### Page 14.4 Ex. 14.1

Q1.

**Answer :**

Let the measure of the fourth angles be *x*°. We know that the sum of the angles of a quadrilateral is 360°.

Therefore,

Hence the measure of the fourth angle is .

Q2.

**Answer :**

We have ,.

So, let ,

,

And

By angle sum property of a quadrilateral, we get:

Also,

And

Similarly,

Hence, the four angles are ,,and .

Q3.

**Answer :**

The quadrilateral can be drawn as follows:

We have CO and DO as the bisectors of angles and respectively.

We need to prove that.

In ,We have,

…… (I)

By angle sum property of a quadrilateral, we have:

Putting in equation (I):

Hence proved.

Q4.

**Answer :**

We have, .

So, let ,

,

and

By angle sum property of a quadrilateral, we get:

Also,

And

Similarly,

Hence, the four angles are , , and .

### Page 14.18 Ex. 14.2

Q1.

**Answer :**

It is given that the two opposite angles of a parallelogram are and .

We know that the opposite angles of a parallelogram are equal.

Therefore,

…… (i)

Thus, the given angles become

Also, .

Therefore the sum of consecutive interior angles must be supplementary.

That is;

Since opposite angles of a parallelogram are equal.

Therefore,

And

Hence the four angles of the parallelogram are , , and .

Q2.

**Answer :**

Let one of the angle of the parallelogram as

Then the adjacent angle becomes

We know that the sum of adjacent angles of the parallelogram is supplementary.

Therefore,

Thus, the angle adjacent to

Since, opposite angles of a parallelogram are equal.

Therefore, the four angles in sequence are ,,and.

Q3.

**Answer :**

Let the smallest angle of the parallelogram be

Therefore, according to the given statement other angle becomes .

Also, the opposite angles of a parallelogram are equal.

Therefore, the four angles become ,,and.

According to the angle sum property of a quadrilateral:

Thus, the other angle becomes

Hence, the four angles of the parallelogram are , , and .

Q4.

**Answer :**

Let the shorter side of the parallelogram be *cm*.

The longer side is given as*cm.*

Perimeter of the parallelogram is given as 22 *cm*

Therefore,

Hence, the measure of the shorter side is * cm.*

Q5.

**Answer :**

It is given that *ABCD* is a parallelogram with

We know that the opposite angles of the parallelogram are equal.

Therefore,

Also, and are adjacent angles, which must be supplementary.

Therefore,

Hence , and .

Q6.

**Answer :**

It is given that *ABCD* is a parallelogram with

We know that the opposite angles of the parallelogram are equal.

Therefore,

Also, and are adjacent angles, which must be supplementary.

Therefore,

Also, and are opposite angles of a parallelogram.

Therefore,

Hence, the angles of a parallelogram are , , and .

Q7.

**Answer :**

The figure is given as follows:

It is given that *ABCD* is a parallelogram.

Thus,

Opposite angles of a parallelogram are equal.

Therefore,

Also, we have *AP* as the bisector of

Therefore,

…… (i)

Similarly,

…… (ii)

We have ,

From (i)

Thus, sides opposite to equal angles are equal.

Similarly,

From (ii)

Thus, sides opposite to equal angles are equal.

Also,

Q8.

**Answer :**

The figure is given as follows:

It is given that *ABCD* is a parallelogram.

Thus

And are alternate interior opposite angles.

Therefore,

…… (i)

We know that the opposite angles of a parallelogram are equal. Therefore,

Also, we have

Therefore,

…… (ii)

In

By angle sum property of a triangle.

From (i) and (ii),we get:

Hence, the required value for is

And is .

### Page 14.19 Ex. 14.2

Q9.

**Answer :**

Figure is given as follows:

It is given that *ABCD* is a parallelogram.

*DE* and *AB* when produced meet at *F*.

We need to prove that

It is given that

Thus, the alternate interior opposite angles must be equal.

