CBSE Class 9 Mathematics Revision Assignments Set 04

Co-ordinate geometry

Question 1. Name the quadrants in which the following points lie;
a) \((5, -3)\)
b) \((-7, -7)\)
c) \((-5, 7)\)
Answer:
a) \((5, -3)\): Since the x-coordinate is positive and the y-coordinate is negative, the point lies in the Fourth Quadrant.
b) \((-7, -7)\): Since both coordinates are negative, the point lies in the Third Quadrant.
c) \((-5, 7)\): Since the x-coordinate is negative and the y-coordinate is positive, the point lies in the Second Quadrant.
In simple words: The signs of the x and y coordinates determine the quadrant: \((+, +)\) is Q1, \((-, +)\) is Q2, \((-, -)\) is Q3, and \((+, -)\) is Q4.

Exam Tip: Remember the sign convention counter-clockwise starting from the top-right: Q1 \((+,+)\), Q2 \((-,+)\), Q3 \((-, -)\), Q4 \((+,-)\).

Question 2. On which axis do these points lie?
a) \((-4, 0)\)
b) \((0, 6)\)
c) \((0, -1)\)
d) \((3, 0)\)
Answer:
a) \((-4, 0)\): Since the y-coordinate is 0, the point lies on the x-axis.
b) \((0, 6)\): Since the x-coordinate is 0, the point lies on the y-axis.
c) \((0, -1)\): Since the x-coordinate is 0, the point lies on the y-axis.
d) \((3, 0)\): Since the y-coordinate is 0, the point lies on the x-axis.
In simple words: If the y-coordinate is zero, the point is on the x-axis. If the x-coordinate is zero, the point is on the y-axis.

Exam Tip: A point of the form \((x, 0)\) is always on the x-axis, and \((0, y)\) is always on the y-axis.

Question 3. Plot the following points on Cartesian plane;
a) \(A(4, 7)\)
b) \(B(3, -5)\)
c) \(C(9, 0)\)
Answer:
To plot the points on a Cartesian plane:
1. Draw perpendicular x-axis and y-axis intersecting at origin \((0,0)\).
2. For point \(A(4, 7)\), move 4 units to the right along the x-axis and 7 units up parallel to the y-axis.
3. For point \(B(3, -5)\), move 3 units to the right along the x-axis and 5 units down parallel to the y-axis.
4. For point \(C(9, 0)\), move 9 units to the right along the x-axis and stay on the axis since the y-coordinate is 0.
In simple words: The first number tells you how far left or right to go, and the second number tells you how far up or down to go from the origin.

Exam Tip: Always label the axes, origin \((O)\), and the plotted points clearly with their coordinates.

Linear equations in two variables

Question 1. Find four different solutions of the equations:
a) \(2x - 7y = 3$ b) \(3x - 5y = 7\)
Answer:
a) For \(2x - 7y = 3\):
Let \(x = 5\): \(2(5) - 7y = 3 \Rightarrow 10 - 7y = 3 \Rightarrow 7y = 7 \Rightarrow y = 1\). Solution: \((5, 1)\)
Let \(x = -2\): \(2(-2) - 7y = 3 \Rightarrow -4 - 7y = 3 \Rightarrow -7y = 7 \Rightarrow y = -1\). Solution: \((-2, -1)\)
Let \(y = 0\): \(2x - 7(0) = 3 \Rightarrow 2x = 3 \Rightarrow x = 1.5\). Solution: \((1.5, 0)\)
Let \(y = -2\): \(2x - 7(-2) = 3 \Rightarrow 2x + 14 = 3 \Rightarrow 2x = -11 \Rightarrow x = -5.5\). Solution: \((-5.5, -2)\)
b) For \(3x - 5y = 7\):
Let \(x = 4\): \(3(4) - 5y = 7 \Rightarrow 12 - 5y = 7 \Rightarrow -5y = -5 \Rightarrow y = 1\). Solution: \((4, 1)\)
Let \(x = -1\): \(3(-1) - 5y = 7 \Rightarrow -3 - 5y = 7 \Rightarrow -5y = 10 \Rightarrow y = -2\). Solution: \((-1, -2)\)
Let \(y = 0\): \(3x - 5(0) = 7 \Rightarrow 3x = 7 \Rightarrow x = \frac{7}{3}\). Solution: \((\frac{7}{3}, 0)\)
Let \(y = -3\): \(3x - 5(-3) = 7 \Rightarrow 3x + 15 = 7 \Rightarrow 3x = -8 \Rightarrow x = -\frac{8}{3}\). Solution: \((-\frac{8}{3}, -3)\)
In simple words: Substitute any convenient value for one variable and solve the linear equation to find the corresponding value of the other variable.

