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

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SECTION B — Very Short Answer Type Questions [2 Marks Each]

• Q26. Prove that the line segments joining the mid-points of the sides of a triangle divide the triangle into four smaller congruent triangles (using similarity/BPT concepts).
  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 $DE \parallel AC$.
  3. This forms four smaller triangles ($\triangle ADF$, $\triangle FDE$, $\triangle DBE$, $\triangle ECF$). Using SSS congruence or similarity and matching side halves, each small triangle is congruent to the others and similar to the outer triangle.
  
  Result: Proved.
• Q27. In $\triangle DEF$, $AB \parallel EF$ meets $DE$ at $A$ and $DF$ at $B$. If $DA = 3\text{ cm}$, $AE = 5\text{ cm}$, and $DB = 4.5\text{ cm}$, find the length of $BF$.
  Answer:
  1. Since $AB \parallel EF$, by Basic Proportionality Theorem (Thales' Theorem): $\frac{DA}{AE} = \frac{DB}{BF}$.
  2. Substitute the given values: $\frac{3}{5} = \frac{4.5}{BF}$.
  3. Solve for $BF$: $BF = \frac{4.5 \times 5}{3} = \frac{22.5}{3} = 7.5\text{ cm}$.
  
  Result: $BF = 7.5\text{ cm}$.
• Q28. If the areas of two similar triangles are equal, prove that they are congruent.
  Answer:
  1. Given $\triangle ABC \sim \triangle PQR$ and $\text{Area}(\triangle ABC) = \text{Area}(\triangle PQR)$.
  2. By area ratio theorem, $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = 1 \implies AB = PQ$.
  3. Similarly, $BC = QR$ and $CA = RP$. By SSS congruence, $\triangle ABC \cong \triangle PQR$.
  
  Result: Proved.
• Q29. In $\triangle ABC$, $\angle B = 90^\circ$ and $BD \perp AC$ at $D$. If $AD = 4\text{ cm}$ and $CD = 9\text{ cm}$, find the length of $BD$.
  Answer:
  1. In a right-angled triangle with an altitude to the hypotenuse, $BD^2 = AD \times CD$.
  2. Substitute the values: $BD^2 = 4 \times 9 = 36$.
  3. Solve for $BD$: $BD = \sqrt{36} = 6\text{ cm}$.
  
  Result: $BD = 6\text{ cm}$.
• Q30. Given $\triangle ABC \sim \triangle PQR$. If $AB = 5\text{ cm}$, area($\triangle ABC$) = $20\text{ cm}^2$, and area($\triangle PQR$) = $45\text{ cm}^2$, find the length of $PQ$.
  Answer:
  1. $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 \implies \frac{20}{45} = \left(\frac{5}{PQ}\right)^2$.
  2. Simplify fraction: $\frac{4}{9} = \left(\frac{5}{PQ}\right)^2$.
  3. Take square root: $\frac{2}{3} = \frac{5}{PQ} \implies PQ = \frac{15}{2} = 7.5\text{ cm}$.
  
  Result: $PQ = 7.5\text{ cm}$.
• Q31. In $\triangle ABC$, $AB = 6\sqrt{3}\text{ cm}$, $AC = 12\text{ cm}$, and $BC = 6\text{ cm}$. Find the measure of $\angle B$.
  Answer:
  1. Find squares of lengths: $AC^2 = 12^2 = 144$, $AB^2 = (6\sqrt{3})^2 = 108$, $BC^2 = 6^2 = 36$.
  2. Observe that $AB^2 + BC^2 = 108 + 36 = 144 = AC^2$.
  3. By the converse of Pythagoras Theorem, $\triangle ABC$ is a right-angled triangle at $B$ (since $AC$ is the hypotenuse). Thus, $\angle B = 90^\circ$.
  
  Result: $\angle B = 90^\circ$.
• Q32. Two poles of height $6\text{ m}$ and $11\text{ m}$ stand vertically on a plane ground. If the distance between their feet is $12\text{ m}$, find the distance between their tops.
  Answer:
  1. Difference in height = $11 - 6 = 5\text{ m}$.
  2. Distance between feet (base) = $12\text{ m}$.
  3. Distance between tops forms a right triangle hypotenuse: $\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ m}$.
  
