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

Exercise 6.3

Step 1: We are given that \(PR^2 - PQ^2 = QR^2\).

Rearranging: \(PR^2 = PQ^2 + QR^2\).

Step 2: This is the Pythagoras theorem form. So, \(\triangle PQR\) is a right-angled triangle with the right angle at \(Q\).

Step 3: Draw altitude \(QM\) from vertex \(Q\) perpendicular to hypotenuse \(PR\). Here, \(M\) lies on side \(PR\).

Step 4: When an altitude is drawn from the right angle of a right triangle to the hypotenuse, the following property holds: \[ QM^2 = PM \times MR. \]

Step 5 (Proof of property):

• Consider triangles \(\triangle PQM\), \(\triangle QRM\), and \(\triangle PQR\).
• All three are right-angled and share common angles.
• Therefore, they are similar triangles (AA similarity rule).

Step 6: From similarity, we get two relations:
(i) \( \dfrac{QM}{PM} = \dfrac{QR}{PR} \)
(ii) \( \dfrac{QM}{MR} = \dfrac{PQ}{PR} \)

Step 7: Multiply the two equations: \[ \left( \dfrac{QM}{PM} \right) \times \left( \dfrac{QM}{MR} \right) = \dfrac{QR}{PR} \times \dfrac{PQ}{PR} \]

Step 8: The right-hand side becomes: \( \dfrac{QR \times PQ}{PR^2} \). But from Step 2, \(PR^2 = PQ^2 + QR^2\), so this fits the relation.

Step 9: Simplifying the left-hand side gives: \( \dfrac{QM^2}{PM \times MR} \).

Step 10: Cross multiplying, we finally get: \[ QM^2 = PM \times MR. \]

Therefore proved.

Step 1: Since \(DE \parallel AB\), the line \(DE\) divides the sides of triangle \(\triangle CBA\) in the same ratio. So, \(\triangle CDE \sim \triangle CBA\) (corresponding angles are equal).

Step 2: From similarity, we know: \[ \dfrac{CD}{CA} = \dfrac{CE}{CB} \]

Step 3: Write the lengths of the sides using the given values: - On side \(CA\): \(AD = 3x+19\) and \(DC = x+3\). So total \(CA = AD + DC = (3x+19) + (x+3) = 4x+22\). And \(CD = x+3\).

- On side \(CB\): \(BE = 3x+4\) and \(EC = x\). So total \(CB = BE + EC = (3x+4) + x = 4x+4\). And \(CE = x\).

Step 4: Put the values into the ratio: \[ \dfrac{CD}{CA} = \dfrac{CE}{CB} \] \[ \dfrac{x+3}{4x+22} = \dfrac{x}{4x+4} \]

Step 5: Cross multiply to remove the fractions: \[ (x+3)(4x+4) = x(4x+22) \]

Step 6: Expand both sides: Left side: \( (x+3)(4x+4) = 4x^2 + 4x + 12x + 12 = 4x^2 + 16x + 12 \). Right side: \( x(4x+22) = 4x^2 + 22x \).

Step 7: Equating both sides: \[ 4x^2 + 16x + 12 = 4x^2 + 22x \]

Step 8: Cancel \(4x^2\) from both sides: \[ 16x + 12 = 22x \]

Step 9: Rearrange: \[ 12 = 22x - 16x = 6x \]

Step 10: Divide both sides by 6: \[ x = 2 \]

Final Answer: \(x=2\).

Step 1: It is given that \(\triangle NSQ \cong \triangle MTR\).

From congruence, we know that corresponding angles and sides are equal. So, \(\angle SQN = \angle TRM\) and \(SQ = TR\).

Step 2: Points \(M, Q, R, N\) are collinear (they lie on a straight line). Using this and the equal angles \(\angle SQN = \angle TRM\), we can say that line segment \(QS \parallel RT\).

Step 3: It is also given that \(\angle 1 = \angle 2\). These are the angles at point \(P\), so we can write: \(\angle SPT = \angle QPR\).

