Q.1 The perimeter of a square is 40 cm. Find the length of its diagonal

Q.2 ABCD is a rectangle with diagonals AC and BD meeting at point O. Find x if OA=5x-7 and OD=4x-5

Q.3 In the adjoining figure, ABCD is a rhombus. The length of its diagonal AC and BD are 24 cm and 10 cm respectively. Find the length of each side of the rhombus. Also, find ∠x if ∠DCO=40º

Q.4 In the adjoining parallelogram ABCD, determine the values of a,b and c. Give a reason, in support of your answer

Solution :

Problem:01 The perimeter of a square is 40 cm. Find the length of its diagonal

Solution: It is given that,

The perimeter of square =40 cm

Each side of square =\frac{40}{4}

Note: All sides of a square are equal.

AB=BC=CD=DA=10 cm

In Δ ABC ,by Pythagoras theorem

(AC)²=(AB)²+(BC)²

(AC)²=(10)²+(10)²

(AC)²=200

AC=10√2 cm

So,Length of the diagonal is 10√2 cm

Problem:02 ABCD is a rectangle with diagonals AC and BD meeting at point O. Find x if OA=5x-7 and OD=4x-5

Solution:

ABCD is a rectangle with diagonals AC and BD meeting at point O

It is given that, OA=5x-7 and OD=4x-5

We know, diagonals of a rectangle bisect each other and are equal in length

AC=BD

OA+OC=OB+OD

OA+OA=OD+OD [OA=OC and OB=OD]

OA=OD

According to Question ,

5x-7=4x-5

x=2

Problem :03 In the adjoining figure, ABCD is a rhombus. The length of its diagonal AC and BD are 24 cm and 10 cm respectively. Find the length of each side of the rhombus. Also, find ∠x if ∠DCO=40º

Solution: As we know ,diagonals of a rhombus bisect each other at right angle

OA= \frac{1}{2}×AC=\frac{1}{2}×24=12 cm

Similarly ,

OB= \frac{1}{2}×BD= \frac{1}{2}×10=5 cm

In ΔAOB ,by pythagoras theorem

(AB)²=(AO)²+(BO)²

(AB)²=(12)²+(5)²

(AB)²=(13)²

AB=13

Hence ,length of each side of rhombus is 13 cm

Now in ΔDOC ,

∠OCD=40º ,∠DOC=90º

∠OCD+∠DOC+∠ODC=180

40º+90º+∠ODC=180

∠ODC=50º

Since ,DC||AB and AC is transversal ,then

∠x=∠CDO

∠x=50º

Problem : 04 In the adjoining parallelogram ABCD, determine the values of a,b and c. Give a reason, in support of your answer

Solution:

We know ,opposite angle of parallelogram are equal

So, ∠BAD=∠BCD

∠a=∠DCA+∠ACB

∠a= 40º+30º

∠a=70º

Since ,AB||CD and AC is transvesal

∠BCA=∠DAC [alternate angle]

∠DAC=30º

Now ,in ΔADC

∠CAD+∠ADC+∠ACD=180º

30º+∠ADC+40º=180º

∠ADC=180º-70º

∠ADC=110º i.e ∠c=110º

So,∠b=110º [opposite angle are equal ]