In and , we have

(Proved above)

(Given)

(Vertically opposite angles)

Therefore,

(By ASA Congruency )

By corresponding parts of congruent triangles property, we get

DC = BF …… (i)

It is given that *ABCD* is a parallelogram. Thus, the opposite sides should be equal. Therefore,

…… (ii)

But,

From (i), we get:

From (ii), we get:

Hence proved.

Q10.

Answer :

(i) Statement: In a parallelogram, the diagonals are equal.

False

(ii) Statement: In a parallelogram, the diagonals bisect each other.

True

(iii) Statement: In a parallelogram, the diagonals intersect each other at right angles.

False

(iv) Statement: In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.

False

(v) Statement: If all the angles of a quadrilateral are equal, then it is a parallelogram.

True

(vi) Statement: If three sides of a quadrilateral are equal, then it is not necessarily a parallelogram.

False

(vii) Statement: If three angles of a quadrilateral are equal, then it is no necessarily a parallelogram.

False

(viii) Statement: If all sides of a quadrilateral are equal, then it is a parallelogram.

True

### Page 14.38 Ex. 14.3

Q1.

**Answer :**

The parallelogram can be drawn as:

We have ,thus and are consecutive interior angles.

These must be supplementary.

Therefore,

Q2.

**Answer :**

Since *ABCD* is a parallelogram with .

Opposite angles of a parallelogram are equal.

Therefore,

Also, let

Similarly,

We know that the sum of the angles of a quadrilateral is .

Hence the measure of other angles are , and .

Q3.

**Answer :**

The figure can be drawn as follows:

In and,

(Sides of a square are equal)

(Diagonals of a parallelogram bisect each other)

(Common)

So, by SSS Congruence rule, we have

Also,

(Corresponding parts of congruent triangles are equal)

But, (Linear pairs)

We have,

Hence, the required measure of is .

Q4.

**Answer :**

The rectangle is given as follows with

We have to find .

An angle of a rectangle is equal to .

Therefore,

Hence, the measure for is .

Q5.

**Answer :**

Figure is given as follows:

It is given that *ABCD* is a parallelogram.

*E *is the mid point of *AB*

Thus,

,

…… (i)

Similarly,

……(ii)

From (i) and (ii)

Also,

Thus,

Therefore, *EBFD* is a parallelogram.

Q6.

**Answer :**

Figure can be drawn as follows:

We have *P* and *Q* as the points of trisection of the diagonal *BD* of parallelogram ABCD.

We need to prove that *AC* bisects *PQ.* That is, .

Since diagonals of a parallelogram bisect each other.

Therefore, we get:

and

*P* and *Q* as the points of trisection of the diagonal *BD.*

Therefore*,*

and

Now, and

Thus,

Hence proved.

Q7.

**Answer :**

Square *ABCD* is given:

*E, F, G* and *H* are the points on *AB, BC, CD* and *DA* respectively, such that :

We need to prove that *EFGH* is a square.

Say,

As sides of a square are equal. Then, we can also say that:

In and ,we have:

(Given)

(Each equal to 90°)

(Each equal to *y* )

By SAS Congruence criteria, we have:

Therefore, EH = EF

Similarly, EF= FG, FG= HG and HG= HE

Thus, HE=EF=FG=HG

Also,

and

But,

and

Therefore,

i.e.,

Similarly,

Thus,* EFGH* is a square.

Hence proved.

Q8.

**Answer :**

Rhombus *ABCD* is given:

We have

We need to prove that

We know that the diagonals of a rhombus bisect each other at right angle.

Therefore,

,,

In *A* and *O *are the mid-points of *BE* and *BD* respectively.

By using mid-point theorem, we get:

Therefore,

In *A* and *O *are the mid-points of *BE* and *BD* respectively.

By using mid-point theorem, we get:

Therefore,

Thus, in quadrilateral *DOCG*,we have:

and

Therefore,* DOCG *is a parallelogram.

