# Maharashtra Board 9th Class Maths Part 2 Practice Set 5.5 Solutions Chapter 5 Quadrilaterals

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.5 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

## Practice Set 5.5 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals

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
$$\frac { PM }{ PR }$$ = $$\frac { 1 }{ 3 }$$