This page provides comprehensive Chapter 6: Triangles - HOTS Worksheet - SJMaths. High Order Thinking Skills (HOTS) worksheet for Class 10 Triangles. Advanced problems for CBSE Board Exams aligned with rationalised syllabus.
Question 1: Two poles of height $a$ metres and $b$ metres are $p$ metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by $\frac{ab}{a+b}$ metres.
Solution: Step 1: Let poles be $AB = a$ and $CD = b$. Distance $BD = p$. Intersection point $P$ at height $h$ ($PM \perp BD$). Step 2: In $\Delta ABD$, $PM || AB \Rightarrow \frac{h}{a} = \frac{MD}{BD}$. Step 3: In $\Delta CDB$, $PM || CD \Rightarrow \frac{h}{b} = \frac{BM}{BD}$. Step 4: Adding: $\frac{h}{a} + \frac{h}{b} = \frac{MD + BM}{BD} = \frac{BD}{BD} = 1$. Step 5: $h(\frac{1}{a} + \frac{1}{b}) = 1 \Rightarrow h(\frac{a+b}{ab}) = 1 \Rightarrow h = \frac{ab}{a+b}$.
Question 2: $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.
Solution: Step 1: In $\Delta ABC$ and $\Delta DAC$: $\angle BAC = \angle ADC$ (Given) $\angle C = \angle C$ (Common) Step 2: By AA Similarity, $\Delta ABC \sim \Delta DAC$. Step 3: $\frac{CA}{CD} = \frac{CB}{CA}$ (Corresponding sides). Step 4: Cross multiplying: $CA^2 = CB \cdot CD$.
Question 3: A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution: Step 1: Height of lamp $AB = 3.6$ m. Height of girl $CD = 0.9$ m. Speed = 1.2 m/s. Step 2: Distance covered in 4s ($BD$) = $1.2 \times 4 = 4.8$ m. Let shadow $DE = x$. Step 3: $\Delta ABE \sim \Delta CDE$. So $\frac{AB}{CD} = \frac{BE}{DE} \Rightarrow \frac{3.6}{0.9} = \frac{4.8 + x}{x}$. Step 4: $4 = \frac{4.8 + x}{x} \Rightarrow 4x = 4.8 + x \Rightarrow 3x = 4.8 \Rightarrow x = 1.6$ m. Answer: 1.6 m.
Question 4: The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that $ABCD$ is a trapezium.
Solution: Step 1: Given $\frac{AO}{BO} = \frac{CO}{DO} \Rightarrow \frac{AO}{CO} = \frac{BO}{DO}$. Also $\angle AOB = \angle COD$ (Vertically opposite). Step 2: By SAS similarity, $\Delta AOB \sim \Delta COD$. Step 3: Therefore, $\angle OAB = \angle OCD$ (Alternate interior angles). Step 4: Since alternate angles are equal, $AB || DC$. Hence $ABCD$ is a trapezium.
Question 5: Through the mid-point $M$ of the side $CD$ of a parallelogram $ABCD$, the line $BM$ is drawn intersecting $AC$ in $L$ and $AD$ produced in $E$. Prove that $EL = 2BL$.
Solution: Step 1: In $\Delta BMC$ and $\Delta EMD$: $MC=MD$ (M is midpoint), $\angle BMC = \angle EMD$ (Vert. opp.), $\angle BCM = \angle EDM$ (Alt. int.). Step 2: $\Delta BMC \cong \Delta EMD$ (ASA). So $BC = DE$. Also $BC = AD$ (Parallelogram). So $AE = AD + DE = 2BC$. Step 3: In $\Delta AEL$ and $\Delta CBL$: $\angle ALE = \angle CLB$ (Vert. opp.), $\angle EAL = \angle BCL$ (Alt. int.). Step 4: $\Delta AEL \sim \Delta CBL$. So $\frac{EL}{BL} = \frac{AE}{BC} = \frac{2BC}{BC} = 2$. Thus $EL = 2BL$.
Question 6: In $\Delta ABC$, $AD \perp BC$. If $AD^2 = BD \cdot DC$, prove that $\angle BAC = 90^\circ$.
Solution: Step 1: Given $AD^2 = BD \cdot DC \Rightarrow \frac{AD}{BD} = \frac{DC}{AD}$. Step 2: In $\Delta ABD$ and $\Delta CAD$: $\angle ADB = \angle CDA = 90^\circ$ and sides are proportional. Step 3: By SAS similarity, $\Delta ABD \sim \Delta CAD$. Step 4: So $\angle BAD = \angle ACD$ and $\angle ABD = \angle CAD$. Step 5: In $\Delta ABC$, $\angle A + \angle B + \angle C = 180^\circ \Rightarrow \angle A + \angle CAD + \angle BAD = 180^\circ \Rightarrow 2\angle A = 180^\circ \Rightarrow \angle A = 90^\circ$.
Question 7: If the areas of two similar triangles are equal, prove that they are congruent.
Solution: Step 1: Let $\Delta ABC \sim \Delta PQR$. Area ratio = Square of ratio of corresponding sides. Step 2: $\frac{\text{Area}(ABC)}{\text{Area}(PQR)} = (\frac{AB}{PQ})^2 = 1$ (Since areas are equal). Step 3: $\frac{AB}{PQ} = 1 \Rightarrow AB = PQ$. Similarly $BC = QR$ and $AC = PR$. Step 4: By SSS congruence, $\Delta ABC \cong \Delta PQR$.
Question 8: In Figure, $PA, QB, RC$ are perpendiculars to $AC$. If $AP = x, QB = z, RC = y, AB = a, BC = b$, prove that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$.
Question 9: 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 $\Delta ABC \sim \Delta PQR$.
Solution: Step 1: Given $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$. Extend $AD$ to $E$ such that $AD=DE$ and $PM$ to $N$ such that $PM=MN$. Join $CE$ and $RN$. Step 2: $\Delta ABD \cong \Delta ECD$ (SAS). So $AB = CE$. Similarly $PQ = RN$. Step 3: Ratio becomes $\frac{CE}{RN} = \frac{AC}{PR} = \frac{AE}{PN}$. By SSS, $\Delta ACE \sim \Delta PRN$. Step 4: $\angle CAE = \angle RPN$. Similarly $\angle BAE = \angle QPM$. Adding gives $\angle BAC = \angle QPR$. Step 5: By SAS similarity ($\frac{AB}{PQ} = \frac{AC}{PR}$ and included angle), $\Delta ABC \sim \Delta PQR$.
Question 10: In an equilateral triangle $ABC$, $D$ is a point on side $BC$ such that $BD = \frac{1}{3}BC$. Prove that $9AD^2 = 7AB^2$.