Maths M.C.Q

2. II and III

Solution:

Consider I.

We know that, in a rectangle the diagonals are not bisectors of each other, since the adjacent side.

Thus, I is false.

Consider II.

We know that, in a square the diagonals bisect the opposite angles.

So, in a square ABCD, the diagonals AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$

Thus, II is true.

Consider III.

We know that, in a rhombus the diagonals bisect the opposite angles.

So, in a rhombus ABCD, the diagonals AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$

Thus, III is true.

3. 60°.

Solution:

Since ABCD is a parallelogram, AB || CD since opposite angles of a parallelogram are equal.

$\Rightarrow\angle\text{ABD}=\angle\text{BCD}=45^{\circ}$ ...(Alternate angles)

In $\triangle\text{ADB},$

$\angle\text{ABD}+\angle\text{BDA}+\angle\text{DAB}=180^{\circ}$ ...(Angle sum property)

$\Rightarrow45+\angle\text{BDA}+75=180$

$\Rightarrow\angle\text{BDA}+120=180$

$\Rightarrow\angle\text{BDA}=60^{\circ}$

$\Rightarrow\angle\text{CBD}=\angle\text{BDA}=60^{\circ}$ ...(Alternate angles)

$\Rightarrow\angle\text{CBD}=60^{\circ}$

2. 12cm.

Solution:

Let ABCD be the rhombus with diagonals AB = 10cm and Ac = 16cm.

Since the diagonals of a rhombus are perpendicular bisectors of each other,

$\Rightarrow\text{OA}=8\text{cm},\text{BD = 2OB}$ and $\angle\text{AOB}=90^{\circ}$

In right $\triangle\text{AOB},$

By Pythagoras theorem,

$\text{AB}^2=\text{OA}^2+\text{OB}^2$

$\Rightarrow(10)^{2}=(8)^2+\text{OB}^2$

$\Rightarrow100=64+\text{OB}^2$

$\Rightarrow\text{OB}^2=36$

$\Rightarrow\text{OB}=\sqrt{36}$

$\Rightarrow\text{OB}=6\text{cm}$

$\Rightarrow\text{BD}=2\times\text{OB}$

$\Rightarrow\text{BD}=2\times6$

$\Rightarrow\text{BD}=12\text{cm}$

Hence, the lenght of the other diagonal is 12cm.

4. Diagonals of ABCD are equal and perpendicular.

Solution:

In $\triangle\text{ABC},$ P and Q are the mid-points of sides AB and BC respectively.

$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC ...(i)}$

In $\triangle\text{BCD},$ Q and R are the mid-points of sides BC and CD respectively.

$\therefore\text{QR || BD}$ and $\text{QR}=\frac{1}{2}\text{BD ...(ii)}$

In $\triangle\text{ADC},$ S and R are the mid-points of sides AD and CD respectively.

$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC ...(iii)}$

In $\triangle\text{ABD},$ P and S are the mid-points of sides AB and AD respectively.

$\therefore\text{SP || BD}$ and $\text{SP}=\frac{1}{2}\text{BD ...(iv)}$

$\Rightarrow\text{PQ || RS}$ and $\text{QR || SP }$ [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, $\text{AC = BD}$ (given)

$\Rightarrow\frac{1}{2}\text{AC}=\frac{1}{2}\text{BD}$

$\Rightarrow\text{PQ = QR = RS = SP}$ [From (i), (ii), (iii) and (iv)]

Let the diagonals AC and BD intersect at O.

Now,

$\text{PS || BD}$

$\Rightarrow\text{PN || MO}$

Also, from (i), $\text{PQ || AC}$

$\Rightarrow\text{PM || NO}$

Thus, in quadrilateral PMON, $\text{PM || NO}$ and $\text{PN || MO}$

$\Rightarrow\text{PMON}$ is a parallelogram.

