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.