WBBSE Solutions For Class 7 Maths Geometry Chapter 8 Construction Of Quadrilateral Exercise 8 Solved Problems

Class 7 Math Solution Geometry Chapter 8 Construction Of Quadrilateral Exercise 8 Solved Problems

For the construction of a quadrilateral five of its parts must be known.

We can construct quadrilaterals in the following cases:

1. If the lengths of four sides and measure of one angle is given
2.  If the lengths of four sides and are diagonal is given
3. If the lengths of three sides and measures of two included angles is given
5.  If the lengths of two adjacent sides and measures of three angles is given
6. If the length of a side of a square is given
7. If two adjacent sides and their included angle of a parallelogram is given
8. If the length of a side of a rhombus and the measure of its angle is given.

Read and Learn More WBBSE Solutions for Class 7 Maths

Question 1. Construct a quadrilateral ABCD in which AB = 5 cm, BC= 5.8 cm, CD = 4.1 cm, AD = 6.7 cm and AC = 9.4 cm.

Solution:

 

 

Method Of Contruction:

1. We take a ray AX of any length.
2.  With centre A and radius 9-4 cm an arc is drawn at C on AX.
3. With A and C as centres and radii 5 cm and 5.8 cm respectively, two arcs are drawn, which intersect each other at point B.
4. Again, with A and C as centres and radii 6.7 cm and 4.1 cm respectively, two arcs are drawn on the other side of AC opposite to B, which intersect each other at point D.
5.  I join A, B; B, C; A, D and C, D.

Thus ABCD is the required quadrilateral whose AB = 5 cm, BC = 5.8 cm, CD = 4.1 cm, AD = 6.7 cm and AC = 9.4 cm.

“WBBSE class 7 maths geometry chapter 8 solutions”

Question 2. Construct a quadrilateral PQRS in which PQ = 6.5 cm, QR = 6 cm, RS = 5 cm, SP 4 cm and ∠PQR = 60°

Solution:

 

 

 

Method Of Construction:

1. A ray PX is drawn and from it cut off a line segment PQ of length 6.5 cm.
2. Using pencil compass and scale an angle ∠PQY = 60° is drawn at Q.
3. From QY cut off a line segment QR of length 6 cm.
4. With P and R as centres and with radii equal to 4 cm and 5 cm respectively two arcs are drawn; to intersect each other at S. Then P, S and S, R are joined.

Thus PQRS is the required quadrilateral in which PQ = 6.5 cm, QR = 6 cm, RS = 5
cm, SP = 4 cm and ∠PQR = 60°

“construction of quadrilateral WBBSE class 7 maths notes”

Question 3. Construct a parallelogram ABCD in which AB = 4 cm, BC = 6 cm, and ∠ABC = 45°

Solution:

 

 

 

Method Of Construction :

1. A ray BX is drawn and from it cut off a line segment BC of length 6 cm.

2.At B on BX a ∠YBX = 45° is drawn.

3. From BY cut off a line segment AB of length 4 cm.

4. With centres A and C and radii equal to 6 cm and 4 cm respectively two arcs are drawn; to intersect each other at D. Then A, D and C, D are joined.

Thus ABCD is the required parallelogram in which AB = DC = 4 cm, AD = BC = 6 cm and ∠ABC = 45°.

“exercise 8 solved problems on quadrilateral construction class 7”

Question 4. Construct a rectangle PQRS in which PQ = 4.5 cm and QR = 6.5 cm

Solution:

 

Method Of Construction:

1. A ray QX is drawn and at Q on QX the angle ∠XQY = 90° is drawn.
2. The lengths QR = 6-5 cm and is cut down QX and the length QP = 4.5 cm is cut from QY.
3. With centres P and R and radii equal to 6.5 cm and 4.5 cm respectively two arcs are drawn to intersect each other at S. Then R, S and P, S are joined. Thus PQRS is the required rectangle.

Question 5. Construct a square ABCD on a side of length 6 cm.

Solution:

 

 

Method Of Construction:

1. A ray AX is drawn and from it cut off a line segment AB of length 6 cm.
2. A perpendicular AY is drawn on AX at A and from it cut off a line segment AD of length 6 cm.
3.  With B and D as centres and radii equal to 6 cm two arcs are drawn, which intersect each other at C. Then D, C and B, C are joined.

Thus ABCD is a required square.

Question 6. Construct a rhombus ABCD in which AC = 4 cm and BD = 6 cm

Solution:

 

 

 

 

⇒ [The diagonals of a rhombus bisect each other perpendiculerly]

Method Of Construction:

1. I take a line segment AC of length 4 cm and bisect it.

2. A ray BX is drawn and from it cut off a line segment BD of length 6 cm and bisect it by YZ at O.

3. With centre O and with a radius 2 cm (half of AC) draw an arc which cuts OY and OZ at A and C respectively. I join A, B; B, C; C, D and A, D.