Thus, opposite angles of a parallelogram should be equal.

Also, it is given that

Therefore,

Or,

Hence proved.

Q9.

**Answer :**

*ABCD *is a parallelogram, *AD* produced to *E *such that .

Also , *AB* produced to *F.*

We need to prove that

In , *D* and *O* are the mid-points of *AE* and *AC* respectively.

By using Mid-point Theorem, we get:

Since, *BD* is a straight line and *O *lies on *AC*.

And, *C* lies on *EF*

Therefore,

…… (i)

Also, is a parallelogram with .

Thus,

In and ,we have:

So, by ASA Congruence criterion, we have:

By corresponding parts of congruence triangles property, we get:

Hence proved.

### Page 14.55 Ex. 14.4

Q1.

**Answer :**

is given with *D,E* and *F *as the mid-points of *BC , CA* and *AB* respectively as shown below:

Also, , and .

We need to find the perimeter of

In , *E *and *F *are the mid-points of *CA* and *AB* respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get:

Similarly, we get

And

Perimeter of

Hence, the perimeter of is .

Q2.

**Answer :**

It is given that *D, E* and *F *be the mid-points of *BC , CA* and *AB* respectively.

Then,

, and .

Now, and transversal *CB* and *CA* intersect them at *D* and *E* respectively.

Therefore,

[(Given)]

and

[(Given)]

Similarly,

Therefore,

[(Given)]

and

[(Given)]

Similarly,

Therefore,

[(Given)] and

[(Given)]

Now *BC* is a straight line.

Similarly,

and

Hence the measure of angles are , and.

Q3.

**Answer :**

It is given that *P, Q* and *R *are the mid-points of *BC, CA* and *AB* respectively.

Also, we have , and

We need to find the perimeter of quadrilateral *ARPQ*

In , *P *and *R *are the mid-points of *CB* and *AB* respectively.

Theorem states, the line segment joining the mid-points of any two sides of a traingle is parallel to the third side and equal to half of it.

Therefore, we get:

Similarly, we get

We have *Q* and *R* as the mid points of *AC* and *AB* respectively.

Therefore,

And

Perimeter of

Hence, the perimeter of quadrilateral *ARPQ *is.

Q4.

**Answer :**

is given with *AD* as the median extended to point *X* such that .

Join *BX* and *CX*.

We get a quadrilateral ABXC, we need to prove that it’s a parallelogram.

We know that *AD* is the median.

By definition of median we get:

Also, it is given that

Thus, the diagonals of the quadrilateral *ABCX* bisect each other.

Therefore, quadrilateral *ABXC *is a parallelogram.

Hence proved.

Q5.

**Answer :**

is given with *E* and *F* as the mid points of sides *AB* and *AC.*

Also, intersecting *EF* at *Q*.

We need to prove that

In , *E* and *F *are the mid-points of *AB* and *AC* respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get:

Since, *Q* lies on *EF*.

Therefore,

This means,

*Q* is the mid-point of *AP*.

Thus, (Because, F is the mid point of AC and)

Hence proved.

Q6.

**Answer :**

In , *BM* and *CN* are perpendiculars on any line passing through *A*.

Also.

We need to prove that

From point *L* let us draw

It is given that , and

Therefore,

Since, L is the mid points of BC,

Therefore intercepts made by these parallel lines on MN will also be equal

Thus,

Now in ,

And . Thus, perpendicular bisects the opposite sides.

Therefore, is isosceles.

Hence

Hence proved.

Q7.

**Answer :**

We have right angled at B.

It is given that and

*D* and *E* are the mid-points of sides *AB* and *AC* respectively.

(i) We need to calculate length of *BC.*

In right angled at B:

By Pythagoras theorem,

Hence the length of *BC* is .

(ii) We need to calculate area of *.*

In right angled at *B*, *D* and *E *are the mid-points of *AB* and *AC *respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, .

