Comprehensive Study Guide to Similar Triangles, Tangents to Circles, and Mensuration

SIMILAR TRIANGLES: FUNDAMENTAL THEOREMS, CRITERIA, AND PROPERTIES

• Definition of Similar Triangles: Two triangles are said to be similar if:
  • (i) Their corresponding angles are equal.
  • (ii) Their corresponding sides are proportional (in the same ratio).
• Congruence versus Similarity:
  • All congruent triangles are similar, but similar triangles need not be congruent.
• Similarity of Polygons:
  • Two polygons having the same number of sides are similar if:
  • (i) Their corresponding angles are equal.
  • (ii) Their corresponding sides are in the same ratio.
• Basic Proportionality Theorem (Thales' Theorem):
  • If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, it divides the two sides in the same ratio.
  • For $\Delta ABC$, if $DE \parallel BC$ where $D$ is on $AB$ and $E$ is on $AC$, then:     $\frac{AD}{DB} = \frac{AE}{EC}$
  • Corollaries of the theorem include:     $\frac{AB}{AD} = \frac{AC}{AE} \quad \text{and} \quad \frac{AB}{DB} = \frac{AC}{EC}$
• Converse of Basic Proportionality Theorem:
  • If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
• Criteria for Similarity of Triangles:
  • AAA-Similarity Criterion: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio, and consequently, the two triangles are similar.
  • AA-Similarity Corollary: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
  • SSS-Similarity Criterion: If the corresponding sides of two triangles are proportional, then their corresponding angles are equal, and the triangles are similar.
  • SAS-Similarity Criterion: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the triangles are similar.
• Ratio of Areas of Similar Triangles:
  • The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides:     $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}$
  • The ratio of areas is also equal to the ratio of the squares of their corresponding altitudes, medians, and angle bisectors.
• Internal Angle Bisector Theorem:
  • The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle:     $\frac{BD}{DC} = \frac{AB}{AC}$
• External Angle Bisector Theorem:
  • The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle:     $\frac{BE}{CE} = \frac{AB}{AC}$
• 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:     $AC^2 = AB^2 + BC^2$
• Converse of Pythagoras Theorem:
  • In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

SIMILAR TRIANGLES: WORKED THEOREMS AND STEP-BY-STEP PROOFS

• Midpoint Transversal Bisecting the Third Side:
  • Problem: Using similar triangles, prove that the line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side.
  • Given: In $\Delta ABC$, $D$ is the mid-point of $AB$ ($AD = DB$), and $DE \parallel BC$.
  • To Prove: $AE = EC$.
  • Proof:
  • In $\Delta ADE$ and $\Delta ABC$:       $\angle 1 = \angle 2 \quad [\text{Corresponding } \angle s]$$\angle 3 = \angle 4 \quad [\text{Corresponding } \angle s]$$[\because DE \parallel BC]$$\therefore \Delta ABC \sim \Delta ADE$
  • By Basic Proportionality Theorem:       $\frac{AD}{DB} = \frac{AE}{EC}$
  • Since $AD = DB$, $\frac{AD}{DB} = 1$, which yields:       $\frac{AE}{EC} = 1 \implies AE = EC$
• Trisection of Side $AC$ ($AL = \frac{1}{3}AC$):
  • Given: In $\Delta ABC$, $P$ is the mid-point of side $BC$ ($BP = PC$), and $Q$ is the mid-point of $AP$ ($AQ = QP$). Line $BQ$ is extended to meet side $AC$ at point $L$.
  • To Prove: $AL = \frac{1}{3}AC$.
  • Construction: Draw line segment $PM \parallel BL$ meeting $AC$ at point $M$.
  • Proof:
  • In $\Delta BLC$, $PM \parallel BL$. By Basic Proportionality Theorem:       $\frac{BP}{PC} = \frac{LM}{MC}$
  • Since $BP = PC$:       $\frac{LM}{MC} = 1 \implies LM = MC \quad \dots (i)$
  • In $\Delta APM$, $QL \parallel PM$ and $Q$ is the mid-point of $AP$. By Basic Proportionality Theorem:       $\frac{AQ}{QP} = \frac{AL}{LM}$
  • Since $AQ = QP$:       $\frac{AL}{LM} = 1 \implies AL = LM \quad \dots (ii)$
  • Combining equations $(i)$ and $(ii)$:       $AL = LM = MC$
  • Since $AC = AL + LM + MC = 3AL$:       $AL = \frac{1}{3}AC$

• Intersection Height of Lines Joining Tops and Bases of Two Poles:
  • Problem: Two poles of heights $a$ and $b$ ($b > a$) are $c$ metres apart. Prove that the height $h$ in metres of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is $\frac{ab}{a + b}$.
