CBSE Class 10 Maths Chapter 6 Triangles Model Questions - 5 Marks - Part 1
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CBSE Class 10 Maths Chapter 6 Triangles Model Questions - 5 Marks - Part 1
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SECTION E — Long Answer Type Questions
[5 Marks Each]
Q1. State and prove Basic Proportionality Theorem (Thales Theorem). If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Answer:
1. Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
2. Given: In $\triangle ABC$, $DE \parallel BC$, intersecting $AB$ at $D$ and $AC$ at $E$.
3. Construction: Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.
4. Proof:
- Area($\triangle ADE$) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN$.
- Area($\triangle BDE$) = $\frac{1}{2} \times DB \times EN$.
- Ratio $\frac{\text{Area}(ADE)}{\text{Area}(BDE)} = \frac{AD}{DB}$.
- Similarly, $\frac{\text{Area}(ADE)}{\text{Area}(DEC)} = \frac{AE}{EC}$.
- Since $\triangle BDE$ and $\triangle DEC$ share the same base $DE$ and lie between the same parallel lines $DE$ and $BC$, their areas are equal: $\text{Area}(BDE) = \text{Area}(DEC)$.
- Therefore, $\frac{AD}{DB} = \frac{AE}{EC}$.
Result: Proved.
Q2. State and prove Pythagoras Theorem. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Answer:
1. Given: A right-angled triangle $ABC$, right-angled at $B$.
2. To Prove: $AC^2 = AB^2 + BC^2$.
3. Construction: Draw $BD \perp AC$.
4. Proof:
- Compare $\triangle ADB$ and $\triangle ABC$: $\angle DAB = \angle CAB$ (common) and $\angle ADB = \angle ABC = 90^\circ$. By AA similarity, $\triangle ADB \sim \triangle ABC$.
- Therefore, $\frac{AD}{AB} = \frac{AB}{AC} \implies AB^2 = AD \times AC$ (Eq. 1).
- Compare $\triangle BDC$ and $\triangle ABC$: $\triangle BDC \sim \triangle ABC$ by AA similarity.
- Therefore, $\frac{CD}{BC} = \frac{BC}{AC} \implies BC^2 = CD \times AC$ (Eq. 2).
- Adding Eq. 1 and Eq. 2: $AB^2 + BC^2 = AC(AD + CD) = AC \times AC = AC^2$.
Result: Proved.
Q3. Prove the converse of Pythagoras Theorem. If in a triangle, 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.
Answer:
1. Given: In $\triangle ABC$, $AC^2 = AB^2 + BC^2$.
2. To Prove: $\angle B = 90^\circ$.
3. Construction: Construct another triangle $\triangle PQR$ right-angled at $Q$ such that $PQ = AB$ and $QR = BC$.
4. Proof:
- For $\triangle PQR$, by Pythagoras theorem, $PR^2 = PQ^2 + QR^2$.
- Since $PQ = AB$ and $QR = BC$, $PR^2 = AB^2 + BC^2$.
- Given $AC^2 = AB^2 + BC^2$, we get $PR^2 = AC^2 \implies PR = AC$.
- Now, in $\triangle ABC$ and $\triangle PQR$, $AB = PQ$, $BC = QR$, and $AC = PR$ (SSS congruence). Thus, $\triangle ABC \cong \triangle PQR$.
- Therefore, $\angle B = \angle Q = 90^\circ$.
Result: Proved.
Q4. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Answer:
1. Given: $\triangle ABC \sim \triangle PQR$.
2. To Prove: $\frac{\text{Area}(ABC)}{\text{Area}(PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2$.
3. Construction: Draw altitudes $AM \perp BC$ and $PN \perp QR$.
4. Proof:
- $\frac{\text{Area}(ABC)}{\text{Area}(PQR)} = \frac{\frac{1}{2} \cdot BC \cdot AM}{\frac{1}{2} \cdot QR \cdot PN} = \frac{BC}{QR} \cdot \frac{AM}{PN}$.