Thus ABCD is the required rhombus.

Class 7 Math Solution Construction Of Quadrilateral Construction Of Quadrilateral Exercise 8.1

Question 1. Let me draw a quadrilateral REST in which RE = 6 cm, ES = 4.5 cm, ST = 5 cm TR = 5.5 cm & ET = 7.5 cm
Solution :

1. From there ray RX, a line segment Re (= 6cm) is cut off.
2. With centre E two arcs of radius 4.5 cm & 7.5 cm is drawn.
3. Again with centre R, a arc of radius 5.5 cm is drawn.
4. By joining ES, TS, TR we get the required quadrilateral REST is obtained.

“WBBSE class 7 maths chapter 8 important questions”

Question  2. Let me draw a parallelogram whose adjacent sides are 5 cm. & 8 cm and the length of the diagonal is 10 cm.
Solution :

1. From the ray GX, aline segment Gl (= 8 cm.) is taken, with centre I, two arcs of radius 5 cm. & radius 5 cm. is drawn which cuts the previous arc atL
2. By joining Gl, IR, RL & LG we get the required parallelo gram GIRL is obtained.

Question 3. Let me draw a rough sketch of a quadrilateral LAND & try to find it will be possible to draw a quadrilateral, in which LA = 4 cm, AN = 5 cm, ND = 4 cm, DL = 6.5 cm & AD = 1 0 cm.
Solution :

Given

• From the ray LX, a line segment LA = 4 cm is taken.
• With centre A & OL, two arcs of radius 10 cm & 6.5 cm is drawn, they intersect at D.
• Now with centre A & D two arcs of radius 5 cm & 4 cm respectively arc drawn. Bu they do not intersect each other.
• So it is not possible to draw the quadrilateral LAND.

“step-by-step solutions for quadrilateral construction class 7”

Question 4. 1 try to draw a rhombus whose one side is 3 cm and its diagonal is 5 cm.
Solution :

• From the ray KX, a line segment Kl (= 3 cm.) is cut off. Now with centre
• I two arcs of radius 3 cm & 5 cm. arc drawn.
• Now with centre K an arc of radius = 3 cm is drawn which cuts the previous arc at E.
• By joining IT, TE & EK, we get the required rhombus KITE is obtained.

Class 7 Math Solution Construction Of Quadrilateral Exercise 8.2

Question 1. Let us draw a quadrilateral GOLD, where two diagonals are GL = 8cm and DO = 10cm. and the other three sides are GO = 6cm, OL = 5 and DL = 5.6 cm. Let’s Measure the length GD and let’s write.
Solution :

1. Draw a ray GX, from this ray a line segment GO (6cm) is cut off.
2. Now with centre G & O two arcs of radius 8cm and 5cm is cut off. They meet at L, Join OL.
3. Now with O & L, two arcs of radius 10cm and 5 6cm is cut off. They meet at D, join GD.
4. Thus a quadrilateral GOLD is formed.

∴ GD = 95cm

Question 2. Let us draw a rhombus REST, whose two diagonals are, RS = 6.8cm and ET = 7.2cm. [Hints: The diagonals of a rhombas bisect each other at right angles. Hence let us draw a perpendicular bisector of diagonal ET with the mid point of the diagonal ET as centre side of the arcs are drawn with hold the length of the other diagonal RS, On either side of the perpendicular biscetor at R & S. Hence joining the points R, E, S and T, rhombus is formed.]
Solution :

1. Take a ray RS, from this ray a line segment RS. ( = 6.8cm) is cut off.
2. Now RS-. is bisected at O. At O, draw a perpendicular PQ. From OP &
3. OQ two parts OE & OT each equal 3.6cm (Total ET = 7.2cm.)
4. Join ER, RT, TS & SE.

∴ The rhombus REST is drawn.

Question 3. Let’s draw a quadrilateral LAND such that LA = 6.5cm AN = 5.4cm, ND = 4cm, DL = 3.5cm, LAN = 45°.
Solution :

Froim the ray AX, AN (=5.4cm) is cut off.

• Now at A draw an angle L AN (= 45° ) from the line sent LA (6.5cm is cut off.
• With centre N & L two arcs equal to 4cm & 3.5cm drawn they meet at D.
• Joining LD & ND, we get the required quadrilateral LAND is obtained.

Question 4. Let’s draw a parallelogram GATE whose adjacent sides are, GA = 7cm and AT – 5cm and ∠GAT = 45°
Solution :

Take a ray GX. From GX, take a line segment GA equal to 7cm.

• Now at A, draw an angle GAY equal to 45°. From AY, AT equal to 5cm is cut off.
• Now, at G & at T draw two arcs equal to radius 5cm and 7cm respectively.
• By joining GE & ET we get the required parallelogram GATE.

