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: A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.

Area of park \$=l \times b= 45 \times 30=1350\ cm^2\$Area of park with path is \$=45+2.5+2.5 \times 30+2.5+2.5 = 1750\ cm^2\$Therefore,Area of path \$=\$ Area of park with path \$-\$ Area of rectangular park \$= 1750-1350=400\ cm^2\$Hence, this is the answer.

Q: Show that a diagonal of a parallelogram divides into two congruent triangles and hence prove that the opposite sides of a parallelogram are equal.

REF. Image.consider \$\Delta \$ ABC and \$\Delta \$ ACDSince the line segments AB+CD are parallelto each other and AC is a transversal\$\angle \$ ACB = \$\angle \$ CAD.AC = AC (common side)\$\angle \$CAB = \$\angle \$ ACD.Thus, by ASA criteria\$ \Delta \$ABC \$\cong \$ \$ \Delta \$ACDThe corresponding part of the congruenttriangle are congruentAB = CD + AD = BC

Q: Through the mid point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC at L and AD produced at E. Prove EL=2BL.

In \$\bigtriangleup BMC\$ and \$\bigtriangleup EMD\$\$\angle BMC=\angle EMD\$ [Vertically opposite]\$MC=DM\$ [Given]\$\angle BCM=\angle EDM\$ [Alternate angles]\$\therefore \bigtriangleup BMC\cong \bigtriangleup EMD\$ [By ASA]Hence, \$BC=DE\$ [By CPCT] \$\rightarrow (1)\$\$AE=AD+DE=BC+BC=2BC\$ \$\rightarrow (2)\$Now, \$\bigtriangleup BLC\sim \bigtriangleup ELA\$ (AA Similarly)\$\dfrac{BL}{EL}=\dfrac{BC}{AE}\$ [By CPCT]\$\dfrac{BL}{EL}=\dfrac{BC}{2BC}\$\$\dfrac{BL}{EL}=\dfrac{1}{2}\$\$EL=2BL\$

Q: Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If \$OA=3\$cm and \$OD=2\$cm determine the lengths of AC and BD.

Here, \$ABCD\$ is a parallelogram and \$AC\$ and \$BD\$ are diagonals of parallelogram intersect each other at \$O.\$\$OA=3\,cm\$ and \$OD=2\,cm\$We know that, diagonals of a parallelogram bisect each other.\$\therefore\$ \$AO=OC\$ and \$BO=OD\$\$\Rightarrow\$ \$AC=2\times OA=2\times 3\,cm=6\,cm\$\$\Rightarrow\$ \$BD=2\times OD=2\times 2\,cm=4\,cm\$\$\therefore\$ The length of the diagonals \$AC\$ and \$BD\$ are \$6\,cm\$ and \$4\,cm.\$

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}}.\$

Q: The diagonals of a rhombus measure \$16\$cm and \$30\$cm. Find its perimeter.

Given: Diagonals AC\$=30\$cm and DB\$=16\$cm.Since the diagonals of the rhombus bisect at right angle to each other.Therefore, OD\$=\dfrac{DB}{2}=\dfrac{16}{2}=8\$cmand OC\$=\dfrac{AC}{2}=\dfrac{30}{2}=15\$cmNow, In right angle triangle DOC,\$(DC)^2=(OD)^2+(CO)^2\$\$\Rightarrow (DC)^2=(8)^2+(15)^2\$\$\Rightarrow (DC)^2=64+225=289\$\$\Rightarrow DC=\sqrt{289}=17\$cmPerimeter of the rhombus\$=4\times\$ side\$=4\times 17=68\$cmThus, the perimeter of rhombus is \$68\$ cm.

Q: How many sides does a pentagon have?

A polygon has as many angles as sides.Like, triangle has \$3\$ angles and \$3\$ sides.A pentagon means a polygon which has \$5\$ sides and \$5\$ angles.Hence, option B is correct.

Question 1

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 2

A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.

Accepted Answer

Area of park $=l \times b= 45 \times 30=1350\ cm^2$Area of park with path is $=45+2.5+2.5 \times 30+2.5+2.5 = 1750\ cm^2$Therefore,Area of path $=$ Area of park with path $-$ Area of rectangular park             $= 1750-1350=400\ cm^2$Hence, this is the answer.

Question 3

The diagonals $AC$ and $BD$ of a parallelogram $ABCD$ intersect each other at the point $O$.If $\angle DAC = 32^o$ and $\angle AOB = 70^o$, then $\angle DBC$ is equal to?

