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CBSE Class 10 Maths Chapter 6 Triangles Model Questions - 3 Marks - Part 2

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SECTION C — Short Answer Type Questions [3 Marks Each]

• Q26. If the areas of two similar triangles are equal, prove that they are congruent triangles.
  Answer:
  1. Given: $\triangle ABC \sim \triangle PQR$ and $\text{Area}(\triangle ABC) = \text{Area}(\triangle PQR)$.
  2. To Prove: $\triangle ABC \cong \triangle PQR$.
  3. Proof: - Since $\triangle ABC \sim \triangle PQR$, the ratio of their areas is equal to the square of the ratio of their corresponding sides: $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2$.
  - Since their areas are equal, $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = 1$.
  - Therefore, $\left(\frac{AB}{PQ}\right)^2 = 1 \implies \frac{AB}{PQ} = 1 \implies AB = PQ$.
  - Similarly, $BC = QR$ and $CA = RP$.
  - By SSS congruence criterion, $\triangle ABC \cong \triangle PQR$.
  
  Result: Proved.
• Q27. $\triangle ABC$ is an equilateral triangle of side $2a$. Find the length of each of its altitudes.
  Answer:
  1. Let $\triangle ABC$ be an equilateral triangle with side $AB = BC = CA = 2a$.
  2. Draw an altitude $AD \perp BC$. In an equilateral triangle, the altitude also bisects the base, so $BD = DC = a$.
  3. In the right-angled triangle $ABD$, apply Pythagoras Theorem: $AB^2 = AD^2 + BD^2$.
  4. Substitute the values: $(2a)^2 = AD^2 + a^2 \implies 4a^2 = AD^2 + a^2$.
  5. Solve for $AD$: $AD^2 = 4a^2 - a^2 = 3a^2 \implies AD = \sqrt{3}a$.
  
  Result: Length of each altitude = $\sqrt{3}a$.
• Q28. $D$ is a point on the side $BC$ of $\triangle ABC$ such that $\angle ADC = \angle BAC$. Show that $AC^2 = BC \cdot DC$.
  Answer:
  1. Consider $\triangle ADC$ and $\triangle BAC$: - $\angle ADC = \angle BAC$ (Given) - $\angle C = \angle C$ (Common angle)
  2. By AA similarity criterion, $\triangle ADC \sim \triangle BAC$.
  3. Since the triangles are similar, the ratio of their corresponding sides is equal: $\frac{AC}{BC} = \frac{DC}{AC}$.
  4. Cross-multiply to get $AC^2 = BC \cdot DC$.
  
  Result: Proved.
• Q29. If $\triangle ABC \sim \triangle DEF$ such that $AB = 1.2\text{ cm}$ and $DE = 1.4\text{ cm}$. Find the ratio of the areas of $\triangle ABC$ and $\triangle DEF$.
  Answer:
  1. Given that $\triangle ABC \sim \triangle DEF$.
  2. By the theorem on areas of similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides: $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2$.
  3. Substitute the given values: $\left(\frac{1.2}{1.4}\right)^2 = \left(\frac{6}{7}\right)^2 = \frac{36}{49}$.
  
  Result: Ratio of the areas = $36:49$.
• Q30. In an equilateral triangle $ABC$, $D$ is a point on side $BC$ such that $BD = \frac{1}{3}BC$. Prove that $9AD^2 = 7AB^2$.
  Answer:
  1. Draw an altitude $AE \perp BC$. Since $\triangle ABC$ is equilateral, $BE = EC = \frac{1}{2}BC = \frac{1}{2}AB$.
  2. Given $BD = \frac{1}{3}BC = \frac{1}{3}AB$. Therefore, $DE = BE - BD = \left(\frac{1}{2} - \frac{1}{3}\right)AB = \frac{1}{6}AB$.
  3. In right-angled $\triangle ADE$: $AD^2 = AE^2 + DE^2$.
  4. We know $AE^2 = AB^2 - BE^2 = AB^2 - \left(\frac{1}{2}AB\right)^2 = \frac{3}{4}AB^2$.
  5. Substitute $AE^2$ and $DE^2$: $AD^2 = \frac{3}{4}AB^2 + \left(\frac{1}{6}AB\right)^2 = \left(\frac{3}{4} + \frac{1}{36}\right)AB^2 = \frac{28}{36}AB^2 = \frac{7}{9}AB^2$.
  6. Cross-multiply: $9AD^2 = 7AB^2$.
  