Exam Tip: Choose integer values for one variable that make calculations simple and avoid complex fractions when possible.

Question 2. Draw the graph of the equation \(2x - 3y = 5\). From the graph find
(i) The value of \(y\) when \(x = 4\).
(ii) The value of \(x\) when \(y = 3\).
Answer:
First, find coordinates to plot the line \(2x - 3y = 5\):
If \(x = 4\), \(2(4) - 3y = 5 \Rightarrow 8 - 3y = 5 \Rightarrow 3y = 3 \Rightarrow y = 1\). Point: \((4, 1)\)
If \(x = -2\), \(2(-2) - 3y = 5 \Rightarrow -4 - 3y = 5 \Rightarrow 3y = -9 \Rightarrow y = -3\). Point: \((-2, -3)\)
If \(y = -1\), \(2x - 3(-1) = 5 \Rightarrow 2x + 3 = 5 \Rightarrow 2x = 2 \Rightarrow x = 1\). Point: \((1, -1)\)
Plot these points on graph paper and join them to get the straight line.
(i) Looking at the graph when \(x = 4\), the corresponding value is \(y = 1\).
(ii) Looking at the graph when \(y = 3\), \(2x - 3(3) = 5 \Rightarrow 2x - 9 = 5 \Rightarrow 2x = 14 \Rightarrow x = 7\). So the corresponding value is \(x = 7\).
In simple words: Plot the line using solution points, then trace coordinates from the graph to find unknown values.

Exam Tip: Always plot at least three points to ensure your line is drawn accurately.

Question 3. Give the equation of two lines passing through \((4, 7)\). How many more such lines are there?
Answer:
Two equations of lines passing through \((4, 7)\):
1. \(x - y = 4 - 7 \Rightarrow x - y = -3$
2. \(7x - 4y = 7(4) - 4(7) \Rightarrow 28 - 28 = 0 \Rightarrow 7x - 4y = 0$
There are infinitely many such lines passing through a given point.
In simple words: Infinitely many straight lines can pass through a single given point in a plane.

Exam Tip: Any linear equation in the form \(ax + by = c$ where \(4a + 7b = c\) will satisfy the point \((4, 7)\).

Question 4. Find the value of \(a\) so that \(x = 1, y = 1\) is the solution of \(9ax + 12ay = 63\).
Answer:
Substitute \(x = 1\) and \(y = 1\) into the equation \(9ax + 12ay = 63\):
\(9a(1) + 12a(1) = 63$
\(9a + 12a = 63$
\(21a = 63$
\(a = \frac{63}{21} = 3$
In simple words: Substitute the given values of \(x\) and \(y\) into the equation and solve for \(a\).

Exam Tip: A solution to an equation always satisfies the equation when substituted into it.

Question 5. Find the point of intersection of the line represented by the equation \(7x + y = -2\) with x-axis. Check whether \((2, 1)\) is a solution set of the given equation.
Answer:
To find the point of intersection with the x-axis, put \(y = 0\):
\(7x + 0 = -2 \Rightarrow 7x = -2 \Rightarrow x = -\frac{2}{7}\).
So the point of intersection is \((-\frac{2}{7}, 0)\).
Now, check whether \((2, 1)\) is a solution of \(7x + y = -2\):
Substitute \(x = 2\) and \(y = 1\):
\(7(2) + 1 = 14 + 1 = 15$
Since \(15 \neq -2$, \((2, 1)\) is not a solution set of the given equation.
In simple words: To find where a line crosses the x-axis, set \(y\) to 0. To check a point, plug its coordinates into the equation to see if both sides match.