  • Given: Two vertical poles $AB = a$ and $CD = b$, separated by distance $BD = c$. Line segments $AD$ and $BC$ intersect at point $E$. Let $EF = h$ be the perpendicular dropped from $E$ to $BD$, and let $BF = x$, so $FD = c - x$.
  • Proof:
  • In $\Delta ABD$ and $\Delta EFD$:       $\angle ABD = \angle EFD = 90^\circ$$\angle ADB = \angle EDF \quad [\text{Common}]$$\therefore \Delta ABD \sim \Delta EFD \quad [\text{AA-Similarity}]$$\frac{EF}{AB} = \frac{FD}{BD} \implies \frac{h}{a} = \frac{c - x}{c} \quad \dots (i)$
  • In $\Delta CDB$ and $\Delta EFB$:       $\angle CDB = \angle EFB = 90^\circ$$\angle CBD = \angle EBF \quad [\text{Common}]$$\therefore \Delta CDB \sim \Delta EFB \quad [\text{AA-Similarity}]$$\frac{EF}{CD} = \frac{BF}{BD} \implies \frac{h}{b} = \frac{x}{c} \implies x = \frac{hc}{b}$
  • Substituting $x = \frac{hc}{b}$ into equation $(i)$:       $\frac{h}{a} = \frac{c - \frac{hc}{b}}{c} = \frac{c\left(1 - \frac{h}{b}\right)}{c} = \frac{b - h}{b}$$bh = a(b - h) \implies bh = ab - ah$$(a + b)h = ab \implies h = \frac{ab}{a + b}$
• Shadow of a Girl Walking Away from a Lamp Post:
  • Problem: A girl of height $90\,cm$ ($0.9\,m$) 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 $4\,s$.
  • Solution:
  • Let $AB = 3.6\,m$ be the lamp post, and let $CD = 0.9\,m$ be the girl.
  • Distance covered by the girl in $4\,s$:       $BD = 1.2\,m/s \times 4\,s = 4.8\,m$
  • Let the shadow length $DE = x\,m$, so $BE = BD + DE = 4.8 + x$.
  • In $\Delta ABE$ and $\Delta CDE$:       $\angle B = \angle D = 90^\circ$$\angle E = \angle E \quad [\text{Common}]$$\therefore \Delta ABE \sim \Delta CDE \quad [\text{AA-Similarity}]$$\frac{BE}{DE} = \frac{AB}{CD} \implies \frac{4.8 + x}{x} = \frac{3.6}{0.9} = 4$$4.8 + x = 4x \implies 3x = 4.8 \implies x = 1.6\,m$
  • The shadow of the girl is $1.6\,m$ long.
• Parallel Segment from Quadrilateral Angle Bisectors:
  • Problem: $ABCD$ is a quadrilateral in which $AB = AD$. The bisectors of $\angle BAC$ and $\angle CAD$ intersect sides $BC$ and $CD$ at points $E$ and $F$ respectively. Prove that segment $EF \parallel BD$.
  • Proof:
  • In $\Delta ABC$, $AE$ bisects $\angle BAC$. By Internal Angle Bisector Theorem:       $\frac{AB}{AC} = \frac{BE}{EC} \quad \dots (i)$
  • In $\Delta CAD$, $AF$ bisects $\angle CAD$. By Internal Angle Bisector Theorem:       $\frac{AD}{AC} = \frac{DF}{FC} \quad \dots (ii)$
  • Given $AB = AD$, equating left-hand sides of $(i)$ and $(ii)$:       $\frac{AB}{AC} = \frac{AD}{AC} \implies \frac{BE}{EC} = \frac{DF}{FC}$
  • In $\Delta BCD$, since $\frac{BE}{EC} = \frac{DF}{FC}$, by the Converse of Basic Proportionality Theorem:       $EF \parallel BD$
• Perpendicular Dropped to Hypotenuse in a Right Triangle:
  • Problem: $\Delta ABC$ is a right-angled triangle with $\angle C = 90^\circ$. Let $p$ be the length of the perpendicular from $C$ to $AB$, and let $AB = c$, $BC = a$, and $CA = b$. Prove that:
  • (i) $pc = ab$
  • (ii) $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$
  • Proof:
  • Express area of $\Delta ABC$ in two ways:       $\text{Area}(\Delta ABC) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} c p \quad \dots (i)$$\text{Area}(\Delta ABC) = \frac{1}{2} a b \quad \dots (ii)$$\frac{1}{2} c p = \frac{1}{2} a b \implies p c = a b \implies c = \frac{a b}{p}$
  • In right triangle $\Delta ABC$, by Pythagoras Theorem:       $c^2 = a^2 + b^2$
  • Substituting $c = \frac{ab}{p}$ into the equation:       $\left(\frac{a b}{p}\right)^2 = a^2 + b^2 \implies \frac{a^2 b^2}{p^2} = a^2 + b^2$$\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}$$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$
• Interior Point in a Rectangle:
  • Problem: Point $O$ is inside rectangle $ABCD$. Prove that $OA^2 + OC^2 = OB^2 + OD^2$.
