NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 6: Triangles

Exercise 6.1

Step 1: The triangle \(ABC\) is a right triangle with \(\angle A = 90^\circ\). So, side \(BC\) is the hypotenuse.

Step 2: The line \(AD\) is drawn from the right angle \(A\) perpendicular to the hypotenuse \(BC\). This line is called the altitude on the hypotenuse.

Step 3: There is a special property of right triangles:
When an altitude is drawn from the right angle to the hypotenuse, the square of the altitude = product of the two parts of the hypotenuse.
In symbols: \(AD^2 = BD \times DC\).

Step 4: From the options, this matches option (C): \(BD \cdot CD = AD^2\).

Final Answer: Option (C).

Step 1: A rhombus has diagonals that cut each other at right angles (90°) and also bisect each other (divide each other into two equal parts).

Step 2: The given diagonals are 16 cm and 12 cm.

Step 3: When they are bisected, each half of the diagonals will be:

• Half of 16 cm = 8 cm
• Half of 12 cm = 6 cm

Step 4: These halves form a right-angled triangle with the side of the rhombus as the hypotenuse.

Step 5: Using Pythagoras theorem: \( \text{(side)}^2 = (8)^2 + (6)^2 \) \( \text{(side)}^2 = 64 + 36 = 100 \) \( \text{side} = \sqrt{100} = 10 \; \text{cm} \)

Final Answer: The side length of the rhombus is 10 cm.

Step 1: We are told that \(\triangle ABC \sim \triangle EDF\). This means the triangles are similar, so their corresponding sides are in the same ratio.

Step 2: Write the rule of similarity: \[ \frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF} \]

Step 3: Now, let’s test each option one by one.

(A) \(BC \cdot EF = AC \cdot FD\) From the ratio \(\frac{BC}{DF} = \frac{AC}{EF}\), cross-multiplying gives \(BC \cdot EF = AC \cdot DF\). ✅ This is true.

(B) \(AB \cdot EF = AC \cdot DE\) From the ratio \(\frac{AB}{ED} = \frac{AC}{EF}\), cross-multiplying gives \(AB \cdot EF = AC \cdot ED\). ✅ This is true.

(C) \(BC \cdot DE = AB \cdot EF\) To check this, look at the ratios: \(\frac{AB}{ED}\) and \(\frac{BC}{DF}\). These do not directly give \(BC \cdot DE = AB \cdot EF\). So ❌ this is not always true.

(D) \(BC \cdot DE = AB \cdot FD\) From the ratio \(\frac{AB}{ED} = \frac{BC}{DF}\), cross-multiplying gives \(BC \cdot ED = AB \cdot DF\). ✅ This is true.

Final Answer: Option (C) is not true.

Step 1: In the question, we are given that the sides of the two triangles are in the same ratio:

\(\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}\).

Step 2: For similarity of triangles (by the SSS similarity criterion), the three sides of one triangle must be proportional to the three sides of another triangle.

Step 3: Compare the sides:

• \(AB\) in triangle ABC corresponds to \(QR\) in triangle PQR.
• \(BC\) in triangle ABC corresponds to \(PR\) in triangle PQR.
• \(CA\) in triangle ABC corresponds to \(PQ\) in triangle PQR.

Step 4: Now, write down the order of the vertices according to this correspondence:

\(A \rightarrow Q,\; B \rightarrow R,\; C \rightarrow P\).

Step 5: So the triangle \(BCA\) in the first triangle matches with \(PQR\) in the second triangle.

Final Answer: \(\triangle BCA \sim \triangle PQR\).

Step 1: Look at the lengths of the line segments.

We are given:

• \(PA = 6\,\text{cm}\)
• \(PB = 3\,\text{cm}\)
• \(PC = 2.5\,\text{cm}\)
• \(PD = 5\,\text{cm}\)

So, \(\dfrac{PA}{PB} = \dfrac{6}{3} = 2\).

And, \(\dfrac{PD}{PC} = \dfrac{5}{2.5} = 2\).

Step 2: Since the ratios are equal (both are 2), triangles \(\triangle APB\) and \(\triangle DPC\) are similar.

