GSEB Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.4

Question 1. Show that in a right-angled triangle, the hypotenuse is the longest side.
Answer:Given: Right \( \triangle ABC \) is a right-angled triangle where the right angle is at B, so \( \angle B = 90^\circ \).
In any triangle, the sum of all angles is \( 180^\circ \). Therefore,
\( \angle A + \angle B + \angle C = 180^\circ \)
Substitute the value of \( \angle B \):
\( \angle A + 90^\circ + \angle C = 180^\circ \)
\( \implies \angle A + \angle C = 180^\circ - 90^\circ \)
\( \implies \angle A + \angle C = 90^\circ \)
Since the sum of \( \angle A \) and \( \angle C \) is \( 90^\circ \), it means both \( \angle A \) and \( \angle C \) must be less than \( 90^\circ \). This shows that \( \angle B \) is the largest angle in the triangle. The side opposite the largest angle is always the longest side.
The side opposite to \( \angle B \) is AC, which is the hypotenuse.
Since \( \angle B > \angle A \) and \( \angle B > \angle C \), it follows that:
The side AC is longer than side BC (because the side opposite a greater angle is longer).
Also, the side AC is longer than side AB (because the side opposite a greater angle is longer).
Therefore, AC, which is the hypotenuse, is the longest side in this triangle.
In simple words: In a right triangle, the angle that is 90 degrees is the biggest angle. The side across from this biggest angle is called the hypotenuse, and it is always the longest side of the triangle.

Exam Tip: Remember that in any triangle, the side opposite the largest angle is the longest side, and the side opposite the smallest angle is the shortest side.

Question 2. In figure sides AB and AC of \( \triangle ABC \) are extended to points P and Q respectively. Also, \( \angle PBC < \angle QCB \). Show that AC > AB.
Answer:Given that \( \angle PBC < \angle QCB \).
Since \( \angle ABC \) and \( \angle PBC \) form a linear pair, their sum is \( 180^\circ \). So, \( \angle ABC = 180^\circ - \angle PBC \).
Similarly, \( \angle ACB \) and \( \angle QCB \) form a linear pair, so \( \angle ACB = 180^\circ - \angle QCB \).
Now, if \( \angle PBC < \angle QCB \), then subtracting both from \( 180^\circ \) reverses the inequality:
\( 180^\circ - \angle PBC > 180^\circ - \angle QCB \)
\( \implies \angle ABC > \angle ACB \)
In a triangle, the side opposite to a greater angle is longer.
Since \( \angle ABC > \angle ACB \), the side opposite \( \angle ABC \) (which is AC) must be greater than the side opposite \( \angle ACB \) (which is AB).
Therefore, \( AC > AB \).
In simple words: When two outer angles of a triangle are compared, if one is smaller, its inner angle is larger. The side opposite the larger inner angle is always the longer side.

Exam Tip: Remember that angles forming a linear pair add up to \( 180^\circ \). When an inequality is subtracted from a constant, the inequality sign reverses.

Question 3. In the figure, \( \angle B < \angle A \) and \( \angle C < \angle D \). Show that \( AD < BC \).
Answer:Given that \( \angle B < \angle A \). In \( \triangle AOB \), the side opposite the greater angle is longer. Since \( \angle A > \angle B \), the side opposite \( \angle A \) (OB) must be greater than the side opposite \( \angle B \) (OA).
So, \( OB > OA \) ........(1)
Also given that \( \angle C < \angle D \). In \( \triangle DOC \), since \( \angle D > \angle C \), the side opposite \( \angle D \) (OC) must be greater than the side opposite \( \angle C \) (OD).
So, \( OC > OD \) ........(2)
Now, add equation (1) and equation (2):
\( OB + OC > OA + OD \)
From the figure, \( OB + OC = BC \) and \( OA + OD = AD \).
\( \implies BC > AD \)
Therefore, \( AD < BC \).
In simple words: We are given that certain angles are smaller than others. This means the sides opposite the larger angles are longer. When we add these longer parts, we find that one whole line segment is longer than the other.

Exam Tip: Remember to clearly identify the triangles you are working with and apply the theorem that states the side opposite the greater angle is longer. Break down complex figures into simpler triangles if needed.