  Result: Distance between tops = $13\text{ m}$.
• Q33. In $\triangle ABC$, $AD$ is the bisector of $\angle A$ meeting $BC$ at $D$. If $AB = 5\text{ cm}$, $AC = 7\text{ cm}$, and $BC = 10\text{ cm}$, find the length of $BD$.
  Answer:
  1. By Angle Bisector Theorem, $\frac{BD}{DC} = \frac{AB}{AC} = \frac{5}{7}$.
  2. Let $BD = 5x$ and $DC = 7x$. Since $BC = 10$, $5x + 7x = 10 \implies 12x = 10 \implies x = \frac{5}{6}$.
  3. $BD = 5 \times \frac{5}{6} = \frac{25}{6}\text{ cm}$ (approx $4.17\text{ cm}$).
  
  Result: $BD = \frac{25}{6}\text{ cm}$.
• Q34. If $\triangle ABC \sim \triangle DEF$ such that $2AB = DE$ and $BC = 8\text{ cm}$, find the length of $EF$.
  Answer:
  1. From similarity, $\frac{AB}{DE} = \frac{BC}{EF}$.
  2. Given $2AB = DE$, so $\frac{AB}{DE} = \frac{1}{2}$.
  3. $\frac{1}{2} = \frac{8}{EF} \implies EF = 16\text{ cm}$.
  
  Result: $EF = 16\text{ cm}$.
• Q35. In $\triangle PQR$, $S$ is a point on side $QR$ such that $\angle PSR = \angle QPR$. Show that $PR^2 = QR \times SR$.
  Answer:
  1. In $\triangle PQR$ and $\triangle SPR$: $\angle PSR = \angle QPR$ (given) and $\angle R = \angle R$ (common angle).
  2. By AA similarity, $\triangle PQR \sim \triangle SPR$.
  3. Write corresponding sides proportion: $\frac{PR}{SR} = \frac{QR}{PR} \implies PR^2 = QR \times SR$.
  
  Result: Proved.
• Q36. In $\triangle ABC$, $AD \perp BC$ and $BD = 3CD$. Prove that $2AB^2 = 2AC^2 + BC^2$.
  Answer:
  1. Since $BD = 3CD$, $BC = 4CD$, meaning $CD = \frac{1}{4}BC$ and $BD = \frac{3}{4}BC$.
  2. From right triangles $ABD$ and $ACD$: $AB^2 - BD^2 = AC^2 - CD^2 \implies AB^2 - AC^2 = \left(\frac{3}{4}BC\right)^2 - \left(\frac{1}{4}BC\right)^2 = \frac{1}{2}BC^2$.
  3. Rearranging gives $2AB^2 = 2AC^2 + BC^2$.
  
  Result: Proved.
• Q37. A ladder $10\text{ m}$ long reaches a window $8\text{ m}$ above the ground. Find the distance of the foot of the ladder from the base of the wall.
  Answer:
  1. By Pythagoras theorem: $\text{length}^2 = \text{height}^2 + \text{distance}^2$.
  2. $10^2 = 8^2 + \text{distance}^2 \implies 100 = 64 + \text{distance}^2$.
  3. $\text{distance}^2 = 36 \implies \text{distance} = 6\text{ m}$.
  
  Result: Distance = $6\text{ m}$.
• Q38. In an equilateral triangle of side $2a$, find the length of one of its altitudes.
  Answer:
  1. Altitude bisects the base side $2a$ into segments of length $a$.
  2. Using Pythagoras theorem: $\text{altitude} = \sqrt{(2a)^2 - a^2} = \sqrt{4a^2 - a^2} = \sqrt{3}a$.
  
  Result: Length of altitude = $\sqrt{3}a$.
• Q39. The perimeters of two similar triangles $\triangle ABC$ and $\triangle PQR$ are $30\text{ cm}$ and $20\text{ cm}$ respectively. If $PQ = 8\text{ cm}$, find the length of side $AB$.
  Answer:
  1. Ratio of perimeters equals ratio of corresponding sides: $\frac{\text{Perimeter}(ABC)}{\text{Perimeter}(PQR)} = \frac{AB}{PQ}$.
  2. Substitute values: $\frac{30}{20} = \frac{AB}{8} \implies \frac{3}{2} = \frac{AB}{8}$.
  3. Solve for $AB$: $AB = \frac{3 \times 8}{2} = 12\text{ cm}$.
  
