Quadrilaterals Class 9th Mathematics AP Board Solution

Class 9th Mathematics AP Board Solution
Exercise 8.1
  1. State whether the statements are True or False.(i) Every parallelogram is a…
  2. Complete the following table by writing (YES) if the property holds for the…
  3. ABCD is trapezium in which AB || CD. If AD = BC, show that ∠A = ∠B and ∠C = ∠D.…
  4. The four angles of a quadrilateral are in the ratio 1: 2:3:4. Find the measure…
  5. ABCD is a rectangle AC is diagonal. Find the angles of ∆ACD. Give reasons.…
Exercise 8.2
  1. In the adjacent figure ABCD is a parallelogram ABEF is a rectangle show that…
  2. Show that the diagonals of a rhombus divide it into four congruent triangles.…
  3. In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O.Prove that…
Exercise 8.3
  1. The opposite angles of a parallelogram are (3x-2)^{0} and (x+48)^{0}…
  2. Find the measure of all the angles of a parallelogram, if one angle is 24° less…
  3. In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side…
  4. In the adjacent figure ABCD is a parallelogram P, Q are the midpoints of sides…
  5. ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and…
  6. ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C…
  7. In delta^{3}abc and DEF, AB || DE, AB=DE; BC = EF and BC || EF. Vertices A, B…
  8. ABCD is a parallelogram. AC and BD are the diagonals intersect at O. P and Q are…
  9. ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA…
Exercise 8.4
  1. ABC is a triangle. D is a point on AB such that ad = {1}/{4} ab and E is a…
  2. ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA…
  3. Show that the figure formed by joining the midpoints of sides of a rhombus…
  4. In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC…
  5. Show that the line segments joining the midpoints of the opposite sides of a…
  6. ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse…

Exercise 8.1
Question 1.

State whether the statements are True or False.

(i) Every parallelogram is a trapezium ( )

(ii) All parallelograms are quadrilaterals ( )

(iii) All trapeziums are parallelograms ( )

(iv) A square is a rhombus ( )

(v) Every rhombus is a square ( )

(vi) All parallelograms are rectangles ( )


Answer:

(i) [True]



⇒ The trapezium has a property


One pair of opposite sides are parallel


This is conveyed by parallelogram


as they have 2 pair of parallel sides


∴ Every parallelogram is a trapezium


(ii) [True]



⇒ The quadrilateral is known as


Polygon having four sides


This is conveyed by parallelogram


as they are also four sided polygon.


∴ Every parallelogram is a Quadrilateral


(iii) [False]



⇒ The parallelogram has a property


Both pair of opposite sides are parallel and equal


Which is not conveyed by trapezium


as they have only 1 pair of parallel sides


∴ Every trapezium is not parallelogram


(iv) [True]



⇒ The Rhombus has a property


Diagonal of rhombus bisect each other at 90°


This is conveyed by square


as their diagonal also bisect each other at 90°


∴ A square is a rhombus


(v) [False]



⇒ The square has a property


All angles of square are equal and 90°


And diagonals are equal and perpendicular bisector to each


other


Which is not conveyed by Rhombus


as their diagonal are perpendicular bisector but not equal


And their angles are also not equal to 90°


∴ Every rhombus is not a square.


(vi) [False]



⇒ The rectangle has a property


Every angle of rectangle is 90°


Which is not conveyed by parallelogram


as it have only opposite side angles are equal not 90°


∴ Every parallelogram are not rectangle.



Question 2.

Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.



Answer:




Question 3.

ABCD is trapezium in which AB || CD. If AD = BC, show that ∠A = ∠B and ∠C = ∠D.


Answer:


Given:- ABCD is a isosceles trapezium


Trapezium with one pair of sides is parallel


And other pair of sides is equal[AD=BC].


