Maths 4 Marks Questions

Given in a square ABCD, P Q R and S are the mid-points of AB, BC, CD and DA, respectively.

To show PQRS is a square.

Construction Join AC and BD.

Proof Since, ABCD is a square.

$\therefore$ AB = BC = CD = AD

Also, P. Q. R and S are the mid-points of AB, BC, CD and DA

respectively.

Then, in $\Delta\text{ADC}$

and $\text{SR}=\frac{1}{2}=\text{AC}$ [by mid-point theorem]...(i)

In $\Delta\text{ABC},$ PQ || AC

and $\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{ii})$

From Eqs. (i) and (ii).

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

Similarly, SP || BD and BD || RQ

$\therefore$ SP || RQ and $\text{SP}=\frac{1}{2}\text{BD}$

and $\text{RQ}=\frac{1}{2}\text{BD}$

$\therefore\ \text{SP}=\text{RQ}=\frac{1}{2}\text{BD}$

Since, diagonals of a square bisect each other at right angle.

$\therefore$ AC = BD

$\text{SP}=\text{RQ}=\frac{1}{2}\text{AC}\ ...(\text{iv})$

From Eqs. (i) and (iv). $\text{SR}=\text{PQ}=\text{SP}=\text{RQ}$ [all side are equal]

Now, in quadrilateral OERF,

OE || FR and OF || ER

$\angle\text{EOF}=\angle\text{ERF}=90^\circ$

Hence, PQRS is a square. Hence proved.

Given: $\Delta\text{ABC}$ in which E is the mid-point of a medien AD of $\Delta\text{ABC}$ and BE is produced to meet AC at F.

show that $\text{AF}=\frac{1}{3}\text{AC}$

to prove: $\text{AF}=\frac{1}{3}\text{AC}$

Construction: Draw DG || BF intersecting AC at G.

Proof: In $\Delta\text{ADG},$ E is the mid-point of AD and EF || DG.

$\therefore$ AF = FG ...(1) [Converse of mid-point theorem]

In $\Delta\text{FBC},$ D is the mid-point of BC and DG || BF.

$\therefore$ FG = GC ...(2) (Converse of mid-point theorem)

From equation (1) and (2), we get

AF = FG = GC ...(3)

But, AC = AF + FG + GC

$\Rightarrow$ AC = AF + AF + AF[Using (3)]

$\Rightarrow$ AC = 3AF

$\Rightarrow\ \text{AF}=\frac{1}{3}\text{AC}$

Hence, proved.

(Change D to E in mid points of AD)

Given: A trapezium ABCD in which E and F are repectively the mid -points of the mid poient of the non-parallel sides AD and BC.

 to prove: EF || AB and $\text{EF}=\frac{1}{2}(\text{AB}=\text{CD})$

construction: join DF and produce it to intersect AB produced at G.

proof: in $\Delta\text{CFD}$ and $\Delta\text{BFG},$ we have

DC || AB

$\therefore\ \angle\text{C}=\therefore\ \angle\text{3}$ [Alternate interior angles]

CF = BF

$\therefore\ \angle\text{1}=\therefore\ \angle\text{2}$ [Vertically opposite angles]

[so, by ASA critertion of congruence, we have]

$\Delta\text{CFD}\cong\Delta\text{BFG}$

$\therefore$ CD || BG [CPCT]

In $\Delta\text{DAG},$ EF joins mid-points AD and GD repectively

$\therefore$ EF || AG [$\because$ mid-point therem]

$\Rightarrow$ EF || AB

so, $\text{EF}=\frac{1}{2}\text{AG}$ [mid-point therem]

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

$\Rightarrow\ \text{EF}=\frac{1}{2}(\text{AB}+\text{CD})[\because\ \text{CD}=\text{BG}]$

Hence, proved.

Given in a parallelogram ABCD, P is a mid-point of BC such that $\angle\text{BAP}=\angle\text{DAP}.$

To prove AD = 2CD

Proof Since, ABCD is a parallelogram.

So, AD || BC and AB is transversal, then

$\angle\text{A}+\angle\text{B}=180^\circ$ (sum of cointerior angles is 180°)

$\angle\text{B}=180^\circ-\angle\text{A}\ ...(\text{i})$

in $\Delta\text{ABP}$ $\angle\text{PAB}+2\angle\text{B}+\angle\text{BPA}=180^\circ$[by angle sum property of a triangle)

$\Rightarrow\ \frac{1}{2}\angle\text{A}+180^\circ-\angle\text{A}+\angle\text{BPA}=180^\circ$ [from Eq. (i)]

$\Rightarrow\ \angle\text{BPA}=\frac{\angle\text{A}}{2}=0$

$\Rightarrow\ \angle\text{BPA}=\frac{\angle\text{A}}{2}\ ...\text{(ii)}$

$\Rightarrow\angle\text{BPA}=\angle\text{BAP}$

$\Rightarrow\ \text{AB}=\text{BP}$ [opposite sides of equal angles are equal]

On multiplying both sides by 2, we get

$2\text{AB}=2\text{BP}$

$\Rightarrow\ 2\text{AB}=\text{BP}$ (since Pis the mid-point of BC)

$\Rightarrow\ 2\text{CD}=\text{AD}$

(since, ABCD is a parallelogram, then AB = CD and BC = AD)

Given in a square ABCD, P Q R and S are the mid-points of AB, BC, CD and DA, respectively.

