Utilizing ICSE Class 9 Maths Solutions S Chand Chapter 9 Mid-Point and Intercept Theorems Chapter Test as a study aid can enhance exam preparation.
S Chand Class 9 ICSE Maths Solutions Chapter 9 Mid-Point and Intercept Theorems Chapter Test
Question 1.
Use mid-segment theorem to name following parts of the given triangle.
(a) A mid-segment of ∆ABC
(b) A segment parallel to AC
(c) A segment that has the same length as BD
(d) A segment that has half the length of AC
(e) A segment that has twice the length of EC
Solution:
In the given ∆ABC
AD = DB and BE = EC
∴ D and E are mid points of AB and BC respectively
∴ DE || AC and DE =\(\frac { 1 }{ 2 }\) AC
(a) DE is mid segment of AABC
(b) DE is parallel to AC
(c) AD = BD
(d) DE is half of AC
(e) Twice of EC = BC (∵ BE = EC)
Question 2.
Find each measure:
(a) NM
(b) XZ
(c) NZ
(d) ∠LMN
(e) ∠YXZ
(f) XXLM
Solution:
In the given figure,
∵ YL = LX, YM = MZ and XN = NZ
∴ L, M and N are the mid points of sides XY,
YZ and XZ respectively
∴ LM = \(\frac { 1 }{ 2 }\) XZ, MN = \(\frac { 1 }{ 2 }\) XY
XY = 10 cm, LM = 6 cm, ∠MNZ = 25°‘
(a) NM = \(\frac { 1 }{ 2 }\) XY = \(\frac { 1 }{ 2 }\) x 10 cm = 5 cm
(b) XZ = 2 x LM = 2 x 6 = 12 cm
(c) NZ = \(\frac { 1 }{ 2 }\) x XZ = \(\frac { 1 }{ 2 }\) x 12 = 6 cm
(d) ∠LMN = 25°
(∵ LM || XZ and MN is a transversal)
(Alternate angles)
(e) ∠YXZ = ∠MNZ = 25°
(corresponding angles)
(f) ∠XLM = ∠XNM
(opposite angles of a ||gm) = 180° – 25° = 155°
Question 3.
Find the value of n in each triangle.
Solution:
In ∆ABC,
AL = LB and CM = MB
∴ LM || AC and LM = \(\frac { 1 }{ 2 }\) AC
⇒ 32 = \(\frac { 1 }{ 2 }\) x 4n ⇒ 2n = 32 32
⇒ n = \(\frac { 32 }{ 2 }\) = 16 cm
∴ n = 16 cm
Question 4.
Find the value of n in each triangle.
Solution:
In ∆ABC,
AE = EC and BD = DC
∴ DE || AB and DE = \(\frac { 1 }{ 2 }\) AB
⇒ 6n = 4n + 12
⇒ 6n – 4n = +12
⇒ 2n = 12
n = \(\frac { 12 }{ 2 }\) = 6
Question 5.
In the figure, CG, EH and FJ are mid-segments of ∆ABD, ∆GCD and ∆GHE respectively. Find each measure:
(a) CG
(b) EH
(c) FJ
(d) m∠DCG
(e) m∠GHE
(f) m∠FJH
Solution:
In the given figure,
CG, EH and FJ are mid segments of ∆ABD, ∆GCD and ∆GHE respectively AB = 33 cm, ∠ABC = 57°
(a) CG = \(\frac { 1 }{ 2 }\) AB
= \(\frac { 1 }{ 2 }\) x 33 = 16.5 cm
(b) EH = \(\frac { 1 }{ 2 }\) DC
= \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) DB
= \(\frac { 1 }{ 4 }\) DB = \(\frac { 1 }{ 4 }\) x 44 = 11 cm
(c) FJ = \(\frac { 1 }{ 2 }\)GH = \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) GC
= \(\frac { 1 }{ 4 }\) GC = \(\frac { 1 }{ 4 }\) x 6.5 cm = 4.125 cm
(d) m∠DCG = ∠DBA
(corresponding angles)
= 57°
(e) m∠GHE = ∠GCD
(corresponding angles)
= 57°
(f) m∠FJH = 180° – ∠GHE
= 180° – 57° = 123°
Question 6.
Prove that the perimeter of a mid-segment triangle is half the perimeter of the triangle.
Solution:
In ∆ABC, D, E and F are the mid-points of sides BC, CA and AB respectively, forming mid segment ∆DEF
∵ D, E and F are the mid points of the sides
∴ DE = \(\frac { 1 }{ 2 }\) AB, EF = \(\frac { 1 }{ 2 }\) BC and DF = \(\frac { 1 }{ 2 }\) AC
Now perimeter of ∆DEF = DE + EF + DF
= \(\frac { 1 }{ 2 }\)AB + \(\frac { 1 }{ 2 }\)BC + \(\frac { 1 }{ 2 }\)CA
= \(\frac { 1 }{ 2 }\)(AB + BC + CA)
= Half perimeter of ∆ABC
Hence proved.