Mathematics Part ii Solutions Solutions for Class 8 Math Chapter 1 - Construction Of Quadrilaterals

Answer:

Let the given rhombus be ABCD whose diagonals intersect at a point O.

Given: AB = BC = CD = DA = 13 cm

Diagonal BD = 24 cm

The figure for the given information can be drawn as follows:

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

∴ BO =

Also, ∠BOC = 90°

In ΔBOC, using Pythagoras theorem, we get:

∴ AC = 2 × OC = 2 × 5 cm = 10 cm

Thus, the length of the other diagonal is 10 cm.

Answer:

Let the given rhombus be ABCD whose diagonals intersect at a point O.

Given: Perimeter of the rhombus = 68 cm

Diagonal BD = 30 cm

The figure for the given information can be drawn as follows:

Perimeter of the rhombus ABCD = 4 × AB

⇒ 68 cm = 4 × AB

⇒ AB =

∴ AB = BC = CD = DA = 17 cm

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

∴ BO =

Also, ∠BOC = 90°

In ΔBOC, using Pythagoras theorem, we get:

∴ AC = 2 × OC = 2 × 8 cm = 16 cm

Thus, the length of the other diagonal is 16 cm.

Answer:

Let the given rhombus be ABCD whose diagonals intersect at a point O.

Given: Diagonal BD = 16 cm and diagonal AC = 12 cm

The figure for the given information can be drawn as follows:

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

∴ BO =

OC =

Also, ∠BOC = 90°

In ΔBOC, using Pythagoras theorem, we get:

Since all the sides of a rhombus are equal in length, AB = BC = CD = DA = 10 cm.

Perimeter of the rhombus = 4 × AB

= 4 × 10 cm

= 40 cm

Answer:

Let us consider an isosceles trapezium PQRS such that PQ || SR and PS = QR. 

PR and QS are its diagonals. 

Construction: Draw two perpendiculars, PA and QB on the side SR.

In ΔPSA and ΔQRB, we have:

PS = QR (Given)

∠ PAS = ∠QBR = 90° (By construction)

PA = QB (Perpendicular distance between two parallel lines)

∴ ΔPSA ≅ ΔQRB (By RHS congruence criterion)

⇒ ∠PSA = ∠QRB (By c.p.c.t)

⇒∠PSR = ∠QRS 

In ΔPSR and ΔQRS, we have:

PS = QR (Given)

∠PSR = ∠QRS (Proved above)

SR = RS (Common)

∴ ΔPSR ≅ ΔQRS (By SAS congruence criterion)

⇒ PR = QS (By c.p.c.t)

Hence, the diagonals of an isosceles trapezium are equal.

Answer:

Given: A quadrilateral ABCD in which diagonals AC and BD intersect at point O such that OA = OC and OB = OD.

To prove: ABCD is a parallelogram.

Proof: 

In ΔAOD and ΔBOC:

OA = OC (Given)

OD = OB (Given)

∠AOD = ∠BOC (Vertically opposite angles)

∴ ΔAOD ≅ ΔCOB (By SAS congruence criterion)

⇒ ∠OAD = ∠OCB (By c.p.c.t.)

∴ AD || BC ... (1) (If a transversal intersect two lines in such a way that a pair of alternate interior angles are equal then the two lines are parallel)

Similarly, AB || DC ... (2)

From (1) and (2), we get:

AB || DC and AD || BC

We know that a quadrilateral is a parallelogram, if both the pairs of its opposite sides are parallel.

Hence, ABCD is a parallelogram.

Answer:

There are three types of quadrilaterals in which the diagonals are equal. These are square, rectangle and isosceles trapezium.

A square is a special kind of rectangle. So, a quadrilateral with equal diagonals and which is not a rectangle is an isosceles trapezium.

The below given figure shows an isosceles trapezium, PQRS with equal diagonals, PR = QS.

Answer:

Let ABCD be the required parallelogram.

Given: DC = AB = 6.5 cm

AD = BC = 4 cm

∠ADC = 45°

Here is the rough sketch of the required parallelogram.

