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Chapter 5 Quadrilaterals Practice Set 5.5 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 5 Quadrilaterals Practice Set 5.5. Step-by-step solved exercises, numerical problems, and digest answers.

4 Solved Questions• 706 words

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

Question 1 Maharashtra Board Solution
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 & Step-by-Step Answer:
i. AC = 9 cm [Given] Points X and Y are the midpoints of sides AB and BC respectively. [Given] ∴ XY = AC [Midpoint tfyeprem] = 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 = AB [Midpoint theorem]
= 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 = BC [Midpoint theorem]
= x 11 = 5.5 cm
l(XY) = 4.5 cm, l(YZ) = 2.5 cm, l(XZ) = 5.5 cm

Question 2 Maharashtra Board Solution
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 = SQ. Given: □PQRS and □MNRL are rectangles. M is the midpoint of side PR.
Solution & Step-by-Step Answer:
Toprove: i. SL = LR ii. LN = (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 = SQ [Midpoint theorem]

Question 3 Maharashtra Board Solution
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 & Step-by-Step Answer:
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 = AC …..(ii) [Midpoint theorem]
Points D and E are the midpoints of sides BC and AC respectively.

∴ DE = AB …..(iii) [Midpoint theorem]
Points F and E are the midpoints of sides AB and AC respectively.
∴ FE = BC
∴ FD = DE = FE [From (i), (ii), (iii) and (iv) ]
∴ ∆FED is an equilateral triangle.

Question 4 Maharashtra Board Solution
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 = . [Hint: Draw DN || QM]
Solution & Step-by-Step Answer:
Given: seg PD is a median of ∆PQR. Point T is the midpoint of seg PD. To Prove: = 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 =
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Chapter 5 Quadrilaterals Practice Set 5.4
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Chapter 5 Quadrilaterals Problem Set 5