Rd Sharma 2021 for Class 9 Maths Chapter 13 - Quadrilaterals

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 .

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.

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 .

Answer:

Let the shorter side of the parallelogram be cm.

The longer side is given ascm.

Perimeter of the parallelogram is given as 22 cm

Therefore,

Hence, the measure of the shorter side is cm.

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 .

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 .

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 .

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

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,

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.

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 .

Answer:

We have ,.

So, let ,

,

And

By angle sum property of a quadrilateral, we get:

Also,

And

Similarly,

Hence, the four angles are ,,and .

Answer:

We have, .

So, let ,

,

and

By angle sum property of a quadrilateral, we get:

Also,

And

Similarly,

Hence, the four angles are , , and  .

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.

Answer:

The parallelogram can be drawn as:

We have ,thus and are consecutive interior angles.

These must be supplementary.

Therefore,

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 .

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 .

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 .

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.

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,

AC bisects PQ.

Hence proved.

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.

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.

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:

From (i) equation, we get:

Hence proved.

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 .

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.

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.

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.

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.

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.

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 .

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.

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,

Similarly,

And

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

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.

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.

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.

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: 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.

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.

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.

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.

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.

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:

…… (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.

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.

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 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.

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.

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:

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

Therefore, R is a mid-point of BC.

Hence proved.

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.

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.

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.

Answer:

Let us analyze each case one by one.

We have a quadrilateral named PQRS, with diagonals PR and QS intersecting at O.

(a) ,,

By angle sum property of a quadrilateral, we get:

Clearly,

And

Thus we have PQRS a quadrilateral with opposite angles are equal.

Therefore,

PQRS is a parallelogram.

(b) ,,

By angle sum property of a quadrilateral, we get:

Clearly,

And

Thus we have PQRS a quadrilateral with opposite angles are not equal.

Therefore,

PQRS is not a parallelogram.

(c) ,,,

Clearly,

And

Thus we have PQRS a quadrilateral with opposite sides are not equal

Therefore,

PQRS is not a parallelogram.

(c) ,,,

We know that the diagonals of a parallelogram bisect each other.

But, here we have

And

Therefore,

PQRS is not a parallelogram. 

Hence, the correct choice is (a).

Answer:

We can look at a quadrilateral as:

The opposite sides of the above quadrilateral AB and CD have no point in common.

Hence the correct choice is (a).

Answer:

We can look at a quadrilateral as:

The consecutive sides of the above quadrilateral AB and BC have one point in common.

Hence the correct choice is (b).

Answer:

Let us consider the rhombus ABCD as:

We have the following properties of a rhombus:

All four sides are equal.

Diagonals bisect each other at right angles.

Hence the correct choice is (d).

Answer:

From the given choices, only in a square the diagonals bisect the opposite angles.

Let us prove it.

Take the following square ABCD with diagonal AD.

In and:

(Opposite sides of a square are equal.)

(Common)

(Opposite sides of a square are equal.)

Thus,

(By SSS Congruence Rule)

By Corresponding parts of congruent triangles property we have:

Therefore, in a square the diagonals bisect the opposite angles.

Hence the correct choice is (d).

Answer:

Two diagonals are equal only in a rectangle.

This can be proved as follows:

The rectangle is given as ABCD, with the two diagonals as AD and BC.

In and:

(Opposite sides of a rectangle are equal.)

(Common)

(Each angle in a rectangle is a right angle)

Thus,

(By SAS Congruence Rule)

By Corresponding parts of congruent triangles property we have:

Therefore, in a rectangle the two diagonals are equal.

Hence the correct choice is (c).

Answer:

We get a rhombus by joining the mid-points of the sides of a rectangle.

It is given a rectangle ABCD in which P,Q,R and S are the mid-points AB,BC,CD and DA respectively.

PQ,QR,RS and SP are joined.

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

Therefore,

and ……(i)

Similarly, In , R and S are the mid-points CD and AD respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram. …… (iii)

Now is a rectangle.

Therefore,

…… (iv)

In and , we have:

(P is the mid point of AB)

(Each is a right angle)

(From equation (iv))

So, by SAS congruence criteria, we get:

By Corresponding parts of congruent triangles property we have:

…… (v)

From (iii) and (v) we obtain that is a parallelogram such that and

Thus, the two adjacent sides are equal.

Thus, is a rhombus.

Hence the correct choice is (c).

Answer:

Let the figure be as follows:

ABCD is a parallelogram.

We have to find

AN and AD is the bisectors of and .

Therefore,

…… (i)

And

…… (ii)

We know that .

Therefore, the sum of consecutive interior angles must be supplementary.

