GSEB Class 7 Maths Solutions Chapter 10 Practical Geometry InText Questions

Think, Discuss and Write (Page 195)

Question 1. In the above construction (see Page 193 NCERT Textbook), can you draw any other line through A that would be also parallel to the line l?
Answer: No, because through a specific point, only one line can be drawn that runs parallel to another given line.
In simple words: Only one parallel line can pass through a point outside another line.

Exam Tip: Remember the fundamental postulate of Euclidean geometry which states that through a point not on a given line, there is exactly one line parallel to the given line.

Question 2. Can you slightly modify the above construction (see Page 193 NCERT Textbook), to use the idea of equal corresponding angles instead of equal alternate angles?
Answer: Yes, it is possible to change the construction slightly to use the concept of matching corresponding angles rather than equal alternate angles.
In simple words: Yes, you can use corresponding angles instead of alternate angles for this construction.

Exam Tip: Both alternate interior angles and corresponding angles are crucial for proving lines parallel. Understanding how to use either concept for construction shows a deeper grasp of geometric principles.

Think, Discuss and Write (Page 198)

Question 1. A student attempted to draw a triangle whose rough figure is given here. He drew QR first. Then with as centre, he drew an arc of 3 cm and with R as centre, he drew an arc of 2 cm. But he could not get P. What is the reason? What property of triangle do you know in connection with this problem? Can such a triangle exist? (Remember the property of triangles 'The sum of any two sides of a triangle is always greater than the third side;!)
Answer: Such a triangle cannot exist because a triangle can only be formed if the sum of the lengths of any two sides is greater than the third side. However, in this case, \( 2 \text{ cm} + 3 \text{ cm} \), which is \( 5 \text{ cm} \), is less than \( 6 \text{ cm} \).
In simple words: This triangle cannot be built. The rule is, two sides added together must be longer than the third side, but here, 2cm plus 3cm (5cm) is shorter than 6cm.

Exam Tip: Always check the triangle inequality theorem before attempting to construct a triangle. If \( a+b > c \), \( b+c > a \), and \( a+c > b \) are not all true, the triangle cannot be formed.

Think, Discuss and Write (Page 200)

Question 1. In \( \triangle ABC \) if AB = 3 cm, AC = 5 cm and \( m\angle C = 30^\circ \). Can we draw this triangle?
Answer: We can draw side AC measuring \( 5 \text{ cm} \) and construct \( \angle C = 30^\circ \). Line segment CA serves as one arm of \( \angle C \). Point B needs to lie on the other arm of \( \angle C \). Nevertheless, we notice that point B cannot be precisely located. Thus, this data is not enough to construct \( \triangle ABC \).
In simple words: We can draw parts of it, but we can't find the exact spot for point B, so we cannot completely draw this triangle with the given information.

Exam Tip: For SAS (Side-Angle-Side) or ASA (Angle-Side-Angle) constructions, the angle must be between the two given sides (SAS) or the side must be between the two given angles (ASA). Here, the given angle (C) is not between sides AB and AC, nor is AC between angles. This often leads to ambiguity.

Think, Discuss and Write (Page 202)

Question 1. In the above example (see Page 201, NCERT Textbook), length of a side and measures of two angles were given. Now study the following problem: In \( \triangle ABC \), if AC = 7 cm, \( m\angle A = 60^\circ \) and \( m\angle B = 50^\circ \), can you draw the triangle? (Angle-sum property of a triangle may help you!)
Answer: We are provided with the line segment AC. \( \angle A \) is given, but \( \angle C \) is not. We can determine \( \angle C \) by using the angle-sum property of a triangle.
\( \angle C = 180^\circ - (\angle B + \angle A) \)
\( = 180^\circ - (60^\circ + 50^\circ) \)
\( = 180^\circ - 110^\circ \)
\( = 70^\circ \)
Now, with the help of \( \angle C \), we can certainly construct the triangle.
In simple words: Yes, we can draw this triangle. We find the missing angle C using the rule that all angles in a triangle add up to 180 degrees. Once we have angle C, we have enough information to construct the triangle.

Exam Tip: When given two angles and a non-included side (AAS criteria), first find the third angle using the angle sum property. This converts the problem into an ASA (Angle-Side-Angle) construction, which is always possible.

Miscellaneous Questions (Page 204)

Question 1. Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and, say why you cannot construct them. Construct rest of the triangles.
Answer:
1. For \( \triangle ABC \): \( m\angle A = 85^\circ \); \( m\angle B = 115^\circ \); AB = 5 cm.
We have: \( m\angle A = 85^\circ \) and \( m\angle B = 115^\circ \).
But here, \( \angle A + \angle B = 85^\circ + 115^\circ = 200^\circ \).
This sum is greater than \( 180^\circ \). We know that the sum of angles in any triangle must be exactly \( 180^\circ \). Therefore, a triangle with these angle measures cannot be constructed.
2. For \( \triangle PQR \): \( m\angle Q = 30^\circ \); \( m\angle R = 60^\circ \); QR = 4.7 cm.
We have: \( m\angle Q = 30^\circ \), \( m\angle R = 60^\circ \), and QR = \( 4.7 \text{ cm} \).
Since \( \angle P + \angle Q + \angle R = 180^\circ \), \( \angle P = 180^\circ - (30^\circ + 60^\circ) = 180^\circ - 90^\circ = 90^\circ \).
This triangle can be constructed (ASA criteria).
Steps of construction:
I. Draw a line segment QR = \( 4.7 \text{ cm} \).
II. At Q, construct \( \angle RQX = 30^\circ \).
III. At R, construct \( \angle QRY = 60^\circ \).
IV. Let the rays \( \overrightarrow{QX} \) and \( \overrightarrow{RY} \) intersect at P.
Thus, \( \triangle PQR \) is the required triangle.

