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NCERT Solutions for Class 8 Maths - Practical Geometry - 2

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 for Class 8 Maths - Practical Geometry - 2

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 for Class 8 Maths - Practical Geometry - 2, such that OMX = 60°.
III. At O, draw NCERT Solutions for Class 8 Maths - Practical Geometry - 2 , such that MOY = 105°.
IV. From NCERT Solutions for Class 8 Maths - Practical Geometry - 2 , cut off OR = 4.5 cm.
V. At R, draw NCERT Solutions for Class 8 Maths - Practical Geometry - 2 , such that ORZ = 105°.
Let NCERT Solutions for Class 8 Maths - Practical Geometry - 2 intersectsNCERT Solutions for Class 8 Maths - Practical Geometry - 2 at E.

NCERT Solutions for Class 8 Maths - Practical Geometry - 2
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 for Class 8 Maths - Practical Geometry - 2 such that LAX = 110°.
III. At L, draw NCERT Solutions for Class 8 Maths - Practical Geometry - 2 such that ALY = 75°.
NCERT Solutions for Class 8 Maths - Practical Geometry - 2
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 for Class 8 Maths - Practical Geometry - 2, cut-off LP = 4 cm.
V. At P, draw NCERT Solutions for Class 8 Maths - Practical Geometry - 2 such that LPZ = 90° 
Let PZ and NCERT Solutions for Class 8 Maths - Practical Geometry - 2 intersect at N.
Thus, PLAN is the required quadrilateral.
(iii) Steps of construction: 

NCERT Solutions for Class 8 Maths - Practical Geometry - 2

I. Draw a line segmentNCERT Solutions for Class 8 Maths - Practical Geometry - 2 = 5 cm.
II. At E, drawNCERT Solutions for Class 8 Maths - Practical Geometry - 2 such that HEA = 85°.
III. FromNCERT Solutions for Class 8 Maths - Practical Geometry - 2 , 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 for Class 8 Maths - Practical Geometry - 2

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

IV. At K, draw NCERT Solutions for Class 8 Maths - Practical Geometry - 2 such that OKQ = 90°.

V. From NCERT Solutions for Class 8 Maths - Practical Geometry - 2 cut-off NCERT Solutions for Class 8 Maths - Practical Geometry - 2 = 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 for Class 8 Maths - Practical Geometry - 2 such that ∠DEX = 60°.
III. From NCERT Solutions for Class 8 Maths - Practical Geometry - 2 , cut-off  EA = 5 cm.
IV. At A, draw ray NCERT Solutions for Class 8 Maths - Practical Geometry - 2 such that ∠EAY = 90°.
V. From NCERT Solutions for Class 8 Maths - Practical Geometry - 2 , cut-off AR = 4.5 cm.
VI. Join R and D.

NCERT Solutions for Class 8 Maths - Practical Geometry - 2
Thus, DEAR is the required quadrilateral.

(ii) Steps of construction: 

I. Draw a line segment NCERT Solutions for Class 8 Maths - Practical Geometry - 2 = 3.5 cm.

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

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.

The document NCERT Solutions for Class 8 Maths - Practical Geometry - 2 is a part of the Class 8 Course NCERT Textbooks & Solutions for Class 8.
All you need of Class 8 at this link: Class 8
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FAQs on NCERT Solutions for Class 8 Maths - Practical Geometry - 2

1. What is practical geometry?
Ans. Practical geometry is a branch of mathematics that deals with the construction and measurement of various geometric shapes and figures using physical tools and techniques.
2. How do we construct a perpendicular bisector of a line segment?
Ans. To construct a perpendicular bisector of a line segment, follow these steps: 1. Place the compass on one endpoint of the line segment and draw two arcs that intersect the line segment. 2. Without changing the compass width, place the compass on the other endpoint of the line segment and draw two more arcs that intersect the line segment. 3. Draw a straight line connecting the two points of intersection of the arcs. This line is the perpendicular bisector of the line segment.
3. How do we construct an angle bisector?
Ans. To construct an angle bisector, follow these steps: 1. Draw an angle using a protractor. 2. With the compass, place the needle on the vertex of the angle and draw an arc that intersects both arms of the angle. 3. Without changing the compass width, place the needle on each point of intersection of the arc and the angle's arms and draw two more arcs. 4. Draw a straight line connecting the vertex of the angle to the point where the two arcs intersect. This line is the angle bisector.
4. Can we construct a triangle if we know the lengths of its three sides?
Ans. Yes, we can construct a triangle if we know the lengths of its three sides. This is known as the SSS (side-side-side) criterion. To construct a triangle using this criterion, we can use a ruler and compass to draw three line segments with the given lengths, making sure that the sum of any two sides is greater than the third side.
5. How do we construct a parallelogram if we know the lengths of its adjacent sides?
Ans. To construct a parallelogram if we know the lengths of its adjacent sides, follow these steps: 1. Draw a line segment of the given length to represent one side of the parallelogram. 2. With the compass, place the needle on one endpoint of the line segment and draw an arc that intersects the line segment. 3. Without changing the compass width, place the needle on the other endpoint of the line segment and draw another arc that intersects the line segment. 4. Connect the points of intersection of the arcs with the endpoints of the line segment. This line represents the adjacent side of the parallelogram. 5. Repeat steps 1-4 for the other adjacent side. The figure formed is a parallelogram.
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