Formula used:- SSS congruency property


If all 3 sides of triangle are equal to all 3 sides of


other triangle


Then; Both triangles are congruent


Solution:- In trapezium ABCD


If AB||CD


And; AD=BC


∴ ABCD is a isosceles trapezium


⇒ If ABCD is a isosceles trapezium


Then;


Diagonals of ABCD must be equal


∴ AC=BD


In Δ ABD and Δ ABC


AC=BD [∵ ABCD is isosceles trapezium]


AD=BC [Given]


AB=AB [Common line in both triangle]


∴ Both triangles are congruent by SSS property


∆ABD ≅ ∆ABC


⇒ If both triangles are congruent


Then there were also be equal.


∴ ∠ A=∠ B


As ABCD is trapezium and AB||CD


∠ A+∠ D=180° and ∠ C+∠ B=180°


∠ A =180° - ∠ D and ∠ B=180° - ∠ C


If ∠ A=∠ B;


180° - ∠ D=180° - ∠ C


180° +∠ C - 180° = ∠ D


∴ ∠ C=∠ D;


Conclusion:- If ABCD is trapezium in which AD = BC, then ∠A = ∠B and ∠C = ∠D.



Question 4.

The four angles of a quadrilateral are in the ratio 1: 2:3:4. Find the measure of each angle of the quadrilateral.


Answer:


Given . Angles of a quadrilateral are in the ratio 1: 2:3:4


Solution .


Let any quadrilateral be ABCD


Then


Sum of all angles of quadrilateral is 360°


∴ ∠ A + ∠ B + ∠ C + ∠ D=360°


If angle A,B,C,D are in ratio 1:2:3:4


Then,


1x + 2x + 3x + 4x =360°


10x=360°


x= 


x= 36°


all angles are 1x,2x,3x,4x


Conclusion.


∴ Angles are 36° , 72° , 108° , 144°



Question 5.

ABCD is a rectangle AC is diagonal. Find the angles of ∆ACD. Give reasons.


Answer:


Given:- ABCD is a rectangle


Formula used:- All angles of rectangle are 90°


In right angle triangle


Pythagoras theorem a2+b2=c2


Sin(θ)=


Solution:-


In ΔACD


∠ D=90° [All angles of rectangle are 90° ]


∠ ACD=θ1


And; Sin(θ1) =  =


θ1=arc sin()


∠ DAC=θ2


And; Sin(θ2) =  =


θ1=arc sin()


Conclusion:-


The angles of Δ ACD are :-


∠ D=90°


∠ DAC=arc sin()


∠ ACD=arc sin()




Exercise 8.2
Question 1.

In the adjacent figure ABCD is a parallelogram ABEF is a rectangle show that ∆AFD ≅ ∆BEC.




Answer:

Given :- ABCD is a parallelogram and ABEF is a rectangle


Formula used :-


SAS congruency rule


If two sides of triangle and angle made by the 2 sides are equal then both the triangles are congruent


Solution :-


In ∆AFD and ∆BEC


*AF=BE [opposite sides of rectangle are equal]


*AD=BC [opposite sides of parallelogram are equal]


As angle of rectangle is 90°


∠ DAB +∠ FAD=90°


∠ DAB=90° - ∠ FAD -----1


Sum of corresponding angles of parallelogram is 180°


∠ DAB+∠ ABC =180°


∠ DAB+∠ ABE+∠ EBC=180°


90° - ∠ FAD + 90° +∠ EBC=180° ∵ putting value from 1


180° +∠ EBC-∠ FAD=180°


∠ EBC - ∠ FAD=0


∠ EBC = ∠ FAD


⇒ Hence;


By SAS property both triangles are congruent


Conclusion :-


∆AFD ≅ ∆BEC



Question 2.

Show that the diagonals of a rhombus divide it into four congruent triangles.


