KC Sinha Mathematics Solution Class 9 Chapter 10 Quadrilaterals exercise 10.1

Page 10.20

EXERCISE 10.1


Type 1


QUESTION 1

Fill in the blanks:
(i) A parallelogram will be a rhombus, if and only if its ___are Perpendicular to each other.
(ii) A quadrilateral will be a parallelogram, if and only if its diagonals ___ each other.
(iii) Opposite angles of a parallelogram are __
Sol :


QUESTION 2

Which of the following statements are true (T) and which are false (F)
(i) In any quadrilateral, if a pair or opposite is equal, it is a parallelogram.
(ii) In a parallelogram , diagonals bisect each other.
(iii) If all sides of a quadrilateral are equal, it is a parallelogram.
(iv) In a quadrilateral, if three angles are equal, it is a parallelogram.
(v) In a quadrilateral , if its opposite sides are of equal lengths , it is a parallelogram .
Sol :



QUESTION 3

Answer the following questions :
(i) What is the relation between the diagonals of a rhombus ?
(ii) For a parallelogram to be square , what will be the relation between its diagonals ?
Sol :


QUESTION 4

(i) In a parallelogram ABCD, if ∠D=110° , then write the measure in degree of ∠A and ∠B
(ii) In a parallelogram, sum of two angles is 140° , then find the measure of its acute angle.
(iii) In a parallelogram, one angle is 4/5 of its adjacent angle , then find the values of both the adjacent angles in degree.
(iv) Side BC of a parallelogram is produced to point E. If ∠DCE=65°, find the value of ∠A.
Sol :



QUESTION 5

In a parallelogram ABCD , ∠DAC=40° ,∠ABC=60° , then fill in the blanks .
(a) ∠ACB__
(b) ∠ADC__
(c) ∠BAC__
(d) ∠ACD__
Sol :




QUESTION 6

ABCD is a rhombus whose diagonals meet at a point O , if ∠OAB=30° , then find ∠OBA
Sol :


QUESTION 7

ABCD is a parallelogram whose diagonals AC and BD intersect each other at point O , such that ∠DAC=32° and ∠AOB=70° , find the value of ∠DBC.
Sol :


QUESTION 8

In the adjacent figure , ABCD is a parallelogram in which ∠DAO=35° , ∠OAB=25° , ∠DOC=75° , find

(i) ∠ABO
(ii) ∠ODC
(iii) ∠ACB
(iv) ∠CBD
Sol :


QUESTION 9

In a rhombus, lengths of diagonals are 8 cm and 6 cm. f‌ind the length of its sides
Sol :


QUESTION 10

In a rhombus, length of a side and one diagonal are 13 cm and 10 cm respectively. Find the length of its other diagonal.
Sol :


QUESTION 11

In the adjacent figure, PQ||SR and PS=PR , find α and β
In the adjacent figure, PQ||SR and PS=PR , find α and β
Sol :
[Hint : PS=PR
∠PSR=∠PRS=α
In ΔPSR α+α+30°=180°
⇒$\alpha =\dfrac{180^{\circ}-30^{\circ}}{2}=75^{\circ}$
PQ||SR
β=∠PRS=75°
]

TYPE 3


QUESTION 12

In a parallelogram ABCD , bisectors of consecutive ∠A and ∠B intersect at a point P. Prove that ∠APB=90°

Sol :


QUESTION 13

If diagonals of a parallelogram bisect its angles , then prove that the diagonals will be perpendicular to each other.
Sol :


QUESTION 14

In  a quadrilateral ABCD , ∠A=∠C and ∠B=∠D , prove that ABCD is a parallelogram.
<fig to be added>
[Hint: In quadrilateral ABCD
⇒∠A+∠B+∠C+∠D=360°
⇒∠A+∠B+∠A+∠B=360°
⇒2(∠A+∠B)=360°
⇒∠A+∠B=180°
⇒AD||BC
Now , on joining BD , we have
ΔABD≅ΔCDB [AAS congruency rule]
ABCD is a parallelogram]
Sol :


QUESTION 15

Prove that the bisectors of angles of a parallelogram form a rectangle.
<fig to  be added>
[Hint: Bisectors of ∠A and ∠B meet at point S]
∠BAS+∠ABS=1/2(∠A+∠B)
=1/2 ×180° [AD||BC and AB is a transversal]
So , ∠BAS+∠ABS=90°
In ΔABS , ∠ASB=90° ..(i)
∠PSR=90° [vertically opposite angles]
Similarly , ∠SRQ=90° , ∠RQP=90° and ∠QPS=90°
Hence ,PQRS is a rectangle ]
Sol :


QUESTION 16

In the given figure , ABC is a triangle and through vertices A, B and C  , three lines are drawn which are respectively parallel to opposite sides. Prove that the perimeter of the triangle DEF , so formed by these three lines is two times the perimeter of the triangle ABC.
<fig to be added>
[Hint: FA=BC; FB=AC; AE=BC
EC=AB; DC=AB; BD=AC
∴FA+FB+AE+EC+DC+BD=BC+AC+BC+AB+AB+AC
⇒(FA+AE)+(FBB+BD)+(EC+DC)=2(AB+BC+CA)
⇒EF+FE+ED=2(AB+BC+AC)
]
Sol :


QUESTION 17

ABCD is a parallelogram in which bisectors of ∠A and ∠C meet the diagonals BD at P and Q respectively. Prove that PCQA is a parallelogram.
Sol :


QUESTION 18

ABCD is a parallelogram and EF||BD.  R is mid point of EF. Prove that BE=DF
ABCD is a parallelogram and EF||BD.  R is mid point of EF. Prove that BE=DF
Sol :


 

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