are mutually perpendicular unit vectors.Solution: $\begin{array}{l} \vec{a}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \\ \vec{b}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}) \\ \vec{c}=\frac{1}{7}(6 \hat{\imath}+2...
and 
then find the value of 
so that 
and 
are orthogonal vectors.Solution: $\begin{array}{l} \vec{a}=\hat{\imath}-1 \hat{\jmath}+7 \hat{k} \\ \vec{b}=5 \hat{\imath}-1 \hat{\jmath}+\lambda \hat{k} \\ \vec{a}+\vec{b}=6 \hat{\imath}-2 \hat{\jmath}+(7+\lambda)...
and 
, show that is perpendicular to . 
and 
then show that and are orthogonal.Solution: If two vectors are perpendicular or orthogonal then their dot product is zero. $\begin{array}{l} \text { i) } \quad \vec{a}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \\ \vec{b}=3...
when 
and 
and 
and 
Solution: If $\vec{a}=\mathrm{a} 1 \hat{\imath}+\mathrm{a} 2 \hat{\jmath}+\mathrm{a} 3 \hat{k}$ and $\vec{b}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{k}, \text {...
Correct Answer: (b) 7.5 cm Explanation: ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{Perimeter(\Delta ABC)}}{{Perimeter(\Delta DEF)}} = \frac{{AB}}{{DE}}\\ \frac{{32}}{{24}} = \frac{{10}}{{DE}}\\ DE =...
Correct Answer: (a) 5.6 cm Explanation: DE ‖ BC $\begin{array}{l} \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{{DE}}{{BC}}\\ \frac{{3.5}}{{AB}} = \frac{5}{8}\\ AB = \frac{{3.5 \times 8}}{5}\\ AB =...
Correct Answer: (b)13 m Explanation: Let the poles be and CD AB = 6 m CD = 11 m Let AC be 12 m Draw a perpendicular from CD, meeting CD at E. BE = 12 m Applying...
and 36 respectively. If the altitude of the first triangle is 3.5cm, then the corresponding altitude of the other triangle. (a) 5.6cm (b) 6.3cm (c) 4.2cm (d) 7cmCorrect Answer: (c) 4.2cm Explanation: The ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes. Let ɦ be the altitude of the other triangle....
Answer: ∆ABC ~ ∆ DEF $\begin{array}{l} \frac{{AB}}{{DE}} = \frac{{BC}}{{EF}}\\ \frac{1}{2} = \frac{6}{{EF}}\\ EF = 12cm \end{array}$
Answer: DE || BC $\begin{array}{l} \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\\ \frac{x}{3x+4} = \frac{x+3}{{3x+19}}\\ \end{array}$ ???? (3???? + 19) = (???? + 3)(3???? + 4) 3????2 +...
Answer: Let the ladder be AB and BC be the height of the window from the ground. Given, AB 10 m BC = 8 m Applying theorem in right-angled triangle ACB, ????????2 = ????????2 +...
and ar(∆DEF) = 169. If BC = 4cm, find EF.Answer: ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{ar(\Delta ABC)}}{{ar(\Delta DEF)}} = \frac{{B{C^2}}}{{E{F^2}}}\\ \frac{{64}}{{169}} = \frac{{{4^2}}}{{E{F^2}}}\\ E{F^2} = \frac{{16 \times 169}}{{64}}\\...
. Find ar(∆COD).Answer: In ∆AOB and ∆COD, ∠???????????? = ∠???????????? (???????????????????????????????????????? ???????????????????????????????? ????????????????????????) ∠????????????...
, find the area of the larger triangleAnswer: Given, The triangles are similar and the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides. $\begin{array}{l} \frac{{48}}{{Area\ of\...
Answer: Given, LM || CB and LN || CD Applying Thales’ theorem, $\begin{array}{l} \frac{{AB}}{{AM}} = \frac{{AC}}{{AL}}\\ \end{array}$ And $\begin{array}{l} \frac{{AD}}{{AN}} =...
