CBSE Class 6 Mathematics Practical Geometry Assignment Set 01

Chapter - 14 (Practical Geometry)

1. Fill in the blanks :-

Question 1(i). If diameter of a circle is 18cm, the radius is _____________
Answer: \(9\text{ cm}\)
In simple words: The radius is always half of the diameter.

Exam Tip: Use the formula radius \(r = \frac{\text{Diameter}}{2}\).

Question 1(ii). The longest chord of a circle is _____________
Answer: diameter
In simple words: A chord that passes through the center of the circle is the diameter, which is the longest possible chord.

Exam Tip: Remember that every diameter is a chord, but not every chord is a diameter.

Question 1(iii). Number of circles passing through a given point are ___________
Answer: infinite (or countless)
In simple words: You can draw an endless number of circles of different sizes that all pass through one single point.

Exam Tip: Contrast this with two points, through which an infinite number of circles can also pass, or three non-collinear points through which only one unique circle passes.

Question 1(iv). Circle which have same centre but diffrent radii are called __________ circles.
Answer: concentric
In simple words: Concentric circles share the exact same center point like the rings of a target.

Question 1(v). Angles of set squares are (i) \(45^{\circ}, 45^{\circ}, 90^{\circ}\) (ii) \(30^{\circ}, 60^{\circ}, 90^{\circ}\)
Answer: (i) \(45^{\circ}\), \(45^{\circ}\), \(90^{\circ}\) (ii) \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\)
In simple words: Geometry box set squares have these specific fixed angles.

Exam Tip: Always include the degree symbol when writing angle measurements.

2. Refer to the figure given below, answer the following.

Question 2(i). Name any radius of the circle ____________
Answer: \(OA\) (or \(OC\), \(OD\), \(OE\), \(OF\))
In simple words: A radius connects the center point to any point on the boundary of the circle.

Exam Tip: Use line segment notation or simple capital letters with a bar or just the letters representing the endpoints starting from the center.

Question 2(ii). Name centre of the circle ____________
Answer: \(O$
In simple words: The fixed point right in the middle of the circle is point \(O\).

Exam Tip: The center is always denoted by a single capital letter.

Question 2(iii). Name any segment of the circle ____________
Answer: \(EF\) (or any chord like \(CD\))\
In simple words: A chord is a line segment whose endpoints both lie on the circle.

Exam Tip: Diameter \(EF\) is the longest chord and also a valid line segment of the circle.

Question 2(iv). Name one sector of the circle ____________
Answer: Sector \(AOC\) (bounded by radii \(OA\), \(OC\) and arc \(AC\))
In simple words: A sector is like a slice of pie enclosed by two radii and an arc.

Exam Tip: Always name a sector using its center and two boundary points on the circle.

Question 2(v). Name any point in the interior of the circle ____________
Answer: \(O\) (and point \(D\))\
In simple words: Interior points lie strictly inside the boundary of the circle.

Exam Tip: The center \(O\) is always in the interior.

Question 2(vi). Name any exterior point of circle ____________
Answer: \(L$
In simple words: An exterior point is located completely outside the circle boundary.

Exam Tip: Point \(L\) is clearly labeled outside the circular curve.

Question 2(vii). Name any point on the circle ____________
Answer: \(A\) (or \(B\), \(C\), \(E\), \(F\))\
In simple words: Points on the circle lie directly on the curved boundary line.

Exam Tip: Any letter touching the perimeter ring can be used.

3. Solve

Question 3(i). Given AB = 3cm & CD = 4cm, construct a line segment XY equal to sum of AB & CD.
Answer:
1. Draw a line \(l\) of sufficient length.
2. Mark a point \(X\) on line \(l\).\
3. Using a compass, measure the length of line segment \(AB = 3\text{ cm}\).
4. With center \(X\) and radius equal to \(AB\), cut an arc on line \(l\) at point \(P\), so that \(XP = 3\text{ cm}\).
5. Now, measure the length of line segment \(CD = 4\text{ cm}\) using the compass.
6. With center \(P\) and radius equal to \(CD\), cut another arc further along line \(l\) at point \(Y\), so that \(PY = 4\text{ cm}\).
7. The total line segment \(XY = XP + PY = 3\text{ cm} + 4\text{ cm} = 7\text{ cm}\), which is equal to the sum of \(AB\) and \(CD\).
In simple words: Join two line segments end-to-end on a straight line to get their combined length of 7 cm.

Exam Tip: Always use a sharp pencil and a properly adjusted compass for accurate length transfers.

Question 3(ii). Draw a line segment of length 5cm and construct its perpendicular bisector.
Answer:
1. Draw a line segment \(AB = 5\text{ cm}\) using a ruler.
2. With \(A\) as center and radius more than half of \(AB\) (greater than \(2.5\text{ cm}\)), draw arcs above and below the line segment \(AB\).
3. With \(B\) as center and the same radius, draw arcs above and below the line segment \(AB\) to intersect the previous arcs at points \(P\) and \(Q\).
4. Join points \(P\) and \(Q\) with a line segment \(PQ\).
5. \(PQ\) is the required perpendicular bisector of \(AB\), dividing it into two equal halves of \(2.5\text{ cm}\) each at right angles (\(90^{\circ}\)).
In simple words: Open your compass wider than halfway, draw intersecting arcs from both ends, and connect the crossing points to cut the line right in half at a right angle.

Exam Tip: Do not change the compass radius width when drawing the corresponding arcs from endpoints \(A\) and \(B\).

Question 3(iii). Draw an angle of \(60^{\circ}\) and construct its bisector.
Answer:
1. Draw a ray \(OA\).
2. With \(O\) as center and any convenient radius, draw an arc that cuts ray \(OA\) at point \(B\).
3. With \(B\) as center and the same radius, draw an arc cutting the previous arc at point \(C\).
4. Draw a ray \(OC\) passing through \(C\). Angle \(AOC\) is \(60^{\circ}\).
5. With \(B\) as center and radius more than half of arc length \(BC\), draw an arc in the interior of the angle.
6. With \(C\) as center and the same radius, draw another arc to intersect the previous arc at point \(P\).
7. Draw ray \(OP\). Ray \(OP\) is the required angle bisector, dividing \(60^{\circ}\) into two equal angles of \(30^{\circ}\) each.
In simple words: Construct a standard equilateral triangle corner for 60 degrees, then use compass arcs from both arms to split the angle exactly in half.

Exam Tip: Keep the compass radius identical when creating the \(60^{\circ}\) arc from the baseline.