Telanagana SCERT Solution Class IX (9) Math Chapter 8 Quadrilaterals Exercise 8.4
(1) ABC is a triangle. D is a point on AB such that AD = 1/4AB and E is a point on AC such that AE = 1/4AC. If DE = 2 cm find BC
Solution: Given,
ABC is a triangle
AD = 1/4AB, AE = 1/4AC
DE = 2 cm
Let, Us, consider point F and G mid points of side AB & AC respectively
In △ABC
FG = ½ BC [∵ F and G are midpoints of AB & AC and FG equal to ½ BC which is the third side of the triangle]
Now,
In △AFG
D and E are the midpoints of AF & AG respectively
Since, AD is ¼ of AB & AE is ¼ of AC
And F & G being the mid points of AB & AC respectively
DE = ½ FG [D & E are mid points of AF and AG]
Or, DE = ½ x ½ BC,
Or, DE = ¼ BC (proved)
(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.
Solution: Given,
ABCD is a quadrilateral
E, F, G, H are the mid points of AB, BC, CD and DA respectively
Join,
AC & BD
∴ In △ADC
H & G are mid points of AD & DC respectively
∴ By mid point theorem
HG ∥ AC & HG = ½ AC
Similarly In △ABC
E & F are mid points of AB & BC respectively
∴ By mid point theorem
EF ∥ AC & EF = ½ AC
∴ HG ∥ EF & HG = EF [∵ both are parallel to AC & both are ½ of AC]
Now,
In △ABD
H & E mid points of AD & AB respectively
∴ By Mid points theorem
EH ∥ BD & EH = ½ BD
Similarly, In △BCD
By mid point theorem
FG ∥ BD & FG = ½ BD
∴ EH ∥ FG & EH = FG [∵ both are parallel to BD & both are ½ of BD]
∴ Quad EFGH is a parallelogram since opposite sides are parallel to each other and equal
(3) Show that the figure formed by joining the midpoints of sides of a rhombus successively is a rectangle.
Solution: Given,
ABCD is a rhombus
E, F, G, H are mid points of side AB, DC, CD, DA respectively
In △ABD
E, F are mid points of side AB and AD respectively
By mid points theorem
EF ∥ BD & BF = ½ BD
In, △ABC
E, F are mid points of side AB & BC respectively
By mid points theorem,
EH ∥ AC & EH ½ of AC
We know,
Diagonals of rhombus are perpendicular to each other,
∴ ∠POS = ∠POQ = ∠QOR = ∠SOR = 90o
∠AOB = 90o
∠APE = ∠PES = 90o [alternate angles EH AC]
In △ADC
F & G are mid points of side AD & DC respectively
By mid points theorem
FG ∥ AC & FG = ½ AC
Now, ∠APF = ∠AOD = 90o
∠ADF = ∠PFQ = 90o
[Alternate angles, FG ∥ AC]
[∵ Corresponding angles BF BD & BD & AC are diagonals of rhombus]
Now,
In △BCD
H, G are the mid points of sides BC and DC respectively,
∴ By mid points theorem,
HG ∥ BD & HG = ½ BC
∴ Now,
∠QOR = ∠CRG = 90o [∵corresponding angle HG ∥ BD]
∠QGR = ∠CRG = 90o [alternate angles, FG ∥ AC]
Similarly, ∠SHR = ∠CRH = 90o [alternate angle, EF ∥ AC]
∴ In quad EFGH
∠SHR = ∠QGR = ∠PFQ = PES = 90o
And
Opposite sides EF & GH ∥ BD and equal to ½ BD
Opposite sides FG & EH ∥ AC & equal to ½ AC
∴ Opposite sides of equal EFGH are equal and parallel to each other
And each vertice angles of quad EFGH is are 90o
∴ Quad EFGH is rectangle (proved)
(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: Given,
ABCD is a parallelogram
E and F are mid points of AB & DC respectively
∴ AE = ½ AB
FC = ½ DC
But,
AB = DC [∵ opposite sides of parallelogram ABCD if equal and parallel to each other]
AB ∥ DC
∴ AE = FC [∵ parts of parallel sides are parallel to each other]
& AE ∥ FC
∴ Quad AEFC is a parallelogram
∴ AF ∥ EC
Now,
In △APB
QE ∥ AP [∵ parts of parallel lines are parallel to each other, AF ∥ EC]
E is the mid point of side AB
∴ By mid point theorem,
Q is the mid point of PB since QE ∥ AP
∴ PQ = BQ
Now,
In △DQC
PE ∥ CQ [∵ parts of parallel lines are parallel to each other, EC ∥ AF]
F is the mid point of side DC
∴ By mid point theorem,
P is the mid point of DQ since PE ∥ CQ
∴ DP = PQ
∴ DP = PQ = BQ
Hence AF and CE trisects diagonal BD
(5) Show that the line segments joining the midpoints of the opposite sides of a quadrilateral and bisect each other
ABCD is a quadrilateral
E, F, G, H are the mid points of AD, DC, CB, BA respectively
In △ABD
E and H are mid points of AD & AB respectively
∴ By mid point theorem,
EH ∥ BD & EH = ½ BD – (i)
In △BCD
G and F are mid points of BC & Dc respectively
∴ By mid points theorem,
GF ∥ BD & GF = ½ BD – (ii)
∴ On Comparing equation (i) & (ii) we get,
EH ∥ GF & FH = GF [∵ both equal to ½ BD and parallel to BD]
∴ Quad EFGH is a parallelogram
Hence, EG and FH are diagonals of parallelogram EFGH
∴EG and FH bisects each other
Line segments EG and FH joining mid points of sides AD, BC and AB, DC of quad ABCD bisects each other
(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 = 1/2AB
Solution: Given,
ABC is a triangle right angled at C
A line through mid point of AB which is M lies parallel to BC and meet AC at D
In △ABC
(i) M is the mid point of AB
DM ∥ BC [given]
∴ By mid point theorem,
D is the mid point of AC [∵ mid point when connected are parallel to the third side]
∠MCE = ∠DMC [alternate angles, DM BC]
∠DCM = ∠ACB – ∠MCB [∠ACB = 90o]
Or, ∠DCM = 90 – ∠ACB
Or, ∠DCM = 90 – ∠DMC [∠MCB = ∠DMC] – (i)
In △DCM
∠MDC + ∠DMC + ∠DCM =180o [sum of angles of a triangle is 180o]
Or, ∠MDC + ∠DMC + 90o – ∠DMC = 180o from equation (i)
Or, ∠MDC = 180o – 90o
= 90o
(ii) ∴ Line segment MD is perpendicular to AC
Since angles between the two segments is 90o
(iii) In △ADM & △CDM
(i) DM is common side
(ii) ∠MDC + ∠ADM = 180o [collinear angles]
Or, 90o + ∠ADN = 180o
Or, ∠ADM = 180o – 90o
Or, ∠ADM = 90o
∠ADM = ∠MDC = 90o
(iii) AD = DC [∵ D is the mid point of AC]
∴ △ADM ≅ △CDM by SAS
∴ CM = MA [corresponding sides of congruent triangles are equal]
And
MA = ½ AB [M is the third point of AB]
∴ CM = NA = ½ AB proved