Maths 4 Marks Questions

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

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

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

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

From (i) and (ii), we have

$\text{PQ = SR }$ and $\text{PQ ∥ SR}$

3. Thus, in quadrilateral PQRS, one pair of opposite sides are equal and parallel.

Hence, PQRS is a parallelogram. 

To Prove: BE is a median of $\triangle\text{ABC}.$

Proof: In $\triangle\text{ABC},$

DE || AB [Given]

D is the mid-point of BC.

The line drawn through the midpoint of one side of a triangle, parallel to another side, intersects the third side at its midpoint.

So, by Mid point Theorem, E is the mid-point of AC.

$\therefore$ BE is the median of $\triangle\text{ABC}$ drawn through B.

D is the mid point of BC and DF is parallel to BE.

The line drawn through the midpoint of one side of a triangle, parallel to another side, intersects the third side at its midpoint.

So,by Mid point Theorem Fis the mid point of EC.

$\therefore\text{CF}=\frac{1}{2}\text{EC}$

$=\frac{1}{2}\Big(\frac{1}{2}\text{AC}\Big)$ [BE is the median through B]

$=\frac{1}{4}\text{AC}.$

Thus, $\text{CF}=\frac{1}{4}\text{AC}.$

1. In $\triangle\text{ABC}$ and $\triangle\text{ADC},$ we have:

$\text{AB = AD}$ (Given)

$\text{BC = DC}$ (Given)

$\text{AC}$ is common.

i.e., $\triangle\text{ABC}\cong\triangle\text{ADC}$ (SSS congruence rule)

$\therefore\angle\text{BAC}=\angle\text{DAC}$ and $\angle\text{BCA}=\angle\text{DCA}$ (By C.P.C.T.)

Thus, AC bisects $\angle\text{A}$ and $\angle\text{C}.$

2. Now, in $\triangle\text{ABE}$ and $\triangle\text{ADE},$ we have:

$\text{AB = AD}$ (Given)

$\angle\text{BAE}=\angle\text{DAE}$ (Proven above)

$\text{AE}$ is common.

$\therefore\triangle\text{ABE}\cong\triangle\text{ADE}$ (SAS congruence rule)

$\Rightarrow\text{BE = DE}$ (By C.P.C.T.)

3. $\triangle\text{ABC}\cong\triangle\text{ADC}$ (Proven above)

$\therefore\angle\text{ABC}=\angle\text{ADC}$ (By C.P.C.T.)

Given: ABCD is a parralegram in which AB is produced to E such that BE = AB. DE is joined which cuts BC at O.

To Prove: $\text{OB = OC}$

Proof: In $\triangle\text{OCD}$ and $\triangle\text{OBE},$ we have,

$\angle\text{DOC}=\angle\text{EOB}$ [vertically opposite angles are equal]

$\angle\text{OCD}=\angle\text{OBE}$ [AB || CD, BC is a transversal thus, alternate angles are equal]

$\text{DC = BE}$ [AB = CD and BE = AB]

Thus, by Angle-Angle-Side criterion of congruence, we have

$\therefore\triangle\text{OCD}\cong\triangle\text{OBE}$ [by AAS]

The corresponding parts of the congruent triangles are equal.

$\therefore\text{OC = OB}$

Hence, ED bisect BC.

Given: A parallelogram ABCD, in which diagonals intersect at O. E and F are the points on AB and CD

To Prove: $\text{OE = OF}$

Proof: In $\triangle\text{AOE}$ and $\triangle\text{COF},$ we have

$\angle\text{CAE}=\angle\text{DCA}$ [Alternate angles]

$\text{AO = CO}$ [diagonals are equal and bisect each other]

and, $\angle\text{AOE}=\angle\text{COF}$ [Vertically opposite angles]

Thus by Angle-Side-Angle criterion of congruence, we have,

$\therefore\triangle\text{AOE}\cong\triangle\text{COF}$ [By ASA]

The corresponding parts of the congruent triangles are equal.

$\therefore\text{OE = OF}$ [By C.P.C.T.]

Let the altitude from D to the side AB bisect AB at point P.

Join BD.

In $\triangle\text{AMD}$ and $\triangle\text{BMD},$

$\text{AM = BM}$ (M is the mid-point of AB)

$\angle\text{AMD}=\angle\text{BMD}$ (Each 90°)

$\text{MD = MD}$ (common)

$\therefore\triangle\text{AMD}\cong\triangle\text{BMD}$ (by SAS congruence criterion)

$\Rightarrow\text{AD = BD}$ (C.P.C.T.)

