CBSE Class 7 Mathematics Practical Geometry Assignment Set 04

VII - Mathematics Assignment No - 01 - Practical Geometry

Question 1. Draw a line AB of 5 cm and mark a point P outside it. From P draw a line parallel to AB by using (i) ruler and compass (ii) set square.
Answer: Steps of Construction using Ruler and Compass:
1. Draw a line segment \(AB = 5 \text{ cm}\).
2. Take a point \(P\) outside the line \(AB\).
3. Take any point \(Q\) on \(AB\) and join \(PQ\).
4. With \(Q\) as centre and a convenient radius, draw an arc cutting \(AB\) at \(R\) and \(PQ\) at \(S\).
5. With \(P\) as centre and the same radius, draw an arc \(CD\) cutting \(PQ\) at \(T\).
6. Measure the length \(RS\) using a compass. With \(T\) as centre and radius equal to \(RS\), draw an arc to cut the previous arc \(CD\) at \(U\).
7. Join \(PU\) and produce it to get the required line parallel to \(AB\).

Steps of Construction using Set Square:
1. Place a set square with one of its edges along the line \(AB\).
2. Place a ruler along the edge opposite to the right angle of the set square.
3. Holding the ruler fixed, slide the set square along the ruler until its edge reaches the point \(P\).
4. Draw a line along this edge passing through \(P\), which is parallel to \(AB\).
In simple words: A parallel line is drawn by making alternate interior angles equal or corresponding angles equal.

Exam Tip: Always label all points clearly and show construction arcs with sharp pencil tips for accuracy.

Question 2. Draw a line AB of length 5 cm and draw a line PQ parallel to AB at a distance of 3 cm from AB.
Answer: Steps of Construction:
1. Draw a line segment \(AB = 5 \text{ cm}\).
2. Choose two points, say \(X\) and \(Y\), on the line \(AB\).
3. At \(X\) and \(Y\), construct perpendiculars \(XZ\) and \(YW\) of length \(3 \text{ cm}\) each, using a compass and ruler.
4. Join the points \(Z\) and \(W\) and extend this line in both directions to obtain the line \(PQ\).
In simple words: Construct two perpendiculars of equal height (3 cm) from any two points on the line and join their outer ends.

Exam Tip: Ensure the perpendiculars are drawn accurately at 90 degrees using a compass.

Question 3. Draw a line AB of length 4.5 cm. Take any point P on it and draw perpendicular at P. On this perpendicular take a point Q such that PQ = 3 cm, and from Q, draw a perpendicular QR. What can you say about AB and QR?
Answer: Steps of Construction:
1. Draw a line segment \(AB = 4.5 \text{ cm}\).
2. Take a point \(P\) on \(AB\).
3. Using a compass, draw a perpendicular \(PX\) to \(AB\) at point \(P\).
4. Cut off a line segment \(PQ = 3 \text{ cm}\) along \(PX\).
5. At point \(Q\), draw a perpendicular \(QR\) to \(PQ\).
6. Since \(\angle BPQ = 90^\circ\) and \(\angle PQR = 90^\circ\), the interior angles on the same side of the transversal \(PQ\) are supplementary (\(90^\circ + 90^\circ = 180^\circ\)).
7. Therefore, \(AB \parallel QR\).
In simple words: Lines AB and QR are parallel to each other because consecutive interior angles add up to \(180^\circ\).

Exam Tip: State the geometric reason clearly: sum of co-interior angles is \(180^\circ\).

Question 4. Draw a pair of parallel lines 4 cm apart using ruler and compass.
Answer: Steps of Construction:
1. Draw a line of any convenient length and name it \(l\).
2. Take two points \(A\) and \(B\) on line \(l\).
3. At \(A\) and \(B\), construct perpendiculars to line \(l\) using a compass.
4. Cut off line segments \(AC = 4 \text{ cm}\) and \(BD = 4 \text{ cm}\) along the perpendiculars.
5. Join points \(C\) and \(D\) and extend to form line \(m\).
6. Line \(m\) is parallel to line \(l\) at a distance of \(4 \text{ cm}\).
In simple words: Two parallel lines are created by drawing equal perpendicular heights from a base line.

Exam Tip: Measure the 4 cm distance carefully using a ruler and compass before marking.