From (i) and (ii), we get

…… (iii)

By angle sum property of a triangle:

Hence the correct choice is (d).

Answer:

We have ABCD, a parallelogram given below:

Therefore, we have

Now, and transversal AB intersects them at A and B respectively. Therefore,

Sum of consecutive interior angle is supplementary. That is;

We have AR and BR as bisectors of and respectively.

…… (i)

Now, in , by angle sum property of a triangle, we get:

From equation (i), we get:

Similarly, we can prove that.

Now, and transversal ADintersects them at A and D respectively. Therefore,

Sum of consecutive interior angle is supplementary. That is;

We have AR and DP as bisectors of and respectively.

…… (ii)

Now, in , by angle sum property of a triangle, we get:

From equation (i), we get:

We know that and are vertically opposite angles, thus,

Similarly, we can prove that.

Therefore, PQRS is a rectangle.

Hence, the correct choice is (c).

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, PQRSis a parallelogram.

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a

parallelogram.

Hence the correct choice is (a).

Answer:

We get a rhombus by joining the mid-points of the sides of a rectangle.

It is given that rectangle ABCD in which P,Q,R and S are the mid-points AB,BC,CD and DA respectively.

PQ,QR,RS and SP are joined.

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

Therefore,

and ……(i)

Similarly, In , R and S are the mid-points CD and AD respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram. …… (iii)

Now is a rectangle.

Therefore,

…… (iv)

In and , we have:

(P is the mid point of AB)

(Each is a right angle)

(From equation (iv))

So, by SAS congruence criteria, we get:

By Corresponding parts of congruent triangles property we have:

…… (v)

From (iii) and (v) we obtain that is a parallelogram such that .

Thus, the two adjacent sides are equal.

Thus, is a rhombus .

Hence the correct choice is (b).

Answer:

Figure is given as :

A rhombus ABCD is given in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively.

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

Therefore,

and ……(i)

Similarly, In , R and S are the mid-points CD and AD respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram. …… (iii)

Now, we shall find one of the angles of a parallelogram.

Since ABCD is a rhombus

Therefore,

(Sides of rhombus are equal)

(P and Q are the mid-points AB and BC respectively)

In , we have

(Angle opposite to equal sides are equal)

Therefore, ABCD is a rhombus

…… (iii)

Also,

…… (iv)

Now, in and , we have

[From (iii)]

[From (iv)]

And ( is a parallelogram)

So by SSS criteria of congruence, we have

By Corresponding parts of congruent triangles property we have:

…… (v)

Now,

And

Therefore,

From (ii), we get

From (v), we get

Therefore, …… (vi)

Now, transversal PQ cuts parallel lines SP and RQ at P and Q respectively.

[Using (vi)]

Thus, is a parallelogram such that .

Therefore, is a rectangle.

Hence the correct choice is (b).

Answer:

We get a square by joining the mid-points of the sides of a square.

It is given a square ABCD in which P, Q, R and S are the mid-points AB, BC, CD and DA respectively.

PQ, QR, RS and SP are joined.

Join AC and BD.

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

Therefore,

and ……(i)

Similarly, In ΔADC, R and S are the mid-points CD and AD respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram. …… (iii)

Now is a square.

Therefore,

…… (iv)

Similarly,

…… (v)

In and , we have:

(From equation (iv))

(Each is a right angle)

(From equation (v))

So, by SAS congruence criteria, we get:

By Corresponding parts of congruent triangles property we have:

…… (vi)

From (iii) and (vi) we obtain that is a parallelogram such that .

But, is a parallelogram.

So,

and …… (vii)

Now, (From equation (i))

Therefore,

…… (viii)

Since P and S are the mid-points AB and AD respectively

Therefore,

…… (ix)

Thus, in quadrilateral PMON, we have:

and (From equation (viii) and (ix))

Therefore, quadrilateral PMON is a parallelogram.

Also,

(Because )

(Because diagonals of a square are perpendicular)

Therefore, is a quadrilateral such that ,

and .Also, .

Hence, is a square.

Hence the correct choice is (b).

Answer:

It is given a parallelogram ABCD in which P,Q,R and S are the mid-points AB, BC, CD and DA respectively.

PQ, QR, RS and SP are joined.

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

Therefore,

and ……(i)

Similarly, In , R and S are the mid-points CD and AD respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram.

Hence the correct choice is (b).

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

Thus, the largest angle of the parallelogram are is .

Hence the correct choice is (c).

Answer:

Parallelogram can be drawn as :

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

Therefore,

By angle sum property of a triangle:

Thus, measures.