$\Rightarrow\angle\text{MPN}=\angle\text{MON}$ (opposite angles of a parallelogram are equal) 

$\Rightarrow\angle\text{MPN}=\angle\text{BOA}$ $(\text{Since }\angle\text{BOA}=\angle\text{MON})$

$\Rightarrow\angle\text{MPN}=90^{\circ}$ $(\text{Since }\text{AC}\perp\text{BD},\angle\text{BOA}=90^{\circ})$

$\Rightarrow\angle\text{QPS}=90^{\circ}$

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and $\angle\text{QPS}=90^{\circ}.$

Hence, PQRS is a square if diagonals of ABCD are equal and perpendicular.

4. Quadrilateral whose opposite angles are supplementary.

Solution:

In $\triangle\text{APB},$ by angle sum property,

$\angle\text{APB}+\angle\text{PAB}+\angle\text{PBA}=180^{\circ}$

$\Rightarrow\angle\text{APB}+\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=180^{\circ}$

$\Rightarrow\angle\text{APB}=180^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}\Big)$

In $\triangle\text{CRD},$ by angle sum property,

$\angle\text{CRD}+\angle\text{RDC}+\angle\text{RCD}=180^{\circ}$

$\Rightarrow\angle\text{CRD}+\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}=180^{\circ}$

$\Rightarrow\angle\text{CRD}=180^{\circ}-\Big(\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}\Big)$

Now, $\angle\text{SPQ}+\angle\text{SRQ}=\angle\text{APB}+\angle\text{CRD}$

$=360^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\frac{1}{2}\angle\text{D}\Big)$

$=360^{\circ}=\frac{1}{2}(\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D})$

$=360^{\circ}-\frac{1}{2}\times360^{\circ}$

$=360^{\circ}-180^{\circ}$

$=180^{\circ}$

Now, $\angle\text{PSR}+\angle\text{PQR}=360^{\circ}-(\angle\text{SPQ}+\angle\text{SRQ})$

$=360^{\circ}-180^{\circ}$

$=180^{\circ}$

Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

1. 40°

Solution:

$\text{AD || BC},$

$\Rightarrow\angle\text{DAO}=\angle\text{BCO}=30^{\circ}$ ...(Alternate angles)

$\Rightarrow\angle\text{BCO}=30^{\circ}$

$\angle\text{AOB}+\angle\text{BOC}=180^{\circ}$ ...(Linear pair of angles)

$\Rightarrow70+\angle\text{BOC}=180$

$\Rightarrow\angle\text{BOC}=110^{\circ}$

In $\triangle\text{CBO},$

$\angle\text{BOC}+\angle\text{BCO}+\angle\text{OBC}=180^{\circ}$ ...(Angle sum Property)

$\Rightarrow110+30+\angle\text{OBC}=180$

$\Rightarrow\angle\text{OBC}=40^{\circ}$

$\Rightarrow\angle\text{DBC}=40^{\circ}$ ...(D - O - B)

2. A parallelogram.

Solution:

In $\triangle\text{ABC},$ D and E are the mid-points of sides AB and AC respectively.

$\Rightarrow\text{DE || BC}$ and $\text{DE}=\frac{1}{2}\text{BC }...(\text{i})$

Now, $\text{MN = MP + PN}=\frac{1}{2}\text{BP}+\frac{1}{2}\text{CP}=\frac{1}{2}(\text{BP + CP})$

$\therefore\text{MN}=\frac{1}{2}\text{BC ...(ii)}$

From (i) and (ii),

$\text{DE || MN}$ and $\text{DE = MN}$

Hence, DENM is a parallelogram.

3. AC² + BD² = 4AB²

Solution:

The diagonals of a rhombus bisect each other at right angles.