Answer:


Given :- ABCD is a rhombus


Formula used :-


*SSS congruency rule


If all sides of both triangles are equal then both triangles are congruent


*properties of rhombus


Solution :-


In Δ AOD and Δ COB


AO=OC [diagonal of Rhombus bisect each other]


OD=OB [diagonal of Rhombus bisect each other]


AD=BC [all sides of rhombus are equal]


∴ Δ AOD ≅ Δ COB


In Δ AOB and Δ COD


AO=OC [diagonal of Rhombus bisect each other]


OD=OB [diagonal of Rhombus bisect each other]


CD=BA [all sides of rhombus are equal]


∴ Δ AOB ≅ Δ COD


In Δ AOB and Δ AOD


AO=AO [Common in both triangles]


OD=OB [diagonal of Rhombus bisect each other]


AD=AB [all sides of rhombus are equal]


∴ Δ AOB ≅ Δ AOD


∴ All four triangles divide by diagonals of triangle are congruent



Question 3.

In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O.

Prove that 


Answer:


Given :- ABCD is a quadrilateral


∠ OCB= ∠ OCD=∠C/2 [OC is bisector of ∠ C]


∠ ODA= ∠ ODC=∠D/2 [OD is bisector of ∠ D]


Formula Used:- ∠ A+ ∠ B+∠ C+ ∠ D=360°


Solution :-


In Δ COD


∠ OCD+ ∠ COD +∠ ODC=180°


∠D/2 + ∠C/2 +∠ COD = 180°


(∠ D+∠ C)/2 + ∠ COD = 180°


If;


∠ A+ ∠ B+∠ C+ ∠ D=360°


∠ C+ ∠ D=360° -(∠ A+ ∠ B)


 + ∠ COD=180°


180° - + ∠ COD=180°


∠ COD=




Exercise 8.3
Question 1.

The opposite angles of a parallelogram are  and  Find the measure of each angle of the parallelogram.


Answer:


Given:- Opposite angles of parallelogram  and 


Formula used:- opposite angles of a parallelogram are equal


Corresponding angles sum is 180°


Solution:-


Let ∠ A and ∠ C


∠ A = ∠ C [opposite angles of a parallelogram are equal]


3x-2=x+48


3x – x=48+2


2x=50


x=25


Then;


∠ A=∠ C=x+48=25+ 48 = 73°


∠ A+∠ B=180°


∠ B=180° - 73°


∠ B= 107°


∠ B = ∠ D [opposite angles of a parallelogram are equal]


Conclusion:-


Angles of parallelogram = 73°, 107°, 73°, 107°



Question 2.

Find the measure of all the angles of a parallelogram, if one angle is 24° less than the twice of the smallest angle.


Answer:


Given:- one angle is 24° less than the twice of the smallest angle


Formula used:- opposite angles of a parallelogram are equal


Corresponding angles sum is 180°


Solution:-


Let ∠ A and ∠ C be smallest because


∠ A = ∠ C [opposite angles of a parallelogram are equal]


2∠ A+ 24° =∠ B


∠ A+∠ B=180° [Sum of corresponding angles is 180° ]


∠ B=180°-∠ A


Then,


2∠ A+ 24° =180° -∠ A


3∠ A =180° - 24°


3∠ A=156°


∠ A==52°


∠ B=180° -∠ A


∠ B= 180° -52°


∠ B=128°


∠ B = ∠ D [opposite angles of a parallelogram are equal]


Conclusion:-


∴ Angles of parallelogram = 52°, 128°, 52°, 128°



Question 3.

In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB.




Answer:

Given :- ABCD is a parallelogram


And CE=EB


Formula used:- ASA congruency property


If 2 angles and one side between them in two


triangles are equal then both triangle are congruent


Solutions :-


In Δ DCE and Δ FBE


∠ DEC=∠ BEF [Vertically opposite angles]


As DC || BF


∠ ECD=∠ EBF [Alternate angles]


And;


CE=EB [Given]


∴ Δ DCE ≅ Δ FBE


As Δ DCE ≅ Δ FBE


Then


DC=BF


And DC=AB [opposite sides of parallelogram are equal]


Hence ;


AB=BF


AF=AB+BF


AF=2AB.


