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}$                            $\text{Opposite angles of a parallelogram are equal.}$                   $\text{Therefore the angle opposite to }108^{\circ}\text{ is }108^{\circ}$                     $\text{And , the angle opposite to }72^{\circ}\text{ is }72^{\circ}$  $\textbf{Hence, the angles of the parallelogram are }\mathbf{108^{\circ},\;72^{\circ},\;108^{\circ}\text{ and }\;72^{\circ}}.$

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Rhombus

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Rhombus

Diagonal of Rhombus