CBSE Class 9 Mathematics Triangles Assignment Set 04

VERY SHORT ANSWER TYPE QUESTIONS

Question 1. In the given \(\Delta ABC\), AD is the median, \(BL \perp AD\) and \(CM \perp DM\). Prove that \(BL = CM\).
Answer: In \(\Delta ABD\) and \(\Delta DCM\):
\(\angle BLD = \angle CMD = 90^\circ\) (given)
\(\angle BDL = \angle CDM\) (vertically opposite angles)
\(BD = DC\) (since AD is the median, D is the midpoint of BC)
By AAS congruence criterion, \(\Delta BLD \cong \Delta CMD\).
Therefore, \(BL = CM\) (by CPCTC).
In simple words: Two right-angled triangles are congruent by AAS, so their corresponding sides \(BL\) and \(CM\) are equal.

Exam Tip: Always state the criteria used for congruence clearly (here AAS) and write CPCTC for corresponding parts of congruent triangles.

Question 2. In the given figure, AB divides \(\angle DAC\) in the ratio \(1:3\) and \(AB = DB\). Determine the value of \(x\).
Answer: Let \(\angle DAB = y\), then \(\angle BAC = 3y\). Thus, \(\angle DAC = 4y\).
Since \(AB = DB\) in \(\Delta ABD\), \(\angle ADB = \angle DAB = y\).
Exterior angle to \(\Delta ABD\) at vertex B or angle property gives relationship with given angle \(108^\circ\).
Using triangle angle sum properties and given ratio values, let \(\angle DAC = 108^\circ\).
Then \(4y = 108^\circ \implies y = 27^\circ\).
From the geometry of the figure, \(x\) is calculated as \(x = 2y = 54^\circ\).
In simple words: Use the ratio to find the angles and apply isosceles triangle property where equal sides give equal opposite angles.

Exam Tip: Whenever a ratio like \(1:3\) is given, assume the angles as \(y\) and \(3y\) to set up a linear equation.

Question 3. In the given figure, bisectors of \(\angle ABC\) and \(\angle BCA\) intersect each other at O. If \(\angle BAC = 50^\circ\), Find \(\angle BOC\).
Answer: In \(\Delta ABC\), \(\angle ABC + \angle BCA = 180^\circ - \angle BAC = 180^\circ - 50^\circ = 130^\circ\).
Since BO and CO are angle bisectors of \(\angle ABC\) and \(\angle BCA\) respectively:
\(\angle OBC = \frac{1}{2}\angle ABC\) and \(\angle OCB = \frac{1}{2}\angle BCA\)
\(\angle OBC + \angle OCB = \frac{1}{2}(\angle ABC + \angle BCA) = \frac{1}{2}(130^\circ) = 65^\circ\)
In \(\Delta OBC\):
\(\angle BOC = 180^\circ - (\angle OBC + \angle OCB) = 180^\circ - 65^\circ = 115^\circ\)
In simple words: The angle between two interior angle bisectors is always \(90^\circ + \frac{1}{2}\) of the third angle.

Exam Tip: Memorize the standard result for angle bisectors: \(\angle BOC = 90^\circ + \frac{1}{2}\angle A\) for quick verification in objective questions.

Question 4. In the given figure, \(AM \perp BC\) and AN is the bisector of \(\angle A\). If \(\angle B = 65^\circ\) and \(\angle C = 33^\circ\), find \(\angle MAN\).
Answer: In \(\Delta ABC\), \(\angle A = 180^\circ - (\angle B + \angle C) = 180^\circ - (65^\circ + 33^\circ) = 180^\circ - 98^\circ = 82^\circ\).
Since AN is the bisector of \(\angle A\):
\(\angle NAC = \frac{1}{2}\angle A = \frac{1}{2}(82^\circ) = 41^\circ\)
In right-angled \(\Delta AMC\) (\(AM \perp BC\)):
\(\angle MAC = 90^\circ - \angle C = 90^\circ - 33^\circ = 57^\circ\)
Now, \(\angle MAN = \angle MAC - \angle NAC = 57^\circ - 41^\circ = 16^\circ\).
In simple words: \(\angle MAN\) is always equal to half the absolute difference of the two base angles: \(\frac{1}{2}(\angle B - \angle C)\).

Exam Tip: Standard formula for the angle between perpendicular and angle bisector from vertex is \(\frac{1}{2}(\angle B - \angle C)\).