$\Rightarrow\text{OA}=\frac{1}{2}\text{AC}$ and $\text{OB}=\frac{1}{2}\text{BD}$

Also, $\angle\text{AOB}=90^{\circ}$

In right $\triangle\text{AOB},$

$\therefore\text{AB}^2=\text{OA}^2+\text{OB}^2$ ...(By Pythagoras theorem)

$\Rightarrow\text{AB}^2=\frac{1}{4}\text{AC}^2+\frac{1}{4}\text{BD}^2$

$\Rightarrow\text{AC}^2+\text{BD}^2=4\text{AB}^2$

2. 70°

Solution:

$\angle\text{DAO}+\angle\text{OAB}=\angle\text{DAB}$

$\Rightarrow\angle\text{DAO}+35^{\circ}=90^{\circ}$

$\Rightarrow\angle\text{DAO}=55^{\circ}$

ABCD is a rectangle and diagonals of a rectangle are equal and bisect each other.

$\text{OA = OD}$

$\Rightarrow\angle\text{ODA}=\angle\text{DAO}$ (angles opposte to equal sides are equal)

$\Rightarrow\angle\text{ODA}=55^{\circ}$

In DODA, by angle sum property,

$\angle\text{ODA}+\angle\text{DAO}+\angle\text{AOD}=180^{\circ}$

$\Rightarrow55^{\circ}+\angle55^{\circ}+\angle\text{AOD}=180^{\circ}$

$\Rightarrow\angle\text{AOD}=70^{\circ}$

4. Diagonals of ABCD are perpendicular to each other.

Solution:

In $\triangle\text{ABC},$ P and Q are the mid-point of sides AB and BC respectively.

$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC }...(\text{i})$

In $\triangle\text{ADC},$ R and S are the mid-point of sides CD and AD respectively.

$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC }...(\text{ii})$

From (i) and (ii),

$\text{PQ ∥ RS}$ and $\text{PQ = RS}$

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in $\triangle\text{ABD},$ P and S are the mid-points of sides AB and AD respectively.

$\therefore\text{PS || BD}$

$\Rightarrow\text{PN || MO}$

Also, from (i), $\text{PQ || AC}$

$\Rightarrow\text{PM || NO}$

Thus, in quadrilateral PMON, $\text{PM || NO}$ and $\text{PN || MO},$

⇒ PMON is a parallelogram.

$\Rightarrow\angle\text{MPN}=\angle\text{MON}$ (opposite angles of a parallelogram are equal)

$\Rightarrow\angle\text{MPN}=\angle\text{BOA}$ $(\text{Since }\angle\text{BOA}=\angle\text{MON})$

$\Rightarrow\angle\text{MPN}=90^{\circ}$ $(\text{Since }\text{AC}\perp\text{BD},\angle\text{BOA}=90^{\circ})$

$\Rightarrow\angle\text{QPS}=90^{\circ}$

Thus, PQRS is a parallelogram whose one angle, i.e. $\angle\text{QPS}=90^{\circ}.$

Hence, PQRS is a rectangle if $\text{AC}\perp\text{BD}.$ 

2. AF = 2AB

Solution:

In $\triangle\text{CDE}$ and $\triangle\text{BFE,}$

$\angle\text{DEC}=\angle\text{FEB}$ ...(Vertically opposite angles)

$\angle\text{DCE}=\angle\text{FBE}$ ...(Alternate angles)

and $\text{CE = BE}$ ...(E is the mid-point.)

$\therefore\triangle\text{CDE}\cong\triangle\text{BFE}$ ...(By AA congruence criterion)

$\therefore\text{CD = BF}$ ...(C.P.C.T.)

Now,

$\text{AF = AB + BF}$

$\Rightarrow\text{AF + AB + CD}$ ...(from (i))

$\Rightarrow\text{AF = 2AB}$

3. Rectangle.

Solution:

The bisectors of $\angle\text{APQ},\angle\text{BPQ},\angle\text{CQP}$ and $\angle\text{PQD}$ enclose a rectengle.

4. BC² + AD² + 2AB ⋅ CD

Solution:

Construction: Draw perpendicular from D and C on AB which meets AB at E and F, respectively.

So, DEFC is a parallelogram, since one pair of opposite sides are parallel and equal.

In $\triangle\text{ABC},$

$\angle\text{B}$ is an acute angle.

