9 Math Quadrilaterals

Finish Line & Beyond QUADRILATERALS 1. Sum of the angles of a quadrilateral is 360°. 2. A diagonal of a parallelogram divides it into two congruent triangles. 3. In a parallelogram, (i) opposite sides are equal (ii) opposite angles are equal (iii) diagonals bisect each other 4. A quadrilateral is a parallelogram, if (i) opposite sides are equal or (ii) opposite angles are equal or (iii) diagonals bisect each other or (iv) a pair of opposite sides is equal and parallel 5. Diagonals of a rectangle bisect each other and are equal and vice-versa. 6. Diagonals of a rhombus bisect each other at right angles and vice-versa. 7. Diagonals of a square bisect each other at right angles and are equal, and viceversa. 8. The line-segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it. 9. A line through the mid-point of a side of a triangle parallel to another side bisects the third side. 10. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order, is a parallelogram.

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Finish Line & Beyond EXERCISE 1 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Answer: As you know angle sum of a quadrilateral = 360° SO, 3 x + 5 x + 9 x + 13 x = 360°

⇒ 30x = 360° ⇒ x = 12°

Hence, angles are: 36°, 60°, 108°, 156° 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle. Answer: In the following parallelogram both diagonals are equal:

A

B

D

C

So, ∆ABC ≅ ∆ADC ≅ ∆ABD ≅ ∆BCD Hence, ∠A = ∠B = ∠C = ∠D =90° As all are right angles so the parallelogram is a rectangle. 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.


Finish Line & Beyond

A

D

B O

C

Answer: In the given quadrilateral ABCD diagonals AC and BD bisect each other at right angle. We have to prove that AB=BC=CD=AD In ∆AOB & ∆AOD DO=OB (O is the midpoint) AO=AO (common side) ∠AOB = ∠AOD (Right angle) So, ∆AOB ≅ ∆AOD So, AB=AD Similarly AB=BC=CD=AD can be proved which means that ABCD is a rhombus. 4. Show that the diagonals of a square are equal and bisect each other at right angles. Answer: In the figure given above let us assume that ∠DAB So, ∠DAO = ∠BAO = 45° Hence, ∆AOD DO=AO (Sides opposite equal angles are equal) Similarly AO=OB=OC can be proved This gives the proof of diagonals of square being equal.

= 90°

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Answer: Using the same figure, If DO=AO Then ∠DAO = ∠BAO = 45° (Angles opposite to equal sides are equal) So, all angles of the quadrilateral are right angles making it a square. 6. Diagonal AC of a parallelogram ABCD bisects

∠ A . Show that


Finish Line & Beyond (i) it bisects ∠ C also, (ii) ABCD is a rhombus.

D

A

C

B

Answer: ABCD is a parallelogram where diagonal AC bisects

∠DAB

In ∆ADC & ∆ABC ∠DAC = ∠BAC (diagonal is bisecting the angle) AC=AC (Common side) AD=BC (parallel sides are equal in a parallelogram) Hence, ∆ADC ≅ ∆ABC So, ∠DCA = ∠BCA This proves that AC bisects ∠DCB as well Now let us assume another diagonal DB intersecting AC on O. As it is a parallelogram so DB will bisect AC and vice versa In ∆AOD & ∆BOD ∠DAO = ∠DCO (opposite angles are equal in parallelogram so their halves will be equal) AO=CO DO=DO Hence, ∆AOD ≅ ∆BOD So, ∠DOA = ∠DOB = 90° As diagonals are intersecting at right angles so it is a rhombus



D

C P

Q A

B

7. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that: (i) ∆ APD ≅ ∆ CQB (ii) AP = CQ (iii) ∆ AQB ≅ ∆ CPD (iv) AQ = CP (v) APCQ is a parallelogram Answer: In ∆APD & ∆CQB DP=BQ (given) AD=BC (opposite sides are equal) ∠DAP = ∠BCQ (opposite angles’ halves are equal) Hence, ∆APD ≅ ∆CQB So, AP=CQ Proved In ∆AQB & ∆CPD AB=CD (Opposite sides are equal) DP=BQ (given) ∠BAQ = ∠DCP ( opposite angles’ halves are equal) Hence, ∆AQB ≅ ∆CPD So, AQ=CP Proved ∠DPA = ∠BQC (corresponding angles of congruent triangles APD & CQB)

∆DQP & ∆BQP ∠DPQ = ∠BQP (from previous proof)

In

DP=BQ (given) PQ=PQ (common Side) So, ∆DQP ≈ ∆BQP So,

∠QDP = ∠QBP


Finish Line & Beyond With equal opposite angles and equal opposite sides it is proved that APCQ is a parallelogram 8. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that (i) ∆ APB ≅ ∆ CQD (ii) AP = CQ

D

C

P

Q B

A

∆APB & ∆CQD ∠ABP = ∠CDQ (alternate angles of transversal DB)

Answer: In AB=CD

∠APB = ∠CQD (right angles) Hence, ∆APB ≅ ∆CQD So, AP=CQ 9. In ∆ ABC 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. Show that