Hence Proved;



Question 4.

In the adjacent figure ABCD is a parallelogram P, Q are the midpoints of sides AB and DC respectively. Show that PBCQ is also a parallelogram.




Answer:

Given:-ABCD is a parallelogram


P,Q are the midpoints of sides AB and DC respectively.


Solution:-


In parallelogram ABCD


If P,Q are the midpoints of sides AB and DC respectively


∴ DC=2QC=2QD


∴ AB=2AP=2PB


If DC||AB


Then;


QD||AP and QC||PB


Line PQ divides the parallelogram in 2 equal parts


Because P and Q are midpoints


∴ AD||PQ||CB


If;


PQ||CB and QC||PB


Then, PBCQ is also a parallelogram



Question 5.

ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD || BA as shown in the figure. Show that

(i) ∠DAC = ∠BCA

(ii) ABCD is a parallelogram




Answer:

Given:- AB=AC and ABC is isosceles triangle


∠ QAD=∠ DAC


CD||BA


Formula used:- sum of angles of triangle is 180°


Solution:-


As Δ ABC is a isosceles triangle


∠ ABC=∠ ACB


∠ ABC+∠ ACB+∠ CAB=180°


2∠ ACB+∠ CAB=180°


*As BAQ is a straight line


∠ CAB+∠ DAC+∠ QAD=180°


*∠ QAD=∠ DAC [Given]


∠ CAB=180° - 2∠ DAC


Putting value of ∠ CAB in above equation 1


2∠ ACB+180° - 2∠ DAC=180°


2∠ ACB =2∠ DAC


∠ ACB =∠ DAC


If;


There is ∠ ACB =∠ DAC and AC is the transverse


∴ these are equal by alternate angles


And AD||BC


In ABCD


If;


AD||BC & CD||BA


If both pair of sides of quadrilateral are parallel


Then the quadrilateral is parallelogram


Conclusion:-


ABCD is a parallelogram


And ∠DAC = ∠BCA



Question 6.

ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that

(i) Δ APB ≅ Δ CQD

(ii) AP = CQ




Answer:

Given:- ABCD is parallelogram


Formula used :- AAS property


If 2 angles and any one side is are equal in both triangle


Then both triangles are congruent


Solution:-


In Δ APB and Δ CQD


∠ CQD=∠ APB=90°


If ABCD is a parallelogram


∠ CDB=∠ DBA [alternate angles as DC||AB]


AB=DC [opposite sides of parallelogram are equal]


By AAS property


∴ Δ APB ≅ Δ CQD


If Δ APB ≅ Δ CQD


Then


AP=CQ


Conclusion:-


In parallelogram ABCD; Δ APB ≅ Δ CQD



Question 7.

In  and DEF, AB || DE, AB=DE; BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that

(i) ABED is a parallelogram

(ii) BCFE is a parallelogram

(iii) AC = DF

(iv) ∆ABC ≅ ∆DEF




Answer:

Given:- AB||DE , AB=DE; BC||EF, BC=EF


Formula used:- SSS congruency rule


If all 3 sides of both triangle are equal then


Both triangles are congruent.


Solution:-


(1) In ABED


AB||DE


AB=DE


∴ ABED is a parallelogram


∵ If one pair of side in quadrilateral is equal and parallel


Then the quadrilateral is parallelogram.


(2) In BCFE


BC||EF


BC=EF


∴ BCEF is a parallelogram


∵ If one pair of side in quadrilateral is equal and parallel


Then the quadrilateral is parallelogram.


