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

Exercise 6.2

Step 1: Recall the Pythagoras theorem. It says that in a right-angled triangle,

\(\text{(Hypotenuse)}^2 = (\text{Base})^2 + (\text{Height})^2\).

Step 2: Identify the largest side, because the largest side is always taken as the hypotenuse in the Pythagoras theorem.

Here, the largest side = 25 cm.

Step 3: Now check if

\(25^2 = 24^2 + 5^2\)

Step 4: Calculate each square.

\(25^2 = 25 \times 25 = 625\)

\(24^2 = 24 \times 24 = 576\)

\(5^2 = 5 \times 5 = 25\)

Step 5: Add the squares of the smaller two sides.

\(24^2 + 5^2 = 576 + 25 = 601\)

Step 6: Compare the two values.

Left side = \(25^2 = 625\)

Right side = \(24^2 + 5^2 = 601\)

Since 625 ≠ 601, the Pythagoras theorem is not satisfied.

Final Conclusion: Therefore, the given triangle is not a right triangle.

Step 1: The symbol “\(\sim\)” means the two triangles are similar. Similar triangles have their corresponding angles equal and their sides in proportion.

Step 2: When triangles are written in order, the letters tell us which angles match with which.

• In \(\triangle DEF\), the first letter is D.
• In \(\triangle RPQ\), the first letter is R.
• So, \(\angle D \leftrightarrow \angle R\).

Step 3: Next, the second letters correspond:

• In \(\triangle DEF\), the second letter is E.
• In \(\triangle RPQ\), the second letter is P.
• So, \(\angle E \leftrightarrow \angle P\).

Step 4: Finally, the third letters correspond:

• In \(\triangle DEF\), the third letter is F.
• In \(\triangle RPQ\), the third letter is Q.
• So, \(\angle F \leftrightarrow \angle Q\).

Step 5: Therefore:

• \(\angle D = \angle R\) ✅ (True)
• \(\angle F = \angle P\) ❌ (False, because F matches with Q, not P)

Final Point: The statement is partly true. Only the first equality is correct.

Step 1: Recall the Basic Proportionality Theorem (also called Thales' theorem). It says: If a line drawn through a triangle is parallel to one side, then it divides the other two sides in the same ratio.

Step 2: To check if \(AB \parallel QR\), we compare the ratios:

\[ \frac{PA}{PQ} \quad \text{and} \quad \frac{PB}{PR} \]

Step 3: First, calculate the length of \(PR\):

\(PR = PB + BR = 4 + 6 = 10\,\text{cm}\).

Step 4: Now find the ratios:

\[ \frac{PA}{PQ} = \frac{5}{12.5} \]

Divide numerator and denominator: \(5 \div 12.5 = 0.4\).

\[ \frac{PB}{PR} = \frac{4}{10} \]

Divide numerator and denominator: \(4 \div 10 = 0.4\).

Step 5: Both ratios are equal (0.4 = 0.4).

Step 6: Since the ratios are equal, by the Basic Proportionality Theorem, line \(AB\) is parallel to \(QR\).

Step 1: Note the given lengths from the figure.

• \(PB = 5\,\text{cm}\)
• \(PC = 6\,\text{cm}\)
• \(PD = 10\,\text{cm}\)
• \(PE = 12\,\text{cm}\)

Step 2: Write the ratios of the sides of the two triangles.

For \(\triangle PBC\) and \(\triangle PDE\):

\[ \dfrac{PB}{PD} = \dfrac{5}{10} = \dfrac{1}{2}, \quad \dfrac{PC}{PE} = \dfrac{6}{12} = \dfrac{1}{2} \]

Step 3: Compare the included angle.

At point \(P\), the angle \(\angle BPC\) and the angle \(\angle DPE\) are equal because they are vertically opposite angles.

Step 4: Apply the SAS (Side–Angle–Side) similarity rule.

Since:

• Two pairs of sides are in the same ratio (\(\dfrac{1}{2}\)), and
• The included angle between those sides is equal,

Therefore, \(\triangle PBC \sim \triangle PDE\).

Step 1: Find the missing angle in triangle PQR.

We know the sum of angles in any triangle is \(180^\circ\).

In \(\triangle PQR\):

\(\angle R = 180^\circ - (\angle P + \angle Q)\)

\(= 180^\circ - (55^\circ + 25^\circ)\)

\(= 180^\circ - 80^\circ = 100^\circ\).

So the three angles of \(\triangle PQR\) are: 55°, 25°, and 100°.

Step 2: Find the missing angle in triangle MST.

In \(\triangle MST\):

\(\angle T = 180^\circ - (\angle M + \angle S)\)

\(= 180^\circ - (100^\circ + 25^\circ)\)

\(= 180^\circ - 125^\circ = 55^\circ\).

So the three angles of \(\triangle MST\) are: 100°, 25°, and 55°.

Step 3: Match the equal angles of both triangles.

• \(\angle Q = 25^\circ \leftrightarrow \angle S = 25^\circ\)
• \(\angle P = 55^\circ \leftrightarrow \angle T = 55^\circ\)
• \(\angle R = 100^\circ \leftrightarrow \angle M = 100^\circ\)

Step 4: Write the correct order of similarity.

Corresponding equal angles must come in the same order.

So the correct similarity is: \(\triangle QPR \sim \triangle STM\).

Not \(\triangle TSM\).

Step 1: Recall the meaning of "similar figures".

Two figures are called similar if:

• Their corresponding angles are equal.
• Their corresponding sides are in the same ratio (also called proportional).

Step 2: Check what the statement says.

The statement says that only equal angles are enough to make two quadrilaterals similar.