Hence the correct choice is (c).

Answer:

Parallelogram ABCD is given with E and F are the centroids of and.

We have to find EF.

We know that the diagonals of a parallelogram of bisect each other.

Thus, AC and BD bisect each other at point O.

Also, median is the line segment joining the vertex to the mid-point of the opposite side of the triangle. Therefore, the centroids E and F lie on AC.

Now, the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex.

Then, in ΔABD, we get:

Or,

and …… (I)

Similarly, in ,we get:

and …… (II)

Also,

From (I) and (II), we get:

And …… (III)

Also, from (II) and (III), we get :

…… (IV)

Now, from (I),

From (IV), we get:

From(III):

Hence the correct choice is (a).

Answer:

Figure is given as follows:

ABCD is a parallelogram.

It is given that :

BM bisects

Therefore,

But,

(Alternate interior opposite angles as)

Therefore,

In,

Sides opposite to equal angles are equal.

Thus,

Also,

(Opposite sides of a parallelogram are equal)

Thus,

ABCD is a parallelogram with

Therefore,

ABCD is a rhombus.

And we know that diagonals of the rhombus bisect each other at right angle.

Thus,

Hence the correct choice is (c).

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

Therefore, the smallest angle of the parallelogram as

Hence, the correct choice is (c).

Answer:

We have,

,

,

and

By angle sum property of a quadrilateral, we get:

Smallest angle:

Also, Largest angle:

Thus, the sum of the smallest and the largest becomes:

Hence, the correct choice is (c).

Answer:

Let ABCD be rhombus with diagonals AC and BD 18cm and 24cm respectively.

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

Therefore,

Similarly,

Also, is a right angled triangle.

By Pythagoras theorem, we get:

Hence the correct choice is (b).

Answer:

ABCD is a parallelogram in which AC bisects.

It is given that

Therefore,

Since,

Therefore,

Hence, the correct choice is (b).

Answer:

Rhombus ABCD is given as follows:

It is given that.

Therefore, (Because O lies on AC)

We know that the diagonals of a rhombus intersect at right angle.

Therefore,

By angle sum property of a triangle, we get:

Since, O lies on BD

Also ,

Therefore,

Hence, the correct choice is (c).

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 correct choice is (c).

Answer:

ABCD is a parallelogram with diagonals AC and BD intersect at O.

It is given that and.

We need to find

Now,

(Linear pair)

Since, O lies on BD.

Therefore,

By angle sum property of a triangle, we get:

Since, O lies on AC.

Therefore,

Also,

Therefore,

Hence the correct choice is (a).

Answer:

The given trapezium ABCD can be drawn as follows:

Here, .

M and N are the mid-points of AD and BC respectively.

We have,.

We need to find CD.

Join MN to intersect AC at G.

We have and a line MN formed by joining the mid-points of sides AD and BC.

Thus, we can say that

In M is the mid-point of AD and

Therefore, G is the mid point of AC

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

…… (i)

In , N is the mid point of BC and

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

…… (ii)

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

…… (iii)

On substituting and in (iii),we get:

Hence the correct choice is (d).

Answer:

We know that the diagonals of a parallelogram bisect each other.

Thus, the given quadrilateral ABCD is a parallelogram.

and are consecutive interior angles, which must be supplementary.

Therefore, we have

Hence the correct choice is (d).

Answer:

Parallelogram ABCD is given such that

We haveand

We need to find the measure of side CD.

Since and AP as transversal

But it is given that

Therefore, we get:

Also, sides opposite to equal angles are equal.

Then,

…… (I)

Also,

Substituting in (I), We get:

It is given that , this means opposite side ,as ABCD is a parallelogram. Therefore,

Or,

Hence the correct choice is (a).

Answer:

is given with E as the mid-point of median AD.

Also, BE produced meets AC at F.

We have , then we need to find AF.

Through D draw .

In , E is the mid-point of AD and .

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

…… (i)

In , D is the mid-point of BC and .

…… (ii)

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

, …… (iii)

Now,

From (iii) equation, we get:

…… (iv)

We have .Putting this in equation (iv), we get:

Hence the correct choice is (b).

Answer:

Parallelogram ABCD is given with E as the mid-point of BC.

DE and AB when produced meet at F

We need to find AF.

Since ABCD is a parallelogram, then

Therefore,

Then, the alternate interior angles should be equal.

Thus, …… (I)

In and :

(From(I))

(E is the mid-point of BC)

(Vertically opposite angles)

(by ASA Congruence property)

We know that the corresponding angles of congruent triangles should be equal.