But, $\text{AD = AB}$ (sides of a rhombus)

$\Rightarrow\text{AD = AB = BD}$

$\Rightarrow\triangle\text{ADB}$ is an equilateral triangle.

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

$\Rightarrow\angle\text{C}=\angle\text{A}=60^{\circ}$ (opposite angles are equal)

$\Rightarrow\angle\text{B}=180^{\circ}-\angle\text{A}=180^{\circ}-60^{\circ}=120^{\circ}$

$\angle\text{D}=\angle\text{B}=120^{\circ}$

Hence, in rhombus ABCD, $\angle\text{A}=60^{\circ},\angle\text{B}=120^{\circ},\angle\text{C}=60^{\circ}$ and $\angle\text{D}=120^{\circ}.$

In $\triangle\text{ABC}$ and $\triangle\text{ADC},$

$\text{AB = AD}$ (sides of a rhombus are equal)

$\text{BC = CD}$ (sides of a rhombus are equal)

$\text{AC = AC}$ (common)

$\therefore\triangle\text{ABC}\cong\triangle\text{ADC}$ (by SSS congruence criterion)

$\Rightarrow\angle\text{BAC}=\angle\text{DAC}$ and $\angle\text{BCA}=\angle\text{DCA}$ (C.P.C.T.)

$\Rightarrow\text{AC}$ bisects $\angle\text{A}$ as well as $\angle\text{C}.$

Similarly,

In $\triangle\text{BAD}$ and $\triangle\text{BCD},$

$\text{AB = BC}$ (sides of a rhombus are equal)

$\text{AD = CD}$ (sides of a rhombus are equal)

$\text{BD = BD}$ (common)

$\therefore\triangle\text{BAD}\cong\triangle\text{BCD}$ (by SSS congruence criterion)

$\Rightarrow\angle\text{ABD}=\angle\text{CBD}$ and $\angle\text{ADB}=\angle\text{CDB}$ (C.P.C.T.)

$\Rightarrow\text{BD}$ bisects $\angle\text{B}$ as well as $\angle\text{D}.$

Let $\triangle\text{ABC}$ be an isosceles right triangle, right-angled at B.

$\Rightarrow\text{AB = BC}$

Let PBSR be a square inscribed in $\triangle\text{ABC}$ with common $\angle\text{B}.$

$\Rightarrow\text{PB = BS = SR = RP}$

Now, $\text{AB} - \text{PB = BC} -\text{BS}$

 $\Rightarrow\text{AP = CS ...(i)}$

In $\triangle\text{APR}$ and $\triangle\text{CSR}$

$\text{AP = CS}$ [from (i)]

$\angle\text{APR}=\angle\text{CSR}$ (Each 90°)

$\text{PR = SR}$ (sides of a square)

$\therefore\triangle\text{APR}\cong\triangle\text{CSR}$ (by SAS congruence criterion)

$\Rightarrow\text{AR = CR}$ [C.P.C.T.]

Thus, point R bisects the hypotenuse AC.

$\angle\text{DCM}=\angle\text{DCN}+\angle\text{MCN}$

$\Rightarrow90^{\circ}=\angle\text{DCN}+60^{\circ}$

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

In $\triangle\text{DCN,}$

$\angle\text{DNC}+\angle\text{DCN}+\angle\text{D}=180^{\circ}$

$\Rightarrow90^{\circ}+30^{\circ}+\angle\text{D}=180^{\circ}$

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

$\Rightarrow\angle\text{B}=\angle\text{D}=60^{\circ}$ (opposite angles of parallelogram are equal)

$\Rightarrow\angle\text{A}=180^{\circ}-\angle\text{B}=180^{\circ}-60^{\circ}=120^{\circ}$

$\Rightarrow\angle\text{C}=\angle\text{A}=120^{\circ}$

Thus, the angles of a parallelogram are 60°, 120°, 60° and 120°.

Given: A parrallelogram ABCD in which E is the mid point of side BC, DE and AB when produced meet at F.

To Prove: $\text{AF}=2\text{AB}$

Proof: In $\triangle\text{DEC}$ and $\triangle\text{FEB}$

$\angle\text{DEC}=\angle\text{FEB}$ [vertically opposite angles]

$\angle\text{DCE}=\angle\text{FBE}$ [alternate angles]

$\text{CE = EB}$ [Given]

Thus by Angle-Angle-Side criterion of congruence, we have

$\triangle\text{DEC}\cong\triangle\text{FEB}$ [By AAS]

The corresponding parts of the congruent triangle are equal.

$\therefore\text{DC = FB}$ [By C.P.C.T.]

So, $\text{AF = AB + BF}$

$=\text{AB + DC}$

$=\text{AB + AB}$

$=2\text{AB}$

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