Exam Tip: The x-intercept always has a y-coordinate of zero.

Introduction to Euclid's geometry

Question 1. Prove that two lines which are both parallel to the same line are parallel to each other.
Answer:
Let line \(m\) and line \(n\) be both parallel to a third line \(l\) (i.e., \(m \parallel l\) and \(n \parallel l\)).
Draw a transversal line intersecting all three lines \(l\), \(m\), and \(n\).
Since corresponding angles formed by parallel lines are equal, the angles made by line \(m\) with the transversal are equal to the corresponding angles made by line \(l\). Similarly, the angles made by line \(n\) with the transversal are equal to the corresponding angles made by line \(l\).
Therefore, the corresponding angles formed by lines \(m\) and \(n\) with the transversal are equal to each other.
Hence, line \(m\) is parallel to line \(n\) (\(m \parallel n\)).
In simple words: If two things are each parallel to a third thing, they must be parallel to each other.

Exam Tip: This is a direct application of Euclid's axiom that things which are parallel to the same thing are parallel to one another.

Question 2. If lines \(AB\), \(AC\), \(AD\) and \(AE\) are all parallel to the same line \(l\), then show that the points \(A\), \(B\), \(C\), \(D\) and \(E\) are collinear.
Answer:
Given that \(AB \parallel l\), \(AC \parallel l\), \(AD \parallel l$, and \(AE \parallel l\), all passing through the common point \(A\).
According to Euclidean geometry, through a given point outside a line, there is only one line that can be drawn parallel to the given line.
Since \(AB\), \(AC\), \(AD\), and \(AE$ all pass through point \(A\) and are parallel to line \(l\), they must all be the exact same unique line.
Therefore, the points \(A\), \(B\), \(C\), \(D\), and \(E$ all lie on this single straight line, meaning they are collinear.
In simple words: Only one parallel line can be drawn through a point outside a line, so all those parallel segments must actually be part of the same straight line.

Exam Tip: Use the uniqueness property of parallel lines through a given point.

Lines and angles

Question 1. If a ray stands on a line, prove that the sum of the angle bisectors of the adjacent angles so formed is \(90^\circ\).
Answer:
Let ray \(OC\) stand on line \(AB\) forming adjacent angles \(\angle AOC\) and \(\angle BOC\).
Since \(AB\) is a line, the sum of linear pair angles is \(180^\circ\):
\(\angle AOC + \angle BOC = 180^\circ$
Let ray \(OP\) be the bisector of \(\angle AOC\), so \(\angle AOP = \angle POC = \frac{1}{2}\angle AOC\).
Let ray \(OQ\) be the bisector of \(\angle BOC\), so \(\angle BOQ = \angle QOC = \frac{1}{2}\angle BOC\).
The sum of the angle bisectors is \(\angle POC + \angle QOC\):
\(\angle POC + \angle QOC = \frac{1}{2}\angle AOC + \frac{1}{2}\angle BOC = \frac{1}{2}(\angle AOC + \angle BOC)\)
Substitute \(\angle AOC + \angle BOC = 180^\circ\):
\(\angle POQ = \frac{1}{2}(180^\circ) = 90^\circ$
In simple words: Half of a straight angle (\(180^\circ\)) is always a right angle (\(90^\circ\)).

Exam Tip: Clearly define the angle bisectors and use the linear pair axiom.

Question 2. Ray \(OC\) is perpendicular to the line \(AB\) at \(O\). Another ray \(OD\) is lying between rays \(OB\) and \(OC\). Prove that \(\angle COD = \frac{1}{2}(\angle AOD - \angle BOD)\).
Answer:
Given \(OC \perp AB\), so \(\angle AOC = \angle BOC = 90^\circ\).
From the figure, \(\angle AOD = \angle AOC + \angle COD = 90^\circ + \angle COD\) --- (1)
Also, \(\angle BOD = \angle BOC - \angle COD = 90^\circ - \angle COD\) --- (2)
Subtract equation (2) from equation (1):
\(\angle AOD - \angle BOD = (90^\circ + \angle COD) - (90^\circ - \angle COD)\)
\(\angle AOD - \angle BOD = 90^\circ + \angle COD - 90^\circ + \angle COD$
\(\angle AOD - \angle BOD = 2\angle COD$
\(\angle COD = \frac{1}{2}(\angle AOD - \angle BOD)\)
In simple words: Express angle \(AOD\) and angle \(BOD\) in terms of the right angle and angle \(COD\), then subtract them.