  Result: $AB = 12\text{ cm}$.
• Q40. In $\triangle ABC$, if $AB^2 + AC^2 = BC^2$, state the type of triangle with respect to its angles and name the theorem.
  Answer:
  1. Triangle Type: Right-angled triangle (right-angled at vertex $A$, since $BC$ is the hypotenuse).
  2. Theorem Name: Converse of Pythagoras Theorem.
  
  Result: Right-angled triangle; Converse of Pythagoras Theorem.
• Q41. $D$ is a point on the hypotenuse $AC$ of a right-angled triangle $ABC$ such that $BD \perp AC$. Prove that $BD^2 = AD \times CD$.
  Answer:
  1. By similar triangle properties, $\triangle ADB \sim \triangle BDC$.
  2. Write corresponding side ratios: $\frac{AD}{BD} = \frac{BD}{CD}$.
  3. Cross-multiply to get $BD^2 = AD \times CD$.
  
  Result: Proved.
• Q42. 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 $\triangle ACE$ and $\triangle BCD$: $AE^2 + BD^2 = (AC^2 + CE^2) + (BC^2 + CD^2)$.
  2. Group terms: $(AC^2 + BC^2) + (CE^2 + CD^2) = AB^2 + DE^2$.
  
  Result: Proved.
• Q43. The side of a rhombus is $10\text{ cm}$. If one of its diagonals is $16\text{ cm}$, find the length of the other diagonal.
  Answer:
  1. Half of the first diagonal = $\frac{16}{2} = 8\text{ cm}$.
  2. Using right triangle formed by side and half-diagonals: $10^2 = 8^2 + x^2 \implies 100 = 64 + x^2 \implies x = 6\text{ cm}$.
  3. Other diagonal length = $2 \times 6 = 12\text{ cm}$.
  
  Result: $12\text{ cm}$.
• Q44. 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. Since $\angle B = 90^\circ$, $AC^2 = AB^2 + BC^2$.
  2. Substitute into equation: $AD^2 = AC^2 + CD^2$.
  3. By converse of Pythagoras theorem, $\angle ACD = 90^\circ$.
  
  Result: Proved.
• Q45. In $\triangle ABC$, $BM \perp AC$ and $CN \perp AB$. Prove that $\triangle ABM \sim \triangle ACN$.
  Answer:
  1. In $\triangle ABM$ and $\triangle ACN$: $\angle AMB = \angle ANC = 90^\circ$ and $\angle A = \angle A$ (common).
  2. By AA similarity, $\triangle ABM \sim \triangle ACN$.
  
  Result: Proved.
• Q46. 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. Using right triangle relationships of diagonals bisecting at $90^\circ$: $4AB^2 = AC^2 + BD^2$.
  2. Since all four sides are equal, $AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$.
  
  Result: Proved.
• Q47. 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. From right triangle $AEC$: $AC^2 = AE^2 + EC^2$.
  2. Substitute $AE^2 = AD^2 - DE^2$ and $EC = \frac{BC}{2} + DE$ to expand and simplify terms.
  
  Result: Proved.
• Q48. In $\triangle ABC$, $AB = AC$ and $D$ is a point on $BC$ produced. Prove that $AD^2 - AC^2 = BD \cdot CD$.
  Answer:
  1. Draw altitude $AE \perp BC$. Using right triangles $ADE$ and $ACE$: $AD^2 - AC^2 = ED^2 - EC^2$.
  2. Factorize and simplify using equal segment properties to get $BD \cdot CD$.
  
  Result: Proved.
• Q49. Two triangles have the same base. Prove that the ratio of their areas is equal to the ratio of their corresponding altitudes.
  Answer:
  1. $\text{Area}_1 = \frac{1}{2} \times \text{base} \times h_1$ and $\text{Area}_2 = \frac{1}{2} \times \text{base} \times h_2$.
  2. Taking the ratio: $\frac{\text{Area}_1}{\text{Area}_2} = \frac{h_1}{h_2}$.
  
  Result: Proved.
• Q50. 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 $AC$ and $BD$ intersect at $P$, prove that $AP \cdot PC = BP \cdot PD$.
  Answer:
  1. In $\triangle APB$ and $\triangle DPC$, vertically opposite angles and equal alternate/segment angles give $\triangle APB \sim \triangle DPC$ by AA similarity.
  2. Corresponding sides proportion $\frac{AP}{DP} = \frac{BP}{PC} \implies AP \cdot PC = BP \cdot PD$.
  
  Result: Proved.

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