OP Malhotra Class 9 Maths Solutions Chapter 8 Triangles Chapter Test

Question 1.
(i) Which of the following is not a criterion for congruency of triangles?
(a) SAS
(b) ASA
(c) SSA
(d) SSS
Answer: (c) SSA
In simple words: SSA is not a valid way to prove that two triangles are congruent. This is because SSA can lead to more than one possible triangle, unlike SAS, ASA, or SSS which give a unique triangle.

🎯 Exam Tip: Remember the four main congruence rules: SSS, SAS, ASA, and RHS. SSA is a common trap because it does not guarantee congruence.

Question 1.
(ii) Fom the given figure, AC = FE, AB = FD, BC = DE, which statement correctly indicates that the two given triangles are congruent?
(a) \( \Delta ABC \cong \Delta EFD \)
(b) \( \Delta ABC \cong \Delta DEF \)
(c) \( \Delta ABC \cong \Delta FDE \)
(d) \( \Delta ABC \cong \Delta FED \)
Answer: (c) \( \Delta ABC \cong \Delta FDE \)
In simple words: If all three sides of one triangle are the same length as all three sides of another triangle, then the triangles are exactly alike. This is called the SSS rule. So, triangle ABC is the same as triangle FDE.

🎯 Exam Tip: When matching triangles using SSS, make sure to write the congruence statement with the vertices in the correct corresponding order.

Question 2.
(i) Show that \( \Delta RST \cong \Delta TUR \) when \( x = 18 \).
Answer:
When \( x = 18 \), then:
Side \( RS = 61 \) (given from figure)
Side \( TU = 61 \) (given from figure)
So, \( RS = TU = 61 \)
Angle \( \angle SRT = 2x = 2 \times 18 = 36^\circ \)
Angle \( \angle RTU = 2x = 2 \times 18 = 36^\circ \)
So, \( \angle SRT = \angle RTU \)
Side \( RT \) is common to both triangles (from figure)
Thus, we have two sides and the included angle equal for both triangles: Side (RS/TU), Angle (\( \angle SRT / \angle RTU \)), Side (RT). A bridge connects the ideas here, where \( RT \) serves as the bridge between the two triangles, confirming congruence.
Therefore, \( \Delta RST \cong \Delta TUR \) by the SAS (Side-Angle-Side) criterion.
In simple words: When \( x \) is 18, we find that two sides and the angle between them in triangle RST are exactly the same as in triangle TUR. So, by the SAS rule, both triangles are a perfect match.

🎯 Exam Tip: Always substitute the value of variables into expressions for sides and angles first to get numerical values before applying congruence criteria.

Question 2.
(ii) Show that \( \Delta ABC \cong \Delta DBC \) when \( x = 4 \).
Answer:
In the given figure, ABDC is a kite, which means \( AB = DB \) and \( AC = DC \).
First, we use the condition \( AB = DB \). From the figure, \( AB = 3x - 9 \) and \( DB = 3 \).
So, \( 3x - 9 = 3 \)
\( \implies 3x = 3 + 9 \)
\( \implies 3x = 12 \)
\( \implies x = \frac{12}{3} = 4 \). This value of \( x \) matches the condition given in the question.
Next, we check if \( AC = DC \). From the figure, \( AC = 5 \) and \( DC = 2x - 3 \).
Substitute \( x = 4 \) into the expression for \( DC \): \( DC = 2(4) - 3 = 8 - 3 = 5 \).
So, \( AC = DC = 5 \).
Finally, the side \( BC \) is common to both triangles \( \Delta ABC \) and \( \Delta DBC \). This method shows how algebraic values confirm geometric congruence.
Since all three corresponding sides are equal (\( AB = DB \), \( AC = DC \), \( BC = BC \)), we can conclude that \( \Delta ABC \cong \Delta DBC \) by the SSS (Side-Side-Side) criterion.
In simple words: We use the \( x=4 \) value to check the side lengths. We find that the first side AB is equal to DB, the second side AC is equal to DC, and the third side BC is shared by both triangles. Since all three sides match up, the triangles ABC and DBC are congruent by the SSS rule.

🎯 Exam Tip: Always check if the value of \( x \) (or any other variable) provided in the question satisfies the given conditions in the figure.

Question 3. D is a point on the side 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) \( BA > BD \)
In simple words: When AD cuts angle A into two equal parts, it means that in triangle ABD, the angle at D (\( \angle ADB \)) is bigger than the angle at A (\( \angle BAD \)). In any triangle, the side opposite the bigger angle is longer. So, the side BA is longer than the side BD.

🎯 Exam Tip: Remember the property that in any triangle, the side opposite to the greater angle is longer, and the angle opposite to the longer side is greater.

Question 4. Two sides of a triangle are of lengths 6 cm and 2.6 cm. The lengths of the third side of the triangle cannot be
(a) 4.7 cm
(b) 5 cm
(c) 4.9 cm
(d) 3.2 cm
Answer: (d) 3.2 cm
In simple words: The third side of a triangle must be longer than the difference of the other two sides (6 - 2.6 = 3.4 cm) and shorter than their sum (6 + 2.6 = 8.6 cm). So, the third side must be between 3.4 cm and 8.6 cm. Out of the given options, 3.2 cm is not in this range, so it cannot be the third side.

🎯 Exam Tip: Always apply both parts of the triangle inequality theorem (difference < third side < sum) to find the possible range for the third side.

Question 5. In \( \Delta ABC \), \( \angle B = 30^\circ \), \( \angle C = 80^\circ \) and \( \angle A = 70^\circ \), then
(a) \( AB > BC < AC \)
(b) \( AB < BC > AC \)
(c) \( AB > BC > AC \)
(d) \( AB < BC < AC \)
Answer: (c) \( AB > BC > AC \)
In simple words: First, we find angle A is 70 degrees. Then we look at all angles: angle C (80) is the biggest, then angle A (70), then angle B (30). The side across from the biggest angle is the longest. So, side AB (across from C) is longest, then BC (across from A), then AC (across from B). So, AB > BC > AC.

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

Question 6. Two sides of a triangle are of lengths 4 cm, and 10 cm. If the lengths of the third side is 'a' cm, then
(a) \( a \le 5 \)
(b) \( 6 \le a \le 12 \)
(c) \( a < 6 \)
(d) \( 6 < a < 14 \)
Answer: (d) \( 6 < a < 14 \)
In simple words: For a triangle to be formed, its third side (a) must be bigger than the difference of the other two sides (10 - 4 = 6 cm) and smaller than their sum (10 + 4 = 14 cm). So, 'a' must be between 6 cm and 14 cm.

🎯 Exam Tip: Always remember the two parts of the triangle inequality to define the full possible range for the third side.