  • Construction: Draw line $EF$ through $O$ parallel to $AB$ meeting $AD$ at $E$ and $BC$ at $F$.
  • Proof:
  • Since $EF \parallel AB$ and $AB \perp AD$, $EF \perp AD$ and $EF \perp BC$. Thus, $EABF$ and $DEFC$ are rectangles, which gives $AE = BF$ and $DE = CF$.
  • Applying Pythagoras Theorem to four right triangles formed with vertex $O$:       $OA^2 = OE^2 + AE^2$$OC^2 = OF^2 + CF^2$$OB^2 = OF^2 + BF^2$$OD^2 = OE^2 + DE^2$
  • Adding $OA^2 + OC^2$:       $OA^2 + OC^2 = OE^2 + AE^2 + OF^2 + CF^2$
  • Substitute $AE = BF$ and $CF = DE$:       $OA^2 + OC^2 = OE^2 + BF^2 + OF^2 + DE^2 = (OF^2 + BF^2) + (OE^2 + DE^2) = OB^2 + OD^2$

• Relation Between Sides and Medians of a Triangle:
  • Problem: 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:     $3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$
  • Proof:
  • Let $D, E, F$ be mid-points of $BC, CA, AB$. In $\Delta ABC$, using Apollonius's Theorem derived via coordinate/Pythagorean expansion:       $AB^2 + AC^2 = 2AD^2 + 2BD^2 = 2AD^2 + 2\left(\frac{BC}{2}\right)^2 = 2AD^2 + \frac{1}{2}BC^2 \quad \dots (i)$$BC^2 + CA^2 = 2CF^2 + \frac{1}{2}AB^2 \quad \dots (ii)$$AB^2 + BC^2 = 2BE^2 + \frac{1}{2}AC^2 \quad \dots (iii)$
  • Adding equations $(i)$, $(ii)$, and $(iii)$:       $2(AB^2 + BC^2 + CA^2) = 2(AD^2 + BE^2 + CF^2) + \frac{1}{2}(AB^2 + BC^2 + CA^2)$$\left(2 - \frac{1}{2}\right)(AB^2 + BC^2 + CA^2) = 2(AD^2 + BE^2 + CF^2)$$\frac{3}{2}(AB^2 + BC^2 + CA^2) = 2(AD^2 + BE^2 + CF^2)$$3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$
• Trisection of Side in an Equilateral Triangle:
  • Problem: In an equilateral triangle $ABC$, point $D$ lies on $BC$ such that $BD = \frac{1}{3}BC$. Prove that $9AD^2 = 7AB^2$.
  • Construction: Draw altitude $AL \perp BC$.
  • Proof:
  • In equilateral $\Delta ABC$, the altitude bisects the base: $BL = LC = \frac{1}{2}BC$.
  • Since $BD = \frac{1}{3}BC$:       $DL = BL - BD = \frac{1}{2}BC - \frac{1}{3}BC = \frac{1}{6}BC$
  • In right $\Delta ALD$: $AD^2 = AL^2 + DL^2$.
  • In right $\Delta ALB$: $AB^2 = AL^2 + BL^2$.
