NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev

Class 8 Mathematics by VP Classes

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Class 8 : NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev

The document NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by VP Classes.
All you need of Class 8 at this link: Class 8

Question 1. Can you construct the following quadrilateral MIST if we have 100° at M instead of 75°?

Solution: Yes, the quadrilateral MIST can be constructed with M = 100° instead of 75°.
NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev

Question 2. Can you construct the quadrilateral PLAN if PL = 6 cm, LA = 9.5 cm, P = 75°, L = 150° and A = 140°?

(Hint: Recall angle-sum property)

Solution: Here, P + L + A + N = 75° + 150° + 140° + N
= 365° + 
N

But the sum of all the angles of a quadrilateral is 360°. 

∴ Construction of quadrilateral PLAN is not possible.

Question 3. In a parallelogram, the lengths of adjacent sides are known. Do we still need measures of the angles to construct as in the Q-1 above?
Solution: No, the measures of three angles are not necessary in case of a parallelogram as its opposite sides are parallel.

EXERCISE 4.1
(Question 1.
Construct the following quadrilaterals:

(i) Quadrilateral MORE 
MO = 6 cm
OR = 4.5 cm

M = 60°
O = 105° 
R = 105° 
(ii) Quadrilateral PLAN
PL = 4 cm
LA = 6.5 cm
P = 90°
A = 110°
N = 85°
(iii) Parallelogram HEAR 
HE = 5 cm
EA = 6 cm

R = 85°
(iv) Rectangle OKAY
OK = 7 cm
KA = 5 cm
Solution: (i) Steps of construction: 
I . Draw a line segment MO = 6 cm.  
II. At M, draw  NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev, such that OMX = 60°.
III. At O, draw NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev , such that MOY = 105°.
IV. From NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev , cut off OR = 4.5 cm.
V. At R, draw NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev , such that ORZ = 105°.
Let NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev intersectsNCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev at E.

NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev
Thus, MORE is the required quadrilateral.

(ii) Steps of construction: 
I. Draw a line segment AL = 6.5 cm.
II. At A, draw NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev such that LAX = 110°.
III. At L, draw NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev such that ALY = 75°.
NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev
Note: L = 75° is not given, but we can determine it using angle sum property
∵ Sum of the three given angles = 110° + 90° + 85° = 285°
∴ The fourth angle L = 360° – 285° = 75°.
IV. FromNCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev, cut-off LP = 4 cm.
V. At P, draw NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev such that LPZ = 90° 
Let PZ and NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev intersect at N.
Thus, PLAN is the required quadrilateral.
(iii) Steps of construction: 

NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev

I. Draw a line segmentNCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev = 5 cm.
II. At E, drawNCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev such that HEA = 85°.
III. FromNCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev , cut-off EA = 6 cm.
IV. With centre at A and radius = 5 cm, draw an arc towards H.
V. With centre at H and radius = 6 cm, draw an arc such that it intersects the previous arc at R.
VI. Join RA and RH.
Thus, HEAR is the required quadrilateral.

(iv) Steps of constructions: 

NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev

I. Draw a line segment OK = 7 cm.
II. At O, draw NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev such that KOP = 90°.
III. From NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev , cut-off NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev = 5 cm.

IV. At K, draw NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev such that OKQ = 90°.

V. From NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev cut-off NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev = 5 cm.
VI. Join A and Y.
Thus, OKAY is the required quadrilateral.

 

CONSTRUCTION OF A QUADRILATERAL WHEN THREE SIDES AND TWO INCLUDED ANGLES ARE GIVEN

Question. We used some five measurements to draw quadrilateral so far. Can there be different sets of five measurements (other than seen so far) to draw a quadrilateral? The following problems may help you in answering the question.
(i) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and ∠B = 80°.
(ii) Quadrilateral PQRS with PQ = 4.5 cm, ∠P = 70°,∠Q = 100°, ∠ R = 80° and ∠S = 110°.
Construct a few more examples of your own to find sufficiency/insufficiency of the data for construction of a quadrilateral
Solution: (i) It is possible to construct a quadrilateral ABCD with the given measurements.
(ii) Not possible, because we cannot locate the points R and S with the help of given measurements.

EXERCISE 4.4 
Question 1 Construct the following quadrilaterals.
(i) Quadrilateral DEAR  
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°
(ii) Quadrilateral TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U = 120°
Solution: (i) Steps of construction:

I. Draw a line segment DE = 4 cm.
II. At E, draw NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev such that ∠DEX = 60°.
III. From NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev , cut-off  EA = 5 cm.
IV. At A, draw ray NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev such that ∠EAY = 90°.
V. From NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev , cut-off AR = 4.5 cm.
VI. Join R and D.

NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev
Thus, DEAR is the required quadrilateral.

(ii) Steps of construction: 

I. Draw a line segment NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev = 3.5 cm.

II. At R, draw a ray NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRevsuch that ∠TRX = 75°.
III. From NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev, cut-off RU = 3 cm.
IV. At U, drawNCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev such that ∠RUY = 120°.
V. From UY, cut-offNCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev= 4 cm.
VI. Join E and T.
Thus, TRUE is the required quadrilateral.
NCERT Solutions(Part- 2)- Practical Geometry Class 8 Notes | EduRev

SOME SPECIAL CASES

Using the properties of a square, rectangle, kite and rhombus, etc., we can construct some special quadrilaterals even with less than 5 measurements.

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