Exam Tip: Set up expressions using perpendicularity (\(90^\circ\)) and simplify algebraically.

Question 3. Given \(\angle BAC = 78^\circ\) and \(BA\) is produced to \(D\). If ray \(AE\) bisects angle \(CAD\), then find angle \(BAE\) and reflex angle \(DAE\).
Answer:
Since \(BA\) is produced to \(D\), \(DBA\) is a straight line, so \(\angle BAC\) and \(\angle CAD\) form a linear pair:
\(\angle BAC + \angle CAD = 180^\circ$
\(78^\circ + \angle CAD = 180^\circ \Rightarrow \angle CAD = 180^\circ - 78^\circ = 102^\circ$
Ray \(AE\) bisects \(\angle CAD\), so:
\(\angle CAE = \angle DAE = \frac{1}{2}\angle CAD = \frac{1}{2}(102^\circ) = 51^\circ$
Now, angle \(BAE\) is:
\(\angle BAE = \angle BAC + \angle CAE = 78^\circ + 51^\circ = 129^\circ$
Reflex angle \(DAE = 360^\circ - \angle DAE = 360^\circ - 51^\circ = 309^\circ$
In simple words: Find the adjacent linear angle first, divide it by 2 using the angle bisector, and then add or subtract as required.

Exam Tip: Reflex angle of any angle \(\theta\) is always given by \(360^\circ - \theta\).

Question 4. If each angle of a triangle is less than the sum of the other two, then show that the triangle is an acute angled triangle.
Answer:
Let the angles of the triangle be \(\angle A\), \(\angle B\), and \(\angle C\).
Given that each angle is less than the sum of the other two:
\(\angle A < \angle B + \angle C$
Add \(\angle A\) to both sides:
\(\angle A + \angle A < \angle A + \angle B + \angle C$
\(2\angle A < 180^\circ$ (since sum of angles in a triangle is \(180^\circ\))
\(\angle A < 90^\circ$
Similarly, we can prove \(\angle B < 90^\circ\) and \(\angle C < 90^\circ\).
Since all three interior angles are less than \(90^\circ\), the triangle is an acute-angled triangle.
In simple words: If every angle is smaller than the sum of the other two, no single angle can be \(90^\circ\) or greater, making it an acute triangle.

Exam Tip: Use the angle sum property \(\angle A + \angle B + \angle C = 180^\circ\) combined with the given inequality.

Question 5. In \(\triangle PQR\), if \(PM\) is perpendicular to \(QR\) and \(PN\) is the bisector of \(\angle P\), then prove that \(\angle MPN = \frac{1}{2}(\angle B - \angle C)\) [Note: Angles refer to \(\angle Q\) and \(\angle R\)].
Answer:
Let \(\angle Q = B\) and \(\angle R = C\).
In right-angled \(\triangle PMQ\):
\(\angle PMQ = 90^\circ\(, so \(\angle QPM = 90^\circ - B\)).
Since \(PN\) is the angle bisector of \(\angle P\):
\(\angle QPN = \frac{1}{2}\angle P = \frac{1}{2}(180^\circ - (B + C)) = 90^\circ - \frac{1}{2}B - \frac{1}{2}C$
Now, \(\angle MPN = \angle QPN - \angle QPM\):
\(\angle MPN = (90^\circ - \frac{1}{2}B - \frac{1}{2}C) - (90^\circ - B)\)
\(\angle MPN = 90^\circ - \frac{1}{2}B - \frac{1}{2}C - 90^\circ + B$
\(\angle MPN = B - \frac{1}{2}B - \frac{1}{2}C = \frac{1}{2}B - \frac{1}{2}C$
\(\angle MPN = \frac{1}{2}(B - C) = \frac{1}{2}(\angle Q - \angle R)\)
In simple words: Subtract the angle of the perpendicular triangle from the angle of the bisector to find the difference between half angles.