$\Rightarrow\text{AC}^2=\text{BC}^2+\text{AB}^2-2\text{AB}\times\text{AE}$

In $\triangle\text{ABD},$

$\angle\text{A}$ is an acute angle.

$\Rightarrow\text{BD}^2=\text{AD}^2+\text{AB}^2-2\text{AB}\times\text{AF}$

$\Rightarrow\text{AC}^2+\text{BD}^2=\text{BC}^2+\text{AD}^2+2\text{AB}(\text{AB}-\text{BE}-\text{AF})$

$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{EF}$

$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{CD}$

3. Trapezium.

Solution:

Let the common multiple be x.

$\therefore$ The angle measure 3x, 7x, 6x and 4x.

 Since the sum of the angles of a quadrilateral is 360°, we have

3x + 7x + 6x + 4x = 360

⇒ 20x = 360

⇒ x = 18°

$\therefore$ The angles of the quadrilateral are

3x = 3(18) = 54°

7x = 7(18) = 126°

6x = 6(18) = 108° and

4x = 4(18) = 72°

Now, 54 + 126 = 180° and 108 + 72 = 180°

So, the angles are interior angles and hence we get one pair of parallel sides of ABCD.

Hence, ABCD is a trapezium.

4. 1 : 1

Solution:

Area of a parallelogram = base ⨯ height

Since two parallelogram stand on equal bases and between the same parallel lines,

their heights are same.

$\therefore$ Areas are also same.

$\therefore$ The ratio of their area is 1 : 1.

3. 72°

Solution:

Let one of the angle of the || gm be x°.

According to the given condition,

$\therefore$ the adjacent angle $=\frac{2}{3}\text{x}^{\circ}$

Now,

$\text{x}+\frac{2}{3}\text{x}=180^{\circ}$ ...(Sum of the adjacent angles of || gm is 180°.)

$\Rightarrow\frac{3\text{x}+2\text{x}}{3}=180^{\circ}$

$\Rightarrow\frac{5\text{x}}{3}=180^{\circ}$

$\Rightarrow5\text{x}=540^{\circ}$

$\Rightarrow\text{x}=108^{\circ}$

⇒ the adjacent angles $=\frac{2}{3}(108)=36\times2=72^{\circ}$

Hence, the smallest angle is 72°.

4. 90°

Solution:

Consider parallelogram ABCD,

We know that, the sum if the adjacent angles of a parallelogram is 180°.

$\Rightarrow\angle\text{DAB}+\angle\text{CBA}=180^{\circ}$

$\Rightarrow\frac{1}{2}\angle\text{DAB}+\frac{1}{2}\angle\text{CBA}=\frac{1}{2}(180^{\circ})$

$\Rightarrow\angle\text{OAB}+\angle\text{OBA}=90^{\circ} ...(\text{i})$

Now,

In $\triangle\text{OAB},$

$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^{\circ}$ ...(Angle sum property)

$\Rightarrow90^{\circ}+\angle\text{AOB}=180^{\circ}$ ...(from (i))

$\Rightarrow\angle\text{AOB}=90^{\circ}$

The bisectore of any adjacent angles of a parallelogram intersects at 90°.

2. $\frac{1}{2}(\text{a}+\text{b})$

Solution:

E and F are the given to be the mid-points of AD and BC respectively.

$\therefore\text{EF} = \frac{1}{2}(\text{AB + DC})$

$=\frac{1}{2}(\text{a + b})$

3. Diagonals of ABCD are equal.

Solution:

In $\triangle\text{ABC},$ P and Q are the mid-points of sides AB and BC respectively.

$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC ...(i)}$

In $\triangle\text{BCD},$ Q and R are the mid-points of sides BC and CD respectively.

$\therefore\text{QR || BD}$ and $\text{QR}=\frac{1}{2}\text{BD ...(ii)}$

In $\triangle\text{ADC},$ S and R are the mid-points of sides AD and CD respectively.