Answer: Let the triangle be ABC with AD as the bisector of ∠???? which meets BC at D. Draw CE || DA, meeting BA produced at E. CE || DA ∠2 = ∠3...
Answer: Let ABC be the equilateral triangle with each side equal to a. Let AD be the altitude from A, meeting BC at D. D is the midpoint of BC. Let AD be h. Applying...
Answer: Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O. The diagonals of a rhombus bisect each other at right angles. If AC – 24 cm and BD = 10...
Answer: Let the two triangles be ABC and PQR. BC = a AC = b AB = c PQ = r PR = q QR = p ∆ABC ~ ∆PQR; therefore, their corresponding sides will be proportional. $\begin{array}{l} \frac{a}{p} =...
.Answer: ???????????????? ???????? ⊥ ???????? ???????????? ???????? ⊥ ????O $\begin{array}{l} \frac{{ar(\Delta ABC)}}{{ar(\Delta DBC)}} = \frac{{\frac{1}{2} \times AX \times BC}}{{\frac{1}{2} \times...
Answer: In ∆ ABC and ∆BXY, ∠???? = ∠???? ∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ????????????????????????) ????ℎ????????,...
Answer: Applying Pythagoras theorem in right-angled triangle ADC, ????????2 = ????????2 + ????????2 ????????2 − ????????2 = ????????2 ????????2 = ????????2 − ????????2...
Answer: In ∆ PAC and ∆QBC, ∠???? = ∠???? (????????????ℎ ???????????????????????? ???????????? 900) ∠???? = ∠???? (????????????????????????????????????????????????????...
Column I Column II (a) A man goes 10m due east and then 20m due north. His distance from the starting point is ……m. (p) 25√3 (b) In an equilateral triangle with each side 10cm, the altitude is...
Column I Column II (a) In a given ∆ABC, DE║BC and $\frac{{AD}}{{DB}} = \frac{{3}}{{5}}$. If AC = 5.6cm, then AE = ……..cm. (p) 6 (b) If ∆ABC~∆DEF such that 2AB = 3DE and BC = 6cm, then EF = …….cm....
Correct Answer: (b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides. Explanation: The ratio of the areas of two similar triangles is equal to the...
Correct Answer: (c)Two triangles are similar if their corresponding sides are proportional. Explanation: Given, ∆ABC~ ∆DEF $\frac{{AB}}{{DE}} = \frac{{AC}}{{DF}} = \frac{{BC}}{{EF}}$
Correct Answer: (b) right-angled Explanation: Given, ????????2 + ????????2 = 162 + 122 => 256 + 144 => 400 ????????2 = 202 => 400 ∴ ????????2 + ????????2 = ????????2 ∆ABC is a right-angled...
, then ∠C = ? (a) 
(b) 
(c) 
(d) 
Correct Answer: (d) ${90^0}$ Explanation: Given, AC = BC ????????2 = 2????????2 = ????????2 + ????????2 = ????????2 + ????????2 Applying Pythagoras theorem, ∆ABC is right angled at C ∠???? =...
. Then, ∆OAC and ∆ODB are (a) equilateral and similar (b) equilateral but not similar (c) isosceles and similar (d) isosceles but not similarCorrect Answer: (c) isosceles and similar Explanation: In ∆AOC and ∆ODB, ∠???????????? = ∠???????????? (???????????????????????????????????????? ????????????????????????????????...
, AB = 18cm and BC = 15cm, then PR = ? (a) 18cm (b) 10cm (c) 12 cm (d) 
cmCorrect Answer: (b) 10 cm Explanation: ∆ABC ~ ∆QRP $\begin{array}{l} \frac{{AB}}{{QR}} = \frac{BC}{PR}\\ \end{array}$ $\frac{{ar(\Delta ABC)}}{{ar(\Delta QRP)}} = \frac{9}{4}$ $\begin{array}{l}...
Correct Answer: (b) similar to the original triangle Explanation: The line segments joining the midpoint of the sides of a triangle form four triangles,...