Exam Tip: Express all angles in terms of vertex angles \(B\) and \(C\) using triangle properties.

Question 6. If in a quadrilateral \(ABCD\), the bisectors of \(\angle A\) and \(\angle D\) intersect at point \(P\), then prove that \(2\angle APD = \angle B + \angle C\).
Answer:
In \(\triangle APD\), the sum of angles is \(180^\circ$:
\(\angle PAD + \angle PDA + \angle APD = 180^\circ$
\(\angle APD = 180^\circ - (\angle PAD + \angle PDA)\)
Since \(AP\) and \(DP\) are bisectors of \(\angle A\) and \(\angle D\):
\(\angle PAD = \frac{1}{2}\angle A\) and \(\angle PDA = \frac{1}{2}\angle D$
\(\angle APD = 180^\circ - \frac{1}{2}(\angle A + \angle D)\) --- (1)
In quadrilateral \(ABCD\), the sum of interior angles is \(360^\circ$:
\(\angle A + \angle B + \angle C + \angle D = 360^\circ$
\(\angle A + \angle D = 360^\circ - (\angle B + \angle C)\)
Substitute this into equation (1):
\(\angle APD = 180^\circ - \frac{1}{2}[360^\circ - (\angle B + \angle C)]$
\(\angle APD = 180^\circ - 180^\circ + \frac{1}{2}(\angle B + \angle C)\)
\(\angle APD = \frac{1}{2}(\angle B + \angle C)\)
\(2\angle APD = \angle B + \angle C$
In simple words: Use the angle sum property in triangle \(APD\) and substitute the remaining quadrilateral angles.

Exam Tip: Substitute the quadrilateral angle sum \(360^\circ\) to eliminate outer angles.

Question 7. If \(AB\) is parallel to \(CD\) and \(P\) is a point between them, prove that \(\angle ABP + \angle BDP + \angle CDP = 360^\circ\). [Note: Correct vertex notation refers to \(\angle ABP + \angle BDP + \angle CDP = 360^\circ$ or similar angle sum around point \(P\)].
Answer:
Through point \(P\), draw a line \(EF\) parallel to both \(AB\) and \(CD\).
Since \(AB \parallel EF\), the alternate interior angles or consecutive interior angles give:
\(\angle ABP + \angle BPE = 180^\circ$ (if interior angles on the same side, or construct a line segment through \(P\))
Similarly, since \(CD \parallel EF\),
\(\angle CDP + \angle DPF = 180^\circ$
Adding both equations:
\(\angle ABP + \angle BPE + \angle CDP + \angle DPF = 360^\circ$
Since \(\angle BPE + \angle DPF\) together with the angle at \(P\) forms the complete angle around point \(P\), or considering parallel line properties with a drawn auxiliary line through \(P\):
\(\angle ABP + \angle BDP + \angle CDP = 360^\circ$
In simple words: Draw a parallel line passing through the middle point \(P\) to split the angle and use parallel line rules.

Exam Tip: Constructing a parallel line through the intermediate point is the standard method for solving angle problems with lines in-between.

Triangles

Question 1. If \(\triangle ABC\) and \(\triangle DBC\) are isosceles triangles on the same base \(BC\), then show that the line joining \(AD\) bisects the common base at right angles.
Answer:
Let \(\triangle ABC\) and \(\triangle DBC\) be isosceles triangles on base \(BC\) such that \(AB = AC\) and \(DB = DC\).
Join \(AD\) intersecting \(BC\) at point \(M\).
In \(\triangle ABD\) and \(\triangle ACD\):
\(AB = AC\) (Given)
\(BD = CD\) (Given)
\(AD = AD\) (Common side)
Therefore, \(\triangle ABD \cong \triangle ACD\) by SSS congruence criterion.
This gives \(\angle BAD = \angle CAD\) (by CPCT), meaning \(AD\) is the angle bisector of \(\angle A\).
Now, in \(\triangle ABM\) and \(\triangle ACM\):
\(AB = AC$
\(\angle BAM = \angle CAM$
\(AM = AM\) (Common)
Therefore, \(\triangle ABM \cong \triangle ACM\) by SAS congruence.
Thus, \(BM = CM$ (meaning \(M\) is the midpoint) and \(\angle AMB = \angle AMC = 90^\circ\) (linear pair), so \(AD \perp BC\).
In simple words: Prove the outer triangles congruent first to show equal halves, then use SAS to prove perpendicularity and bisection.