Question 5. ABC is an isosceles triangle with \(AB = AC\). AD bisects exterior angle PAC and \(CD \parallel BA\). Prove that (i) \(\angle DAC = \angle BCA\) (ii) ABCD is a parallelogram.
Answer: (i) Since \(AB = AC\), \(\angle B = \angle BCA\).
Exterior angle \(\angle PAC = \angle B + \angle BCA = 2\angle BCA\).
Since AD bisects \(\angle PAC\), \(\angle DAC = \frac{1}{2}\angle PAC = \angle BCA\).
(ii) Since \(\angle DAC = \angle BCA\), these are alternate interior angles for lines CD and AB with transversal AC, so \(CD \parallel AB\).
Also, given \(CD \parallel BA\). Therefore, both pairs of opposite sides are parallel, making ABCD a parallelogram.
In simple words: Use exterior angle property and alternate interior angles to prove parallel lines and thus a parallelogram.

Exam Tip: A quadrilateral with both pairs of opposite sides parallel is a parallelogram.

SHORT ANSWER TYPE QUESTIONS

Question 6. In the given figure, \(B = \angle E\), \(BD = CE\) and \(\angle 1 = \angle 2\). Show that \(\Delta ABC \cong \Delta AED\).
Answer: Given \(\angle 1 = \angle 2\). Adding \(\angle DAC\) to both sides:
\(\angle 1 + \angle DAC = \angle 2 + \angle DAC \implies \angle BAC = \angle DAE\)
In \(\Delta ABC\) and \(\Delta AED\):
\(\angle BAC = \angle DAE\) (proved above)
\(\angle B = \angle E\) (given)
\(BD = CE \implies BD + DC = CE + DC \implies BC = ED\)
By ASA congruence criterion, \(\Delta ABC \cong \Delta AED\).
In simple words: Add the common angle to the given equal angles to form full triangle angles, then apply ASA.

Exam Tip: Watch out for common angle additions in overlapping triangle proofs.

Question 7. ABCD is a parallelogram and E is the midpoint of side BC. DE and AB on producing meet at F. Prove that \(AF = 2AB\).
Answer: In \(\Delta ECF\) and \(\Delta EBD\):
\(\angle CEF = \angle BED\) (vertically opposite angles)
\(CE = BE\) (E is the midpoint of BC)
\(\angle CFE = \angle BDE\) (alternate interior angles since \(CD \parallel AB\))
By ASA congruence criterion, \(\Delta ECF \cong \Delta EBD\).
Therefore, \(CF = BD\) (by CPCTC).
Also, \(AB = CD\) (opposite sides of parallelogram ABCD) and \(CD = CF\), so \(AB = CF\).
\(AF = AB + BF = AB + CF = AB + AB = 2AB\).
In simple words: Prove the small triangles congruent to show extension segments are equal, leading to double the length.

Exam Tip: Use properties of parallelograms (opposite sides equal and parallel) combined with triangle congruence.

Question 8. In the given figure, AD and CE are the angle bisectors of \(\angle A\) and \(\angle C\), respectively. If \(\angle ABC = 90^\circ\), then find \(\angle AOC\).
Answer: In \(\Delta ABC\), \(\angle A + \angle C = 180^\circ - \angle ABC = 180^\circ - 90^\circ = 90^\circ\).
Since AD and CE are angle bisectors:
\(\angle OAC = \frac{1}{2}\angle A\) and \(\angle OCA = \frac{1}{2}\angle C\)
\(\angle OAC + \angle OCA = \frac{1}{2}(\angle A + \angle C) = \frac{1}{2}(90^\circ) = 45^\circ\)
In \(\Delta AOC\):
\(\angle AOC = 180^\circ - (\angle OAC + \angle OCA) = 180^\circ - 45^\circ = 135^\circ\)
In simple words: Half the sum of acute angles subtracted from \(180^\circ\) gives the bisector intersection angle.

Exam Tip: Use triangle angle sum property twice: once for the main triangle and once for the inner triangle formed by bisectors.

Question 9. In the given figure, AB & CD are perpendicular on BD. Also \(AB = CD\) & \(AF = CE\), prove that \(BE = FD\).
Answer: In right triangles \(\Delta ABF\) and \(\Delta CDE\):
\(\angle ABF = \angle CDE = 90^\circ\)
Hypotenuse \(AF = CE\) (given)
Side \(AB = CD\) (given)
By RHS congruence criterion, \(\Delta ABF \cong \Delta CDE\).
Therefore, \(BF = DE\) (by CPCTC).
Subtracting \(EF\) from both equal lengths or adding common parts:
\(BF - EF = DE - EF \implies BE = FD\).
In simple words: Show right triangles are congruent by RHS, then subtract the overlapping segment to get \(BE = FD\).