Step 3: See why this is not correct.

For polygons with more than 3 sides (like quadrilaterals), equal angles alone are not sufficient. We must also check the ratio of sides.

Step 4: Example to understand better.

Take two rectangles:

• Rectangle A has sides (2 , ext{cm}) and (1 , ext{cm}).
• Rectangle B has sides (3 , ext{cm}) and (1 , ext{cm}).

Both rectangles have all four angles = (90^circ). So, their angles are equal.

But the ratio of sides is different:

• In Rectangle A: ratio = (2:1).
• In Rectangle B: ratio = (3:1).

Since the side ratios are not the same, these two rectangles are not similar.

Final Conclusion: The statement is false. For similarity of quadrilaterals, both equal angles and proportional sides are required.

Step 1: Let the smaller triangle have sides \(x, y, z\).

Step 2: Then, according to the question, the larger triangle will have:

• First side = \(3x\)
• Second side = \(3y\)
• Perimeter = \(3(x + y + z)\)

Step 3: The perimeter of the larger triangle is the sum of its three sides. So, \(3x + 3y + \text{third side} = 3(x + y + z)\).

Step 4: Simplify:

\(3x + 3y + \text{third side} = 3x + 3y + 3z\).

Therefore, the third side = \(3z\).

Step 5: Now the sides of the larger triangle are \(3x, 3y, 3z\).

Step 6: Compare the sides of the two triangles:

\(\dfrac{3x}{x} = \dfrac{3y}{y} = \dfrac{3z}{z} = 3\).

Step 7: Since all three pairs of corresponding sides are in the same ratio, the two triangles are similar by the SSS (Side-Side-Side) similarity criterion.

Let us go step by step:

1. In both triangles, one angle is a right angle (90°).
2. It is given that one acute angle of the first triangle is equal to one acute angle of the second triangle.
3. A triangle has three angles. The sum of all three angles of any triangle is 180°.
4. So, in each triangle we already know two angles: one is 90° (right angle), and the other is the given equal acute angle.
5. That means the third angle in both triangles must also be equal (because 180° − (90° + given angle) will give the same result for both).
6. Therefore, the two triangles have all their angles equal.
7. By the Angle–Angle (AA) similarity rule, if two angles of one triangle are equal to two angles of another triangle, the triangles are similar.

Hence, the two right triangles are similar.

Step 1: In similar triangles, the ratio of any two corresponding linear dimensions (like sides, altitudes, medians, etc.) is the same. Here, the ratio of altitudes is given as \(\dfrac{3}{5}\).

Step 2: When we compare areas of two similar triangles, the ratio of areas is equal to the square of the ratio of their corresponding sides (or altitudes).

Step 3: So, the ratio of areas = \(\left(\dfrac{3}{5}\right)^2\).

Step 4: Calculate the square: \(\left(\dfrac{3}{5}\right)^2 = \dfrac{3^2}{5^2} = \dfrac{9}{25}\).

Step 5: The student’s guess \(\dfrac{6}{5}\) is wrong because areas do not compare in the same ratio as altitudes. They always compare by the square of that ratio.

Final Result: The ratio of their areas is \(\dfrac{9}{25}\), not \(\dfrac{6}{5}\).

Step 1: We are given that \(PD\) is perpendicular to \(QR\). This means \(\angle QDP = 90^\circ\) and \(\angle RDP = 90^\circ\).

Step 2: For two triangles to be similar, we need either:

• Two pairs of corresponding angles equal (AA similarity), or
• The sides to be in the same ratio (SSS or SAS similarity).

Step 3: In \(\triangle PQD\) and \(\triangle RPD\), we already have one common fact: both contain a right angle at point \(D\).

Step 4: But we also need at least one more equal angle (or a clear ratio of sides). For example, we would need \(\angle QPD = \angle RPD\). But this is not guaranteed just from the information given.

Step 5: Because we only know about one equal angle (the right angle), and we have no other angle or side ratio, we cannot apply the similarity rules.

Therefore, we cannot conclude that \(\triangle PQD\) is similar to \(\triangle RPD\).

Step 1: Look at both triangles: \(\triangle ADE\) and \(\triangle ACB\).

Step 2: Notice that \(\angle A\) is present in both triangles. So, they share a common angle.

Step 3: It is given that \(\angle D = \angle C\).

Step 4: Now each triangle has two angles equal:
• In \(\triangle ADE\): angles \(\angle A\) and \(\angle D\).
• In \(\triangle ACB\): angles \(\angle A\) and \(\angle C\).

Step 5: When two angles of one triangle are equal to two angles of another triangle, the two triangles are similar. This rule is called the AA (Angle–Angle) similarity criterion.

Conclusion: Therefore, \(\triangle ADE \sim \triangle ACB\).

Let us go step by step:

Step 1: Recall the condition for SAS similarity of two triangles.

• The angle between two sides of one triangle is equal to the angle between two sides of the other triangle.
• The two sides of one triangle are proportional to the corresponding two sides of the other triangle.

If these two conditions are satisfied, then the triangles are similar.

Step 2: In the question, we are told:

• One angle of the first triangle is equal to one angle of the second triangle.
• Two sides of the first triangle are proportional to the corresponding two sides of the second triangle.

Step 3: This information is not enough unless the equal angle lies between the two proportional sides.

If the equal angle is the included angle between the two proportional sides, then the triangles will definitely be similar (by SAS similarity).

Step 4: But if the equal angle is not the included angle, then the triangles may or may not be similar. In fact, we can even draw counterexamples where similarity fails.

Final Conclusion: Therefore, the statement given in the question is not always true. It is true only when the equal angle is the included angle between the proportional sides.