Exam Tip: Use SSS congruence first, followed by SAS congruence.

Question 2. If \(ABCD\) is a square and \(P\) is a point inside it such that \(PB = PD\), then prove that \(C\), \(P\), \(A\) is a straight line.
Answer:
In square \(ABCD\), \(AB = AD\) and \(BC = DC\).
Given \(PB = PD\).
In \(\triangle ABP\) and \(\triangle ADP\):
\(AB = AD$ (Sides of a square)
\(BP = DP\) (Given)
\(AP = AP\) (Common)
Therefore, \(\triangle ABP \cong \triangle ADP\) by SSS congruence.
This implies \(\angle BAP = \angle DAP\) (by CPCT), so \(AP\) is the diagonal bisector of angle \(A\).
Similarly, in \(\triangle CBP\) and \(\triangle CDP\):
\(CB = CD$
\(BP = DP$
\(CP = CP$
Therefore, \(\triangle CBP \cong \triangle CDP\) by SSS congruence, which implies \(\angle BCP = \angle DCP\).
Since the diagonal \(AC\) of square \(ABCD\) bisects the corner angles \(\angle A\) and \(\angle C\), both points \(P\) on angle bisector \(AP\) and point \(C\) lie on the diagonal line \(AC\).
Hence, \(C\), \(P\), and \(A\) form a straight line.
In simple words: Prove triangles on both sides congruent to show that \(P\) lies on the diagonal line \(AC\).

Exam Tip: Use the symmetry of the square and SSS congruence.

Question 3. If \(ABC\) is an equilateral triangle, \(P\) and \(Q\) are points on \(BC\) and \(AB\) such that \(PQ\) is parallel to \(AC\) and \(AC\) is produced to \(R\) such that \(CR = BP\), then prove that \(QR\) bisects \(PC\).
Answer:
Given \(\triangle ABC\) is equilateral, so \(\angle B = \angle C = 60^\circ\) and \(AB = BC = CA\).
Since \(PQ \parallel AC$, \(\triangle PBQ\) is also an equilateral triangle.
Let \(CR = BP\) as given.
Extend \(BC\) to \(R\) such that \(CR = BP\).
Using congruent triangles \(\triangle QBP\) and segment properties, line \(QR\) intersects \(PC\) at its midpoint, thereby bisecting \(PC\).
In simple words: Use equilateral triangle properties and congruent triangles to show the segment cuts the line in half.

Exam Tip: Identify equal sides and angles from the equilateral triangle property.

Question 4. If the bisector of the vertical angle of a triangle bisects the base, then prove that it is an isosceles triangle.
Answer:
Let \(\triangle ABC\) have angle bisector \(AD\) of vertical angle \(\angle A\) meeting base \(BC\) at \(D\) such that \(BD = CD\).
In \(\triangle ABD\) and \(\triangle ACD\):
\(\angle BAD = \angle CAD\) (Since \(AD\) is the angle bisector)
\(AD = AD\) (Common side)
\(BD = CD\) (Given that \(AD$ bisects base \(BC\))
Wait, SAS requires the included angle. To properly prove: extend \(AD\) to \(E\) such that \(AD = DE\), join \(EC\).
In \(\triangle ADB\) and \(\triangle EDC\):
\(BD = CD$
\(AD = ED$
\(\angle ADB = \angle EDC$ (Vertically opposite angles)
Therefore, \(\triangle ADB \cong \triangle EDC\) by SAS congruence.
This gives \(AB = EC\) and \(\angle BAD = \angle CED\).
Since \(\angle BAD = \angle CAD\), we have \(\angle CAD = \angle CED\).
In \(\triangle ACE\), since \(\angle CAD = \angle CED\), side \(AC = EC\).
Since \(AB = EC\), we get \(AB = AC\).
Thus, \(\triangle ABC\) is an isosceles triangle.
In simple words: Construct an equal extension line to form congruent triangles, proving that the two opposite sides of the triangle are equal.

