GSEB Class 7 Maths Solutions Chapter 10 પ્રાયોગિક ભૂમિતિ Exercise 10.4

Question 1. Construct \( \triangle ABC \) where \( m\angle A = 60^\circ \), \( m\angle B = 30^\circ \) and \( AB = 5.8 \) cm.
Answer:

Construction Steps:
1. First, draw a line segment AB that is \( 5.8 \) cm long.
2. Next, with a compass, make an angle of \( 60^\circ \) at point B on the line segment AB, and draw the ray \( \overrightarrow{BM} \).
3. Then, bisect the angle \( \angle MBA \) to get the ray \( \overrightarrow{BX} \). This makes \( m\angle XBA = 30^\circ \).
4. After that, use a compass to draw a ray \( \overrightarrow{AY} \) from point A on the line segment AB, forming an angle of \( 60^\circ \).
5. Finally, where the rays \( \overrightarrow{BX} \) and \( \overrightarrow{AY} \) cross each other, name that point C. So, \( \triangle ABC \) is the desired triangle.
In simple words: To draw this triangle, first make a line \( AB \) of \( 5.8 \) cm. Then, at point \( A \), draw a \( 60^\circ \) angle, and at point \( B \), draw a \( 30^\circ \) angle. Where these two angle lines meet, that is point \( C \).

Exam Tip: When constructing triangles using ASA (Angle-Side-Angle), accurately measuring and drawing the angles from the endpoints of the given side is crucial for a correct construction.

Question 2. Construct \( \triangle PQR \) where \( PQ = 5 \) cm, \( m\angle PQR = 105^\circ \) and \( m\angle QRP = 40^\circ \). (Hint: Recall the angle sum property of a triangle.)
Answer:
The problem gives us the line segment PQ as \( 5 \) cm. To make a triangle, we typically need the size of two angles and the side between them (ASA criterion). Here, we have the sizes of \( \angle Q \) and \( \angle R \). We know that all three angles in a triangle always add up to \( 180^\circ \). Using this rule, we can easily find the measure of \( \angle P \).
\( m\angle P = 180^\circ – (m\angle Q + m\angle R) \)
\( = 180^\circ - (105^\circ + 40^\circ) \)
\( = 180^\circ – 145^\circ \)
\( = 35^\circ \)
Construction Steps:
1. First, draw a line segment PQ with a length of \( 5 \) cm.
2. Next, at point P on line segment PQ, use a protractor to draw a ray \( \overrightarrow{PM} \) that forms an angle of \( 35^\circ \).
3. Then, at point Q on line segment PQ, use a protractor to draw another ray \( \overrightarrow{QN} \) that forms an angle of \( 105^\circ \).
4. Finally, where ray \( \overrightarrow{PM} \) and ray \( \overrightarrow{QN} \) meet, label that point R. This makes \( \triangle PQR \) the triangle we needed to build.
In simple words: First, find the third angle \( \angle P \) by subtracting the other two angles from \( 180^\circ \). Then, draw the side \( PQ \). From point \( P \), draw the ray for \( \angle P \), and from point \( Q \), draw the ray for \( \angle Q \). The point where these two rays cross is \( R \).

Exam Tip: Always calculate the third angle using the angle sum property of a triangle when only two angles and an included side are given, to apply the ASA construction rule effectively.

Question 3. Check if it is possible to construct \( \triangle DEF \) where \( EF = 7.2 \) cm, \( m\angle E = 110^\circ \) and \( m\angle F = 80^\circ \). Justify your answer.
Answer:
In \( \triangle DEF \), we are given that the measure of \( \angle E \) is \( 110^\circ \) and the measure of \( \angle F \) is \( 80^\circ \).
If we add these two angles together, we get \( m\angle E + m\angle F = 110^\circ + 80^\circ = 190^\circ \).
We know that in any triangle, the sum of all three interior angles must exactly equal \( 180^\circ \).
However, here the sum of just two angles is \( 190^\circ \), which is more than \( 180^\circ \).
Because of this, it is not possible to have a triangle with these specific angle measurements. Therefore, we cannot construct such a triangle.
In simple words: The two angles given for \( \triangle DEF \) add up to \( 190^\circ \). But all three angles in any triangle must only add up to \( 180^\circ \). Since \( 190^\circ \) is more than \( 180^\circ \), you cannot draw a triangle with these angles.

Exam Tip: Before attempting to construct a triangle with given angles, always check if the sum of the angles is \( 180^\circ \). If not, construction is impossible.