Correct Answer: (c) 5:6 Explanation: Let x and y be the corresponding heights of the two triangles. The corresponding angles of the triangles are equal. The triangles are similar. (By AA criterion)...
and ar(∆DEF) = 49 . Then, the ratio of their corresponding sides is (a) 36 : 49 (b) 6 : 7 (c) 7 : 6 (d) √6 : √7Correct Answer: (b) 6:7 Explanation: In ∆ABC ~ ∆DEF, $\begin{array}{l} \frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{AC}}{{DF}}\\ \frac{{ar(\Delta ABC)}}{{ar(\Delta DEF)}} =...
, then ar(∆ABC) : ∆(DEF) = ? (a) 5 : 7 (b) 25 : 49 (c) 49 : 25 (d) 125 : 343Correct Answer: (b) 25 :49 Explanation: In ∆ABC and ∆DEF, $\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{AC}}{{DF}} = \frac{5}{7}$ By SSS criterion, ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{ar(\Delta...
Correct Answer: (b) 4:1 Explanation: In ∆ABC, D is the midpoint of AB and E is the midpoint of AC. By midpoint theorem and Basic Proportionality Theorem,...
, then 
Correct Answer: (d)\frac{9}{4}$ Explanation: Given, ∆ABC ~ ∆PQR $\begin{array}{l} \frac{{BC}}{{QR}} = \frac{2}{3}\\ \end{array}$ $\begin{array}{l} \frac{{ar(\Delta PQR)}}{{ar(\Delta ABC)}} =...
Correct Answer: (d) 16:81 Explanation: If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides. $\begin{array}{l} \frac{{area\...
and ∠CDP = , then ∠PBA = ? (a) (b) (c) (d) 
Correct Answer: (d) ${100^0}$ Explanation: In ∆ APB and ∆ DPC, In ∆ APB and ∆ DPC, ∠???????????? = ∠???????????? = 500 $\begin{array}{l} \frac{{AP}}{{BP}} = \frac{6}{3} = 2\\...
, then (a) ∆PQR ~ ∆CAB (b) ∆PQR ~ ∆ABC (c) ∆CBA ~ ∆PQR (d) ∆BCA ~ ∆PQRCorrect Answer: (a) ∆PQR ~ ∆CAB Explanation: In ∆ABC and ∆PQR, $\begin{array}{l} \frac{{AB}}{{QR}} = \frac{{BC}}{{PR}} = \frac{{CA}}{{PQ}}\\ \end{array}$ ∆ABC ~ ∆QRP
Correct Answer: (b) similar but not congruent Explanation: In ∆ABC and ∆DEF, ∠???? = ∠???? ∠???? = ∠???? Applying AA similarity theorem, ∆ABC - ∆DEF. AB = 3DE AB ≠ DE ∆ABC and ∆DEF are similar but...
Correct Answer: (c) BC. DE = AB. EF Explanation: ∆ABC ~ ∆EDF $\begin{array}{l} \frac{{AB}}{{DE}} = \frac{{AC}}{{EF}} = \frac{{BC}}{{DF}}\\ BC.DE \ne AB.EF \end{array}$
, where 
and 
are the points 
and 
respectively.Solution: $\begin{array}{l} \overrightarrow{O P}=\hat{\imath}+3 \hat{\jmath} \\ \overrightarrow{O Q}=4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k} \\ \overrightarrow{P Q}=\overrightarrow{O...
and 
Solution: $\begin{array}{l} \overrightarrow{O A}=3 \hat{\imath}+2 \hat{\jmath}+6 \hat{k} \\ \overrightarrow{O B}=\hat{\imath}+4 \hat{\jmath}-2 \hat{k} \end{array}$ Let $\mathrm{C}$ be the mid-point...