Accepted Answer

In given figure,Quadrilateral ABCD is a parallelogram.So, AD $\left|\right|$ BC$\therefore$ $\angle$DAC = $\angle$ACB  --- ( Alternate angle)$\therefore$ $\angle$ACB = 32$^\circ$$\angle$AOB + $\angle$BOC = 180$^\circ$  --- (straight angle)$\Rightarrow$70$^\circ$ + $\angle$BOC = 180$^\circ$$\therefore$ $\angle$BOC = 110$^\circ$In $\triangle$BOC,$\angle$OBC + $\angle$BOC + $\angle$OCB = 180$^\circ$$\Rightarrow$$\angle$OBC  + 110$^\circ$ + 32$^\circ$ = 180$^\circ$$\Rightarrow$ $\angle$OBC = 38$^\circ$$\therefore$  $\angle$DBC = 38$^\circ$

Question 4

Show that a diagonal of a parallelogram divides into two congruent triangles and hence prove that the opposite sides of a parallelogram are equal.

Accepted Answer

REF. Image.consider $\Delta $ ABC and $\Delta $ ACDSince the line segments AB+CD are parallelto each other and AC is a transversal$\angle $ ACB = $\angle $ CAD.AC = AC (common side)$\angle $CAB = $\angle $ ACD.Thus, by ASA criteria$ \Delta $ABC $\cong $ $ \Delta $ACDThe corresponding part of the congruenttriangle are congruentAB = CD + AD = BC

Question 5

Through the mid point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC at L and AD produced at E. Prove EL=2BL.

Accepted Answer

In $\bigtriangleup BMC$ and  $\bigtriangleup EMD$$\angle BMC=\angle EMD$  [Vertically opposite]$MC=DM$ [Given]$\angle BCM=\angle EDM$ [Alternate angles]$\therefore \bigtriangleup  BMC\cong \bigtriangleup  EMD$ [By ASA]Hence, $BC=DE$ [By CPCT] $\rightarrow (1)$$AE=AD+DE=BC+BC=2BC$ $\rightarrow (2)$Now,  $\bigtriangleup BLC\sim \bigtriangleup ELA$ (AA Similarly)$\dfrac{BL}{EL}=\dfrac{BC}{AE}$   [By CPCT]$\dfrac{BL}{EL}=\dfrac{BC}{2BC}$$\dfrac{BL}{EL}=\dfrac{1}{2}$$EL=2BL$

Question 6

Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If $OA=3$cm and $OD=2$cm determine the lengths of AC and BD.

Accepted Answer

Here, $ABCD$ is a parallelogram and $AC$ and $BD$ are diagonals of parallelogram intersect each other at $O.$$OA=3\,cm$ and $OD=2\,cm$We know that, diagonals of a parallelogram bisect each other.$\therefore$  $AO=OC$ and $BO=OD$$\Rightarrow$  $AC=2\times OA=2\times 3\,cm=6\,cm$$\Rightarrow$  $BD=2\times OD=2\times 2\,cm=4\,cm$$\therefore$  The length of the diagonals $AC$ and $BD$ are $6\,cm$ and $4\,cm.$

Question 7

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

Question 8

The diagonals of a rhombus measure $16$cm and $30$cm. Find its perimeter.

Accepted Answer

Given: Diagonals AC$=30$cm and DB$=16$cm.Since the diagonals of the rhombus bisect at right angle to each other.Therefore, OD$=\dfrac{DB}{2}=\dfrac{16}{2}=8$cmand OC$=\dfrac{AC}{2}=\dfrac{30}{2}=15$cmNow, In right angle triangle DOC,$(DC)^2=(OD)^2+(CO)^2$$\Rightarrow (DC)^2=(8)^2+(15)^2$$\Rightarrow (DC)^2=64+225=289$$\Rightarrow DC=\sqrt{289}=17$cmPerimeter of the rhombus$=4\times$ side$=4\times 17=68$cmThus, the perimeter of rhombus is $68$ cm.

Question 9

State whether True or False.(a) All rectangles are squares.(b) All rhombuses are parallelograms.(c) All squares are rhombuses and also rectangles.(d) All squares are not parallelograms.(e) All kites are rhombuses.(f) All rhombuses are kites.(g) All parallelograms are trapeziums.(h) All squares are trapeziums.

Accepted Answer

(a) FalseAll squares are rectangles. (All sides of rectangle are not equal).(b) TrueOpposite sides are equal and parallel.(c) TrueAll sides of the rhombus are equal $\therefore$ All squares are rhombuses. All squares are also rectangles as each interval angle measure $90^{\circ}$(d) FalseAll squares are parallelogram as opposite sides are equal and parallel.(e) FalseKite has different lengths.(f) TrueRhombus also has two distinct consecutive pairs of sides of equal length.(g) TrueParallelogram pair of parallel sides.(h) TrueAll squares have a pair of parallel sides.

Question 10

How many sides does a pentagon have?

Accepted Answer

A polygon has as many angles as sides.Like, triangle has $3$ angles and $3$ sides.A pentagon means a polygon which has $5$ sides and $5$ angles.Hence, option B is correct.