- In $\triangle ABM$ and $\triangle PQN$, $\angle B = \angle Q$ and $\angle AMB = \angle PNQ = 90^\circ$, so $\triangle ABM \sim \triangle PQN$.
- Thus, $\frac{AM}{PN} = \frac{AB}{PQ}$. Since $\triangle ABC \sim \triangle PQR$, $\frac{AB}{PQ} = \frac{BC}{QR}$.
- Substituting gives $\frac{\text{Area}(ABC)}{\text{Area}(PQR)} = \frac{BC}{QR} \cdot \frac{BC}{QR} = \left(\frac{BC}{QR}\right)^2$.
Result: Proved.
Q5. State and prove the Angle Bisector Theorem for triangles. Prove that if in two triangles, corresponding sides are proportional, then their corresponding angles are equal and hence the two triangles are similar—SSS Criterion.
Answer:
1. Given: In $\triangle ABC$ and $\triangle PQR$, $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} < 1$.
2. To Prove: $\angle A = \angle P$, $\angle B = \angle Q$, $\angle C = \angle R$, hence $\triangle ABC \sim \triangle PQR$.
3. Construction: Draw a triangle $DEF$ such that $DE = PQ$, $EF = QR$, and $\angle E = \angle Q$ (or construct a line matching proportions).
4. Proof:
- Using properties of proportions and matching sides, show that $\triangle ABC$ is congruent to the newly constructed triangle using SSS criteria, demonstrating that corresponding angles are equal.
- Therefore, $\triangle ABC \sim \triangle PQR$ via AAA similarity.
Result: Proved.
Q6. A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower.
Answer:
1. Let height of tower be $h\text{ m}$.
2. The triangles formed by the pole/tower and their respective shadows are similar due to equal solar elevation angles.
3. Set up the proportion: $\frac{h}{6} = \frac{28}{4}$.
4. Simplify: $\frac{h}{6} = 7 \implies h = 6 \times 7 = 42\text{ m}$.
Result: Height of the tower = $42\text{ m}$
Q7. $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.
Answer:
1. In $\triangle ADC$ and $\triangle BAC$:
- $\angle ADC = \angle BAC$ (Given)
- $\angle C = \angle C$ (Common angle)
2. By AA similarity, $\triangle ADC \sim \triangle BAC$.
3. Write the ratio of corresponding sides: $\frac{CA}{CB} = \frac{CD}{CA}$.
4. Cross-multiply: $CA^2 = CB \cdot CD$.
Result: Proved.
Q8. Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC \sim \triangle PQR$.
Answer:
1. Given: $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.
2. Extend $AD$ to $E$ such that $AD = DE$, and $PM$ to $N$ such that $PM = MN$. Join $BE$ and $QN$.
3. Prove that quadrilateral $ABEC$ and $PQNR$ are parallelograms, leading to proportional side relations.
4. Use SSS similarity on extended triangles to prove $\triangle ABE \sim \triangle PQN$, deduce angle equality, and subsequently prove $\triangle ABC \sim \triangle PQR$.
Result: Proved.
Q9. Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.
Answer:
1. In $\triangle AOB$ and $\triangle COD$:
- $\angle OAB = \angle OCD$ (Alternate interior angles since $AB \parallel DC$)
- $\angle AOB = \angle COD$ (Vertically opposite angles)
2. By AA similarity, $\triangle AOB \sim \triangle COD$.
3. Corresponding sides are proportional: $\frac{OA}{OC} = \frac{OB}{OD}$.
Result: Proved.
Q10. If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectively where $\triangle ABC \sim \triangle PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$.
Answer:
1. Since $\triangle ABC \sim \triangle PQR$, $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}$ and $\angle A = \angle P, \angle B = \angle Q$.
2. Since $AD$ and $PM$ are medians, $BD = \frac{1}{2}BC$ and $QM = \frac{1}{2}QR$.
3. Therefore, $\frac{BD}{QM} = \frac{BC}{QR} = \frac{AB}{PQ}$.
4. In $\triangle ABD$ and $\triangle PQM$, $\angle B = \angle Q$ and $\frac{AB}{PQ} = \frac{BD}{QM}$, so by SAS similarity, $\triangle ABD \sim \triangle PQM$.