The steps of construction are as follows:

(1) Draw a line segment DC of length 6.5 cm.

(2) At point D, draw ∠XDC of measure 45°.

(3) Taking D as the centre and 4 cm as the radius, draw an arc which intersects ray DX at point A.

(4) Taking A as the centre and 6.5 cm as the radius, draw an arc on the right side of point A.

(5) Taking C as the centre and 4 cm as the radius, draw an arc which intersects the previously drawn arc at point B.

(6) Join AB and BC.

ABCD is the required parallelogram.

Answer:

Let ABCD be the required rhombus.

Given: AB = BC = CD = DA = 4 cm

∠ADC = 30°

Here is the rough sketch of the required rhombus.

The steps of construction are as follows:

(1) Draw a line segment DC of length 4 cm.

(2) At point D, draw ∠XDC of measure 30°.

(3) Taking D as the centre and 4 cm as the radius, draw an arc which intersects ray DX at point A.

(4) Taking A as the centre and 4 cm as the radius, draw an arc on the right side of point A.

(5) Taking C as the centre and 4 cm as the radius, draw an arc which intersects the previously drawn arc at point B.

(6) Join AB and BC.

ABCD is the required rhombus.

Answer:

Let ABCD be the required parallelogram.

Given: DC = AB = 7 cm

AD = BC = 4 cm

AC = 8 cm

Here is the rough sketch of the required parallelogram.

The steps of construction are as follows:

(1) Draw a line segment DC of length 7 cm.

(2) Taking D as the centre and 4 cm as the radius, draw an arc on the upper side of line segment DC.

(3) Taking C as the centre and 8 cm as the radius, draw an arc which intersects the previously drawn arc at point A.

(4) Join AD and AC.

(5) Taking A as the centre and 7 cm as the radius, draw an arc on the right side of point A.

(6) Taking C as the centre and 4 cm as the radius, draw an arc which intersects the previously drawn arc at point B.

(7) Join AB and BC.

ABCD is the required parallelogram.

Answer:

Let ABCD be the required parallelogram with diagonals AC and BD intersecting at point O.

Given: DC = 7 cm

AC = 8 cm, BD = 10 cm

We know that diagonals of a parallelogram bisect each other.

∴ OC = OA = = 4 cm

OD = OB = = 5 cm

Here is the rough sketch of the required parallelogram.

The steps of construction are as follows:

(1) Draw a line segment DC of length 7 cm.

(2) Taking D as the centre and 5 cm as the radius, draw an arc on the upper side of line segment DC.

(3) Taking C as the centre and 4 cm as the radius, draw an arc intersecting the previously drawn arc at point O.

(4) Join OD and OC.

(5) Now extend DO and CO to form rays DX and CY respectively.

(6) Taking O as the centre and 5 cm as the radius, draw an arc which intersects ray DX at point B.

(7) Taking O as the centre and 4 cm as the radius, draw an arc which intersects ray CY at point A.

(8) Join AD, AB and BC.

ABCD is the required parallelogram.

Answer:

Let ABCD be the required parallelogram with diagonals AC and BD intersecting at point O.

Given: DC = 5 cm

AC = 6 cm, BD = 8 cm

We know that diagonals of a parallelogram bisect each other.

∴ OC = OA = = 3 cm

OD = OB = = 4 cm

Here is the rough sketch of the required parallelogram.

The steps of construction are as follows:

(1) Draw a line segment DC of length 5 cm.

(2) Taking D as the centre and 4 cm as the radius, draw an arc on the upper side of line segment DC.

(3) Taking C as the centre and 3 cm as the radius, draw an arc intersecting the previously drawn arc at point O.

(4) Join OD and OC.

(5) Now extend DO and CO to form rays DX and CY respectively.

(6) Taking O as the centre and 4 cm as the radius, draw an arc which intersects ray DX at point B.

(7) Taking O as the centre and 3 cm as the radius, draw an arc which intersects ray CY at point A.

(8) Join AD, AB and BC.

ABCD is the required parallelogram.

Answer:

Let ABCD be the required parallelogram with diagonals AC and BD intersecting at point O.