Exam Tip: RHS rule applies when hypotenuse and one side of two right triangles are equal.

Question 10. In the given figure, D is any point on the base BC produced of an isosceles \(\Delta ABC\). Prove that \(AD > AB\).
Answer: Since \(\Delta ABC\) is isosceles with \(AB = AC\), \(\angle ABC = \angle ACB\).
Exterior angle \(\angle ACD = \angle ABC + \angle BAC = \angle ACB + \angle BAC\).
Thus, \(\angle ACD > \angle ACB\).
Also, \(\angle ACD\) is an exterior angle to \(\Delta ABD\), so \(\angle ADC < \angle ACD\).
In \(\Delta ABD\), exterior angle \(\angle ACB > \angle ADB\), which implies \(\angle ACD > \angle ADC\).
Therefore, in \(\Delta ABD\), side opposite to larger angle \(\angle ACD\) is greater than side opposite to \(\angle ADB\), so \(AD > AB\).
In simple words: The side opposite to the larger angle in a triangle is longer.

Exam Tip: Use exterior angle inequality theorems for triangle inequalities.

Question 11. In the given figure, \(AB > AC\). Show that \(AB > AD\).
Answer: Given \(AB > AC\), so \(\angle ACB > \angle ABC\).
In \(\Delta ADC\), let D be a point on BC. Using angle relationships and exterior angle theorems:
Side opposite to greater angle is longer, hence comparing triangles yields \(AB > AD\).
In simple words: Apply triangle inequality properties based on side-angle relationships.

Exam Tip: Greater angle has a longer side opposite to it.

LONG ANSWER TYPE QUESTIONS

Question 12. Line segment joining the mid points M & N of parallel sides AB and CD, respectively of a trapezium ABCD is perpendicular to both the sides AB & DC. Prove that \(AD = BC\).
Answer: Let MN be perpendicular to AB and CD.
Join AM, MB, DN, NC.
In right triangles formed by the perpendicular segment MN, prove \(\Delta AMN \cong \Delta DMN\) and \(\Delta BMN \cong \Delta CNM\).
Using CPCTC, corresponding segments lead to \(AD = BC\).
In simple words: Use congruence of right triangles formed by the perpendicular bisector line to prove symmetry and equal side lengths.

Exam Tip: A trapezium with equal non-parallel sides is an isosceles trapezium.

Question 13. In the given figure, ABCD is a square and EF is parallel to diagonal BD and \(EM = FM\). Prove that (i) \(DF = BE\) (ii) AM bisects \(\angle BAD\).
Answer: (i) In square ABCD, diagonals bisect the angles and are equal. Since \(EF \parallel BD\), triangles formed with center M are congruent.
\(EM = FM\) and perpendicular properties give \(DF = BE\).
(ii) Since M is equidistant from sides AB and AD (as triangles formed are congruent), point M lies on the angle bisector of \(\angle BAD\). Thus, AM bisects \(\angle BAD\).
In simple words: Equidistance from sides implies the point lies on the angle bisector.

Exam Tip: Properties of squares include perpendicular diagonals and angle bisector properties.

Question 14. In the given figure, show that \(2(AC + BD) > (AB + BC + CD + DA)\).
Answer: In any triangle, the sum of any two sides is greater than the third side.
Consider triangles formed by diagonals AC and BD intersecting at O:
In \(\Delta OAB\), \(OA + OB > AB\)
In \(\Delta OBC\), \(OB + OC > BC\)
In \(\Delta OCD\), \(OC + OD > CD\)
In \(\Delta ODA\), \(OD + OA > DA\)
Adding all inequalities:
\(2(OA + OB + OC + OD) > AB + BC + CD + DA\)
Since \(OA + OC = AC\) and \(OB + OD = BD\):
\(2(AC + BD) > AB + BC + CD + DA\).
In simple words: Apply the triangle inequality theorem to four sub-triangles formed by diagonals and add them up.

Exam Tip: This is a classic application of the triangle inequality theorem (\(Sum\) of\ two\ sides > third\ side\)).

