## Maharashtra State Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.5

Question 1.

In the adjoining figure, points X, Y, Z are the midpoints of of ∆ABC respectively, cm. Find the lengths of side AB, side BC and side AC AB = 5 cm, AC = 9 cm and BC = 11c.m. Find the lengths of XY, YZ, XZ.

Solution:

i. AC = 9 cm [Given]

Points X and Y are the midpoints of sides AB and BC respectively. [Given]

∴ XY = \(\frac { 1 }{ 2 }\) AC [Midpoint tfyeprem]

= \(\frac { 1 }{ 2 }\) x 9 = 4.5 cm

ii. AB = 5 cm [Given]

Points Y and Z are the midpoints of sides BC and AC respectively. [Given]

∴ YZ = \(\frac { 1 }{ 2 }\) AB [Midpoint theorem]

= \(\frac { 1 }{ 2 }\) x 5 = 2.5 cm

iii. BC = 11 cm [Given]

Points X and Z are the midpoints of sides AB and AC respectively. [Given]

∴ XZ = \(\frac { 1 }{ 2 }\) BC [Midpoint theorem]

= \(\frac { 1 }{ 2 }\) x 11 = 5.5 cm

l(XY) = 4.5 cm, l(YZ) = 2.5 cm, l(XZ) = 5.5 cm

Question 2.

In the adjoining figure, □PQRS and □MNRL are rectangles. If point M is the midpoint of side PR, then prove that,

i. SL = LR

ii. LN = \(\frac { 1 }{ 2 }\) SQ.

Given: □PQRS and □MNRL are rectangles. M is the midpoint of side PR.

Solution:

Toprove:

i. SL = LR

ii. LN = \(\frac { 1 }{ 2 }\) (SQ)

Proof:

i. □PQRS and □MNRL are rectangles. [Given]

∴ ∠S = ∠L = 90° [Angles of rectangles]

∠S and ∠L form a pair of corresponding angles on sides SP and LM when SR is their transversal.

∴eg ML || seg PS …(i) [Corresponding angles test]

In ∆PRS,

Point M is the midpoint of PR and seg ML || seg PS. [Given] [From (i)]

∴ Point L is the midpoint of seg SR. ……(ii) [Converse of midpoint theorem]

∴ SL = LR

ii. Similarly for ∆PRQ, we can prove that,

Point N is the midpoint of seg QR. ….(iii)

In ∆RSQ,

Points L and N are the midpoints of seg SR and seg QR respectively. [From (ii) and (iii)]

∴ LN = \(\frac { 1 }{ 2 }\)SQ [Midpoint theorem]

Question 3.

In the adjoining figure, ∆ABC is an equilateral triangle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ∆FED is an equilateral triangle.

Given: ∆ABC is an equilateral triangle.

Points F, D and E are midpoints of side AB, side BC, side AC respectively.

To prove: ∆FED is an equilateral triangle.

Solution:

Proof:

∆ABC is an equilateral triangle. [Given]

∴ AB = BC = AC ….(i) [Sides of an equilateral triangle]

Points F, D and E are midpoints of side AB and BC respectively.

∴ FD = \(\frac { 1 }{ 2 }\)AC …..(ii) [Midpoint theorem]

Points D and E are the midpoints of sides BC and AC respectively.

∴ DE = \(\frac { 1 }{ 2 }\)AB …..(iii) [Midpoint theorem]

Points F and E are the midpoints of sides AB and AC respectively.

∴ FE = \(\frac { 1 }{ 2 }\)BC

∴ FD = DE = FE [From (i), (ii), (iii) and (iv) ]

∴ ∆FED is an equilateral triangle.

Question 4.

In the adjoining figure, seg PD is a median of ∆PQR. Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that \(\frac { PM }{ PR }\) = \(\frac { 1 }{ 3 }\). [Hint: Draw DN || QM]

Solution:

Given: seg PD is a median of ∆PQR. Point T is the midpoint of seg PD.

To Prove: \(\frac { PM }{ PR }\) = \(\frac { 1 }{ 3 }\)

Construction: Draw seg DN ||seg QM such that P-M-N and M-N-R.

Proof:

In ∆PDN,

Point T is the midpoint of seg PD and seg TM || seg DN [Given]

∴ Point M is the midpoint of seg PN. [Construction and Q-T-M]

∴ PM = MN [Converse of midpoint theorem]

In ∆QMR,

Point D is the midpoint of seg QR and seg DN || seg QM [Construction]

∴ Point N is the midpoint of seg MR. [Converse of midpoint theorem]

∴ RN = MN …..(ii)

∴ PM = MN = RN …..(iii) [From (i) and (ii)]

Now, PR = PM + MN + RN [ P-M-R-Q-T-M]

∴ PR = PM + PM + PM [From (iii) ]

∴ PR = 3PM

\(\frac { PM }{ PR }\) = \(\frac { 1 }{ 3 }\)