Thus, (Corresponding angles of parallel lines are equal)

And

area of

*D* is the mid-point of side *AB* .

Therefore, area of

Hence the area of is .

Q8.

**Answer :**

We have as follows:

*M, N* and *P* are the mid-points of sides *AB ,AC *and* BC* respectively.

Also, , and

We need to calculate *BC, AB* and *AC.*

In ,* M* and *N *are the mid-points of *AB* and *AC *respectively.

Therefore,

Similarly,

And

Hence, the measure for *BC, AB* and *AC* is , and respectively*.*

### Page 14.56 Ex. 14.4

Q9.

**Answer :**

We have as follows:

Through *A*,*B* and *C* lines are drawn parallel to *BC,CA* and *AB* respectively intersecting at *P,Q* and *R* respectively.

We need to prove that perimeter of is double the perimeter of .

and

Therefore, is a parallelogram.

Thus,

Similarly,

is a parallelogram.

Thus,

Therefore,

Then, we can say that *A* is the mid-point of *QR*.

Similarly, we can say that *B* and* C *are the mid-point of *PR *and* PQ *respectively.

In ,

Theorem states, the line drawn through the mid-point of any one side of a triangle is parallel to the another side, intersects the third side at its mid-point.

Therefore,

Similarly,

Perimeter of is double the perimeter of

Hence proved.

Q10.

**Answer :**

is given with

*AD* is any line from *A* to *BC* intersecting *BE* in *H*.

*P,Q* and *R *respectively are the mid-points of *AH,AB* and *BC*.

We need to prove that

Let us extend *QP* to meet *AC* at *M*.

In , *R *and* Q *are the mid-points of *BC* and *AB* respectively.

Therefore, we get:

…… (i)

Similarly, in,

…… (ii)

From (i) and (ii),we get:

and

We get, is a parallelogram.

Also,

Therefore, is a rectangle.

Thus,

Or,

Hence proved.

Q11.

**Answer :**

We have the following given figure:

We have and and *AP* is the bisector of exterior angle of .

(i) We need to prove that

In ,

We have (Given)

Thus, (Angles opposite to equal sides are equal)

By angle sum property of a triangle, we get:

…… (i)

Now,

(*AP* is the bisector of exterior angle )

(Linear Pair)

…… (ii)

From equation (i) and (ii),we get:

(ii) We need to prove that is a parallelogram.

We have proved that

This means,

Also it is given that

We know that a quadrilateral with opposite sides parallel is a parallelogram.

Therefore, is a parallelogram.

Q12.

**Answer :**

ABCD is a kite such that and

Quadrilateral *PQRS* is formed by joining the mid-points *P,Q,R* and *S *of sides *AB,BC,CD* and *AD* respectively.

We need to prove that Quadrilateral *PQRS *is a rectangle*.*

In , *P* and *Q* are the mid-points of *AB* and *BC* respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, *PQRS *is a parallelogram.

Now,

But, *P* and *S* are the mid-points of *AB* and *AD*

…… (I)

In :* P *and* S *are the mid-point of side *AB *and* AD*

By mid-point Theorem, we get:

Or,

In ,* P *is the mid-point of side *AB* and

By Using the converse of mid-point theorem, we get:

*M* is the mid-point of *AO*

Thus*,*

…… (II)

In and , we have:

(Common)

[From (I)]

[From (II)]

By SSS Congruence theorem, we get:

By corresponding parts of congruent triangles property, we get:

But,

and

Therefore,

(, Corresponding angles should be equal)

Or,

We have proved that

Similarly, .

Then we can say that and

Therefore, is a parallelogram with

Or, we can say that is a rectangle.

We get:

Also,* PQRS *is a parallelogram.

Therefore,* PQRS *is a rectangle.

Hence proved.

Q13.

**Answer :**

, an isosceles triangle is given with *D,E* and *F *as the mid-points of *BC, CA* and *AB* respectively as shown below:

We need to prove that the segment *AD* and *EF* bisect each other at right angle.