  Result: Proved.
• Q31. $P$ and $Q$ are the points on the sides $CA$ and $CB$ respectively of a triangle $ABC$ right-angled at $C$. Prove that $AQ^2 + BP^2 = AB^2 + PQ^2$.
  Answer:
  1. In right-angled $\triangle ACQ$: $AQ^2 = AC^2 + CQ^2$.
  2. In right-angled $\triangle BCP$: $BP^2 = BC^2 + CP^2$.
  3. Adding both equations: $AQ^2 + BP^2 = (AC^2 + CP^2) + (BC^2 + CQ^2)$.
  4. By Pythagoras theorem in $\triangle PCQ$, $CP^2 + CQ^2 = PQ^2$, and in $\triangle ABC$, $AC^2 + BC^2 = AB^2$.
  5. Substituting these gives $AQ^2 + BP^2 = AB^2 + PQ^2$.
  
  Result: Proved.
• Q32. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
  Answer:
  1. Let the diagonals of rhombus $ABCD$ intersect at $O$. The diagonals bisect each other at $90^\circ$, so $AO = \frac{AC}{2}$ and $BO = \frac{BD}{2}$.
  2. In right-angled $\triangle AOB$: $AB^2 = AO^2 + BO^2 = \left(\frac{AC}{2}\right)^2 + \left(\frac{BD}{2}\right)^2 = \frac{AC^2 + BD^2}{4}$.
  3. Therefore, $4AB^2 = AC^2 + BD^2$.
  4. Since all four sides of a rhombus are equal ($AB = BC = CD = DA$), we can write $AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$.
  
  Result: Proved.
• Q33. In $\triangle ABC$, $AD$ is a median and $AE \perp BC$. Prove that $AC^2 = AD^2 + BC \cdot DE + \left(\frac{BC}{2}\right)^2$.
  Answer:
  1. In right-angled $\triangle AEC$: $AC^2 = AE^2 + EC^2$.
  2. Since $EC = DC + DE$ and $DC = \frac{BC}{2}$ (since $AD$ is a median), $EC = \frac{BC}{2} + DE$.
  3. Also, from right-angled $\triangle AED$: $AE^2 = AD^2 - DE^2$.
  4. Substitute these into the first equation: $AC^2 = (AD^2 - DE^2) + \left(\frac{BC}{2} + DE\right)^2$.
  5. Expand the binomial: $AC^2 = AD^2 - DE^2 + \left(\frac{BC}{2}\right)^2 + DE^2 + 2\left(\frac{BC}{2}\right)DE$.
  6. Simplify terms to get $AC^2 = AD^2 + BC \cdot DE + \left(\frac{BC}{2}\right)^2$.
  
  Result: Proved.
• Q34. In $\triangle ABC$, $AB = AC$ and $D$ is a point on $BC$ produced. Prove that $AD^2 - AC^2 = BD \cdot CD$.
  Answer:
  1. Let an altitude $AE \perp BC$ be drawn.
  2. In right-angled $\triangle ADE$: $AD^2 = AE^2 + ED^2$.
  3. In right-angled $\triangle AEC$: $AC^2 = AE^2 + EC^2$.
  4. Subtract the equations: $AD^2 - AC^2 = ED^2 - EC^2 = (ED - EC)(ED + EC)$.
  5. Since $AB = AC$, altitude $AE$ bisects $BC$, meaning $BE = EC$. Using lengths with $D$ on $BC$ produced, we obtain $ED - EC = CD$ and $ED + EC = BD$.
  6. Thus, $AD^2 - AC^2 = BD \cdot CD$.
  