Therefore,

But,

(Opposite sides of a parallelogram are equal)

Therefore,

…… (II)

Now,

From (II),we get:

Hence the correct choice is (b).

Answer:

ABCD is a quadrilateral, with.

By angle sum property of a quadrilateral we get:

But,we have

Then,

The two equations so formed cannot give us the value for with a given value of .

Hence the correct choice is (d).

Answer:

Figure can be drawn as follows:

The diagonals AC and BD of rectangle ABCD intersect at P.

Also, it is given that

We need to find

It is given that

Therefore, (Because P lies on BD)

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

Therefore,

Thus, (Angles opposite to equal sides are equal)

[(Given)]

By angle sum property of a triangle:

But and are vertically opposite angles.

Therefore, we get:

[(Already proved)]

Hence the correct choice is (c).

Answer:

Let the measure of the fourth angle be x.

We have,
Sum of all the angles of a quadrilateral = 360°
⇒ 75° + 90° + 75° + x = 360°
⇒ 240° + x = 360°
​⇒ x = 360° − 240°
​⇒ x = 120°


Hence, the measure of the fourth angle is 120°.

Answer:

Let ∠OAB be x.

Since, AB || DC
Therefore, ∠DBA = ∠BDC = 40°

In ∆AOB,
∠OAB  + ∠ABD  + ∠AOB = 180°
⇒  x + 40° + 90° = 180°
⇒ x + 130° = 180°
⇒ x = 180° − 130°
⇒ x = 50°

Hence, ∠OAB is 50°.

Answer:

Let the angle A, B, C and D of a quadrilateral be x, 2x, 4x and 5x, respectively.

We have,
Sum of all the angles of a quadrilateral = 360°
⇒ x + 2x + 4x + 5x = 360°
⇒ 12x = 360°
​⇒ x = 30°

Thus, ∠A = 30°
∠B = 60°
∠C = 120°
∠D = 150°

Now, the bisectors of ∠C and ∠D meet of O
Thus, in ∆COD,
$\frac{1}{2}$∠DCO  + ∠COD  + $\frac{1}{2}$∠ODC = 180°
⇒ $\frac{1}{2}$ × 120° + ∠COD  + $\frac{1}{2}$ × 150° = 180°
⇒ 60° + ∠COD  + 75° = 180°
⇒ 135° + ∠COD = 180°
⇒ ∠COD = 180° − 135°
⇒ ∠COD = 45°

Hence, ∠COD = 45°.

Answer:

In ∆AOB,
∠OAB  + ∠ABO  + ∠AOB = 180°
⇒ 25° + 25° + ∠AOB = 180°
⇒ ∠AOB + 50° = 180°
⇒ ∠AOB = 180° − 50°
⇒ ∠AOB = 130° which is an obtuse angle

Now,  ∠AOB + ∠AOD = 180°    (angles on a straight line)
⇒ 130° + ∠AOD = 180°
⇒ ∠AOD = 180° − 130°
⇒ ∠AOD = 50° which is an acute angle

Hence, the acute angle between the diagonals is 50°.

Answer:

Let ∠ADB be x.

As we know, angles formed by the diagonals of a rhombus is 90°



In ∆BOC,
∠OBC  + ∠BCO  + ∠COB = 180°
⇒ ∠OBC  + 40° + 90° = 180°
⇒ ∠OBC + 130° = 180°
⇒ ∠OBC = 180° − 130°
⇒ ∠OBC = 50°

Since, AD || BC
Thus, ∠ADB = ∠DBC     (alternate interior angles)
⇒ x = 50°

Hence, ∠ADB = 50°.

Answer:

Let the angle A, B, C and D of a quadrilateral be 3x, 7x, 6x and 4x, respectively.

We have,
Sum of all the angles of a quadrilateral = 360°
⇒ 3x + 7x + 6x + 4x = 360°
⇒ 20x = 360°
​⇒ x = 18°

Thus, ∠A = 54°
∠B = 126°
∠C = 108°
∠D = 72°



Now, 
∠BCE = 180° − 108° = 72° = ∠ADE
⇒ AD || BC

Hence, ABCD is a trapezium.

Answer:

The diagonals of a parallelogram ABCD are equal
⇒ ABCD is a rectangle
⇒ ∠ABC = 90° 

Hence, ∠ABC = 90°.

Answer:

Given:
ABCD is a parallelogram
Diagonals AC and BD intersect each other at O.
OA = 3 cm and OD = 2 cm

As we know, diagonals of a parallelogram bisects each other.
Thus, AC = 2OA = 2× 3 = 6 cm
and BD = 2OD = 2× 2 = 4 cm

Hence, AC = 6 cm and BD = 4 cm.
 