Given: AC = 6 cm, BD = 7 cm

∠DOC = 70°

We know that diagonals of a parallelogram bisect each other.

∴ OC = OA = = 3 cm

OD = OB = = 3.5 cm

Here is the rough sketch of the required parallelogram.

The steps of construction are as follows:

(1) Draw line segment AO of length 3 cm and produce AO to a point C such that OC = OA.

(2) At point O, draw ∠COP of measure 70°.

(3) Taking O as the centre and 3.5 cm as the radius, draw an arc which intersects ray OP at point D.

(4) Produce OD to a point B such that OD = OB.

(5) Join AD, DC, CB and BA.

ABCD is the required parallelogram.

Answer:

(i)

Let ABCD be the required trapezium.

Given: AB = 5 cm and DC = 7.5 cm

AD = 5.5 cm and BC = 6 cm.

Here is the rough sketch of the required trapezium.

The steps of construction are as follows:

(1) Draw a line segment DC of length 7.5 cm as the bottom side of the trapezium and then mark a point E at a distance of 7 cm − 5 cm = 2.5 cm from its right end such that EC is 2.5 cm long.

(2) Taking E and C as centres with radius 5.5 cm and 6 cm respectively, draw arcs intersecting each other at point B. 

(3) Join EB and BC.

(4) Draw a line segment AB of length 5 cm parallel to the bottom line DC through the top vertex of the triangle ECB, using a set square.

(5) Join AD.

ABCD is the required trapezium.

(ii)

Let ABCD be the required trapezium.

Given: AB = 5.5 cm and DC = 8 cm

∠ADC = 55° and ∠BCD = 70°

Here is the rough sketch of the required trapezium.

The steps of construction are as follows:

(1) Draw a line segment DC of length 8 cm as the bottom side of the trapezium and then mark a point E at a distance of 8 cm − 5.5 cm = 2.5 cm from its right end such that EC is 2.5 cm long.

(2) At points D and C, draw ∠PDC and ∠RCD of measures 55° and 70° respectively.

(3) Draw a ray EQ parallel to DP at point E, which intersects the ray CR at point B.

(4) Draw a line segment from point B parallel to the bottom line DC which intersects the ray DP at point A, using a set square.

ABCD is the required trapezium.

Answer:

(i) Here is the rough sketch of the required quadrilateral.

The steps of construction are as follows:

(1) Draw a line segment BC of length 5 cm.

(2) At point B, draw ∠PBC of measure 60°.

(3) At point C, draw ∠QCB of measure 140°.

(4) Taking B as the centre and 7 cm as the radius, draw an arc which intersects ray BP at point A.

(5) Taking C as the centre and 4 cm as the radius, draw an arc which intersects ray CQ at point D.

(6) Join AD.

ABCD is the required quadrilateral.

(ii) Here is the rough sketch of the required quadrilateral.

The steps of construction are as follows:

(1) Draw a line segment AB of length 8 cm.

(2) At point B, draw ∠PBA of measure 50°.

(3) Taking B as the centre and 6 cm as the radius, draw an arc which intersects BP at point C.

(4) Taking C as the centre and 5.5 cm as the radius, draw an arc on the left side of ray BP.

(5) Taking A as the centre and 3 cm as the radius, draw an arc which intersects the previously drawn arc at point D.

(6) Join AD and CD.

ABCD is the required quadrilateral.

(iii) Here is the rough sketch of the required quadrilateral.

The steps of construction are as follows:

(1) Draw a line segment DC of length 5 cm.

(2) Taking D as the centre and 7.5 cm as the radius, draw an arc on the upper side of line segment CD.

(3) Taking C as the centre and 4.5 cm as the radius, draw an arc which intersects the previously drawn arc at point B.

(4) Join BD and BC.

(5) Taking B as the centre and 8.5 cm as the radius, draw an arc on the left side of point B.

(6) Taking D as the centre and 6 cm as the radius, draw an arc which intersects the previously drawn arc at point A.

(7) Join AD and AB.

ABCD is the required quadrilateral.