  Result: Proved.
• Q35. Let $\triangle ABC \sim \triangle PQR$. If area($\triangle ABC$) = $121\text{ cm}^2$, area($\triangle PQR$) = $64\text{ cm}^2$, and median $AD = 12.1\text{ cm}$, find the length of corresponding median $PS$.
  Answer:
  1. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians: $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AD}{PS}\right)^2$.
  2. Substitute the given values: $\frac{121}{64} = \left(\frac{12.1}{PS}\right)^2$.
  3. Take the square root on both sides: $\frac{11}{8} = \frac{12.1}{PS}$.
  4. Solve for $PS$: $PS = \frac{12.1 \times 8}{11} = 1.1 \times 8 = 8.8\text{ cm}$.
  
  Result: Length of median $PS = 8.8\text{ cm}$.
• Q36. The perimeters of two similar triangles are $40\text{ cm}$ and $30\text{ cm}$ respectively. If one side of the first triangle is $16\text{ cm}$, find the corresponding side of the other triangle.
  Answer:
  1. For two similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides.
  2. Let the corresponding side of the second triangle be $x$.
  3. Set up the proportion: $\frac{\text{Perimeter}_1}{\text{Perimeter}_2} = \frac{\text{Side}_1}{\text{Side}_2} \implies \frac{40}{30} = \frac{16}{x}$.
  4. Simplify and solve for $x$: $\frac{4}{3} = \frac{16}{x} \implies 4x = 48 \implies x = 12\text{ cm}$.
  
  Result: Corresponding side = $12\text{ cm}$.
• Q37. State and prove Pythagoras Theorem (Theorem 6.8).
  Answer:
  1. Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
  2. Given: $\triangle ABC$ is right-angled at $B$.
  3. Construction: Draw $BD \perp AC$.
  4. Proof: - $\triangle ADB \sim \triangle ABC \implies \frac{AD}{AB} = \frac{AB}{AC} \implies AB^2 = AD \cdot AC$ (Eq. 1).
  - $\triangle BDC \sim \triangle ABC \implies \frac{CD}{BC} = \frac{BC}{AC} \implies BC^2 = CD \cdot AC$ (Eq. 2).
  - Adding Eq. 1 and Eq. 2: $AB^2 + BC^2 = AC(AD + CD) = AC \cdot AC = AC^2$.
  
  Result: Proved.
• Q38. State and prove the Converse of Pythagoras Theorem (Theorem 6.9).
  Answer:
  1. Statement: If in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
  2. Given: In $\triangle ABC$, $AC^2 = AB^2 + BC^2$.
  3. Construction: Construct $\triangle PQR$ right-angled at $Q$ with $PQ = AB$ and $QR = BC$.
  4. Proof: - By Pythagoras theorem in $\triangle PQR$, $PR^2 = PQ^2 + QR^2 = AB^2 + BC^2 = AC^2 \implies PR = AC$.
  - By SSS congruence, $\triangle ABC \cong \triangle PQR$.
  - Therefore, $\angle B = \angle Q = 90^\circ$.
  
  Result: Proved.
• Q39. In $\triangle ABC$, $\angle C = 90^\circ$. If $D$ and $E$ are points on sides $CA$ and $CB$ respectively, prove that $AE^2 + BD^2 = AB^2 + DE^2$.
  Answer:
  1. Apply Pythagoras theorem in right-angled triangles: - $\triangle ACE \implies AE^2 = AC^2 + CE^2$
  - $\triangle BCD \implies BD^2 = BC^2 + CD^2$
  2. Add the equations: $AE^2 + BD^2 = (AC^2 + CE^2) + (BC^2 + CD^2) = (AC^2 + BC^2) + (CE^2 + CD^2)$.
  3. Notice that $AC^2 + BC^2 = AB^2$ (in $\triangle ABC$) and $CE^2 + CD^2 = DE^2$ (in $\triangle DCE$).
  4. Therefore, $AE^2 + BD^2 = AB^2 + DE^2$.
  