A

D B

C

E

F


Finish Line & Beyond (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ∆ ABC ≅ ∆ DEF. Answer: In ∠ABC & ∆DEF AB=DE (given) BC=EF (given) ∠ABC = ∠DEF (AB||DE & BC||EF) Hence, ∠ABC ≅ ∆DEF In quadrilateral ABED AB= ED AB||ED So, ABED is a parallelogram (opposite sides are equal and parallel) So, BE||AD ------------ (1) Similarly quadrilateral ACFD can be proven to be a parallelogram So, BE||CF ------------ (2) From equations (1) & (2) It is proved that AD||CF So, AD=CF Similarly AC=DF and AC||DF can be proved 10. ABCD is a trapezium in which AB || CD and AD = BC. Show that (i) ∠ A = ∠ B (ii) ∠ C = ∠ D (iii) ∆ ABC ≅ ∆ BAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

A

D

B

E

C


Finish Line & Beyond Answer: In ∆BCE EC=AD (Opposite sides are equal in parallelogram) AD=BC (given) So, BC=EC

⇒ ∠CBE = ∠CEB ∠CBE + ∠CBA = 180° (angles on the same side of a straight line) ∠CEB + ∠DAB = 180° (adjacent angles of parallelogram are complementary) Substituting ∠CBE = ∠CEB it is clear that ∠DBA = ∠CBA Now, ∠DAB + ∠CDA = 180° (adjacent angles of parallelogram) And, ∠CBA + ∠DCB = 180° ( adjacent angles of a parallelogram) As ∠DBA = ∠CBA , so it is clear that ∠CDA = ∠DCB In ∆ABC & ∆BAD AB=AB (common side) AD=BC (given)

∠DBA = ∠CBA ∆ABC ≅ ∆BAD

Hence,

EXERCISE 2 1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that : (i) SR || AC and SR =

1 AC 2

(ii) PQ = SR (iii) PQRS is a parallelogram.

T

D R C

S

Q A P

B

Answer: Let us extend the line SR to T so that CT is parallel to AS In ∆DSR & ∆CRT DR=RC (R is the mid point of side DC) ∠DRS = ∠TRS (Opposite angles)


Finish Line & Beyond ∠DSR = ∠RTC (alternate angle of transversal ST when DA||CT) Hence, ∆DSR ≅ ∆CRT So, SR=RT ST=AC (Opposite sides of parallelogram) So, SR=

1 AC 2

As SR is touching the mid points of DA and DC so as per mid point theorem SR||AC Similarly AC||PQ can be proven which will prove that PQRS is a parallelogram. 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

D R

S

A

C

Q

P B

Answer: Following the method used in the previous question it can be proved that PQRS is a parallelogram. To prove it to be a rectangle we need to prove that

∠S = ∠R = ∠Q = ∠P = 90° In ∆DSR, ∆CRQ, ∆BQP & ∆APS

DS=CR=BQ=AP=DR=CQ=BP=AS (All sides of rhombus are equal and PQRS are midpoints)

∠DSR = ∠DRS = ∠CRQ = ∠CQR = ∠BQP = ∠BPQ = ∠APS = ∠ASP So, ∆DSR ≅ ∆CRQ ≅ ∆BQP ≅ ∆APS So, ∠SDR = ∠RCQ = ∠QBP = ∠PAS = 90° Hence, ∠DSR + ∠DRS = 90° Or, ∠DSR = ∠DRS = ∠CRQ = ∠CQR = ∠BQP = ∠BPQ = ∠APS = ∠ASP = 45° As, ∠ASP + ∠PSR + ∠DSR = 180° ⇒ ∠PSR = 180° − (45° + 45°) = 90° Similarly ∠S = ∠R = ∠Q = ∠P = 90° Hence, PQRS is a rectangle.


Finish Line & Beyond 3. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F. Show that F is the mid-point of BC.

C

D

E

A

G

F

B

Answer: In ∆ADB DG=GB A parallel line to the base originating from mid point of second side will intersect at the midpoint of the third side. AB||DC AB||EF So, EF||DC So, In ∆ABD EG||AB E is the mid point of AD So, G is the mid point of DB Now, in ∆DCB GF||DC G is the mid point of BD So, F will be mid point of BC ( Mid point theorem) 4. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD. Answer: In ∆ADE & ∆CBF AD=BC (Opposite sides are equal in parallelogram) BF=DE (Half of opposite sides of parallelogram) ∠ADE = ∠CBF (Opposite angles are equal) So, ∆ADE ≅ ∆CBF Hence, AE=CF In quadrilateral AECF EC||AF & EC=AF AE=CF So, AE||CF So, AECF is a parallelogram.



D

C

E

P

Q

A

F

B

In ∆DQC PE||QC (proved earlier by proving AE||CF) E is the mid point of DC So, P is the mid point of DQ So, DP=PQ

∆APB

FQ||AP F is the mid point of AB So, PQ=QB So, DP=PQ=QB proved 5. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. Answer: ABCD is a quadrilateral in which P, Q, R, & S are mid points of AB, BC, CD & AD In ∆ACD SR is touching mid points of CD and AD So, SR||AC Similarly following can be proved PQ||AC QR||BD PS||BD So, PQRS is a parallelogram. PR and QS are diagonals of the parallelogram PQRS, so they will bisect each other.



D R C S Q

A

P

B

6. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA =

1 AB 2

B

M

C

D

A

Answer: DM||BC M is the mid point of AB


Finish Line & Beyond So, D is the mid point of AC (Mid point theorem) ∠ACD = ∠MDA = 90° (alternate angle to transversal MD) Now in ∆CDM & ∆ADM CD=AD MD=MD

∠MDC = ∠MDA So, ∆CDM ≅ ∆ADM (SAS Theorem) So, MC=MA MA=

1 AB 2

So, MC=MA=

1 AB 2


9 Math Quadrilaterals

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