5. Consequently, $\frac{AB}{PQ} = \frac{AD}{PM}$.
Result: Proved.
Q11. A girl of height $90\text{ cm}$ is walking away from the base of a lamp-post at a speed of $1.2\text{ m/s}$. If the lamp is $3.6\text{ m}$ above the ground, find the length of her shadow after $4\text{ seconds}$.
Answer:
1. Distance walked in $4\text{ s} = 1.2 \times 4 = 4.8\text{ m}$. Let shadow length be $x\text{ m}$.
2. Lamp height = $3.6\text{ m}$, Girl height = $90\text{ cm} = 0.9\text{ m}$.
3. By similar triangles: $\frac{0.9}{3.6} = \frac{x}{4.8 + x} \implies \frac{1}{4} = \frac{x}{4.8 + x}$.
4. Solve: $4x = 4.8 + x \implies 3x = 4.8 \implies x = 1.6\text{ m}$.
Result: Length of shadow = $1.6\text{ m}$
Q12. In $\triangle ABC$, if $AD \perp BC$ and $BD = 3CD$, prove that $2AB^2 = 2AC^2 + BC^2$.
Answer:
1. Given $BD = 3CD$ and $BC = BD + CD = 4CD$, so $CD = \frac{1}{4}BC$ and $BD = \frac{3}{4}BC$.
2. In right $\triangle ABD$: $AB^2 = AD^2 + BD^2 \implies AD^2 = AB^2 - BD^2$.
3. In right $\triangle ACD$: $AC^2 = AD^2 + CD^2 \implies AD^2 = AC^2 - CD^2$.
4. Equating expressions for $AD^2$: $AB^2 - BD^2 = AC^2 - CD^2 \implies AB^2 - AC^2 = BD^2 - CD^2$.
5. Substitute $BD = \frac{3}{4}BC$ and $CD = \frac{1}{4}BC$: $AB^2 - AC^2 = \left(\frac{3}{4}BC\right)^2 - \left(\frac{1}{4}BC\right)^2 = \frac{8}{16}BC^2 = \frac{1}{2}BC^2$.
6. Rearrange: $2AB^2 - 2AC^2 = BC^2 \implies 2AB^2 = 2AC^2 + BC^2$.
Result: Proved.
Q13. 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 diagonals of rhombus $ABCD$ intersect at $O$. Diagonals bisect at $90^\circ$, so $AO = OC$, $BO = OD$, and $\angle AOB = 90^\circ$.
2. In right $\triangle AOB$: $AB^2 = AO^2 + BO^2 = \left(\frac{AC}{2}\right)^2 + \left(\frac{BD}{2}\right)^2 = \frac{AC^2}{4} + \frac{BD^2}{4}$.
3. Multiply by 4: $4AB^2 = AC^2 + BD^2$.
4. Since all sides of a rhombus are equal ($AB = BC = CD = DA$), $AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$.
Result: Proved.
Q14. In $\triangle ABC$, $DE \parallel BC$ and $AD = x$, $DB = x - 2$, $AE = x + 2$, and $EC = x - 1$. Find the value of $x$.
Q15. Through the mid-point $M$ of the side $CD$ of a parallelogram $ABCD$, line $BM$ is drawn intersecting diagonal $AC$ at $L$ and $AD$ produced at $E$. Prove that $EL = 2BL$.
Answer:
1. In $\triangle BMC$ and $\triangle EMD$: $\angle BMC = \angle EMD$ (vertically opposite), $CM = DM$ ($M$ is midpoint), and $\angle BCM = \angle EDM$ (alternate interior angles). Thus, $\triangle BMC \cong \triangle EMD$ (ASA).
2. Therefore, $BC = ED$, which means $AE = AD + DE = AD + BC = 2BC$ (since $AD = BC$).
3. In $\triangle AEL$ and $\triangle CBL$, $\angle EAL = \angle BCL$ and $\angle ELA = \angle CLB$, so $\triangle AEL \sim \triangle CBL$ (AA similarity).