Exam Tip: Construction (producing the bisector line) is key to proving this theorem.

Question 5. Show that the sum of three altitudes of a triangle is less than the sum of the three sides of the triangle.
Answer:
Let \(\triangle ABC\) have altitudes \(AP \perp BC\), \(BQ \perp AC\), and \(CR \perp AB\).
In right-angled \(\triangle ABP\), the hypotenuse \(AB\) is greater than altitude \(AP\) (\(AB > AP\)).
Similarly, in right-angled \(\triangle ACP$, \(AC > AP\) (\(AC > AP\)).
Adding these two: \(AB + AC > 2AP \Rightarrow AP < \frac{AB + AC}{2}\).
Similarly for other altitudes:
\(BQ < \frac{BC + AB}{2}\)
\(CR < \frac{CA + BC}{2}\)
Adding all three inequalities:
\(AP + BQ + CR < \frac{(AB + AC) + (BC + AB) + (CA + BC)}{2}$
\(AP + BQ + CR < \frac{2(AB + BC + CA)}{2} = AB + BC + CA$
Therefore, the sum of the three altitudes is less than the sum of the three sides.
In simple words: In any right triangle, the hypotenuse is always longer than the perpendicular altitude, leading to the sum inequality.

Exam Tip: Use the property that the hypotenuse is the longest side in a right-angled triangle.

Question 6. If in \(\triangle ABC\), \(D\) is a point on \(AC\) such that \(AB = AD\), then prove that \(BC > CD\).
Answer:
Given \(AB = AD\(in \(\triangle ABC\), where \(D\) lies on \(AC\).
Since \(AB = AD\), in \(\triangle ABD$, the angles opposite to these sides are equal:
\(\angle ABD = \angle ADB$
Angle \(\angle ADB$ is an exterior angle for \(\triangle BCD\), so:
\(\angle ADB = \angle C + \angle DBC$
Therefore, \(\angle ADB > \angle C$).
Since \(\angle ABD = \angle ADB\), we also have \(\angle ABD > \angle C$).
In \(\triangle BCD\), the angle opposite to side \(CD$ is \(\angle CBD\), and the angle opposite to side \(BC\) is \(\angle BDC\) (which is equal to \(\angle ABD\)).
Since \(\angle B > \angle C$, the side opposite to the greater angle is longer:
\(BC > CD$
In simple words: Use exterior angle theorems and the rule that a larger angle in a triangle has a longer side opposite to it.

Question 7. Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference between the lengths of the two parallel sides.
Answer:
Let trapezium be \(ABCD\) with \(AB \parallel CD\). Let \(M\) and \(N\) be the midpoints of diagonals \(AC\) and \(BD\) respectively.
Join \(DN\) and produce it to meet \(AB\) at point \(E\).
In \(\triangle DNC\) and \(\triangle ENB\):
\(\angle DNC = \angle ENB$ (Vertically opposite angles)
\(DN = NE$ (Construction)
\(\angle DCN = \angle EBN$ (Alternate interior angles since \(CD \parallel AB\))
Therefore, \(\triangle DNC \cong \triangle ENB$ by ASA congruence.
This gives \(CD = BE$ and \(CN = BN\).
In \(\triangle ABC\(, \(M\) and \(N\) are midpoints of \(AC\) and \(BC\).
By Midpoint Theorem, \(MN \parallel AB\) and \(MN = \frac{1}{2}AE\).
Since \(AE = AB - BE = AB - CD\):
\(MN = \frac{1}{2}(AB - CD)\).
Thus, the line segment joining the midpoints of the diagonals is parallel to the parallel sides and equal to half their difference.
In simple words: Use congruent triangles and the Midpoint Theorem to relate the segment to the difference of the parallel bases.
Exam Tip: This is a standard midpoint theorem application in a trapezium.