This means their angles are equal in pairs: \(\angle A = \angle D\), \(\angle B = \angle C\), etc.

Step 3: From the question, we know \(\angle CDP = 30^\circ\).

Because of similarity, \(\angle A = \angle D = 30^\circ\).

Step 4: In triangle \(APB\):

• \(\angle APB = 50^\circ\) (given)
• \(\angle PAB = 30^\circ\) (from similarity)
• \(\angle PBA = ?\)

Step 5: The sum of angles in any triangle is \(180^\circ\).

So,

\(\angle PBA = 180^\circ - (\angle APB + \angle PAB)\)

\(= 180^\circ - (50^\circ + 30^\circ)\)

\(= 180^\circ - 80^\circ\)

\(= 100^\circ\).

Final Answer: \(\angle PBA = 100^\circ\). ✅

Step 1: We are told that \(\angle D = \angle Q\) and \(\angle R = \angle E\).

Step 2: From this, we know the triangles are similar by the AA (Angle-Angle) similarity rule.

Step 3: In similar triangles, equal angles tell us which vertices correspond:

• \(D \leftrightarrow Q\)
• \(E \leftrightarrow R\)
• \(F \leftrightarrow P\)

Step 4: So the corresponding sides are:

• \(DE \leftrightarrow QR\)
• \(EF \leftrightarrow RP\)
• \(DF \leftrightarrow QP\)

Step 5: Now check each option:

• (A) \(\tfrac{EF}{PR} = \tfrac{DF}{PQ}\) → Correct, sides match.
• (B) \(\tfrac{DE}{PQ} = \tfrac{EF}{RP}\) → Wrong, because \(DE\) corresponds to \(QR\), not to \(PQ\).
• (C) \(\tfrac{DE}{QR} = \tfrac{DF}{PQ}\) → Correct.
• (D) \(\tfrac{EF}{RP} = \tfrac{DE}{QR}\) → Correct.

Step 6: Therefore, option (B) is not true.

Step 1: We are told that \(\angle B = \angle E\) and \(\angle F = \angle C\).

This means two angles of triangle \(ABC\) are equal to two angles of triangle \(DEF\).

Step 2: If two angles of one triangle are equal to two angles of another triangle, the triangles are similar by the AA (Angle–Angle) similarity rule.

Step 3: To be congruent, all sides of the two triangles must be exactly the same length (or have the same ratio of 1:1).

But here, \(AB = 3 \times DE\). So the ratio of sides is \(3:1\), not \(1:1\).

Step 4: Therefore, the triangles are similar (because of equal angles), but not congruent (because the sides are not equal in length).

Final Answer: Similar but not congruent.

Step 1: We are told that \(\triangle ABC \sim \triangle PQR\). This means both triangles are similar, so their corresponding sides are in the same ratio and their areas are related by the square of that ratio.

Step 2: From the question, \(\dfrac{BC}{QR} = \dfrac{1}{3}\). This means side \(BC\) of triangle ABC is one-third the length of side \(QR\) of triangle PQR.

Step 3: The rule is: \[ \dfrac{\text{Area of one triangle}}{\text{Area of other triangle}} = \left(\dfrac{\text{corresponding side of one}}{\text{corresponding side of other}}\right)^2 \]

Step 4: Apply the rule: \[ \dfrac{\operatorname{ar}(BCA)}{\operatorname{ar}(PRQ)} = \left(\dfrac{BC}{QR}\right)^2 \]

Step 5: Substitute the values: \[ \dfrac{\operatorname{ar}(BCA)}{\operatorname{ar}(PRQ)} = \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9} \]

Step 6: Take reciprocal (since question asks for \(\dfrac{\operatorname{ar}(PRQ)}{\operatorname{ar}(BCA)}\)): \[ \dfrac{\operatorname{ar}(PRQ)}{\operatorname{ar}(BCA)} = 9 \]

Final Answer: Option A (9)

Step 1: In similar triangles, the order of letters shows the matching corners.
So, \(A \leftrightarrow D\), \(B \leftrightarrow F\), and \(C \leftrightarrow E\).