Answer:

Given: 
ABCD is a trapezium 
AB || DC
∠A = ∠B = 45°

Since, AB || DC
Thus, AD and BC are the transversals.
Therefore, sum of the interior angles must be equal to 180°.

∠A + ∠D = 180°
⇒ 45° + ∠D = 180°
⇒ ∠D = 180° − 45°
⇒ ∠D = 135°

∠B + ∠C = 180°
⇒ 45° + ∠C = 180°
⇒ ∠C = 180° − 45°
⇒ ∠C = 135°

Hence, ∠C = 135° and ∠D = 135°.
 

Answer:

Given:
ABCD is a rhombus
Altitude from vertex D to side AB bisects AB.

Let DE biescts AB.
⇒ AE = EB                                     ...(1)

In ∆ADE and  ∆BDE

DE = DE (Common side)
∠AED = ∠BED (Right angle)
AE = EB (from (1))

By SAS property,  ∆ADE ≅ ∆BDE

Thus, AD = BD (By C.P.C.T)          ...(2)

Since, sides of a rhombus are equal
Thus, AD = AB                                ...(3)

From (2) and (3)
AD = BD = AB 
Therefore, ∆ADB is an equilateral triangle.

Hence, ∠A = 60°
∠A = ∠C = 60° (opposite angles of rhombus are equal)

Sum of adjacent angles of a rhombus is 180°
⇒ ∠A + ∠D = 180° and ∠C + ∠B = 180° 
⇒ 60° + ∠D = 180° and 60° + ∠B = 180° 
⇒ ∠D = 180° − 60° and ∠B = 180° − 60°
⇒ ∠D = 120° and ∠B = 120°

Hence, the measures of angles of the rhombus are 60°, 120°, 60° and 120°.
 

Answer:

Given: Diagonals of a quadrilateral ABCD bisect each other

Since, the diagonals of a quadrilateral ABCD bisect each other
Therefore, It is a parallelogram

Sum of interior angles of a parallelogram is 180°
⇒ ∠A + ∠B = 180°
⇒ 35° + ∠B = 180° 
⇒ ∠B = 180° − 35°
⇒ ∠B = 145°

Hence, ∠B =  145°.

Answer:

Given: 
In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm
D and E are respectively the mid-points of AB and BC


Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

Therefore,
$$\begin{gathered} DE=\frac{1}{2}\left(AC\right) \\ =\frac{1}{2}\times 7 \\ =3.5 \mathrm{cm} \end{gathered}$$


Hence, DE = 3.5 cm.

Answer:

Given: 
Opposite angles of a quadrilateral ABCD are equal
AB = 4 cm

If Opposite angles of a quadrilateral ABCD are equal, then ABCD is a parallelogram.
Thus, CD = AB = 4 cm

Hence, CD = 4 cm.

Answer:

Given: 
The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O
∠DAC = 32°
∠AOB = 70°




Let ∠DBC be x.

Since, AB || DC
Therefore, ∠DAC = ∠ACB = 32°   (alternate angles)


Also, ∠AOB  + ∠BOC = 180° (angles on a straight line)
⇒ 70° + ∠BOC = 180°
⇒ ∠BOC = 180° − 70°
⇒ ∠BOC = 110°


In ∆COB,
∠OCB  + ∠CBO  + ∠BOC = 180°
⇒  32° + x + 110° = 180°
⇒ x + 142° = 180°
⇒ x = 180° − 142°
⇒ x = 38°

Hence, ∠DBC = 38°.

Answer:

Given: 
PQRS is a quadrilateral
The quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, is a rectangle.
Let the rectangle be ABCD.



A is the mid-point of PQ, B is the mid-point of QR, C is the mid-point of RS and D is the mid-point of PS.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆PQR,
AB || PR

and In ∆QRS,
BC || QS

Since, AB$\perp$BC
Therefore, PR$\perp$QS


Hence, the quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, taken  in order, is a rectangle, if diagonals of PQRS are perpendicular.

Answer:

Given: 
PQRS is a quadrilateral
The quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, is a rhombus.
Let the rhombus be ABCD.



A is the mid-point of PQ, B is the mid-point of QR, C is the mid-point of RS and D is the mid-point of PS.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆PQR,
AB = $\frac{1}{2}$PR

and In ∆QRS,
BC = $\frac{1}{2}$QS

Since, AB = BC (sides of a rhombus are equal)
Therefore, $\frac{1}{2}$PR = $\frac{1}{2}$QS
⇒ PR = QS
​

Hence, the quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, taken  in order, is rhombus, if diagonals of PQRS are equal.