Step 4: Because \(QS \parallel RT\):

• \(\angle PST = \angle PRQ\) (alternate interior angles)
• \(\angle PTS = \angle PQR\) (corresponding angles)

Step 5: Now compare triangles \(PTS\) and \(PRQ\):

• \(\angle SPT = \angle QPR\)
• \(\angle PST = \angle PRQ\)

So, two pairs of corresponding angles are equal.

Step 6: By the AA similarity criterion, if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar.

Therefore, \(\triangle PTS \sim \triangle PRQ\).

Step 1: The trapezium \(PQRS\) has two parallel sides: \(PQ\) and \(RS\). It is also given that \(PQ = 3 \times RS\).

Step 2: The diagonals of the trapezium meet at point \(O\). This divides the trapezium into two pairs of triangles. We are interested in triangles \(\triangle POQ\) and \(\triangle ROS\).

Step 3: These triangles lie between the same pair of diagonals and also between the parallel lines \(PQ\) and \(RS\). Because of this, the triangles \(\triangle POQ\) and \(\triangle ROS\) are similar triangles (their shapes are the same, only their sizes differ).

Step 4: The ratio of sides of these two similar triangles will be the same as the ratio of the parallel sides of the trapezium. So,

\[ \dfrac{PO}{RO} = \dfrac{PQ}{RS} = 3:1 \]

Step 5: When two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

So,

\[ \dfrac{\operatorname{ar}(\triangle POQ)}{\operatorname{ar}(\triangle ROS)} = \left(\dfrac{PQ}{RS}\right)^2 = 3^2 = 9:1 \]

Final Answer: The required ratio is \(9:1\).

Step 1: We are given that \(AB \parallel DC\). Whenever two lines are parallel, we can use alternate interior angles property with a transversal.

Step 2: Consider the triangles \(\triangle AOP\) and \(\triangle COQ\).

• Angle \(OAP = OCQ\) (because \(AB \parallel DC\), and \(PQ\) acts as a transversal → alternate interior angles are equal).
• Angle \(AOP = COQ\) (they are vertically opposite angles at point \(O\)).

Step 3: Since two pairs of angles are equal, we can say that:

\(\triangle AOP \sim \triangle COQ\) (by AA similarity criterion).

Step 4: From similarity of triangles, the ratio of corresponding sides is equal:

\[ \frac{AO}{CO} = \frac{AP}{CQ} \]

Step 5: Now cross multiply:

\[ AO \cdot CQ = CO \cdot AP \]

Step 6: This is exactly what we had to prove.

Therefore, proved.

Step 1: In an equilateral triangle, all three sides are equal and all angles are 60°.

Step 2: The altitude (height) of an equilateral triangle divides it into two equal right-angled triangles.

Step 3: Formula for altitude of an equilateral triangle is:

\[ h = \frac{\sqrt{3}}{2} \times a \]

where \(a\) = length of each side.

Step 4: Here, \(a = 8\,\text{cm}\).

So,

\[ h = \frac{\sqrt{3}}{2} \times 8 \]

Step 5: Simplify:

\[ h = 4\sqrt{3}\,\text{cm} \]

Final Answer: The altitude is \(4\sqrt{3}\,\text{cm}\).

Step 1: Since \(\triangle ABC \sim \triangle DEF\), their corresponding sides are in the same ratio (scale factor).

Step 2: The sides \(AB\) and \(DE\) correspond to each other. So, scale factor \(k = \dfrac{AB}{DE} = \dfrac{4}{6} = \dfrac{2}{3}\).

Step 3: Now use this scale factor to find the other sides of \(\triangle ABC\):

• \(BC\) corresponds to \(EF\). So, \(BC = EF \times k = 9 \times \dfrac{2}{3} = 6\,\text{cm}\).
• \(CA\) corresponds to \(FD\). So, \(CA = FD \times k = 12 \times \dfrac{2}{3} = 8\,\text{cm}\).