Let’s join *DF* and *DE*.

In , *D* and *E *are the mid-points of *BC* and *AC* respectively.

Therefore, we get: Or

Similarly, we can get

Therefore, *AEDF* is a parallelogram

We know that opposite sides of a parallelogram are equal.

and

Also, from the theorem above we get

Thus,

Similarly,

It is given that , an isosceles triangle

Thus,

Therefore,

Also,

Then, *AEDF* is a rhombus.

We know that the diagonals of a rhombus bisect each other at right angle.

Therefore, *M* is the mid-point of *EF *and

Hence proved.

Q14.

**Answer :**

is given with *D* a point on *AB* such that .

Also, *E* is point on *AC* such that.

We need to prove that

Let *P* and *Q* be the mid points of *AB* and *AC* respectively.

It is given that

and

But, we have taken *P* and *Q* as the mid points of *AB* and *AC* respectively.

Therefore,* D *and *E* are the mid-points of *AP* and *AQ* respectively.

In , *P* and *Q* are the mid-points of *AB* and *AC* respectively.

Therefore, we get and …… (i)

In , *D *and *E* are the mid-points of *AP* and *AQ* respectively.

Therefore, we get and …… (ii)

From (i) and (ii),we get:

Hence proved.

Q15.

**Answer :**

Figure is given as follows:

*ABCD* is a parallelogram, where *P* is the mid-point of *DC* and *Q* is a point on *AC* such that

.

*PQ* produced meets *BC* at *R*.

We need to prove that *R* is a mid-point of *BC.*

Let us join *BD* to meet *AC* at *O.*

It is given that *ABCD* is a parallelogram.

Therefore, (Because diagonals of a parallelogram bisect each other)

Also,

Therefore,

In , *P* and *Q *are the mid-points of *CD* and *OC* respectively.

Therefore, we get:

Also, in , Q is the mid-point of *OC *and

Therefore, *R* is a mid-point of *BC.*

Hence proved.

Q16.

**Answer :**

Rectangles *ABCD* and *PQRC *are given as follows:

*Q* is the mid-point of *AC*.

In , *Q* is the mid-point of *AC* such that

Using the converse of mid-point theorem, we get:

*P *is the mid-point of *DC*

That is;

Similarly, *R* is the mid-point of *BC*.

Now, in , *P *and* R *are the mid-points of *DC *and* BC *respectively.

Then, by mid-point theorem, we get:

Now, diagonals of a rectangle are equal.

Therefore putting ,we get:

Hence Proved.

Q17.

**Answer :**

*ABCD* is a parallelogram with *E *and *F* as the mid-points of *AB* and *CD* respectively.

We need to prove that

Since *E* and *F* are the mid-points of *AB* and *CD* respectively.

Therefore,

,

And

,

Also,* ABCD* is a parallelogram. Therefore, the opposite sides should be equal.

Thus,

Also, (Because )

Therefore, *BEFC *is a parallelogram

Then, and …… (i)

Now,

Thus, (Because as *ABCD* is a parallelogram)

We get,

*AEFD *is a parallelogram

Then, we get:

…… (ii)

But, *E* is the mid-point of *AB.*

Therefore,

Using (i) and (ii), we get:

Hence proved.

### Page 14.57 Ex. 14.4

Q18.

**Answer :**

In *BM* and *CN* are perpendiculars on any line passing through *A*.

Also.

We need to prove that

From point *L* let us draw

It is given that , and

Therefore,

Since, L is the mid points of BC,

Therefore intercepts made by these parallel lines on MN will also be equal

Thus,

Now in ,

And . Thus, perpendicular bisects the opposite sides.

Therefore, is isosceles.

Hence

Hence proved.

Q19.

**Answer :**

Figure can be drawn as:

Let *ABCD* be a quadrilateral such that *P,Q ,R* and *S* are the mid-points of side *AB,BC,CD* and *DA* respectively.

