GSEB Class 7 Maths Solutions Chapter 10 Practical Geometry Exercise 10.4

Question 1. Construct \( \Delta ABC \), given \( m\angle A = 60^\circ \), \( m\angle B = 30^\circ \) and \( AB = 5.8 \) cm.
Answer:
Steps of construction:
I. Draw a line segment \( AB = 5.8 \) cm.
II. Construct \( \angle BAX = 60^\circ \) at A.
III. At B, construct \( \angle ABY = 30^\circ \).
IV. Let the rays AX and BY intersect at C.
Thus, \( \Delta ABC \) is the required triangle.
In simple words: First, draw the base line. Then, from each end of this line, draw rays at the given angles. The point where these rays cross becomes the third corner of your triangle, forming the complete shape you need.

Exam Tip: When constructing triangles, always start with the given side, then use a protractor to draw the angles from the endpoints of that side. The intersection of the two rays will complete your triangle.

Question 2. Construct \( \Delta PQR \) if \( PQ = 5 \) cm, \( m\angle PQR = 105^\circ \) and \( m\angle QRP = 40^\circ \). (Hint: Recall angle-sum property of a triangle).
Answer:
Here, the line segment PQ is provided. We can construct the triangle if the measure of \( \angle QPR \) is known. To find this, we use the angle sum property of a triangle.
Here, \( m\angle QPR = 180^\circ - (m\angle PQR + m\angle QRP) \)
\( \implies m\angle QPR = 180^\circ - (105^\circ + 40^\circ) \)
\( \implies m\angle QPR = 180^\circ - 145^\circ \)
\( \implies m\angle QPR = 35^\circ \)
Steps of construction:
I. Draw a line segment \( PQ = 5 \) cm.
II. At P, construct \( \angle QPX = 35^\circ \).
III. At Q, construct \( \angle PQY = 105^\circ \).
IV. Let the rays \( \overrightarrow{PX} \) and \( \overrightarrow{QY} \) intersect at R.
Thus, \( \Delta PQR \) is the required triangle.
In simple words: First, find the missing angle using the rule that all angles in a triangle add up to 180 degrees. Then, draw the given side, and from each end, draw rays using the known angles. Where these rays meet is the last point of your triangle.

Exam Tip: Always calculate the third angle using the angle-sum property before attempting construction if two angles and one side are given. This ensures you know all necessary parameters.

Question 3. Examine whether you can construct \( \Delta DEF \) such that \( EF = 7.2 \) cm, \( m\angle E = 110^\circ \) and \( m\angle F = 80^\circ \). Justify your answer.
Answer:
Here, the given angles are \( m\angle E = 110^\circ \) and \( m\angle F = 80^\circ \).
The sum of these two angles is \( m\angle E + m\angle F = 110^\circ + 80^\circ = 190^\circ \).
Since the sum of the three angles of a triangle is always \( 180^\circ \), and here the sum of just two angles is \( 190^\circ \), which is greater than \( 180^\circ \).
Therefore, \( \Delta DEF \) cannot be formed because the total of its two specified angles is more than \( 180^\circ \). A triangle cannot exist with such angle measures.
In simple words: A triangle's three angles must always add up to exactly 180 degrees. If only two angles already add up to more than 180 degrees, then it's impossible to make a triangle with those angles.

Exam Tip: Before starting any triangle construction with given angles, always check if the sum of the angles is consistent with the angle-sum property of triangles (i.e., sums to 180°).