  • Subtracting the two equations:       $AB^2 - AD^2 = BL^2 - DL^2 = (BL + DL)(BL - DL)$$BL + DL = \frac{1}{2}BC + \frac{1}{6}BC = \frac{2}{3}BC$$BL - DL = BD = \frac{1}{3}BC$$AB^2 - AD^2 = \left(\frac{2}{3}BC\right)\left(\frac{1}{3}BC\right) = \frac{2}{9}BC^2$
  • Since $AB = BC$:       $AB^2 - AD^2 = \frac{2}{9}AB^2 \implies 9(AB^2 - AD^2) = 2AB^2$$9AB^2 - 9AD^2 = 2AB^2 \implies 7AB^2 = 9AD^2$

SIMILAR TRIANGLES: ADDITIONAL WORKED EXAMPLES

• Basic Proportionality Computations:
  • Example: In $\Delta ABC$, $DE \parallel BC$. Given $AD = 3.6\,cm$, $AB = 10\,cm$, and $AE = 4.5\,cm$. Find $EC$ and $AC$.     $\frac{AB}{AD} = \frac{AC}{AE} \implies \frac{10}{3.6} = \frac{AC}{4.5} \implies AC = \frac{4.5 \times 10}{3.6} = 12.5\,cm$$EC = AC - AE = 12.5 - 4.5 = 8\,cm$
  • Example: Find $x$ if $DE \parallel BC$ where $AD = 4$, $DB = x - 4$, $AE = 8$, and $EC = 3x - 19$.     $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{4}{x - 4} = \frac{8}{3x - 19}$$4(3x - 19) = 8(x - 4) \implies 12x - 76 = 8x - 32$$4x = 44 \implies x = 11$
• Angle Bisector Calculations:
  • Internal Bisector Example: In $\Delta ABC$, $AD$ is the bisector of $\angle A$. $AB = 10\,cm$, $AC = 14\,cm$, $BC = 6\,cm$. Let $BD = x$, $DC = 6 - x$.     $\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{10}{14} = \frac{x}{6 - x} \implies \frac{5}{7} = \frac{x}{6 - x}$$5(6 - x) = 7x \implies 30 - 5x = 7x \implies 12x = 30 \implies x = 2.5\,cm$$BD = 2.5\,cm \quad \text{and} \quad DC = 3.5\,cm$
  • External Bisector Example: In $\Delta ABC$, $AE$ is the external bisector of $\angle CAD$ meeting $BC$ extended at $E$. $AB = 10\,cm$, $AC = 6\,cm$, $BC = 12\,cm$. Let $CE = x$.     $\frac{BE}{CE} = \frac{AB}{AC} \implies \frac{12 + x}{x} = \frac{10}{6} = \frac{5}{3}$$3(12 + x) = 5x \implies 36 + 3x = 5x \implies 2x = 36 \implies x = 18\,cm$$CE = 18\,cm$
• Similar Triangle Proportions and Altitudes:
  • Example: Given $\Delta ACB \sim \Delta APQ$. $BC = 10\,cm$, $PQ = 5\,cm$, $BA = 7.4\,cm$, $AP = 2.8\,cm$. Find $CA$ and $AQ$.     $\frac{AC}{AP} = \frac{CB}{PQ} = \frac{AB}{AQ} \implies \frac{AC}{2.8} = \frac{10}{5} = 2 \implies AC = 5.6\,cm$$\frac{7.4}{AQ} = 2 \implies AQ = \frac{7.4}{2} = 3.7\,cm$
  • Right-Angled Triangle Altitude Example: In $\Delta ABC$, $\angle ABC = 90^\circ$, $BD \perp AC$. $AB = 5.7\,cm$, $BD = 3.8\,cm$, $CD = 5.4\,cm$. Find $BC$.
  • Since $\Delta BDC \sim \Delta ABC$:       $\frac{AB}{BD} = \frac{BC}{DC} \implies \frac{5.7}{3.8} = \frac{BC}{5.4} \implies BC = \frac{5.7 \times 5.4}{3.8} = 8.1\,cm$
• Similar Triangles Areas and Heights:
  • Example: In $\Delta ABC$, $DE \parallel BC$. $DE = 4\,cm$, $BC = 6\,cm$, $\text{Area}(\Delta ADE) = 16\,cm^2$. Find $\text{Area}(\Delta ABC)$.     $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta ABC)} = \frac{DE^2}{BC^2} = \frac{4^2}{6^2} = \frac{16}{36}$$\frac{16}{\text{Area}(\Delta ABC)} = \frac{16}{36} \implies \text{Area}(\Delta ABC) = 36\,cm^2$
  • Heights of Isosceles Triangles: Two isosceles triangles have equal vertical angles and area ratio $16 : 25$. Find the ratio of their corresponding heights.