which divides the line joining 
and 
in the ratio 
(i) internally (ii) externally.Solution: $\begin{array}{l} \overrightarrow{O A}=-2 \hat{\imath}+\hat{\jmath}+3 \hat{k} \\ \overrightarrow{O B}=3 \hat{\imath}+5 \hat{\jmath}-2 \hat{k} \end{array}$ (i) $\mathrm{R}$ divides...
and 
are 
and 
respectively. Find the position vector of a point 
which divides 
externally in the ratio 
. Also, show that 
is the mid-point of the line segment 
.Solution: Given: $\overrightarrow{O A}=(2 \vec{a}+\vec{b})$ $\overrightarrow{O B}=(\vec{a}-3 \vec{b})$ Position vector of $\mathrm{C}$ which divides $\mathrm{AB}$ in the ratio $1: 2$ externally is...
and 
(i) internally and (ii) externally in the ratio 
Solution: Given: $\overrightarrow{O A}=(2 \vec{a}-3 \vec{b})$ $\overrightarrow{O B}=(3 \vec{a}-2 \vec{b})$ (i) Let $\mathrm{P}$ be the point that divides $\mathrm{A}, \mathrm{B}$ internally in the...
and 
are the vertices of a rightangled triangle.Solution: $\begin{array}{l} \overrightarrow{O A}=\hat{\imath}-\hat{\jmath} \\ \overrightarrow{O B}=4 \hat{\imath}-3 \hat{\jmath}+\hat{k} \\ \overrightarrow{O C}=2 \hat{\imath}-4 \hat{\jmath}+5...
and of a 
be 
and 
, respectively, prove that is equilateral.Solution: $\begin{array}{l} \overrightarrow{O A}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} \\ \overrightarrow{O B}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k} \\ \overrightarrow{O C}=3...
and 
having position vectors 
and 
respectively, are collinear.Solution: $\begin{array}{l} \overrightarrow{O A}=\hat{\imath}+2 \hat{\jmath}+7 \hat{k} \\ \overrightarrow{O B}=2 \hat{\imath}+6 \hat{\jmath}+2 \hat{k} \\ \overrightarrow{O C}=3 \hat{\imath}+10...
and 
, find 
Solution: $\begin{array}{l} \vec{a}=\hat{\imath}-2 \hat{\jmath} \\ \vec{b}=2 \hat{\imath}-3 \hat{\jmath} \\ \vec{c}=2 \hat{\imath}+3 \hat{k} \\ \vec{a}+\vec{b}+\vec{c}=(\hat{\imath}-2...
Correct Answer: (b) $\frac{{DE}}{{PQ}} = \frac{{EF}}{{RP}}$ Explanation: In ∆DEF and ∆PQR, ∠???? = ∠???? ∠???? = ∠???? Applying AA similarity theorem, ∆DEF ~ ∆QRP $\frac{{DE}}{{PQ}} =...
, then (a) ∠B = ∠E (b) ∠A = ∠D (c) ∠B = ∠D (d) ∠A = ∠FCorrect Answer: (c)∠???? = ∠D Explanation: ∆ ABC − EDF The corresponding angles, ∠???? ???????????? ∠???? ???????????????? ???????? ????????????????????. ∠???? = ∠D
Answer: Given, ???????? = 6√3???????? ????????2 = 108 ????????2 AC = 12 cm ????????2 = 144 ????????2 BC = 6 cm ????????2 = 36 ???????? ∴ ????????2 = ????????2 + ????????2 The square of the longest...
.Solution: $\vec{A}=5 \hat{\imath}-\hat{\jmath}+2 \hat{k}$ Magnitude of required vector is 8 units $\mathrm{I} \vec{A} \mathrm{I}=\sqrt{5^{2}+1^{2}+2^{2}}=\sqrt{25+1+4}=\sqrt{30}$ $\begin{array}{l}...
and AD⊥BC. Then, (a) BC.CD = 
(b) AB.AC = (c) BD.CD = 
(d) AB.AC = Correct Answer: (c) BD.CD = $A{D^2}$ Explanation: In ∆ BDA and ∆ADC, ∠???????????? = ∠???????????? = 900 ∠???????????? = 900 − ∠???????????? => 900 − (900 − ∠????????????)...