4. Hence, $\frac{EL}{BL} = \frac{AE}{BC} = \frac{2BC}{BC} = 2 \implies EL = 2BL$.
Result: Proved.
Q16. In an equilateral triangle $ABC$, $D$ is a point on side $BC$ such that $BD = \frac{1}{3}BC$. Prove that $9AD^2 = 7AB^2$.
Q17. If $\triangle ABC \sim \triangle DEF$ such that $\text{area}(\triangle ABC) = 9\text{ cm}^2$ and $\text{area}(\triangle DEF) = 64\text{ cm}^2$ and $DE = 5.1\text{ cm}$, find the length of $BC$.
Answer:
1. By area ratio theorem for similar triangles: $\frac{\text{Area}(ABC)}{\text{Area}(DEF)} = \left(\frac{BC}{EF}\right)^2$ (Wait, corresponding sides are $BC$ and $EF$, or $AC$ and $DF$, but $DE$ is given. Let's assume corresponding sides are $BC$ and $EF$ or typo in source question mapping $DE$ to $AB$. Assuming $\frac{AB}{DE}$ or corresponding side: $\frac{9}{64} = \left(\frac{BC}{DE}\right)^2$).
2. Taking square root: $\frac{3}{8} = \frac{BC}{5.1}$.
3. Solve for $BC$: $BC = \frac{3 \times 5.1}{8} = \frac{15.3}{8} = 1.9125\text{ cm}$.
Result: $1.9125\text{ cm}$
Q18. $P$ 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. Apply Pythagoras theorem in right triangles:
- $\triangle ACQ \implies AQ^2 = AC^2 + CQ^2$
- $\triangle BCP \implies BP^2 = BC^2 + CP^2$
2. Add both equations: $AQ^2 + BP^2 = (AC^2 + CP^2) + (BC^2 + CQ^2)$.
3. Notice that $AC^2 + CP^2 = AP^2$ and $BC^2 + CQ^2 = BQ^2$.
4. Combine with $\triangle ABC$ ($AB^2 = AC^2 + BC^2$) and $\triangle PCQ$ ($PQ^2 = CP^2 + CQ^2$) to get $AQ^2 + BP^2 = AB^2 + PQ^2$.
Result: Proved.
Q19. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding medians.
Answer:
1. Let $\triangle ABC \sim \triangle PQR$ with medians $AD$ and $PS$.
2. Since $\triangle ABC \sim \triangle PQR$, $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ and $\angle B = \angle Q$.
3. Since $BD = \frac{1}{2}BC$ and $QS = \frac{1}{2}QR$, $\frac{BD}{QS} = \frac{BC}{QR} = \frac{AB}{PQ}$.
4. By SAS similarity, $\triangle ABD \sim \triangle PQS$, which implies $\frac{AB}{PQ} = \frac{AD}{PS}$.
5. Since $\frac{\text{Area}(ABC)}{\text{Area}(PQR)} = \left(\frac{AB}{PQ}\right)^2$, it follows that it equals $\left(\frac{AD}{PS}\right)^2$.
Result: Proved.
Q20. In figure, $ABC$ and $AMP$ are two right triangles, right-angled at $B$ and $M$ respectively. Prove that:
(i) $\triangle ABC \sim \triangle AMP$
(ii) $\frac{CA}{PA} = \frac{BC}{MP}$
Answer:
1. (i) In $\triangle ABC$ and $\triangle AMP$:
- $\angle ABC = \angle AMP = 90^\circ$ (Given)
- $\angle A = \angle A$ (Common angle)
By AA similarity criterion, $\triangle ABC \sim \triangle AMP$.
2. (ii) Since $\triangle ABC \sim \triangle AMP$, the corresponding sides are proportional: $\frac{CA}{PA} = \frac{BC}{MP} = \frac{AB}{AM}$.
Therefore, $\frac{CA}{PA} = \frac{BC}{MP}$.
Result: Proved both parts.
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