# 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

(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__

(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

Sol :

[Hint : PS=PR

∠PSR=∠PRS=α

In ΔPSR α+α+30°=180°

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.

⇒∠A+∠B+∠C+∠D=360°

⇒∠A+∠B+∠A+∠B=360°

⇒2(∠A+∠B)=360°

⇒∠A+∠B=180°

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.

[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.

[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

Sol :

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