(3) As ABED is a parallelogram


BE||AD , BE=AD ;


As BCFE is a parallelogram


BE||CF , BE=CF ;


∴ By concluding both above statements


AD||CF and AD=CF


∴ ACFD is a parallelogram


∵ If one pair of side in quadrilateral is equal and parallel


Then the quadrilateral is parallelogram


If ACDF is a parallelogram ;


Then


AC=DF [opposite sides of parallelogram are equal ]


(4) In Δ ABC and Δ DEF


AB=DE [Given]


BC=EF [Given]


AC=DF [Proved above]


Δ ABC ≅ Δ DEF [SSS congruency rule]



Question 8.

ABCD is a parallelogram. AC and BD are the diagonals intersect at O. P and Q are the points of tri section of the diagonal BD. Prove that CQ || AP and also AC bisects PQ (see figure).




Answer:

Given:- ABCD is parallelogram


BP=BD/3


DQ=BD/3


Formula used:- SAS congruency property


If 2 sides and angle between the two sides of both


the triangle are equal, then both triangle are congruent


Solution:-


⇒ In parallelogram ABCD


AO=OC and DO=OB [diagonal of parallelogram bisect each other]


As DO=OB


Where DO=DQ+OQ


OB=OP+PB


∴ DQ+OQ=OP+PB


⇒ +OQ=OP+


⇒ OQ=OP+ - 


⇒ OQ=OP


PQ = OP+OQ = 2(OQ) = 2(OP)


∴ AC diagonal bisect PQ


⇒ In Δ QOC and Δ POA


OQ=PO [Proven above]


AO=OC [Diagonal of parallelogram bisect each other]


∠ QOC=∠ AOP [vertically opposite angles]


Hence Δ QOC ≅ Δ POA


∴ ∠ CQO=∠ OPA


If QC and PA are 2 lines


And QP is the transverse


And ∠ CQO=∠ OPA by Alternate angles


∴ QC||PA


Conclusion:-


CQ||PA and CA is bisector of PQ.



Question 9.

ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square.




Answer:

Given:- *ABCD is a square


*E, F, G and H are the mid points of AB, BC, CD and DA


respectively


*AE = BF = CG = DH


Formula used:- *Isosceles Δ property


⇒ If 2 sides of triangle are equal then


Corresponding angle will also be equal


*Properties of quadrilateral to be square


⇒ All sides are equal


⇒ All angles are 90°


Solution:-


In Δ AHE, Δ EBF, Δ FCG, Δ DHG


⇒ AE = BF = CG = DH [Given]


If;


⇒ AE=EB [E is midpoint of AB]


⇒ BF=FC [F is midpoint of BC]


⇒ CG=GD [G is midpoint of CD]


⇒ DH=HA [H is midpoint of DA]


⇒ AE = BF = CG = DH


On replacing every part we get;


⇒ EB=FC=GD=HA;


⇒ ∠ A+∠ B+∠ C+∠ D=90° [All angles of square are 90°]


Hence;


All triangles Δ AHE, Δ EBF, Δ FCG, Δ DHG


are congruent by SAS property


∴ Δ AHE≅ Δ EBF≅ Δ FCG ≅ Δ DHG


⇒ HE=EF=FG=GH [All triangles are congruent]


In Δ AHE, Δ EBF, Δ FCG, Δ DHG


∵ all sides of square are equal and after the midpoint of each sides


Every half side of square are equal to half of other sides.


HA=AE , EB=FB ,FC=GC ,HD=DG


∴ All Δ AHE, Δ EBF, Δ FCG, Δ DHG are isosceles


⇒ as central angle of all triangle is 90°


It makes all Δ AHE, Δ EBF, Δ FCG, Δ DHG are right angle isosceles Δ


∴ all corresponding angles of equal side will be 45°


∠ AHE=∠ BEF=∠ CFG=∠ DHG=∠ AEH=∠ BFE=∠ CGF=∠ DGH=45°


⇒ as AB is straight line


Then; ∠ AEH+∠ HEF+∠ BEF=180°


45°+∠ HEF+45° =180°


∠ HEF=180° -90° =90°


Similarly ;


∠ EFG=90°


∠ FGH=90°


∠ GHE=90°


Conclusion:-


All angles are 90° and all sides are equal of quadrilateral


Hence quadrilateral is square




Exercise 8.4
Question 1.

