Ex.4.3 Q1 Practical Geometry Solution - NCERT Maths Class 8

Go back to  'Ex.4.3'

Question

Construct the following quadrilaterals.

(i) Quadrilateral \( MORE \)

\(MO = 6\,\rm{cm}\)

\(OR = 4.5\,\rm{cm}\)

\(\angle {M = 60 }^\circ\)

\(\angle {O = 105 }^\circ\)

\(\angle R = 105^\circ\)

(ii) Quadrilateral \( PLAN\)

\(PL=4\,\rm{cm}\)

\(LA=6.5\,\rm{cm}\)

\(\angle P=90^\circ\)

\(\angle A=110^\circ\)

\(\angle N=85^\circ\)

(iii) Parallelogram \(HEAR\)

\(HE = 5\,\rm{cm}\)

\(EA = 6\,\rm{cm} \)

\(\angle R = 85^\circ\)

(iv) Rectangle \(OKAY\)

\(OK=7\,\rm{cm}\)

\(KA=5\,\rm{cm}\)

 Video Solution
Practical Geometry
Ex 4.3 | Question 3

Text Solution

What is known?

Measurements of two sides and three angles

What is unknown?

Construction of a Quadrilateral

Reasoning:

As you are aware, we need five measurements to draw a quadrilateral .

The measurements of two adjacent sides and three angles 

Steps:

Let us draw a rough diagram of the quadrilateral

Let us construct the quadrilateral

Step 1: Draw a line segment \(MO=6 \,\rm{cm}.\)

Step 2: With \(M\) as center draw angle of measure \(60^\circ.\)

Step 3: Draw \( 105^\circ\) from \(O.\)

Step 4: With \(O\) as center and radius \(4.5\,\rm{cm}\) draw an arc cutting the ray from \(O\) at \(R.\) \(OR = 4.5\,\rm{cm}\)

Step 5: Construct an angle of \(105^{\circ}\) from \(R\) as above. The ray from \(R\) meets the ray from \(M\) at a point. Mark the intersection point as \(E.\)

Step 6: \(MORE\)  is the required quadrilateral.

  
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school