, ∠C = , , AB = 5cm, AC = 8cm and DF = 7.5cm, then which of the following is true? (a) DE = 12cm, ∠F = , (b) DE = 12cm, ∠F = , (c) DE = 12cm, ∠D = , (d) EF = 12cm, ∠D = ,Correct Answer: (b) DE = 12cm, ∠F = ${100^0}$ Explanation: Given, In triangle ABC, ∠???? + ∠???? + ∠???? = 1800 ∠???? = 180 − 30 − 50 => 1000 ∵ ∆ABC ~ ∆DFE ∠???? = ∠???? = 300 ∠???? = ∠???? =...
.Solution: $\vec{A}=-2 \hat{\imath}+\hat{\jmath}+2 \hat{k}$ $\mathrm{I} \vec{A} \mathrm{I}=\sqrt{(-2)^{2}+1^{2}+2^{2}}=\sqrt{4+1+4}=\sqrt{9}=3$ Note: Any vector $\vec{X}$ is given by $\vec{X}=$...
Correct Answer: (d) 4 : 1 Explanation: Given, ABC and BDE are two equilateral triangles D is the midpoint of BC and BDE is also an equilateral triangle. E is also the midpoint of AB. D and E are the...
Correct Answer: (d) 30 cm Explanation: Perimeter of ∆ABC = AB + BC + CA => 9 + 6 + 7.5 => 22.5 cm $\begin{array}{l} \frac{{Perimeter(\Delta ABC)}}{{Perimeter(\Delta DEF)}} =...
and 
then find a unit vector parallel to .Solution: $\begin{array}{l} \vec{a}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \\ \vec{b}=2 \hat{\imath}+4 \hat{\jmath}+9 \hat{k} \end{array}$ Then $\vec{a}+\vec{b}=(\hat{\imath}+2 \hat{\jmath}-3...
Correct Answer: (a) 35 cm Explanation: Given, ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{Perimeter(\Delta ABC)}}{{Perimeter(\Delta DEF)}} = \frac{{AB}}{{DE}}\\ \frac{{Perimeter(\Delta ABC)}}{{25}} =...
Correct Answer: (b) 5.4 cm Explanation: Given, ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{Perimeter(\Delta ABC)}}{{Perimeter(\Delta DEF)}} = \frac{{BC}}{{EF}}\\ \frac{{30}}{{18}} = \frac{9}{{EF}}\\ EF =...
and 
then find a unit vector in the direction of .Solution: $\begin{array}{l} \vec{a}=3 \hat{\imath}+\hat{\jmath}-5 \hat{k} \\ \vec{b}=\hat{\imath}+2 \hat{\jmath}-\hat{k} \end{array}$ Then $\vec{a}-\vec{b}=(3 \hat{\imath}+\hat{\jmath}-5...
. If AC = 5.6cm, then AE = ? (a) 4.2cm (b) 3.1cm (c) 2.8cm (d) 2.1cmCorrect Answer: (d) 2.1 cm Explanation: Given, DE || BC. Applying Thales’ theorem, $\begin{array}{l} \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\\ \end{array}$ AE be x cm. EC = (5.6 –...
and 
then find the unit vector in the direction of .Solution: $\vec{a}=-\hat{\imath}+\hat{\jmath}-\hat{k}$ $\vec{b}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k}$ Then $(\vec{a}+\vec{b})=(-\hat{\imath}+\hat{\jmath}-\hat{k})+(2 \hat{\imath}-\hat{\jmath}+2...
Correct Answer: (c) x = 4 Explanation: Given, DE || BC Applying Thales’ theorem, $\begin{array}{l} \frac{{AD}}{{BD}} = \frac{{AE}}{{EC}}\\ \frac{{7x-4}}{{3x+4}} =...