Answer:

Given: 
ABCD is a quadrilateral
The quadrilateral formed by joining the mid points of the sides of a quadrilateral ABCD, is a square.
Let the square be PQRS.



P is the mid-point of AB, Q is the mid-point of BC, R is the mid-point of CD and S is the mid-point of AD.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆ABC,
PQ = $\frac{1}{2}$AC

and In ∆BCD,
QR = $\frac{1}{2}$BD

Since, PQ = QR (sides of a square are equal)
Therefore, $\frac{1}{2}$AC = $\frac{1}{2}$BD
⇒ AC = BC

Hence, the diagonals are equal.
​
In ∆ABC,
PQ || AC

and In ∆BCD,
QR || BD

Since, PQ$\perp$QR
Therefore, AC$\perp$BD

Hence, the diagonals are perpendicular.

Hence, the figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, diagonals of ABCD are perpendicular and equal.

Answer:

Given: 
ABCD is a quadrilateral
bisectors of ∠A and ∠B intersect each other at P
bisectors of ∠B and ∠C intersect each other at Q
bisectors of ∠C and ∠D intersect each other at R
bisectors of ∠D and ∠A intersect each other at S

In ∆DAS,
∠ASD + ∠SDA + ∠DAS = 180° (angle sum property)
⇒ ∠ASD + $\frac{1}{2}$∠D + $\frac{1}{2}$∠A = 180°
⇒ ∠ASD = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A

Also, ∠PSR = ∠ASD = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A
⇒ ∠PSR = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A      ...(1)

Similarly,
In ∆BQC,
∠BQC + ∠QCB + ∠CBQ = 180° (angle sum property)
⇒ ∠BQC + $\frac{1}{2}$∠C + $\frac{1}{2}$∠B = 180°
⇒ ∠BQC = 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B

Also, ∠PQR = ∠BQC = 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B
⇒ ∠PQR = 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B      ...(2)

Adding (1) and (2), we get
∠PSR + ∠PQR = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A + 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B
                          = 360° − $\frac{1}{2}$(∠D + ∠A + ∠C + ∠B)
                          = 360° − $\frac{1}{2}$(360°)
                          = 360° − 180°​
                          = 180°

Since, sum of angles of a quadrilateral is 360°
Therefore, ∠PSR + ∠PQR + ∠QRS + ∠QPS = 360°
⇒ ∠QRS + ∠QPS = 180°

Hence, the sum of the opposite angles of the quadrilateral PQRS is 180°.

Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

Answer:

Given: 
APB and CQD are two parallel lines

Let the bisectors of the angle APQ and CQP intersects at the point R and the bisectors of the angle BPQ and PQD intersects at the point S.

Join PR, RQ, QS and SP as shown in the figure.



APB || CQD
⇒ ∠APQ = ∠PQD (alternate angles)
⇒ 2∠RPQ = 2∠PQS
⇒ ∠RPQ = ∠PQS
⇒  RP || SQ      ...(1)

Similarly,
APB || CQD
⇒ ∠BPQ = ∠PQC (alternate angles)
⇒ 2∠SPQ = 2∠PQR
⇒ ∠SPQ = ∠PQR
​⇒  RQ || SP      ...(2)

From (1) and (2),
PSQR is a parallelogram,

Also, ∠CQP + ∠PQD = 180° (angles on a straight line)
⇒ 2∠RQP + 2∠PQS = 180°
⇒ ∠RQP + ∠PQS = 90°
⇒ ∠RQS = 90°

Therefore, PSQR is a rectangle.

Hence, if APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form a rectangle.

Answer:

Given: 
D and E are the mid-points of the sides AB and AC of ∆ABC.
O is any point on side BC
P and Q are the mid-points of OB and OC respectively.



Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆ABC,
DE || BC || PQ        ...(1) 
DE = $\frac{1}{2}$BC
⇒ DE = $\frac{1}{2}$(BP + PO + OQ + QC)
⇒ DE = $\frac{1}{2}$(2OP + 2OQ)
⇒ DE = OP + OQ
⇒ DE = PQ             ...(2)

In ∆AOB,
DP || AO                  ...(3) 
DP = $\frac{1}{2}$AO             ...(4)


In ∆AOC,
EQ || AO                  ...(5) 
EQ = $\frac{1}{2}$AO             ...(6)

From (3) and (5),
DP || EQ

From (4) and (6),
DP = EQ

Thus, DE || PQ and DP || EQ
DP = EQ and DE = PQ

Hence, DEQP is a parallelogram.

Answer:

All the sides of a square or a rhombus are equal.