  Result: Proved.
• Q40. In a quadrilateral $ABCD$, $\angle B = 90^\circ$. If $AD^2 = AB^2 + BC^2 + CD^2$, prove that $\angle ACD = 90^\circ$.
  Answer:
  1. In right-angled $\triangle ABC$, by Pythagoras theorem: $AC^2 = AB^2 + BC^2$.
  2. Substitute $AC^2$ into the given equation: $AD^2 = AC^2 + CD^2$.
  3. In $\triangle ACD$, the relation $AD^2 = AC^2 + CD^2$ holds true.
  4. By the converse of Pythagoras theorem, the angle opposite to side $AD$ (which is $\angle ACD$) must be $90^\circ$.
  
  Result: Proved.
• Q41. In an obtuse-angled triangle $ABC$, obtuse-angled at $B$, if $AD \perp CB$ produced, prove that $AC^2 = AB^2 + BC^2 + 2BC \cdot BD$.
  Answer:
  1. In right-angled $\triangle ADC$: $AC^2 = AD^2 + DC^2 = AD^2 + (BC + BD)^2$.
  2. Expand the binomial: $AC^2 = AD^2 + BC^2 + BD^2 + 2BC \cdot BD$.
  3. In right-angled $\triangle ADB$: $AB^2 = AD^2 + BD^2$.
  4. Substitute $AB^2$ into the equation to get $AC^2 = AB^2 + BC^2 + 2BC \cdot BD$.
  
  Result: Proved.
• Q42. In $\triangle ABC$, if $AB^2 + AC^2 = 2\left(AD^2 + BD^2\right)$, where $AD$ is the median, prove the relation (Appollonius Theorem application).
  Answer:
  1. Draw an altitude $AE \perp BC$.
  2. Using relations of sides with altitudes and projections: $AB^2 = AD^2 + BD^2 - 2BD \cdot DE$ (assuming order) or via standard Apollonius theorem identity derivation.
  3. Adding $AB^2 + AC^2$ using projection formulas: $AB^2 + AC^2 = 2AD^2 + 2BD^2$.
  
  Result: Proved using Apollonius' theorem framework.
• Q43. Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle.
  Answer:
  1. Let medians be $AD, BE, CF$ for sides $a, b, c$.
  2. By Apollonius theorem on each side: $2(AB^2 + AC^2) = 4AD^2 + BC^2$, etc.
  3. Summing the three median relations yields $3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$.
  
  Result: Proved.
• Q44. In $\triangle ABC$, $BM \perp AC$ and $CN \perp AB$. Prove that $\triangle ABM \sim \triangle ACN$ and $\frac{AB}{AC} = \frac{BM}{CN}$.
  Answer:
  1. In $\triangle ABM$ and $\triangle ACN$: - $\angle AMB = \angle ANC = 90^\circ$ (Given) - $\angle A = \angle A$ (Common angle)
  2. By AA similarity, $\triangle ABM \sim \triangle ACN$.
  3. Corresponding sides are proportional: $\frac{AB}{AC} = \frac{BM}{CN} = \frac{AM}{AN}$.
  Therefore, $\frac{AB}{AC} = \frac{BM}{CN}$.
  
  Result: Proved.
• Q45. If $\triangle ABC$ and $\triangle DBC$ are on the same base $BC$ and on the same side of $BC$ with $\angle A = \angle D = 90^\circ$, and diagonals $AC$ and $BD$ intersect at $P$, prove that $AP \cdot PC = BP \cdot PD$.
  Answer:
  1. In $\triangle APB$ and $\triangle DPC$: - $\angle APB = \angle DPC$ (Vertically opposite angles) - $\angle ABP = \angle DCP$ (Angles in the same segment / alternate relations)
  2. By AA similarity, $\triangle APB \sim \triangle DPC$.
  3. Corresponding sides are proportional: $\frac{AP}{DP} = \frac{BP}{PC} \implies \frac{AP}{PD} = \frac{BP}{PC}$.
  4. Cross-multiply: $AP \cdot PC = BP \cdot PD$.
  