Question 5. Draw a line AB of 3.5 cm. Mark a point P 4 cm above AB. Through P, draw PQ parallel to AB.
Answer: Steps of Construction:
1. Draw a line segment \(AB = 3.5 \text{ cm}\).
2. Take any point \(X\) on \(AB\) and construct a perpendicular \(XY\) of length \(4 \text{ cm}\). Let the end point be \(P\).
3. Take another point \(Z\) on \(AB\) and construct a perpendicular of length \(4 \text{ cm}\).
4. Join \(P\) and the top of the second perpendicular, extending it to form line \(PQ\).
In simple words: Construct a height of 4 cm from the line AB to locate point P, then draw a line through P parallel to AB.

Exam Tip: Verify the distance of 4 cm at multiple points to ensure the line is truly parallel.

Question 6. Construct a triangle ABC, having AB = 4 cm, \(\angle B = 30^\circ\) and BC = 3 cm.
Answer: Steps of Construction:
1. Draw a line segment \(AB = 4 \text{ cm}\).
2. At point \(B\), construct an angle of \(30^\circ\) using a compass (\(60^\circ\) bisected).
3. With \(B\) as centre and radius \(3 \text{ cm}\), draw an arc intersecting the ray of the angle at point \(C\).
4. Join \(A\) to \(C\). \(\triangle ABC\) is the required triangle.
In simple words: Draw side AB, make a \(30^\circ\) angle at B, cut off BC = 3 cm, and join AC.

Exam Tip: To make \(30^\circ\), first construct \(60^\circ\) and then bisect it.

Question 7. Construct a triangle ABC in which AB = 2 cm, BC = 4.5 cm and AC = 4 cm.
Answer: Steps of Construction:
1. Draw a line segment \(BC = 4.5 \text{ cm}\).
2. With \(B\) as centre and radius \(2 \text{ cm}\), draw an arc.
3. With \(C\) as centre and radius \(4 \text{ cm}\), draw another arc intersecting the first arc at point \(A\).
4. Join \(AB\) and \(AC\). \(\triangle ABC\) is the required triangle.
In simple words: Use the SSS (Side-Side-Side) criterion by drawing arcs of given side lengths from the endpoints of the base.

Question 8. Construct a triangle ABC in which AC = 4 cm, \(\angle A = 70^\circ\) and \(\angle C = 65^\circ\).
Answer: Steps of Construction:
1. Draw a line segment \(AC = 4 \text{ cm}\).
2. At point \(A\), construct an angle of \(70^\circ\) using a protractor (or compass if standard).
3. At point \(C\), construct an angle of \(65^\circ\).
4. Extend the arms of the two angles until they intersect each other at point \(B\).
5. \(\triangle ABC\) is the required triangle.
In simple words: Draw the base, construct the given angles at both ends of the base, and mark their intersection point as the third vertex.

Exam Tip: This is an ASA (Angle-Side-Angle) construction. Ensure the side is included between the two given angles.

Question 9. Construct a right angled triangle ABC in which hypotenuse AC = 4.5 cm, side AB = 3.5 cm.
Answer: Steps of Construction:
1. Draw a line segment \(AB = 3.5 \text{ cm}\).
2. At point \(A\), construct a right angle (\(90^\circ\)) using a compass.
3. With \(B\) as centre and radius equal to the hypotenuse \(4.5 \text{ cm}\), draw an arc cutting the perpendicular ray at point \(C\).
4. Join \(B\) to \(C\). \(\triangle ABC\) is the required right-angled triangle.
In simple words: Draw side AB, construct a \(90^\circ\) angle at A, and swing an arc of length 4.5 cm from B to intersect the perpendicular at C.

Exam Tip: For RHS (Right angle-Hypotenuse-Side) construction, always draw the right angle first, then use the compass for the hypotenuse.

Question 10. Is it possible to construct \(\triangle ABC\) in which AB = 6 cm, \(\angle A = 105^\circ\), \(\angle B = 85^\circ\)? If not, why not?
Answer: No, it is not possible to construct such a triangle.
Reason: Sum of all three interior angles of a triangle must always be equal to \(180^\circ\).
Here, given angles are \(\angle A = 105^\circ\) and \(\angle B = 85^\circ\).
Sum of these two angles = \(105^\circ + 85^\circ = 190^\circ\).
Since \(190^\circ > 180^\circ\), the angle sum property is violated.
In simple words: The triangle cannot be formed because the sum of just two angles exceeds the total angle sum of a triangle (\(180^\circ\)).

Exam Tip: Always check the angle sum property (\(\angle A + \angle B + \angle C = 180^\circ\)) before attempting any triangle construction.