  • Since the vertical angles are equal and legs are equal in each ($AB = AC$ and $DE = DF$), base angles are equal:       $\angle B = \frac{180^\circ - \angle A}{2} = \frac{180^\circ - \angle D}{2} = \angle E$
  • Therefore, $\Delta ABC \sim \Delta DEF$ by AA-similarity.     $\frac{\text{Height}_1^2}{\text{Height}_2^2} = \frac{\text{Area}_1}{\text{Area}_2} = \frac{16}{25} \implies \frac{\text{Height}_1}{\text{Height}_2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

SIMILAR TRIANGLES: PRACTICE EXERCISES

• Exercise 1 Solutions and Problems:
  • Q1: $\Delta ABC \sim \Delta DEF$, $\text{Area}(\Delta ABC) = 9\,cm^2$, $\text{Area}(\Delta DEF) = 16\,cm^2$, $BC = 2.1\,cm$. Length of $EF$:     $\frac{9}{16} = \frac{(2.1)^2}{EF^2} \implies \frac{3}{4} = \frac{2.1}{EF} \implies EF = \frac{4 \times 2.1}{3} = 2.8\,cm$
  • Q2: $\Delta ABC \sim \Delta DEF$, $BC = 3\,cm$, $EF = 4\,cm$, $\text{Area}(\Delta ABC) = 54\,cm^2$. Area of $\Delta DEF$:     $\frac{54}{\text{Area}(\Delta DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \implies \text{Area}(\Delta DEF) = \frac{54 \times 16}{9} = 96\,cm^2$
  • Q3: Right triangle has hypotenuse $q\,cm$ and side $p\,cm$. If $q - p = 2$, third side in terms of $q$:     $\text{Third side} = \sqrt{q^2 - p^2} = \sqrt{q^2 - (q - 2)^2} = \sqrt{q^2 - (q^2 - 4q + 4)} = \sqrt{4q - 4} = 2\sqrt{q - 1}$
  • Q4: $\Delta ABC \sim \Delta DEF$, $AB = 2DE$, $\text{Area}(\Delta ABC) = 56\,cm^2$. Area of $\Delta DEF$:     $\frac{56}{\text{Area}(\Delta DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{2}{1}\right)^2 = 4 \implies \text{Area}(\Delta DEF) = \frac{56}{4} = 14\,cm^2$
  • Q5: Two poles of heights $7\,m$ and $12\,m$ stand on plane ground with bases $12\,m$ apart. Distance between tips:     $\text{Height difference} = 12 - 7 = 5\,m$$\text{Distance between tips} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\,m$
  • Q6: $DE \parallel BC$ and $AD : DB = 5 : 4$. Area ratio evaluation for $\Delta DEF$ and $\Delta CFB$.
  • Q7: $DE \parallel BC$ and $AD : DB = 2 : 3$. Ratio $\text{Area}(\Delta ADE) : \text{Area}(\Delta ABC)$.     $\frac{AD}{AB} = \frac{2}{2 + 3} = \frac{2}{5} \implies \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta ABC)} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}$
  • Q8: Similar triangles areas $81\,cm^2$ and $49\,cm^2$. Altitude of first is $6.3\,cm$. Corresponding altitude $h_2$:     $\frac{81}{49} = \frac{(6.3)^2}{h_2^2} \implies \frac{9}{7} = \frac{6.3}{h_2} \implies h_2 = \frac{7 \times 6.3}{9} = 4.9\,cm$
  • Q9: Areas $121\,cm^2$ and $64\,cm^2$. Median of first is $12.1\,cm$. Median of second $m_2$:     $\frac{11}{8} = \frac{12.1}{m_2} \implies m_2 = \frac{8 \times 12.1}{11} = 8.8\,cm$
  • Q10: Areas $100\,cm^2$ and $49\,cm^2$. Altitude of bigger is $5\,cm$. Other altitude:     $\frac{10}{7} = \frac{5}{h} \implies h = 3.5\,cm$
  • Q11: Sum of squares of sides of rhombus equals sum of squares of its diagonals:     $4a^2 = d_1^2 + d_2^2$
  • Q12: Points $S$ and $T$ trisect side $QR$ of right $\Delta PQR$. Prove $8PT^2 = 3PR^2 + 5PS^2$.
  • Q13: $BL$ and $CM$ are medians of $\Delta ABC$ right-angled at $A$. Prove $4(BL^2 + CM^2) = 5BC^2$.
  • Q14: $\Delta ABC \sim \Delta PQR$ and Quad $ABCD \sim$ Quad $PQRS$. Find unknown values $x, y, z$.
  • Q15: Values of $x$ that make $DE \parallel AB$ when segments are given as $8x + 9$, $x + 3$, $3x + 4$, and $x$.
