SR\parallel PQ$ and $SR=PQ=\dfrac{1}{2}AC$    ---- ( 3 )Similarly, $SP\parallel BD$ and $BD\parallel RQ$$\therefore$  $SP\parallel RQ$ and $SP=\dfrac{1}{2}BD$and $RQ=\dfrac{1}{2}BD$$\therefore$  $SP=RQ=\dfrac{1}{2}BD$Since, diagonals of a square bisect each other at right angle.$\therefore$  $AC=BD$$\Rightarrow$  $SP=RQ=\dfrac{1}{2}AC$          ----- ( 4 )From ( 3 ) and ( 4 )$SR=PQ=SP=RQ$We know that the diagonals of a square bisect each other at right angles.$\angle EOF=90^o.$Now, $RQ\parallel DB$$RE\parallel FO$Also, $SR \parallel AC$$\Rightarrow$  $FR \parallel OE$$\therefore$  $OERF$ is a parallelogram.So, $\angle FRE =\angle EOF=90^o$  (Opposite angles are equal)Thus, $PQRS$ is a parallelogram with $\angle R=90^o$ and $SR=PQ=SP=RQ$$\therefore$  $PQRS$ is a square.

Question 10

The measures of two adjacent angles of a parallelogram are in the ratio $3 : 2$. Find the measure of each of the angles of the parallelogram

Accepted Answer

$\textbf{Step - 1: Finding the adjacent angles of the parallelogram}$                   $\text{Since the adjacent angles are in the ratio }3:2\text{, let the angles be }3x\text{ and }2x$                   $\text{Adjacent angles of a parallelogram are supplementary}$                   $\therefore 3x+2x=180^\circ$                   $\Rightarrow5x=180^\circ$                   $\Rightarrow x=36^{\circ}$                   $\therefore 3x=108^{\circ} \text{ and }2x=72^{\circ}$$\textbf{Step - 2: Finding the opposite angles of the parallelogram}$                            $\text{Opposite angles of a parallelogram are equal.}$                   $\text{Therefore the angle opposite to }108^{\circ}\text{ is }108^{\circ}$                     $\text{And , the angle opposite to }72^{\circ}\text{ is }72^{\circ}$  $\textbf{Hence, the angles of the parallelogram are }\mathbf{108^{\circ},\;72^{\circ},\;108^{\circ}\text{ and }\;72^{\circ}}.$

Revise with Concepts

Rhombus

Rectangles

Squares

Quick Summary With Stories

Rhombus

Diagonal of Rhombus

Applications of properties of Rhombus

Rectangle

Diagonals of a Rectangle

Square

Diagonals of a Square

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

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

$A B C D$ is a rhombus and $P, Q, R$ and $S$ are the mid point of the sides $A B, B C, C D$ and $D A$ respectively.show that the quadrilateral $P Q R S$ is a rectangle

ABCD is a rhombus. If $∠ A C B = 4 0^{o}$ , then $∠ A D B$ is

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

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

A diagonal of a rectangle is inclined to one side of the rectangle at $2 5^{o}$ . The acute angle between the diagonals is:

Sum of the area of two squares is $640$ m $^{2}$ . If the difference of their perimeters is $64$ m, find the sides of the two squares.

Show that the quadrilateral, formed by joining the mid-points of the sides of a square is also a square.

The measures of two adjacent angles of a parallelogram are in the ratio $3 : 2$ . Find the measure of each of the angles of the parallelogram

Revise with Concepts

Rhombus

Rectangles

Squares

Quick Summary With Stories

Rhombus

Diagonal of Rhombus

Applications of properties of Rhombus

Rectangle

Diagonals of a Rectangle

Square

Diagonals of a Square

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

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

ABCD is a rhombus and P,Q,R and S are the mid point of the sides AB,BC,CD and DA respectively.show that the quadrilateral PQRS is a rectangle

ABCD is a rhombus. If ∠ACB=40o, then ∠ADB is

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

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

A diagonal of a rectangle is inclined to one side of the rectangle at 25o. The acute angle between the diagonals is:

Sum of the area of two squares is 640 m2. If the difference of their perimeters is 64 m, find the sides of the two squares.

Show that the quadrilateral, formed by joining the mid-points of the sides of a square is also a square.

The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram

$A B C D$ is a rhombus and $P, Q, R$ and $S$ are the mid point of the sides $A B, B C, C D$ and $D A$ respectively.show that the quadrilateral $P Q R S$ is a rectangle

ABCD is a rhombus. If $∠ A C B = 4 0^{o}$ , then $∠ A D B$ is

A diagonal of a rectangle is inclined to one side of the rectangle at $2 5^{o}$ . The acute angle between the diagonals is:

Sum of the area of two squares is $640$ m $^{2}$ . If the difference of their perimeters is $64$ m, find the sides of the two squares.

The measures of two adjacent angles of a parallelogram are in the ratio $3 : 2$ . Find the measure of each of the angles of the parallelogram