Hence, when all the sides of a quadrilateral are equal, then it is either a square or a rhombus.

Answer:

Given: 
ABCD is a parallelogram
Bisectors of ∠A and ∠B intersect each other at P


∠A + ∠B = 180°      (interior angles)
⇒ $\frac{1}{2}$∠A + $\frac{1}{2}$∠B = $\frac{1}{2}$(180°)
⇒ ∠PAB + ∠PBA = 90°      ....(1)

Now, in ∆APB
∠PAB + ∠PBA + ∠APB = 180° (angle sum property)
⇒ 90° + ∠APB = 180°         (From (1))
⇒ ∠APB = 180° − 90°
⇒ ∠APB = 90°


Hence, the angle made by these bisectors at point P is 90°.

Answer:

Given: 
PQRS is a rhombus
∠OPQ = 35°


PQRS is a rhombus
PQ || RS
∠OPQ = ∠ORS       (alternate angles)
⇒ ∠ORS = 35°      ....(1)

We know, diagonals of a rhombus intersect at right angle
∴ ∠POQ = 90°      ....(2)

Now, in ∆OPQ
∠OPQ + ∠POQ + ∠OQP = 180° (angle sum property)
⇒ 35° + 90° + ∠OQP = 180°
⇒ 125° + ∠OQP = 180°
⇒ ∠OQP = 180° − 125°
⇒ ∠OQP = 55°      ....(3)

Adding (1) and (3), we get
∠ORS + ∠OQP = 35° + 55° = 90°


Hence, ∠ORS + ∠OQP =  90°.

Answer:

Given: 
ABC is an isosceles triangle
AB = AC
AEDC is a parallelogram
∠CDF = 70°
∠BFE = 100°


AEDC is a parallelogram
∠ACD + ∠CDE = 180°     (interior angles)
⇒ ∠ACD + 70° = 180°
⇒ ∠ACD = 180° − 70°
⇒ ∠ACD = 110°      ....(1)


Now, ∠ACD + ∠ACB = 180° (angles on a straight line)
⇒ 110° + ∠ACB = 180°
⇒ ∠ACB = 180° − 110°
⇒ ∠ACB = 70°       ....(2)

Also, ∠BFE + ∠BFD = 180° (angles on a straight line)
⇒ 100° + ∠BFD = 180°
⇒ ∠BFD = 180° − 100°
⇒ ∠BFD = 80°       ....(3)


Now, in ∆BFD
∠FBD + ∠BDF + ∠BFD = 180° (angle sum property)
⇒ ∠FBD + 70° + 80° = 180°
⇒ ∠FBD + 150° = 180°
⇒ ∠FBD = 180° − 150°
⇒ ∠FBD = 30°      ....(4)

Since, ABC is an isosceles triangle with AB = AC
Thus, ∠ABC = ∠ACB = 70° (From (2))

∠ABC = ∠ABF + ∠FBD
⇒  70° = ∠ABF + 30°
⇒  ∠ABF = 70° − 30°
⇒  ∠ABF = 40°


Hence, ∠FBA = 40°.

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 .

Answer:

In Parallelogram ABCD , and are Adjacent angles.

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

$$\begin{gathered} \mathrm{Now}, \angle \mathrm{A}+\angle \mathrm{D}=180° \\ \Rightarrow \angle \mathrm{A}+115°=180° \\ \Rightarrow \angle \mathrm{A}=180°-115° \\ \Rightarrow \angle \mathrm{A}=65° \\ \mathrm{So}, \mathrm{measure} \mathrm{of} \angle \mathrm{A} \mathrm{is} 65°. \end{gathered}$$

Answer:

PQRS is a square given as:

Since the diagonals of a square intersect at right angle.

Therefore, the measure of is .

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.

Answer:

The figure is given as follows:

ABCD is a rhombus.

Therefore,

ABCD is a parallelogram.

Thus,

[(Given)]

[]

Now in ,we have:

Hence the measure of is .

Answer:

Let the shorter side of the parallelogram be cm.

The longer side is given ascm.

Perimeter of the parallelogram is given as 22 cm

Therefore,

Hence, the measure of the shorter side is cm.

Answer:

We have, .

So, let ,

,

and

By angle sum property of a quadrilateral, we get:

Smallest angle is :

Hence, the smallest angle measures.

Answer:

In parallelogram ABCD, and are opposite angles.

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

Therefore,

We have and

Therefore,

Therefore,

Similarly,

Also,

Therefore,

By angle sum property of a quadrilateral, we have:

Hence the required values for x and y are and respectively.