“different methods of constructing quadrilaterals WBBSE”

Question 5. Let me draw a rectangle RICH, So that Rl = 4cm and Cl = 7.5cm. [Hints : Opposite sides of a rectangle are equal and in each angle is degree.]
Solution :

To draw a rectangle RICH, take a ray RX, from RX, cut a line segment RH equal to 7.5cm.

Now, at R, draw perpendicular RY. From RY cut a line segment Rl equal to 4cm.

• Now, with centre H & I draw two arcs equal to radius 4cm & 7.5cm respectively.
• The two arcs intersects at C, join. Cl & CH, we get required rectangle RICH.

Question 6. Let’s construct a rhombus whose one angle is 60° and its equal sides are 6.5cm each.
Solution :

From the AX, a line segment AB (= 6.5) is cut off.

• At draw an angle YAX equal to 60° . From YA, a line segment DA = 6.5cm is cut off.
• Now with centre B & D two arcs of length 6.5cm is drawn. They meet at G. Joining BC & CD we get the required Rhombus ABCD.

Question 7. Let’s draw a square PATH whose sides are 8cm each.
Solution :

From the ray PX, PA = 8cm is cut off. Now, at P drawn angle 99° , YPA.

• From PY, PH = 8cm is taken,.Now with centre A & H two arcs equal to 8cm is drawn, let they meet at T.
• Join TA & TH. Thus, we get the required square

Question 8. Let us draw a quadrilateral PLAN is which PL = 4.6cm LA = 5.5cm, AN = 5cm, AN = 5cm & ∠PLA = 60° & ∠LAN = 90°
Solution :

• Draw 3 line segments PL (4.6cm), LA (5,5cm) & AN (5cm)
• Take a ray AX, from AX, AL (= 5.5cm) is cut off.
• At A & L dras two angles equal to 90° & 60° ie ∠LAY(=90°)& ∠ALK(= 60°) drawn.
• From AY, AN = 5cm & from LK, LP = 4.6cm is cut off.
• Join PN.

∴ Quadrilateral PLAN is obtained.

Question 9. Let us draw a quadrilateral HEAR, in which HE = 5cm RH = 6.8 cm, ∠EHR = 90°, ∠HEA = 1 20° and ∠HRA = 70°
Solution:

To construct a quardritateral HEAR

• Take a ray HP from HP cut off HF = 5cm. On HE at H & at E draw an angle = 90° and angle 120° respectively. From HY cut off HR = 6.8 cm. Now, at
• R, draw an angle = 70°  Which cut EQ at A. JoinÿR & we get the required quadrilateral HEAR.

Exercise

Question 1. Let’s think and write answers of the following: If four sides of a quardrilateral are given, is it possible to construct it. It we can’t draw, then what more data should be given so that the quardrilateral can be constructed, let’s find.

1. To draw specific parallelogram, what least data are required, let’s find.
Solution: 

Two adjacent side and an angle between two sides are required.

2. Let’s find, the least number of data required to construct a square.
Solution:   One side is required.

3. Let’s find if it is possible to construct a rhombus when the length of its two diagonals are given.
Solution: To construct a rhombus length of two diagrams is required.

4.  Let’s find the least number of data required to construct a rectangle.
Solution:  To the adjacent side’s length is required.

“properties of quadrilaterals used in construction class 7”

Question 2. Let’s construct a quadrilateral ABCD in which AB = 5.2cm, BC = 6cm, CD = 4.4cm, AD = 7cm and AC = 1 0cm. If in this quadrilateral, AC= 1 8cm, could it be possible to construct the quadrilateral, let’s find.
Solution :

To constract a quadrilateral ABCD, take a line AX, From AX, AB = 5.2cm is cut off.

• Now, with centre A & B two arcs of radius 10cm and 6cm are drawn, let they meet at C.
• Again with centre A & C two arcs 7cm and 4.4cm are drawn, they meet each other at D.
• By joining BC, Cd & DA we get the required quadrilateral ABCD.
• In AC = 1 2cm, it is not possible to draw the quadrilateral, as arcs AC & BC will never meet each other.

Question 3. Let construct a parallelogram JUMP in which JU = 5.2cm, UM = 4.8cm and JM = 7cm.
Solution :

From the ray JX, JU = 5.2cm is cut off.

• Now, with centres J & U two arcs of radius 7cm, 4.8cm are drawn, they meet at M.
• Again with centre J & M two arcs of radius 4.8cm & 5.2 cm are drawn.
• They meet at P. By joining JU, UM & MU we get the required parallelogram JUMP.

Question 4. Let’s contruct a rhombus PQRS in which PQ = 5.4cm and PR = 8cm.
Solution :

From the ray PX, take a line segment PQ equal to 7cm.