Step 2: We are given \(\angle A = 30^\circ\) and \(\angle C = 50^\circ\).
The sum of angles in a triangle is \(180^\circ\).
So, \(\angle B = 180^\circ - (30^\circ + 50^\circ) = 100^\circ\).

Step 3: Since \(B \leftrightarrow F\), we know \(\angle F = 100^\circ\).
Also, \(A \leftrightarrow D\), so \(\angle D = 30^\circ\).

Step 4: Find the scale factor between the triangles.
\(AB = 5\,\text{cm}\) and \(DF = 7.5\,\text{cm}\).
So, scale factor \(k = \dfrac{DF}{AB} = \dfrac{7.5}{5} = 1.5\).

Step 5: To find \(DE\), use the matching side of \(AC\).
\(AC = 8\,\text{cm}\). Multiply by the scale factor:
\(DE = AC \times k = 8 \times 1.5 = 12\,\text{cm}\).

Final Answer: \(DE = 12\,\text{cm}\) and \(\angle F = 100^\circ\).
So, the correct option is B.

Step 1: To prove two triangles are similar, one common method is the SAS (Side-Angle-Side) similarity rule. It says: If two sides of one triangle are in the same ratio as two sides of another triangle, and the angle between those sides is equal, then the triangles are similar.

Step 2: Here we are given that \(\dfrac{AB}{DE} = \dfrac{BC}{FD}\). So, side \(AB\) of triangle \(ABC\) is proportional to side \(DE\) of triangle \(DEF\), and side \(BC\) is proportional to side \(FD\).

Step 3: Notice the position of these sides. In triangle \(ABC\), the angle between sides \(AB\) and \(BC\) is \(\angle B\). In triangle \(DEF\), the angle between sides \(DE\) and \(FD\) is \(\angle D\).

Step 4: For SAS similarity, we need these included angles to be equal. That means \(\angle B = \angle D\).

Final Answer: Therefore, the triangles will be similar when \(\angle B = \angle D\).

Step 1: We are told \(\triangle ABC \sim \triangle QRP\). That means the two triangles are similar. In similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Step 2: The ratio of the areas is given as: \[ \dfrac{\operatorname{ar}(ABC)}{\operatorname{ar}(PQR)} = \dfrac{9}{4} \]

Step 3: Take square root to find the ratio of the sides (scale factor): \[ \dfrac{AB}{QR} = \dfrac{BC}{RP} = \dfrac{CA}{PQ} = \sqrt{\dfrac{9}{4}} = \dfrac{3}{2} \]

Step 4: From the correspondence in the question, side \(BC\) of \(\triangle ABC\) matches with side \(PR\) of \(\triangle QRP\).

Step 5: Use the ratio of sides: \[ \dfrac{BC}{PR} = \dfrac{3}{2} \]

Step 6: Put \(BC = 15\,\text{cm}\). \[ \dfrac{15}{PR} = \dfrac{3}{2} \]

Step 7: Cross-multiply: \[ 15 \times 2 = 3 \times PR \] \[ 30 = 3 \times PR \]

Step 8: Divide both sides by 3: \[ PR = \dfrac{30}{3} = 10\,\text{cm} \]

Final Answer: \(PR = 10\,\text{cm}\). So the correct option is A.

Step 1: It is given that \(PS = QS = RS\). This means the point \(S\) is the same distance from all three vertices \(P, Q, R\).

Step 2: A point that is equidistant from all three vertices of a triangle is called the circumcentre of the triangle.

Step 3: Here, \(S\) lies on the side \(PQ\). Normally, the circumcentre of a triangle lies inside (for an acute triangle), outside (for an obtuse triangle), or on the midpoint of the hypotenuse (for a right triangle).

Step 4: Since \(S\) lies on side \(PQ\), the only possibility is that the triangle \(PQR\) is a right-angled triangle with the right angle at \(R\).

Step 5: In a right-angled triangle, the side opposite the right angle is called the hypotenuse. Here, side \(PQ\) is the hypotenuse.

Step 6: By the Pythagoras theorem, we know that:

\[ PR^2 + QR^2 = PQ^2 \]

Final Answer: Option (C) is correct.