  • Q16: Perimeters of two similar triangles are $30\,cm$ and $20\,cm$. If one side of the first is $15\,cm$, the corresponding side of the second is:     $\frac{30}{20} = \frac{15}{s_2} \implies s_2 = 10\,cm$
  • Q17: Sides ratio $1 : 3$. Area of smaller is $40\,cm^2$. Area of larger:     $\frac{40}{A_2} = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \implies A_2 = 360\,cm^2$
  • Q18: $P, Q$ on $AB, AC$ of $\Delta ABC$. $AP = 2\,cm$, $PB = 4\,cm$, $AQ = 3\,cm$, $QC = 6\,cm$. Prove $BC = 3PQ$.
  • Q19: Vertical stick $12\,cm$ casts shadow $8\,cm$. Tower casts shadow $40\,m$. Height of tower:     $\frac{h}{40} = \frac{12}{8} = 1.5 \implies h = 60\,m$
  • Q20: In figure, $AP = 3\,cm$, $AR = 4.5\,cm$, $AQ = 6\,cm$, $AB = 5\,cm$, and $AC = 10\,cm$. Length of $AD$.
  • Q21: $\angle A = \angle CED$, $AB = 9\,cm$, $AD = 7\,cm$, $CD = 8\,cm$, $CE = 10\,cm$. Find $DE$.
  • Q22: In $\Delta ABC$, if $AD$ is median, show Apollonius's theorem: $AB^2 + AC^2 = 2(AD^2 + BD^2)$.
  • Q23: Parallelogram $ABCD$, $M$ is mid-point of $CD$. $BM$ meets $AC$ in $L$ and $AD$ produced in $E$. Prove $EL = 2BL$.
  • Q24: Right $\Delta ABC$ with perpendicular $p$ to hypotenuse: prove $pc = ab$ and $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.
  • Q25: Equilateral $\Delta PQR$ with $QR$ trisected at $S$. Prove $9PS^2 = 7PQ^2$.
  • Q26: If diagonals of a quadrilateral divide each other proportionally, prove it is a trapezium.
• Exercise 2 Problems:
  • Q1: Basic proportionality questions with variants: (i) $BD = 3.6\,cm$, (ii) $AE = 3\,cm$, (iii) $AC = 20.25\,cm$.
  • Q2: Solve for $x$ given $LM \parallel AB$, $AL = x - 3$, $AC = 2x$, $BM = x - 2$, $BC = 2x + 3$.
  • Q3: Verification whether $DE \parallel BC$ given side segment measures.
  • Q4: Determine $\angle BAD$ given $\frac{AB}{AC} = \frac{BD}{DC}$, $\angle B = 70^\circ$, $\angle C = 50^\circ$.
  • Q5 & Q6: Verification and numerical determination using angle bisector theorem.
  • Q7-Q14: Trapezium proportionality, equiangular angle bisector segment properties, and numerical proportions.
  • Q15-Q27: Right triangle relations, altitude calculations, coordinate walking ($10\,m$ south, $24\,m$ west yields $\sqrt{10^2 + 24^2} = 26\,m$), and inscribed circle radii.

SIMILAR TRIANGLES: MULTIPLE CHOICE QUESTIONS

• Q1: In $\Delta ABC$, $DE \parallel BC$, $AD = 2.4\,cm$, $AE = 3.2\,cm$, $EC = 4.8\,cm$. Value of $AB$:   $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{2.4}{DB} = \frac{3.2}{4.8} = \frac{2}{3} \implies DB = 3.6\,cm$$AB = AD + DB = 2.4 + 3.6 = 6\,cm \quad [\text{Option b}]$
• Q2: In $\Delta ABC$, $DE \parallel BC$, $AD = 7x - 4$, $AE = 5x - 2$, $DB = 3x + 4$, $EC = 3x$. Value of $x$:   $\frac{7x - 4}{3x + 4} = \frac{5x - 2}{3x} \implies 3x(7x - 4) = (5x - 2)(3x + 4)$$21x^2 - 12x = 15x^2 + 14x - 8 \implies 6x^2 - 26x + 8 = 0 \implies 3x^2 - 13x + 4 = 0$$(3x - 1)(x - 4) = 0 \implies x = 4 \quad [\text{Option c}]$