• Now at Q draw an angle YQP = 60°  From Qy, QR = 6.5cm taken, now with centre R and P two arcs of radius 5.2cm & 4.4 cm are dran, they meet at S by joining PS &
• RS we get the required quadrilateral PQRS.

Question 5. Let’s construct a quadrilateral PQRS in which PQ = 7cm, QR = 6.5cm, RS = 5.2cm, SP = 4.40cm, = 60°
Solution :

From the ray PX, PQ = 5.4cm is cut off.

• Now, with centre P &Q drawn two arcs of radius 8cm and 5.4cm are drawn.
• They meet at R. Now again with centre P & R two arcs are drawn with a radius 5.4cm, The meet at S, By joining SR & PS we get the required quadrilateral PQRS.

Question 6. Let’s construct a rhombus BEST in which BS = 6.8cm, and ET = 5.8cm.
Solution :

⇒ From the ray BL, BS = 6.8cm is taken, Now, we draw the perpendicular bisector PQ of BS, which cuts BS at O.

⇒ From OP & OQ, OE & OT equal to 2.9cm are taken, Now Joint BE, ES, ST & TB and we get the rhombus BEST.

“how to draw a parallelogram, rhombus, and trapezium class 7”

Question 7. Let’s construct a square DEAR in which DE = 5.2cm.
Solution :

From the ray DX, De = 5.2cm is cut off.

• Now, at D draw an angle ZYDE = 90° From Dy, Dr = 5.2cm is taken. Now, with centres E and R two arcs equal to drawn.
• Let them meet at A. After joining AR & AE we getting square DEAR.

Question 8. Let’s draw a rectangle READ in which RE = 6cm & EA = 5cm.
Solution :

From the ray EX, ER = 6cm is cut off.

• Now at E draw a perpendicular EY, from EY cut-off.
• EA = 5cm. Now, with centre A & R draw two arcs equal to radius 6cm & 5cm respectively.
• They meet each other at DF joining DA & DR, we get the required rectangle READ.

“real-life applications of quadrilateral construction WBBSE”

Question 9. Let’s draw a quadrilateral SAND, such that SA = 5.6cm, AN = 4.5cm, ∠ASD = 75°, ∠SAN = 75° and ∠AND = 110°.
Solution :

From the ray SX, SA = 5.6cm is cut off.

• At A draw the angle ∠SAY = 75° > form AY cut off AN = 4.5. Then at N draw angle ∠PNA = 110°  at S draw ∠QSA = 75°
• They meet at D, by joining DA we get the required quadrilateral SAND.

Question 10. Let’s draw a parallelogram LAND such that LA = 6.6cm, AN = 4cm and ∠LAN = 45°
Solution :

• Draw two line segments LA (6.6 cm) & AN (5.4 cm).
• Now from the ray AX, AN = 5.4 cm is cut off.
• At A, ∠YAN = 45° is drawn, and from AY, AL = (6.6 cm) is cut off.
• Now at L & N draw two arcs with radius AN & AL arc drawn, There two arcs intersects at D. Join LD & DN.
• Thus the required parallelo gram LAND is obtained.

Question 11. Let’s construct a rhombus HOME such that ∠HOM 60 and HO = 6cm.
Solution :

From the ray HX, HO = 6cm is taken.

• Now at O draw an angle ∠YOH = 60° From OY , OM = 6cm is taken (As al1 the side of a rhombus are equal).
• Now, centre H & M two arcs equal to 8cm are drawn, then by joining EM & EH, we get the required rhombus HOME.

Question 12. Let’s construct a rhombus ROAD in which RA = 8cm and OD = 6cm.
Solution :

From the ray PX, take a line segment PQ equal to 7cm.

• Now, at Q draw an angle YQR = 60° From QY, QR = 6.5cm taken, now with centre R and P two arcs of radius 5.2cm & 4.4cm are drawn.
• They meet at S by joining PS & RS we get the required quadrilateral PQRS.

“chapter 8 construction of quadrilateral long and short questions”

Question 13. Let’s draw a square TRAM such that TA = 6cm.
Solution :

⇒ A square TRAM

⇒ A line segment TA = 6cm is drawn.

⇒ Now, draw the perpendicular bisector (PQ) of TA at O, from OP & OQ OM & OR each equal = 3cm is taken, by joining TM, MA, AR, RT we got the required square TRAM.

“common mistakes in quadrilateral construction class 7”

Question 14.  Let’s construct a rectangle ABCD such that AC = 5cm and ∠BAC = 30°.
Solution :

On the time AX at A draw an angle ∠YAX = 30°.

• From AY, ACT = 5cm is cut off. From C draw the perpendicular CB on AX.
• Now, with centres C & A draw two arcs equal to AB & BC respectively.
• They meet at D joining CD & AD, we get the required rectangle ABCD.