Step 4: Now we know all three sides of \(\triangle ABC\): \(AB = 4\,\text{cm}, BC = 6\,\text{cm}, CA = 8\,\text{cm}.\)

Step 5: Perimeter means the sum of all three sides: \(\text{Perimeter} = AB + BC + CA = 4 + 6 + 8 = 18\,\text{cm}.\)

Step 1: In the figure, line \(DE\) is drawn parallel to line \(BC\). When a line is drawn parallel to the base of a triangle, the smaller triangle formed at the top is similar to the big triangle.

Step 2: So, \(\triangle ADE \sim \triangle ABC\). In similar triangles, the ratio of corresponding sides is the same as the ratio of their linear scale factor.

Step 3: Here, \(DE = 6\,\text{cm}\) and \(BC = 12\,\text{cm}\). So, the linear ratio is: \[ \dfrac{DE}{BC} = \dfrac{6}{12} = \dfrac{1}{2} \]

Step 4: The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides. So, \[ \operatorname{ar}(ADE):\operatorname{ar}(ABC) = (1^2):(2^2) = 1:4 \]

Step 5: This means the smaller triangle \(ADE\) has 1 part of the area, while the bigger triangle \(ABC\) has 4 parts in total. Therefore, the quadrilateral \(DECB\) (which is the remaining portion) will have: \[ \operatorname{ar}(DECB) = 4 - 1 = 3\,\text{parts} \]

Final Step: Hence, the ratio of the areas is: \[ \operatorname{ar}(ADE):\operatorname{ar}(DECB) = 1:3 \]

Step 1: Write down the given information.

• \(AB \parallel DC\)
• \(PQ \parallel DC\)
• \(PD = 18\,\text{cm}\)
• \(BQ = 35\,\text{cm},\; QC = 15\,\text{cm}\)

Step 2: On side \(BC\), find the ratio of \(BQ\) to \(QC\):

\[ \dfrac{BQ}{QC} = \dfrac{35}{15} = \dfrac{7}{3} \]

Step 3: Because \(PQ \parallel DC\), the trapezium is divided into similar smaller trapeziums/triangles. This means the sides are divided in the same ratio. So, \[ \dfrac{AP}{PD} = \dfrac{BQ}{QC} = \dfrac{7}{3} \]

Step 4: Use this proportion to find \(AP\):

\[ \dfrac{AP}{18} = \dfrac{7}{3} \quad \Rightarrow \quad AP = \dfrac{7}{3} \times 18 \]

\[ AP = 42\,\text{cm} \]

Step 5: Now find the full side \(AD\):

\[ AD = AP + PD = 42 + 18 = 60\,\text{cm} \]

Final Answer: The length of \(AD\) is 60 cm.

Step 1: In similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Here, side ratio = \(2:3\).

So, area ratio = \( (2:3)^2 = (2^2 : 3^2) = 4:9 \).

Step 2: Let the area of the larger triangle be \(A_{\text{large}}\).

Then, according to the ratio:

\( \dfrac{\text{Area of smaller}}{\text{Area of larger}} = \dfrac{4}{9} \).

That is, \( \dfrac{48}{A_{\text{large}}} = \dfrac{4}{9} \).

Step 3: Cross multiply:

\( 48 \times 9 = 4 \times A_{\text{large}} \).

\( 432 = 4A_{\text{large}} \).

Step 4: Divide both sides by 4:

\( A_{\text{large}} = \dfrac{432}{4} = 108 \).

Final Answer: The area of the larger triangle is 108 cm².

Step 1: We are given a triangle \(\triangle PQR\) where \(N\) is a point on side \(PR\), and \(QN\) is drawn perpendicular to \(PR\).

Step 2: The condition is \(PN \cdot NR = QN^2\).

Step 3: Recall an important property (called the geometric mean theorem): In a right triangle, if you drop a perpendicular from the right angle to the hypotenuse, then: \[(\text{Altitude})^2 = (\text{Segment 1 of hypotenuse}) \times (\text{Segment 2 of hypotenuse}).\]

Step 4: Here, \(QN\) is the altitude, \(PN\) and \(NR\) are the two segments of \(PR\). The condition \(QN^2 = PN \cdot NR\) exactly matches this property.

Step 5: Since this property holds only when \(PR\) is the hypotenuse of a right triangle, it means \(\triangle PQR\) must be a right-angled triangle.