ABC is a triangle. D is a point on AB such that  and E is a point on AC such that  If DE = 2 cm find BC.


Answer:


Given:- DE=2cm




Formula used:- Line joining midpoints of 2 sides of triangle will be


half of its 3rd side.


Solution:-


Assume mid points of line AB and AC is M and N


The formation of new triangle will be there Δ AMN


As;


 ⇒ AC=4AE


and 


∴ AM=2AE


Hence;


E is the midpoint of AM


Similarly


D is the midpoint of AN


→ In Δ AMN


If E,D are midpoints


ED=MN


MN=2ED


MN=2×2cm=4cm


→ In Δ ABC


If M,N are midpoints


AB=MN


AB=2MN


AB=2×4cm=8cm


Conclusion:-


BC=8cm



Question 2.

ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram.


Answer:


Given:- ABCD is quadrilateral E, F, G and H are the midpoints of


AB, BC, CD and DA respectively


Formula used:- Line joining midpoints of 2 sides of triangle


Is parallel to 3rd side


Solution:-


BD is diagonal of quadrilateral


EH is the line joined by midpoints of triangle ABD,


∴EH is parallel to BD


GF is line joined by midpoints of side BC&BD of triangle BCD


∴GF is parallel to BD


⇒ If HE is parallel to BD and BD is parallel to GF


∴ It gives HE is parallel to GF


⇒ AC is another diagonal of quadrilateral


GH is the line joined by midpoints of triangle ADC,


∴GH is parallel to AC


FE is line joined by midpoints of side BC&AB of triangle ABC


∴FE is parallel to AC


⇒ If GH is parallel to AC and AC is parallel to FE


∴ It gives GH is parallel to FE


As HE||GF and GH||FE


∴ EFGH is a parallelogram


Conclusion:-


EFGH is a parallelogram



Question 3.

Show that the figure formed by joining the midpoints of sides of a rhombus successively is a rectangle.


Answer:


Given:- ABCD is a rhombus


E,F,G,H are the mid points of AB,BC,CD,DA


Formula used:- Line joining midpoints of 2 sides of triangle


Is parallel and half of 3rd side


Solution:-


BD is diagonal of rhombus


EH is the line joined by midpoints of triangle ABD,


∴EH is parallel and half of BD


GF is line joined by midpoints of side BC&BD of triangle BCD


∴GF is parallel and half BD


⇒ If HE is parallel to BD and BD is parallel to GF


∴ It gives HE is parallel to GF


⇒ If HE is half of BD and GF is also half of BD


∴ It gives HE is equal to GF


⇒ AC is another diagonal of rhombus


GH is the line joined by midpoints of triangle ADC,


∴GH is parallel to AC


∴GH is half of AC


FE is line joined by midpoints of side BC&AB of triangle ABC


∴FE is parallel to AC


∴ FE is half of AC


⇒ If GH is parallel to AC and AC is parallel to FE


∴ It gives GH is parallel to FE


⇒ If GH is half of AC and FE is also half of AC


∴ It gives GH is equal to FE


As diagonals of rhombus intersect at 90°


And AC is parallel to FE and GH


All angles of EFGH is 90°


All angles are 90° and opposite sides are equal and parallel


Conclusion:-


EFGH is a rectangle



Question 4.

In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC respectively. Show that the line segments AF and EC trisect the diagonal BD.




Answer:

Given:- ABCD is a parallelogram


E and F are the midpoints of the sides AB and DC respectively


Formula used:- Line drawn through midpoint of one side of triangle


Parallel to other side , bisect the 3rd side.