3. For \( \triangle ABC \): \( m\angle A = 70^\circ \); \( m\angle B = 50^\circ \); AC = 3 cm.
We have: \( m\angle A = 70^\circ \), \( m\angle B = 50^\circ \), and AC = \( 3 \text{ cm} \).
Since \( \angle A + \angle B + \angle C = 180^\circ \), \( \angle C = 180^\circ - (70^\circ + 50^\circ) = 180^\circ - 120^\circ = 60^\circ \).
This triangle can be constructed (ASA criteria).
Steps of construction:
I. Draw a line segment AC = \( 3 \text{ cm} \).
II. At A, construct \( \angle CAX = 70^\circ \).
III. At C, construct \( \angle ACY = 60^\circ \).
IV. Let the rays \( \overrightarrow{AX} \) and \( \overrightarrow{CY} \) meet at B.
Thus, \( \triangle ABC \) is the required triangle.

4. For \( \triangle LMN \): \( m\angle L = 60^\circ \); \( m\angle N = 120^\circ \); LM = 5 cm.
We have: \( m\angle L = 60^\circ \) and \( m\angle N = 120^\circ \).
Since, in \( \triangle LMN \), \( \angle L + \angle M + \angle N = 180^\circ \).
But here, \( \angle L + \angle N = 60^\circ + 120^\circ = 180^\circ \).
This implies that \( \angle L + \angle N + \angle M = 180^\circ + \angle M \). For the sum to be \( 180^\circ \), \( \angle M \) would have to be \( 0^\circ \), which is impossible for a triangle.
Therefore, \( \triangle LMN \) cannot be constructed.
5. For \( \triangle ABC \): BC = 2 cm; AB = 4 cm; AC = 2 cm.
We have: \( BC = 2 \text{ cm, } AB = 4 \text{ cm, } AC = 2 \text{ cm} \).
We know that a triangle is only possible when the total length of any two sides is greater than the length of the third side.
Here, BC + AC = \( 2 \text{ cm} + 2 \text{ cm} = 4 \text{ cm} \).
This sum is equal to AB (which is \( 4 \text{ cm} \)). Since \( BC + AC \) is not greater than AB, \( \triangle ABC \) is not possible.
6. For \( \triangle PQR \): PQ = 3.5 cm; QR = 4 cm; PR = 3.5 cm.
We have: \( PQ = 3.5 \text{ cm, } QR = 4 \text{ cm, } PR = 3.5 \text{ cm} \).
This triangle can be constructed (SSS criteria), as all side lengths satisfy the triangle inequality (e.g., \( 3.5 + 3.5 > 4 \), \( 3.5 + 4 > 3.5 \)).
Steps of construction:
I. Draw a line segment PQ = \( 3.5 \text{ cm} \).
II. With center P and radius \( 3.5 \text{ cm} \), draw an arc.
III. With center Q and radius \( 4 \text{ cm} \), draw another arc, such that it cuts the previous arc at R.
IV. Join RP and RQ.
Thus, \( \triangle PQR \) is the required triangle.

7. For \( \triangle XYZ \): XY = 3 cm; YZ = 4 cm; XZ = 5 cm.
We have: \( XY = 3 \text{ cm, } YZ = 4 \text{ cm, } XZ = 5 \text{ cm} \).
This is a right-angled triangle (since \( 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \)). It can be constructed (SSS criteria).
Steps of construction:
I. Draw a line segment YZ = \( 4 \text{ cm} \).
II. With center Y and radius \( 3 \text{ cm} \), draw an arc.
III. With center Z and radius \( 5 \text{ cm} \), draw another arc, such that it cuts the previous arc at X.
IV. Join XY and XZ.
Thus, \( \triangle XYZ \) is the required triangle.

8. For \( \triangle DEF \): DE = 4.5 cm; EF = 5.5 cm; DF = 4 cm.
We have: \( DE = 4.5 \text{ cm, } EF = 5.5 \text{ cm, } DF = 4 \text{ cm} \).
This triangle can be constructed (SSS criteria), as all side lengths satisfy the triangle inequality (e.g., \( 4.5 + 4 > 5.5 \), \( 4.5 + 5.5 > 4 \), \( 4 + 5.5 > 4.5 \)).
Steps of construction:
I. Draw a line segment EF = \( 5.5 \text{ cm} \).
II. With center E and radius \( 4.5 \text{ cm} \), draw an arc.
III. With center F and radius \( 4 \text{ cm} \), draw another arc such that it intersects the previous arc at D.
IV. Join DE and DF.
Thus, \( \triangle DEF \) is the required triangle.

Exam Tip: For problems involving triangle construction, first evaluate if the given conditions satisfy the triangle inequality theorem (sum of any two sides is greater than the third side) or angle sum property (sum of angles is 180°). This preliminary check helps determine if construction is possible.