Correct Answer: (b) 4cm Explanation: Given, DE || BC Applying basic proportionality theorem, $\begin{array}{l} \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}}\\ \frac{{4.5}}{{AB}} =...
Solution: If $\vec{a}=\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{k}$, then Unit vector in the direction of $\vec{a}$ can be given by $\hat{a}=\frac{\vec{a}}{I...
Correct Answer: (b) 6 cm Explanation: Given, DE || BC Applying basic proportionality theorem, $\begin{array}{l} \frac{{AD}}{{BD}} = \frac{{AE}}{{EC}}\\ \frac{{2.4}}{{BD}} =...
. If ∠B = 
and ∠C = , then ∠BAD = ? (a) (b) 
(c) (d) Correct Answer: (a) ${30^0}$ Explanation: Given, $\begin{array}{l} \frac{{AB}}{{AC}} = \frac{{BD}}{{DC}}\\ \end{array}$ Applying angle bisector theorem, AD bisects ∠????. In...
Correct Answer: (a) 2 Explanation: Given, The diagonals of a trapezium are proportional. $\begin{array}{l} \frac{{OA}}{{OC}} = \frac{{OB}}{{OD}}\\ \frac{{3X-1}}{{5X-3}} =...
Correct Answer: (c) isosceles Explanation: Let AD be the angle bisector of angle A in triangle ABC. Applying angle bisector theorem, $\begin{array}{l} \frac{{AB}}{{AC}} =...
Solution: A. $\vec{a}=\hat{\imath}+2 \hat{\jmath}+5 \hat{k}$ If $\vec{a}=\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{k}$, then $\mathrm{I} \vec{a}...
Correct Answer: (a) parallelogram Explanation: The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Correct Answer: (b) trapezium Explanation: Diagonals of a trapezium divide each other proportionally.
Correct Answer: (b) 13 cm Explanation: Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O. AC = 24 cm BD = 10 cm Diagonals of a rhombus bisect each...
Correct Answer: (c) 16 cm Explanation: Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O. Also, diagonals of a rhombus bisect each other at...
Correct Answer: (c) 3???????? 2 = 4???????? 2 Explanation: Applying Pythagoras theorem, In right-angled triangles ABD and ADC, $\begin{array}{l} A{B^2} = A{D^2} + B{D^2}\\ A{B^2} = {\left(...
Correct Answer: (b) Isosceles Explanation: In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.
Correct Answer: (b) 3.5cm Explanation: Using angle bisector in ∆ABC, $\begin{array}{l} \frac{{AB}}{{AC}} = \frac{{BD}}{{DC}}\\ \frac{10}{14} = \frac{6-x}{x}\\...
Correct Answer: (d) 7.5 cm Explanation: Given, AD bisects angle A Applying angle bisector theorem, $\begin{array}{l} \frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}\\ \frac{4}{5} =...
Correct Answer: (a) 3 : 4 Explanation: In ∆ ABD and ∆ACD, ∠???????????? = ∠???????????? $\begin{array}{l} \frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}\\ \frac{6}{8} = \frac{3}{4}\\ BD:DC =...
Correct Answer: (d) 24 cm Explanation: In triangle ABC, Let the altitude from A on BC meets BC at D. AD = 5 cm AB = 13 cm D is the midpoint of BC Applying Pythagoras theorem in...
Correct Answer: (b) 6√3????m Explanation: Let ABC be the equilateral triangle with AD as its altitude from A. In right-angled triangle ABD, ????????2 = ????????2 + ????????2...
Correct Answer: (b) 15 cm, 20 cm Explanation: Given, Length of hypotenuse = 25 cm Let the other two sides be x cm and (x−5) cm. Applying Pythagoras theorem, 252 = ????2 + (???? − 5 ) 2 625 = ????2 +...
.OM = 16 cm and OP = 12 cm if MN = 21cm and ∆NMP = then NP=?Answer: In right triangle MOP, By using Pythagoras theorem, ????????2 = ????????2 + ????????2 => 122 + 162 => 144 + 256 => 400 MO = 20 cm In right triangle MPN, By using...