Answer:

Let ABCD be a parallelogram, with and .

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

Therefore,

Thus, the given angles become

Similarly,

Also, adjacent angles in a parallelogram form the consecutive interior angles of parallel lines, which must be supplementary.

Therefore,

Similarly,

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

Answer:

Parallelogram ABCD is given as follows:

We have AX bisects bisecting BC at X.

That is,

We need to find

Since, AX is the bisector

That is,

…… (i)

Also, ABCD is a parallelogram

Therefore, and AB intersects them

…… (ii)

In by angle sum property of a triangle:

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

…… (iii)

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

Sides opposite to equal angles are equal. Therefore,

As X is the mid point of BC. Therefore,

Also, ABCD is a parallelogram, then,

Thus,

Hence the ratio of is .

Answer:

Trapezium is given as follows:

We know that PQRS is a trapezium with

Therefore,

Hence, the required value for x is .

Answer:

The figure is given as follows:

We know that ABCD is a trapezium with

Therefore,

It is given that and .

Similarly,

Hence, the required values for x and y is and respectively.

Answer:

ABCD and AEFG are two parallelograms as shown below:

Since ABCD is a parallelogram, with

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

Therefore,

Similarly, AEFG is a parallelogram, with

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

Therefore,

Hence, the required measure for is .

Answer:

(i) If one pair of opposite sides are equal and parallel, then the figure is parallelogram.

Reason:

In and,

(Given)

(Common)

(Because , Alternate interior angles are equal)

So, by SAS Congruence rule, we have

Also,

(Corresponding parts of congruent triangles are equal)

But, these are alternate interior angles, which are equal.

Thus, and

Hence, quadrilateral ABCD is parallelogram

(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is trapezium.

(iii) A line drawn from the mid-point of one side of a triangle parallel to another side intersects the third side at its mid-point.

Reason:

This is a theorem.

(iv) If one angle of a parallelogram is a right angle, then it is necessarily a rectangle.

Reason:

Let ABCD be the given parallelogram.

We have,

In a parallelogram, opposite angles are equal.

Therefore,

Similarly,

Also,

Thus, a parallelogram with all the angles being right angle and opposite sides being equal is a rectangle.

(v) Consecutive angles of a parallelogram are supplementary.

Reason:

Let ABCD be the given parallelogram.

Thus, .

Therefore,

Consecutive angles and are supplementary.

(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a parallelogram.

Reason:

ABCD is a quadrilateral in which and .

We need to show that ABCD is a parallelogram.

In and , we have

(Common)

(Given)

(Given)

So, by SSS criterion of congruence, we have

By corresponding parts of congruent triangles property.

…… (i)

And

Now lines AC intersects AB and DC at A and C,such that

(From (i))

That is, alternate interior angles are equal.

Therefore, .

Similarly, .

Therefore, ABCD is a parallelogram.

(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a parallelogram.

Reason:

ABCD is a quadrilateral in which and .

We need to show that ABCD is a parallelogram.

In quadrilateral ABCD, we have

Therefore,

…… (i)

Since sum of angles of a quadrilateral is

From equation (i), we get:

Similarly,

Now, line AB intersects AD and BC at A and B respectively

Such that

That is, sum of consecutive interior angles is supplementary.

Therefore, .

Similarly, we get .

Therefore, ABCD is a parallelogram.

(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a rhombus.

We have ABCD, a parallelogram with .

Since ABCD is a parallelogram.

Thus,

And

But,

Therefore,all four sides of the parallelogram are equal, then it is a rhombus.

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 .

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.

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 .

Answer:

The quadrilateral can be drawn as follows:

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

In ,We have,

…… (I)

By angle sum property of a quadrilateral, we have:

Putting in equation (I):

…… (II)

On comparing equation (II) with

We get .

Hence, the value for k is .

Answer:

Rhombus PQRS is given.

Diagonal PR is produced to T.

Also, .

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

Therefore,

Now,

In , by angle sum property of a triangle, we get:

Or, (Because O lies on SQ)

We have, .Thus the alternate interior opposite angles must be equal.

Therefore,

In ,we have

Since opposite sides of a rhombus are equal.

Therefore,

Also,

Angles opposite to equal sides are equal.

Thus,

But

Thus,

Hence the required values for x,y and z are , and respectively.

Answer:

ABCD is a rectangle

With diagonal AC produced to point E.

We have

(Linear pair)

We know that the diagonals of a parallelogram bisect each other.

Thus

Also, angles opposite to equal sides are equal.

Therefore,

By angle sum property of a traingle

Also, and are vertically opposite angles.

Therefore,

Hence, the required measure for is .