Question 4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that \( \angle A > \angle C \) and \( \angle B > \angle D \).
Answer:Construction: Join AC.
Proof: Consider \( \triangle ABC \). Given that AB is the smallest side of the quadrilateral. Therefore, in \( \triangle ABC \), \( AB < BC \) and \( AB < AC \).
The angle opposite to a longer side is greater.
So, since \( BC > AB \), \( \angle BAC > \angle BCA \) ........(1)
Now consider \( \triangle ACD \). Given that CD is the longest side of the quadrilateral. Therefore, in \( \triangle ACD \), \( CD > AD \) and \( CD > AC \).
Since \( CD > AD \), \( \angle CAD > \angle ACD \) ........(2)
Add equation (1) and equation (2):
\( \angle BAC + \angle CAD > \angle BCA + \angle ACD \)
From the figure, \( \angle BAC + \angle CAD = \angle A \) and \( \angle BCA + \angle ACD = \angle C \).
\( \implies \angle A > \angle C \)
Similarly, by joining diagonal BD, we can prove that \( \angle B > \angle D \).
In simple words: When a diagonal is drawn in a quadrilateral, it creates two triangles. By comparing sides and angles in these triangles, we can show that the angle at the vertex where the smallest side meets is larger than the angle at the vertex where the longest side meets. The same logic applies to the other pair of angles.

Exam Tip: For problems involving quadrilaterals, drawing a diagonal often helps to break it down into two triangles, allowing you to apply triangle properties. Remember to apply the property that the angle opposite the longer side is greater.

Question 5. In figure, PR > PQ and PS bisects \( \angle QPR \). Prove that \( \angle PSR > \angle PSQ \).
Answer:Given: In \( \triangle PQR \), \( PR > PQ \) and PS bisects \( \angle QPR \).
To Prove: \( \angle PSR > \angle PSQ \)
Proof: In \( \triangle PQR \), we are given \( PR > PQ \).
According to the angle-side relationship in a triangle, the angle opposite the longer side is greater.
\( \implies \angle PQR > \angle PRQ \) (or \( \angle PQS > \angle PRS \)) ........(1)
Since PS bisects \( \angle QPR \), it divides the angle into two equal parts:
\( \angle QPS = \angle RPS \) ........(2)
Now, consider \( \triangle PQS \). The exterior angle \( \angle PSR \) is equal to the sum of the two opposite interior angles.
\( \angle PSR = \angle PQS + \angle QPS \) ........(3)
Similarly, consider \( \triangle PRS \). The exterior angle \( \angle PSQ \) is equal to the sum of the two opposite interior angles.
\( \angle PSQ = \angle PRS + \angle RPS \) ........(4)
From (1), we know \( \angle PQS > \angle PRS \).
From (2), we know \( \angle QPS = \angle RPS \).
Add the two inequalities (treating the equality as an inequality):
\( \angle PQS + \angle QPS > \angle PRS + \angle RPS \)
Substitute (3) and (4) into this inequality:
\( \implies \angle PSR > \angle PSQ \)
Hence proved.
In simple words: We know one side is longer, so the angle opposite it is bigger. Since the line PS cuts the top angle in half, both halves are equal. Using the rule that an outside angle of a triangle equals the sum of the two opposite inside angles, we can show that the angle we need to prove is indeed larger.

Exam Tip: Remember the exterior angle theorem, which states that an exterior angle of a triangle is equal to the sum of its two opposite interior angles. This theorem is crucial for solving problems involving angle relationships like this one.

Question 6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Answer:Given: Let `l` be a line and P be a point not lying on line `l`. Let PQ be the perpendicular line segment from P to `l`, so \( PQ \perp l \) at Q. Let R be any other point on line `l`, different from Q.
To Prove: \( PQ < PR \)
Proof: Consider the triangle \( \triangle PQR \).
Since PQ is perpendicular to `l`, \( \angle PQR = 90^\circ \).
In any triangle, the sum of all angles is \( 180^\circ \). So, in \( \triangle PQR \):
\( \angle QPR + \angle PQR + \angle PRQ = 180^\circ \)
Substitute \( \angle PQR = 90^\circ \):
\( \angle QPR + 90^\circ + \angle PRQ = 180^\circ \)
\( \implies \angle QPR + \angle PRQ = 90^\circ \)
This implies that both \( \angle QPR \) and \( \angle PRQ \) are acute angles (less than \( 90^\circ \)).
Since \( \angle PQR = 90^\circ \), it is the largest angle in \( \triangle PQR \).
In a triangle, the side opposite the largest angle is the longest side.
Therefore, the side opposite \( \angle PQR \) (which is PR) must be longer than the side opposite \( \angle PRQ \) (which is PQ).
So, \( PR > PQ \)
This can also be written as \( PQ < PR \).
Since R was any arbitrary point on line `l` (other than Q), this proves that the perpendicular line segment (PQ) is the shortest distance from point P to line `l`.
In simple words: When you drop a straight line down to another line so it makes a perfect square corner (90 degrees), that line is always the shortest way to get from the point to the line. Any other line you draw will be longer because it forms a triangle where the 90-degree angle is the biggest angle.

Exam Tip: This is a fundamental concept in geometry. When asked to find the shortest distance from a point to a line, always remember it's the length of the perpendicular segment.