• Q3: In $\Delta ABC$, $AD$ is internal bisector of $\angle A$. $BD = 5\,cm$, $BC = 7.5\,cm$. Then $DC = 2.5\,cm$. Ratio $AB : AC$:   $\frac{AB}{AC} = \frac{BD}{DC} = \frac{5}{2.5} = \frac{2}{1} = 2 : 1 \quad [\text{Option a}]$
• Q4: In trapezium $ABCD$, $AB \parallel DC$, diagonals intersect at $O$. $AO = 3x - 1$, $BO = 2x + 1$, $OC = 5x - 3$, $OD = 6x - 5$. Then $x$:   $\frac{AO}{OC} = \frac{BO}{OD} \implies \frac{3x - 1}{5x - 3} = \frac{2x + 1}{6x - 5}$$(3x - 1)(6x - 5) = (2x + 1)(5x - 3) \implies 18x^2 - 21x + 5 = 10x^2 - x - 3$$8x^2 - 20x + 8 = 0 \implies 2x^2 - 5x + 2 = 0 \implies (2x - 1)(x - 2) = 0 \implies x = 2 \quad [\text{Option a}]$
• Q5: In $\Delta ABC$, $\frac{AB}{AC} = \frac{BD}{DC}$, $\angle B = 70^\circ$, $\angle C = 50^\circ$. $\angle A = 180^\circ - (70^\circ + 50^\circ) = 60^\circ$. Since $AD$ bisects $\angle A$:   $\angle BAD = \frac{60^\circ}{2} = 30^\circ \quad [\text{Option a}]$
• Q6: In $\Delta ABC$, $AD$ is bisector of $\angle A$. Then $\frac{\text{Area}(\Delta ABD)}{\text{Area}(\Delta ACD)} = \frac{BD}{DC} = \frac{AB}{AC} \quad [\text{Option b}]$.
• Q7: $\Delta ABC \sim \Delta DEF$, perimeters are $30\,cm$ and $18\,cm$. If $BC = 9\,cm$, then $EF$:   $\frac{30}{18} = \frac{9}{EF} \implies EF = \frac{9 \times 18}{30} = 5.4\,cm \quad [\text{Option b}]$
• Q8: $\Delta ABC \sim \Delta DEF$, $AB = 9.1\,cm$, $DE = 6.5\,cm$, perimeter of $\Delta DEF = 25\,cm$. Perimeter of $\Delta ABC$:   $\frac{\text{Perimeter}(\Delta ABC)}{25} = \frac{9.1}{6.5} = 1.4 \implies \text{Perimeter} = 35\,cm \quad [\text{Option a}]$
• Q9: Vertical stick $20\,m$ casts shadow $10\,m$. Tower casts shadow $50\,m$. Height of tower:   $\frac{h}{50} = \frac{20}{10} = 2 \implies h = 100\,m \quad [\text{Option a}]$
• Q10: $\Delta ABC$ and $\Delta EBD$ are equilateral triangles such that $D$ is mid-point of $BC$. Ratio of areas:   $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta EBD)} = \left(\frac{BC}{BD}\right)^2 = \left(\frac{2}{1}\right)^2 = 4 : 1 \quad [\text{Option c}]$
• Q11: Two isosceles triangles have equal vertical angles and area ratio $16 : 25$. Ratio of corresponding heights:   $\sqrt{\frac{16}{25}} = 4 : 5 \quad [\text{Option a}]$
• Q12: $\Delta ABC \sim \Delta DEF$, $2AB = DE$, $BC = 8\,cm$. Then $EF$:   $\frac{AB}{DE} = \frac{1}{2} = \frac{BC}{EF} \implies EF = 2 \times 8 = 16\,cm \quad [\text{Option a}]$
• Q13: Poles of height $6\,m$ and $11\,m$, distance between feet $12\,m$. Distance between tops:   $\sqrt{12^2 + (11 - 6)^2} = \sqrt{144 + 25} = 13\,m \quad [\text{Option c}]$
• Q14: In $\Delta ABC$, line $XY \parallel BC$. If $BY$ bisects $\angle XYC$, then $BC = CY \quad [\text{Option c}]$.
• Q15: In $\Delta ABC$, $DE : BC = 3 : 5$. Ratio $\text{Area}(\Delta ADE) : \text{Area}(\square BCED)$.   $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta ABC)} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$$\text{Area}(\square BCED) = 25 - 9 = 16 \implies \text{Ratio} = 9 : 16 \quad [\text{Option b}]$
• Q16: In trapezium $ABCD$, $BC \parallel AD$, $AD = 4\,cm$, $\frac{AO}{OC} = \frac{DO}{OB} = \frac{1}{2}$. Then $BC = 8\,cm \quad [\text{Option b}]$.