To show PQRS is a square.

Construction Join AC and BD.

Proof Since, ABCD is a square.

$\therefore$ AB = BC = CD = AD

Also, P. Q. R and S are the mid-points of AB, BC, CD and DA

respectively.

Then, in $\Delta\text{ADC}$

and $\text{SR}=\frac{1}{2}=\text{AC}$ [by mid-point theorem]...(i)

In $\Delta\text{ABC},$ PQ || AC

and $\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{ii})$

From Eqs. (i) and (ii).

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

Similarly, SP || BD and BD || RQ

$\therefore$ SP || RQ and $\text{SP}=\frac{1}{2}\text{BD}$

and $\text{RQ}=\frac{1}{2}\text{BD}$

$\therefore\ \text{SP}=\text{RQ}=\frac{1}{2}\text{BD}$

Since, diagonals of a square bisect each other at right angle.

$\therefore$ AC = BD

$\text{SP}=\text{RQ}=\frac{1}{2}\text{AC}\ ...(\text{iv})$

From Eqs. (i) and (iv). $\text{SR}=\text{PQ}=\text{SP}=\text{RQ}$ [all side are equal]

Now, in quadrilateral OERF,

OE || FR and OF || ER

$\angle\text{EOF}=\angle\text{ERF}=90^\circ$

Hence, PQRS is a square. Hence proved.

To prove: ABCD is a square.

Proof: DC || AB [Opposite sides of a rectangle are parallel]

$\Rightarrow\ \angle4=\angle1\ ...(\text{i})$ [Alternate interior angles]

Similarly, $\angle3=\angle2\ ...(\text{ii})$ [Alternate interior angles]

And $\angle1=\angle2\ ...(\text{iii})$ [Given]

From equation (1), (2) and (3), we get

$\angle3=\angle4$

In $\Delta\text{BAD}$ and $\Delta\text{BDC}$ we have

$\angle1=\angle2$[Given]

BD = BD[Common side)

$\angle3=\angle4$ [proved above]

So, By ASA criterion of congruence, we have

$\Delta\text{BAD}\cong\Delta\text{BCD}$

$\therefore\ \text{AB}=\text{BC}$ [CPCT]

As, adjacent sides of rectangle are equal. So, ABCD is a square.

To prove: PRQS is a parallelogram.

Proof: DC||AB [$\therefore$ Opposite sides of a parallelogram are parallel]

$\Rightarrow$ AP || QC

DC = AB [$\because$ Opposite sides of a parallelogram are equal]

$\Rightarrow\ \frac{1}{2}\text{DC}=\frac{1}{2}\text{AB}$ 

$\Rightarrow$ QC = AP [$\because$ P is the mid-point of AB and Q is mid-point of CD]

$\Rightarrow$ APCQ is a parallelogram. [$\because$ AP || CQ and QC = AP]

$\therefore$ AQ||PC [$\because$ Opposite sides of a || gm are parallelogram)

$\Rightarrow$ SQ || PR

Similarly, SP || QR

$\therefore$ Qudrilateral PRQS is a parallogram.

Hence. proved

Proof In $\Delta\text{ADC},$ S and R are the mid-points of AD and DC respectively. Then, by mid-point theorem.

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

In $\Delta\text{ABC},$ P and Q are the mid-points of AB and BC respectively. Then, by mid-point theorem.

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

From Eqs. (i) and (ii). $\text{SR}=\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{iii})$

Similarly, in $\Delta\text{BCD}$ RQ|| BD and $\text{RQ}=\frac{1}{2}\text{BD}\ ...(\text{iv})$

and in $\Delta\text{BAD}$ SP || BD and $\text{SP}=\frac{1}{2}\text{BD}\ ...(\text{v})$

From Eqs. (iv) and (v). $\text{SP}=\text{RQ}=\frac{1}{2}\text{BD}=\frac{1}{2}\text{AC}$ (given, AC = BD]...(vi)

From Egs. (ii) and (vi). SR = PQ = SP = RQ

It shows that all sides of a quadrilateral PQRS are equal. Hence, PQRS is a rhombus.

Hence proved.

Then, $\text{AD}=\text{BD}=\frac{1}{2}\text{AB},\ \text{BE}=\text{EC}=\frac{1}{2}\text{BC}$

$\therefore$ OD = OB...(1) [$\therefore$ Diagonals of parallelogram bisect each other]

OA = OC...(2)($\because$ Diagonals of parallelogram bisect each other]

AE = CF...(3)[Given]

Subtracting (3) from (2), we get

OA - AE = OC - CF

$\Rightarrow$ OE = OF...(4)

$\therefore$ BFDE is parallelogram. [$\therefore$ OD = OB and OE = OF]

Hence, proved.