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Problem Set 5, Chapter 5 - Quadrilaterals Mathematics Part II Solutions for Class 9 Math

Problem Set 5
Question 1.

Choose the correct alternative answer and fill in the blanks.

If all pairs of adjacent sides of a quadrilateral are congruent then it is called ....
A. rectangle

B. parallelogram

C. trapezium

D. rhombus


Answer:

As per the properties of a rhombus:- A rhombus is a parallelogram in which adjacent sides are equal(congruent).


Question 2.

Choose the correct alternative answer and fill in the blanks.

If the diagonal of a square is 12√2 cm then the perimeter of square is ......
A. 24 cm

B. 24√2 cm

C. 48 cm

D. 48√2 cm


Answer:

Here d= 12√2 = √2 s where s is side of square

Given diagonal = 20 cm


⇒ s = 


Therefore, perimeter of the square is 4s = 4 x 12


= 48cm. (C)


Question 3.

Choose the correct alternative answer and fill in the blanks.

If opposite angles of a rhombus are (2x)° and (3x - 40)° then value of x is .......
A. 100°

B. 80°

C. 160°

D. 40°


Answer:

As rhombus is a parallelogram with opposite angles equal

⇒ 2x = 3x -40


x= 40°


Question 4.

Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.


Answer:

Adjacents sides are 7cm and 24 cm

In a rectangle angle between the adjacent sides is 90°


⇒ the diagonal is hypotenuse of right Δ


By pythagorus theorem


Hypotenuse2 = side2 + side2


Hypotenuse2 = 49 + 576 = 


length of the diagonal = 25cm



Question 5.

If diagonal of a square is 13 cm then find its side.


Answer:

given Diagonal of the Square = 13cm

The angle between each side of the square is 90°


Using Pythagoras theorem


Hypotenuse2 = side2 + side2










Side =  cm


Question 6.

Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of its each side.


Answer:

In a parallelogram opposite sides are equal

Let the sides of parallelogram be x and y


2x+ 2y =112 and given 


⇒ 


⇒7y = 224


y= 32


x= 24


four sides of the parallelogram are 24cm , 32 cm, 24cm, 32cm.



Question 7.

Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.


Answer:

According to the properties of Rhombus diagonals of the rhombus bisects each other at 90°



In the rhombus PQRS


SO = OQ = 10 cm


PO= OR = 12cm


So in ΔPOQ


∠ POQ = 90°


⇒ PQ is hypotenuse

By Pythagoras theorem,

102 + 242 =PQ2


100 + 576 = PPQ2


676 = PQ2


26cm = PQ Ans


Question 8.

Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50° then find the measure of ∠MPS.


Answer:

The figure is given below:



Given PQRS is a rectangle


⇒ PS ∥ QR (opposite sides are equal and parallel)


QS and PR are transversal


So ∠ QMR = ∠ MPS (vertically opposite angles)


Given ∠ QMR = 50°


∴ ∠ MPS = 50°


Question 9.

In the adjacent Figure 5.42, if seg AB || seg PQ, seg AB ≅ seg PQ, seg AC || seg PR, seg AC  seg PR then prove that, seg BC || seg QR and seg BC  seg QR.



Answer:

Given

AB ∥ PQ


AB ≅ PQ ( or AB = PQ )


⇒ ABPQ is a parallelogram (pair of opposite sides is equal and parallel)


⇒ AP ∥ BQ and AP ≅ BQ……………..1


Similarly given,


AC ∥ PR and AC ≅ PR


⇒ACPR is a parallelogram (pair of opposite sides is equal and parallel)


⇒ AP ∥ CR and AP ≅ CR ……………………2


From 1 and 2 we get


BQ ∥ CR and BQ ≅ CR


Hence BCRQ is a parallelogram with a pair of opposite sides equal and parallel.


Hence proved.



Question 10.

In the Figure 5.43, ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively.

Then prove that, PQ || AB and 



Answer:

Given AB ∥ DC

P and Q are mid points of AD and BC respectively.


Construction :- Join AC


The figure is given below:



In Δ ADC


P is mid point of AD and PQ is ∥ DC the part of PQ which is PO is also ∥ DC


By mid=point theorem


A line from the mid-point of a side of Δ parallel to third side, meets the other side in the mid-point


⇒ O is mid-point of AC


⇒ PO = 1/2 DC…………..1


Similarly in Δ ACB


Q id mid-point of BC and O is mid –point of AC


⇒ OQ∥ AB and OQ = 1/2 AB………………2


Adding 1 and 2


PO + OQ = 1/2 (DC+ AB)


PQ = 1/2 (AB +DC)


And PQ ∥ AB


Hence proved.




Question 11.

In the adjacent figure 5.44, 􀀍ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.



Answer:

Given AB ∥ DC

M is mid-point of AC and N is mid-point of DB


Given ABCD is a trapezium with AB ∥ DC


P and Q are the mid-points of the diagonals AC and BD respectively


The figure is given below:



To Prove:- MN ∥ AB or DC and


In ΔAB


AB || CD and AC cuts them at A and C, then


∠1 = ∠2 (alternate angles)


Again, from ΔAMR and ΔDMC,


∠1 = ∠2 (alternate angles)


AM = CM (since M is the mid=point of AC)


∠3 = ∠4 (vertically opposite angles)


From ASA congruent rule,


ΔAMR ≅ ΔDMC


Then from CPCT,


AR = CD and MR = DM


Again in ΔDRB, M and N are the mid points of the sides DR and DB,


then PQ || RB


⇒ PQ || AB


⇒ PQ || AB and CD ( ∵ AB ∥ DC)


Hence proved.