• Q17: Equilateral $\Delta ABC$, altitude $BE \perp CA$. Then:   $AB^2 + BC^2 + CA^2 = 3a^2 = 3\left(\frac{4}{3}BE^2\right) = 4BE^2 \quad [\text{Option c}]$
• Q18: Right triangle hypotenuse $p$, side $q$, with $p - q = 1$. Third side is $\sqrt{p^2 - q^2} = \sqrt{(p - q)(p + q)} = \sqrt{1(2q + 1)} = \sqrt{2q + 1} \quad [\text{Option a}]$.
• Q19: Triangle $ABC$ with $AC = BC$ and $AB^2 = 2AC^2 = AC^2 + BC^2$. Angle is right at $\angle C \quad [\text{Option c}]$.
• Q20: Right $\Delta ABC$ at $A$, $AD \perp BC$. Side $AB = c$, $AC = b$. Then hypotenuse is $\sqrt{b^2 + c^2}$, giving:   $AD = \frac{bc}{\sqrt{b^2 + c^2}} \quad [\text{Option a}]$
• Q21: Right $\Delta ABC$ at $A$, $AD \perp BC$. $BC = 13\,cm$, $AC = 5\,cm$. Ratio $\text{Area}(\Delta ABC) : \text{Area}(\Delta ADC) = \left(\frac{BC}{AC}\right)^2 = \frac{169}{25} = 169 : 25 \quad [\text{Option b}]$.
• Q22: Median property in right triangle at $B$: $4LC^2 = AB^2 + 4BC^2 \quad [\text{Option a}]$.
• Q23: Similar right triangles transversal expression: $x = \frac{ac}{b + c} \quad [\text{Option a}]$.
• Q24: Medial triangle area ratio: $\text{Area}(\Delta EFD) : \text{Area}(\Delta ABC) = 1 : 4 \quad [\text{Option d}]$.
• Q25: Aeroplane north at $1000\,km/h$ and west at $1200\,km/h$ for $1.5\,h$ ($\frac{3}{2}\,h$):   $D_1 = 1500\,km, \quad D_2 = 1800\,km$$\text{Distance} = \sqrt{1500^2 + 1800^2} = 300\sqrt{5^2 + 6^2} = 300\sqrt{61}\,km \quad [\text{Option a}]$
• Q26: Standard perpendicular relation: $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} \quad [\text{Option d}]$.
• Q27: Given triangle sides and angle relation calculation yielding $x = 4\,cm \quad [\text{Option b}]$.
• Q28: Angle calculation using bisector property yielding $\angle ABD = 40^\circ \quad [\text{Option b}]$.
• Q29: Equilateral triangle and square of same perimeter $P$:
  • Triangle side $s = \frac{P}{3}$, Area $X = \frac{\sqrt{3}}{4}\left(\frac{P}{3}\right)^2 = \frac{\sqrt{3}}{36}P^2 \approx 0.0481 P^2$.
  • Square side $a = \frac{P}{4}$, Area $Y = \left(\frac{P}{4}\right)^2 = \frac{1}{16}P^2 = 0.0625 P^2$.
  • Therefore, $X < Y \quad [\text{Option b}]$.
• Q30: Point equidistant from all three sides of a triangle: Four points exist (1 incenter and 3 excenters) $[\text{Option c}]$.
• Q31: Triangle $OSD$ formed by bisector and perpendicular bisector is isosceles right-angled $[\text{Option c}]$.
• Q32: Right triangle area $40\,cm^2$, perimeter $40\,cm$:   $\frac{1}{2}ab = 40 \implies ab = 80$$a + b + c = 40 \implies a + b = 40 - c$$(a + b)^2 = a^2 + b^2 + 2ab \implies (40 - c)^2 = c^2 + 160$$1600 - 80c + c^2 = c^2 + 160 \implies 80c = 1440 \implies c = 18\,cm \quad [\text{Option b}]$
• Q33: Isosceles triangle base $10$, legs $13$. Height $h = \sqrt{13^2 - 5^2} = 12$. Area $= \frac{1}{2} \times 10 \times 12 = 60$. Other base yielding area $60$ with legs $13$ is $24\,inch \quad [\text{Option c}]$.
• Q34: In right $\Delta ABC$ at $B$, with $P, Q$ on $AB, BC$: $AQ^2 + CP^2 = AC^2 + PQ^2 \quad [\text{Option c}]$.
• Q35: Right $\Delta ABC$ at $A$, $AB = 5\,cm$, $AC = 12\,cm$, hypotenuse $BC = 13\,cm$. Perpendicular to hypotenuse:   $AD = \frac{AB \times AC}{BC} = \frac{5 \times 12}{13} = \frac{60}{13}\,cm \quad [\text{Option b}]$