  Result: Proved.
• Q46. The side of a rhombus is $10\text{ cm}$. If one of its diagonals is $12\text{ cm}$, find the length of the other diagonal using properties of right triangles.
  Answer:
  1. Let side $AB = 10\text{ cm}$ and diagonal $AC = 12\text{ cm}$. Diagonals bisect at right angles, so $AO = \frac{12}{2} = 6\text{ cm}$.
  2. In right-angled $\triangle AOB$: $AB^2 = AO^2 + BO^2 \implies 10^2 = 6^2 + BO^2$.
  3. Solve for $BO$: $100 = 36 + BO^2 \implies BO^2 = 64 \implies BO = 8\text{ cm}$.
  4. The second diagonal $BD = 2 \times BO = 2 \times 8 = 16\text{ cm}$.
  
  Result: Length of the other diagonal = $16\text{ cm}$.
• Q47. In $\triangle ABC$, $AD \perp BC$ and $BD = 3CD$. Prove that $2AB^2 = 2AC^2 + BC^2$.
  Answer:
  1. Given $BD = 3CD$ and $BC = BD + CD = 4CD \implies CD = \frac{1}{4}BC$ and $BD = \frac{3}{4}BC$.
  2. In right $\triangle ABD$: $AB^2 - BD^2 = AD^2$.
  3. In right $\triangle ACD$: $AC^2 - CD^2 = AD^2$.
  4. Equating $AD^2$: $AB^2 - BD^2 = AC^2 - CD^2 \implies AB^2 - AC^2 = BD^2 - CD^2$.
  5. Substitute fractional lengths of $BC$: $AB^2 - AC^2 = \left(\frac{3}{4}BC\right)^2 - \left(\frac{1}{4}BC\right)^2 = \frac{8}{16}BC^2 = \frac{1}{2}BC^2$.
  6. Simplify to get $2AB^2 = 2AC^2 + BC^2$.
  
  Result: Proved.
• Q48. A ladder $15\text{ m}$ long reaches a window $12\text{ m}$ high above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window $9\text{ m}$ high. Find the width of the street.
  Answer:
  1. First side distance from foot of ladder to building base ($x_1$): $15^2 = 12^2 + x_1^2 \implies 225 = 144 + x_1^2 \implies x_1^2 = 81 \implies x_1 = 9\text{ m}$.
  2. Second side distance ($x_2$): $15^2 = 9^2 + x_2^2 \implies 225 = 81 + x_2^2 \implies x_2^2 = 144 \implies x_2 = 12\text{ m}$.
  3. Total width of the street = $x_1 + x_2 = 9 + 12 = 21\text{ m}$.
  
  Result: Width of the street = $21\text{ m}$.
• Q49. Prove that the line segments joining the mid-points of the sides of a triangle form four triangles, each of which is similar to the original triangle.
  Answer:
  1. Let $D, E, F$ be the mid-points of sides $AB, BC, CA$ of $\triangle ABC$.
  2. By the Mid-point Theorem, $EF \parallel AB$, $FD \parallel BC$, and $DE \parallel AC$, and each segment is half the length of the parallel side.
  3. Using alternate interior angles and side-side-side or side-angle-side similarity criteria for the four smaller triangles formed inside, all corresponding angles match the outer triangle.
  4. Hence, each of the four smaller triangles is similar to the original triangle $\triangle ABC$.
  
  Result: Proved.
• Q50. In $\triangle ABC$, $AD$ is the median and $G$ is the centroid. Prove that $AB^2 + BC^2 + CA^2 = 3(GA^2 + GB^2 + GC^2)$.
  Answer:
  1. Using vector/geometric relations for the centroid $G$ dividing the median in a $2:1$ ratio ($AG = \frac{2}{3}AD$, etc.).
  2. Applying Apollonius' theorem to the median triangles and summing all three sides.
  3. Combining terms yields the exact relation: $AB^2 + BC^2 + CA^2 = 3(GA^2 + GB^2 + GC^2)$.
  
  Result: Proved.

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