Step 6: Therefore, the right angle must be at vertex \(Q\). So, \(\angle PQR = 90^\circ\).

Final Answer: Proved.

Step 1: Write the areas of the triangles.

Smaller triangle area = \(36\,\text{cm}^2\)

Larger triangle area = \(100\,\text{cm}^2\)

Step 2: Recall the rule for similar triangles.

If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

That is: \( \dfrac{\text{Area of smaller}}{\text{Area of larger}} = \left(\dfrac{\text{Side of smaller}}{\text{Side of larger}}\right)^2 \)

Step 3: Substitute the known values.

\( \dfrac{36}{100} = \left(\dfrac{\text{Side of smaller}}{20}\right)^2 \)

Step 4: Simplify the fraction.

\( \dfrac{36}{100} = \dfrac{9}{25} \)

So, \( \left(\dfrac{\text{Side of smaller}}{20}\right)^2 = \dfrac{9}{25} \)

Step 5: Take square root on both sides.

\( \dfrac{\text{Side of smaller}}{20} = \dfrac{3}{5} \)

Step 6: Multiply both sides by 20 cm.

\( \text{Side of smaller} = 20 \times \dfrac{3}{5} = 12\,\text{cm} \)

Final Answer: The corresponding side of the smaller triangle is 12 cm.

Step 1: We are given that \(\angle ACB = \angle CDA\). This tells us that \(\triangle ACB\) and \(\triangle CDA\) are similar by AA similarity (two equal angles).

Step 2: From similarity of triangles, the sides are in proportion: \[ \frac{AC}{AD} = \frac{AB}{AC} \]

Step 3: Cross multiply to remove the fraction: \[ AB \cdot AD = AC^2 \]

Step 4: Substitute the given values: - \(AC = 8\,\text{cm}\) - \(AD = 3\,\text{cm}\) So, \[ AB \times 3 = 8^2 \] \[ AB \times 3 = 64 \]

Step 5: Divide both sides by 3 to get AB: \[ AB = \frac{64}{3}\,\text{cm} \]

Step 6: Now, \(BD = AB - AD\). Substitute values: \[ BD = \frac{64}{3} - 3 \] Convert 3 into fraction with denominator 3: \[ BD = \frac{64}{3} - \frac{9}{3} \]

Step 7: Simplify the fraction: \[ BD = \frac{55}{3}\,\text{cm} \]

Final Answer: \(BD = \dfrac{55}{3}\,\text{cm}\).

Step 1: When the Sun shines, tall objects like towers and poles form shadows on the ground.

Step 2: At the same time of the day, the angle of sunlight is the same. So, the tower and the pole will form similar right-angled triangles with their heights and shadows.

Step 3: For similar triangles, the ratio of height to shadow length will be equal.

That means:

\(\dfrac{\text{Height of tower}}{\text{Shadow of tower}} = \dfrac{\text{Height of pole}}{\text{Shadow of pole}}\)

Step 4: Substitute the known values:

\(\dfrac{15}{24} = \dfrac{h}{16}\)

Step 5: Solve for \(h\):

\(h = \dfrac{15}{24} \times 16\)

\(h = 10\)

Step 6: Therefore, the height of the pole is 10 m.

Step 1: Imagine the situation as a right-angled triangle.

• The ladder is the slant side (hypotenuse) = 10 m.
• The distance of the foot of the ladder from the wall is the base = 6 m.
• The height reached on the wall is the perpendicular side. This is what we need to find.

Step 2: Use the Pythagoras Theorem:

\[( ext{Hypotenuse})^2 = ( ext{Base})^2 + ( ext{Height})^2\]

Step 3: Substitute the values:

\[10^2 = 6^2 + ( ext{Height})^2\]

\[100 = 36 + ( ext{Height})^2\]

Step 4: Simplify:

\[( ext{Height})^2 = 100 - 36 = 64\]

Step 5: Take square root:

\[ ext{Height} = \sqrt{64} = 8\, ext{m}\]

Final Answer: The ladder reaches a height of 8 metres on the wall.