AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.4 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.4

Question 1.

ABC is a triangle. D is a point on AB such that AD = \(\frac { 1 }{ 4 }\) AB and E is a point on AC such that AE = \(\frac { 1 }{ 4 }\) AC. If DE = 2 cm find BC.

Solution:

Given that D and E are points on AB and AC.

Such that AD = \(\frac { 1 }{ 4 }\) AB and AE = \(\frac { 1 }{ 4 }\) AC

Let X, Y be midpoints of AB and AC.

Joint D, E and X, Y.

Now in ΔAXY; D, E are the midpoints of sides AX and AY.

∴ DE // XY and DE = \(\frac { 1 }{ 2 }\) XY

⇒ 2 cm = \(\frac { 1 }{2 }\) XY

⇒ XY = 2 x 2 = 4cm

Also in ΔABC; X, Y are the midpoints of AB and AC.

∴ XY//BC and XY = \(\frac { 1 }{2 }\) BC

4 cm = \(\frac { 1 }{2 }\) BC

⇒ BC = 4 x 2 = 8 cm

Question 2.

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

Solution:

Given that E, F, G and H are the midpoints of the sides of quad. ABCD.

In ΔABC; E, F are the midpoints of the sides AB and BC.

∴ EF//AC and EF = \(\frac { 1 }{2 }\) AC

Also in ΔACD; HG // AC

and HG = \(\frac { 1 }{ 2 }\) AC

∴ EF // HG and EF = HG

Now in □EFGH; EF = HG and EF // HG

∴ □EFGH is a parallelogram.

Question 3.

Show that the figure formed by joining the midpoints of sides of a rhom¬bus successively is a rectangle.

Solution:

Let □ABCD be a rhombus.

P, Q, R and S be the midpoints of sides of □ABCD

In ΔABC,

P, Q are the midpoints of AB and BC.

∴ PQ//AC and PQ = \(\frac { 1 }{2 }\)AC …………………..(1)

Also in ΔADC, ,

S, R are the midpoints of AD and CD.

∴ SR//AC and SR = \(\frac { 1 }{2 }\)AC ………………(2)

From (1) and (2);

PQ // SR and PQ = SR

Similarly QR // PS and QR = PS

∴ □PQRS is a parallelogram.

As the diagonals of a rhombus bisect at right angles.

∠AOB – 90°

∴ ∠P = ∠AOB = 90°

[opp. angles of //gm PYOX] Hence □PQRS is a rectangle as both pairs of opp. sides are equal and parallel, one angle being 90°.

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.

Solution:

□ABCD is a parallelogram. E and F are the mid points of AB and CD.

∴ AE = \(\frac { 1 }{2 }\)AB and CF = \(\frac { 1 }{2 }\)CD

Thus AE = CF [∵ AB – CD]

Now in □AECF, AE = CF and AE ||CF

Thus □AECF is a parallelogram.

Now in ΔEQB and ΔFDP

EB = FD [Half of equal sides of a //gm]

∠EBQ = ∠FDP[alt. int.angles of EB//FD]

∠QEB = ∠PFD

[∵∠QED = ∠QCF = ∠PFD]

∴ ΔEQB ≅ ΔFPD [A.S.A. congruence]

∴ BQ = DP [CPCT] ……………… (1)

Now in ΔDQC; PF // QC and F is the midpoint of DC.

Hence P must be the midpoint of DQ

Thus DP = PQ …………….. (2)

From (1) and (2), DP = PQ = QB

Hence AF and CE trisect the diagonal BD.

Question 5.

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

Solution:

Let ABCD be a quadrilateral.

P, Q, R, S are the midpoints of sides of □ABCD.

Join (P, Q), (Q, R), (R, S) and (S, P).

In ΔABC; P, Q are the midpoints of AB and BC.

∴ PQ // AC and PQ = \(\frac { 1 }{2 }\)AC ………….(1)

Also from ΔADC

S, R are the midpoints of AD and CD

SR // AC and SR = \(\frac { 1 }{2 }\) AC …………………(2)

∴ From (1) & (2)

PQ = SR and PQ //SR

∴ □PQRS is a parallelogram.

Now PR and QS are the diagonals of □ PQRS.

∴ PR and QS 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= \(\frac { 1 }{2 }\)AB

Solution:

Given that in ΔABC; ∠C = 90°

M is the midpoint of AB.

i) If ‘D’ is the midpoints of AC.

The proof is trivial.

Let us suppose D is not the mid point of AC.

Then there exists D’ such that AD’ = D’C

Then D’M is a line parallel to BC through M.

Also DM is a line parallel to BC through M.

There exist two lines parallel to same line through a point M.

This is a contradiction.

There exists only one line parallel to a given line through a point not on the line.

∴ D’ must coincides with D

∴ D is the midpoint of AC

ii) From (i) DM // BC

Thus ∠ADM = ∠ACB = 90°

[corresponding angles]

⇒ MD ⊥ AC

iii) In ∆ADM and ∆CDM

AD = CD [ ∵ D is midpoint from (i)]

∠ADM = ∠MDC (∵ 90° each)

DM = DM (Common side)

∴ ∆ADM = ∆CDM (SAS congruence)

⇒ CM = MA (CPCT)

CM = \(\frac { 1 }{2 }\) AB (∵ M is the midpoint of AB)

∴ CM = MA = \(\frac { 1 }{2 }\)AB