In , *P* and *Q* are the mid-points of *AB* and *BC* respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, *PQRS *is a parallelogram.

Since, diagonals of a parallelogram bisect each other.

Therefore, *PR* and *QS* bisect each other.

Hence proved.

Q20.

**Answer :**

(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is isosceles.

Explanation:

Figure can be drawn as: A

, an isosceles triangle is given.

*F* and *E* are the mid-points of *AB* and *AC* respectively.

Therefore,

…… (I)

Similarly,

…… (II)

And

…… (III)

Now, is an isosceles triangle.

From equation (II) and (III), we get:

Therefore, in two sides are equal.

Therefore, it is an isosceles triangle.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is right triangle.

Explanation:

Figure can be drawn as: A

right angle at *B* is given.

*F* and *E* are the mid-points of *AB* and *AC* respectively.

Therefore,

…… (I)

Similarly,

…… (II)

And

…… (III)

Now, *DE || AB* and transversal *CB* and *CA* intersect them at *D* and *E* respectively.

Therefore,

and

Similarly,

Therefore,

and

Similarly,

Therefore,

Now *AC* is a straight line.

Now, by angle sum property of ,we get:

Therefore,

But,

Then we have:

(iii) The figure formed by joining the mid-points of the consecutive sides of a quadrilateral is **parallelogram**.

Explanation:

Figure can be drawn as:

Let *ABCD* be a quadrilateral such that *P, Q, R* and *S* are the mid-points of side *AB, BC, CD* and *DA* respectively.

In , *P* and *Q* are the mid-points of *AB* and *BC* respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, *PQRS *is a parallelogram.

### Page 14.62 Formative Assessment_VSA

Q1.

**Answer :**

In Parallelogram *ABCD*, and are adjacent angles.

Thus, .

Then, we have and as consecutive interior angles which must be supplementary.

Hence, the sum of and is .

Q2.

**Answer :**

In Parallelogram *ABCD* , and are Adjacent angles.

We know that in a parallelogram, adjacent angles are supplementary.

$\mathrm{Now},?\mathrm{A}+?\mathrm{D}=180\xb0\phantom{\rule{0ex}{0ex}}??\mathrm{A}+115\xb0=180\xb0\phantom{\rule{0ex}{0ex}}??\mathrm{A}=180\xb0\u2013115\xb0\phantom{\rule{0ex}{0ex}}??\mathrm{A}=65\xb0\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{measure}\mathrm{of}?\mathrm{A}\mathrm{is}65\xb0.$Q3.

**Answer :**

PQRS is a square given as:

Since the diagonals of a square intersect at right angle.

Therefore, the measure of is .

Q4.

**Answer :**

The quadrilateral can be drawn as follows:

We have *AO* and *BO* as the bisectors of angles and respectively.

In ,We have,

…… (1)

By angle sum property of a quadrilateral, we have:

Putting in equation (1):

……(2)

It is given that in equation (2), we get:

Hence, the sum of and is .

Q5.

**Answer :**

The rectangle *ABCD* is given as:

We have,

(Linear pair)

Since, diagonals of a rectangle are equal and they bisect each other. Therefore, in , we have

(Angles opposite to equal sides are equal.)

Therefore,

Now,in , we have

Since, each angle of a rectangle is a right angle.

Therefore,

Thus,

Hence, the measure of is.

Q6.

**Answer :**

The square *PQRS* is given as:

Since *PQRS* is a square.

Therefore,

and

Now, in , we have

That is, (Angles opposite to equal sides are equal)

By angle sum property of a triangle.

()

Hence, the measure of is.

Q7.

**Answer :**

Figure is given as :

Suppose the diagonals AC and BD intersect at O.

Since, diagonals of a rectangle are equal and they bisect each other.

Therefore, in , we have

Angles opposite to equal sides are equal.

Therefore,

But,

Now,

Hence, the measure of is .

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