**Chapter 5 Quadrilaterals Practice Set 5.5**

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:

Solution:

**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 = LRGiven: □PQRS and □MNRL are rectangles. M is the midpoint of side PR.Solution:**

Toprove:

i. SL = LR

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)]

**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.

**Question 4.Solution:**
Given: seg PD is a median of ∆PQR. Point T is the midpoint of seg PD.

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