$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC ...(iii)}$

In $\triangle\text{ABD},$ P and S are the mid-points of sides AB and AD respectively.

$\therefore\text{SP || BD}$ and $\text{SP}=\frac{1}{2}\text{BD ... (iv)}$

$\Rightarrow\text{PQ ∥ RS}$ and $\text{QR ∥ SP}$ [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

$\Rightarrow\frac{1}{2}\text{AC}=\frac{1}{2}\text{BD}$

$\Rightarrow\text{PQ = QR = RS = SP }$ [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus if diagonals of ABCD are equal.

2. $\frac{1}{3}\text{AC}$

Solution:

Construction: Join DG and G be the mid-point of FC.

Now,

In $\triangle\text{BCF, D}$ is the mid-point of BC and G is the mid-point of FC and F is the mid-point of AG.

$\Rightarrow\text{DG || BF}$

$\Rightarrow\text{DG || EF}$ ...(B - E - F)

$\Rightarrow\text{AF}=\text{FG}=\text{GC}$ ...(Since G is the mid-point.)

$\Rightarrow\text{AF}=\frac{1}{2}\text{AC}$

3. 45°

Solution:

We know that, the opposite angles of a parallelogram are equal.

$\therefore\angle\text{C}=\angle\text{A}=75^{\circ}$

In $\triangle\text{BCD},$

$\angle\text{BCD}+\angle\text{BDC}+\angle\text{CBD}=180^{\circ}$ ...(Angle sum property)

$\Rightarrow75^{\circ}+\angle\text{BDC}+60^{\circ}=180^{\circ}$

$\Rightarrow\angle\text{BDC}=45^{\circ}$

1. 10cm.

Solution:

We konw that, the diagonals of a rhombus bisect each other at right angles.

So, Ac = 16cm and BD = 12cm

⇒ OA = 8cm and OB = 6cm

Also, $\angle\text{OAB}=90^{\circ}$

In right $\triangle\text{OAB},$

By Pythagoras theorem,

$\text{AB}^2=\text{OA}^2+\text{OB}^2$

$\Rightarrow\text{AB}^2=(8)^2+(6)^2$

$\Rightarrow\text{AB}^2=64+36$

$\Rightarrow\text{AB}^2=100$

$\Rightarrow\text{AB}=\sqrt{100}$

$\Rightarrow\text{AB}=10\text{cm}$

Hence, the length of each side of the rhombus is 10cm.

3. Rectangle.

Solution:

Given that ABCD is a parallelogram.

We konw that, opposite sides of a parallelogram are parallel.

$\Rightarrow\angle\text{A}+\angle\text{B}=180^{\circ}$ ...(interior angles)

Also, $\angle\text{A}=\angle\text{B}=90^{\circ}$ ...(Given)

Since opposite angles of a parallelogram are equal,

$\angle\text{A}=\angle\text{C}$ and $\angle\text{B}=\angle\text{D}$

So, $\angle\text{A}=\angle\text{C}=\angle\text{B}=\angle\text{D}=90^{\circ}$

$\therefore$ ABCD is a rectangle.

2. 50°

Solution:

We know that, sum of the angles of a quadrilateral is 360°.

$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}$

$\Rightarrow\angle\text{A}+\angle\text{B}+70^{\circ}+30^{\circ}=360^{\circ}$

$\Rightarrow\angle\text{A}+\angle\text{B}=260^{\circ}$

$\Rightarrow\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=\frac{1}{2}(260)^{\circ}$

$\Rightarrow\angle\text{BAO}+\angle\text{ABO}=130^{\circ}...(\text{i})$

In $\triangle\text{AOB},$

$\angle\text{BAO}+\angle\text{ABO}+\angle\text{AOB}=180^{\circ}$ ...(Angle sum Property)

$\Rightarrow130^{\circ}+\angle\text{AOB}=180^{\circ}$ ...(from (i))

$\Rightarrow\angle\text{AOB}=50^{\circ}$