Solution:-


As ABCD is parallelogram


AB=CD and AB||CD;


⇒ AE=CF and AE||CF [E and F are the midpoints of AB and CD]


In quadrilateral AECF


⇒ AE=CF and AE||CF


AECF is a parallelogram.


∵ Quadrilateral having one pair of side equal and parallel are parallelogram


If AECF is a parallelogram


∴ AF||CE


Hence ;


PF||CQ and AP||QE [As AF=AP+PF and CE=CQ+QE ]


In Δ DQC


DF=FC [F is midpoint]


PF||CQ


Then;


P is midpoint of DQ


DP=PQ


∵ Line drawn through midpoint of one side of triangle Parallel to


other side , bisect the 3rd side


In Δ APB


AE=EB [E is midpoint]


AP||QE


Then;


Q is midpoint of PB


PQ=QB


∵ Line drawn through midpoint of one side of triangle Parallel to


other side , bisect the 3rd side


If DP=PQ and PQ=QB


Then DP=PQ=QB


Conclusion:-


Line segments AF and EC trisect the diagonal BD.



Question 5.

Show that the line segments joining the midpoints of the opposite sides of a quadrilateral and bisect each other.


Answer:


Given:- ABCD is a quadrilateral


Formula used:- Line joining midpoints of 2 sides of triangle


Is parallel and half of 3rd side


Solution:-


BD is diagonal of quadrilateral


EH is the line joined by midpoints of triangle ABD,


∴EH is parallel and half of BD


GF is line joined by midpoints of side BC&BD of triangle BCD


∴GF is parallel and half BD


⇒ If HE is parallel to BD and BD is parallel to GF


∴ It gives HE is parallel to GF


⇒ If HE is half of BD and GF is also half of BD


∴ It gives HE is equal to GF


⇒ AC is another diagonal of quadrilateral


GH is the line joined by midpoints of triangle ADC,


∴GH is parallel to AC


∴GH is half of AC


FE is line joined by midpoints of side BC&AB of triangle ABC


∴FE is parallel to AC


∴ FE is half of AC


⇒ If GH is parallel to AC and AC is parallel to FE


∴ It gives GH is parallel to FE


⇒ If GH is half of AC and FE is also half of AC


∴ It gives GH is equal to FE


If both opposite sides are parallel and equal


Then, the quadrilateral is parallelogram


If EFGH is parallelogram


Then their diagonal bisect each other


If diagonal of parallelogram is the line joining midpoint of opposite sides of quadrilateral


Then;


Line joining midpoints of opposite sides of quadrilateral bisect each other.


Conclusion:-


Lines joining midpoints of opposite sides of quadrilateral bisect each other



Question 6.

ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and Parallel to BC intersects AC at D. Show that

(i) D is the midpoint of AC

(ii) MD ⊥ AC

(iii) CM = MA = AB




Answer:

Given:- ACB is right angle triangle


M is midpoint of AB


DM||CB


Formula used:- Line drawn through midpoint of one side of triangle


Parallel to other side , bisect the 3rd side


* SAS congruency property


If 2 sides and angle between the two sides of both


the triangle are equal, then both triangle are congruent


Solution:-


(1) In Δ ABC


M is midpoint of AB


And DM||CB


∴ D is midpoint of AC


AD=DC


∵ Line drawn through midpoint of one side of triangle


Parallel to other side , bisect the 3rd side


(2) In Δ ABC


DM||CB


∠ ADM=∠ ACB [Corresponding angles]


∠ ACB =90°


∠ ADM=90°


(3) In Δ ADM and Δ DMC


⇒ DM=DM [Common in both triangles ]


⇒ AD=DC [D is the midpoint]


As ADC is straight line


∠ ADM+∠ MDC=180°


∠ MDC=180° - ∠ ADM


∠ MDC=90°


⇒ ∠ ADM=∠ MDC


Hence;


In Δ ADM ≅ Δ DMC


∴ CM=MA


⇒ MA=AB [M is midpoint of AB]


∴ CM=MA=AB


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