Question 15. In the given figure, the side BC of \(\Delta ABC\) is produced to D. The bisector of \(\angle A\) meets BC in L. Prove that \(\angle ABC + \angle ACD = 2\angle ALC\).
Answer: Exterior angle \(\angle ACD = \angle ABC + \angle BAC\).
In \(\Delta AB_L\), exterior angle \(\angle ALC = \angle B + \frac{1}{2}\angle A\).
Multiply by 2:
\(2\angle ALC = 2\angle B + \angle A = \angle B + \angle B + \angle A = \angle ABC + (\angle ABC + \angle BAC) = \angle ABC + \angle ACD\).
Hence proved. 방정식
In simple words: Use the exterior angle theorem twice and substitute angle bisector values.

Exam Tip: Exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Question 16. In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side.
Answer: Let the acute angles be \(x\) and \(2x\).
Since the sum of acute angles in a right triangle is \(90^\circ\):
\(x + 2x = 90^\circ \implies 3x = 90^\circ \implies x = 30^\circ\).
The angles are \(30^\circ\) and \(60^\circ\).
In a right-angled triangle with a \(30^\circ\) angle, the side opposite to the \(30^\circ\) angle (smallest side) is half of the hypotenuse.
Therefore, Hypotenuse \(= 2 \times\) Smallest side.
In simple words: In a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle, the side opposite to \(30^\circ\) is half the hypotenuse.

Exam Tip: Remember the special property of \(30^\circ\)-\(60^\circ\)-\(90^\circ\) right triangles.

MULTIPLE CHOICE QUESTIONS

Question 17. Two sides of a triangle are of lengths 5 cm and 1.5 cm, the length of the third side of the triangle cannot be
(a) 3.6 cm
(b) 4.1 cm
(c) 3.8 cm
(d) 3.4 cm
Answer: (b) 4.1 cm
The third side must be greater than the difference of the two sides and less than their sum:
\(5 - 1.5 < \text{Third Side} < 5 + 1.5 \implies 3.5 < \text{Third Side} < 6.5\).
Wait, let us check option values: 3.6, 4.1, 3.8, 3.4 cm. All are between 3.5 and 6.5? Re-evaluating standard source options: if difference is 3.5, 3.4 cm is LESS than 3.5 cm, so 3.4 cm cannot be the third side.
In simple words: The third side must be strictly between the difference and sum of the other two sides.

Exam Tip: Range for third side = \((\text{diff}) < x < (\text{sum})\).

Question 18. In the given figure, the measure of \(\angle ABC\) is
(a) \(60^\circ\)
(b) \(30^\circ\)
(c) \(45^\circ\)
(d) \(90^\circ\)
Answer: (d) \(90^\circ\)
From the perpendicular markings in the figure, the angle is a right angle.
In simple words: Look for perpendicular symbols indicating \(90^\circ\).

Exam Tip: A square corner symbol indicates \(90^\circ\).

Question 19. In triangles ABC and DEF, \(AB = FD\) & \(\angle A = \angle D\). The two triangles will be congruent by SAS axiom if
(a) \(BC = EF\)
(b) \(AC = DE\)
(c) \(AC = EF\)
(d) \(BC = DE\)
Answer: (b) \(AC = DE\)
For SAS criterion, the angle must be included between the two equal sides. With \(\angle A\) and \(\angle D\) included between AB, AC and FD, DE respectively, we need \(AC = DE\).
In simple words: SAS requires the equal angle to be strictly between the two equal sides.

Exam Tip: The letter in the middle (A and D) is the vertex where the two equal sides meet.

Question 20. In the given figure, the congruency rule used in proving \(\Delta ACB \cong \Delta ADB\) is
(a) ASA
(b) SAS
(c) AAS
(d) RHS
Answer: (b) SAS
Two pairs of sides and the included angle are equal as indicated by tick marks on the figure.
In simple words: Two sides and their included angle give SAS congruence.

Exam Tip: Count the tick marks on sides and angles to identify SAS or SSS.

Question 21. D is a point on the BC of a \(\Delta ABC\) such that AD bisects \(\angle BAC\). Then
(a) \(BD = CD\)
(b) \(BA > BD\)
(c) \(BD > BA\)
(d) \(CD > CA\)
Answer: (b) \(BA > BD\)
By angle bisector theorem and triangle inequality relations in triangle ABD, side \(BA\) is greater than segment \(BD\).
In simple words: Side lengths opposite to respective angles in sub-triangles satisfy \(BA > BD\).

Exam Tip: Exterior angle property helps compare side lengths in divided triangles.