Correct Answer: (a) 7 m Explanation: Let the ladder BC reaches the building at C. Let the height of building where the ladder reaches be AC. BC = 25 m AC = 24 m In right-angled...
Correct Answer: (d) 5.0 Explanation: Suppose DE is a 5 m long stick and BC is a 12.5 m high tree. Suppose DA and BA are the shadows of DE and BC. In ∆ABC and ∆ADE...
Correct Answer: (d) 30m Explanation: Let AB and AC be the vertical pole and its shadow, AB = 6 m AC = 3.6 m Let DE and DF be the tower and its shadow. DF = 18 m DE =? In...
Correct Answer: (c) 1.5m Explanation: Let AB and AC be the vertical stick and its shadow, AB = 1.8 m AC = 45 cm => 0.45 m Let DE and DF be the pole and its shadow, DE = 6 m...
Correct Answer: (b) 10 m Explanation: Let AB and DE be the two poles. AB = 13 m DE = 7 m Distance between their bottoms = BE = 8 m Draw a perpendicular DC to AB from D,...
Correct Answer: (c) 26 m Explanation: The man starts from point A and goes 24 m due west to point B. From here, he goes 10 m due north and stops at C. In right triangle...
Answer: ABCD is a rhombus having AC and BD its diagonals. The diagonals of a rhombus perpendicular bisect each other. AOC is a right-angled triangle. In right triangle AOC, By using Pythagoras...
Answers: (i) True Given, ∆ABC ~ ∆DEF ∠???????????? = ∠???????????? ∠???? = ∠???? (∠???????????? ~ ∆????????????) By AA criterion, ∆ABP and ∆DEQ $\frac{A B}{D E}=\frac{A P}{D Q}$...
and AC = 8 cm and in a DEF , DF = 9 cm D = and DE= 12 cm, then ABC ~ DEF. (ii) the polygon formed by joining the midpoints of the sides of a quadrilateral is a rhombus.Answers: (i) False In ∆ABC, AB = 6 cm ∠???? = 450 ???????? = 8 ???????? I???? ∆????????????, ???????? = 9 ???????? ∠???? = 450 ???????? = 12 ???????? ∆???????????? ~ ∆???????????? (ii) False...
Answers: (i) False If two triangles are similar, their corresponding angles are equal and their corresponding sides are proportional. (ii) True ABC is a triangle with M, N DE is...
Answers: (i) False Two rectangles are similar if their corresponding sides are proportional. (ii) True Two circles of any radii are similar to each other.
Answer: ABCD is a rhombus. The diagonals of a rhombus perpendicularly bisect each other. ∠???????????? = 900 ???????? = 20 ???????? ???????? = 21 ???????? In right...
. If p, q and r are the lengths of AM, MB and BC respectively then express the length of MN of terms of P, q and r.Answer: In ∆AMN and ∆ABC, ∠???????????? = ∠???????????? =$76^{\circ}$ ∠???? = ∠???? (????????????????????????) By AA similarity criterion, ∆AMN ~ ∆ABC If two triangles...
Answer: Let, DC = x BD = a - x Using angle bisector there in ∆ ABC, $\frac{A B}{A C}=\frac{B D}{D C}$ $\frac{c}{b}=\frac{a-x}{x}$ $c x=a b-b x$ $x(b+c)=a b$ $x=\frac{a...
Answer: In right-angled triangle SOW, Using Pythagoras theorem, ????????2 = ????????2 + ????????2 => 352 + 122 => 1225 + 144 => 1369 ???????? = 37 ???? Hence,...
Answer: The altitude drawn from the vertex opposite to the non-equal side bisects the non-equal side. ABC is an isosceles triangle having equal sides AB and BC. The altitude drawn from the vertex...
Answer: When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional. ∆BMP ~ ∆CNR $\frac{B M}{C N}=\frac{B P}{C R}=\frac{M P}{N R}$ $\quad \frac{B...
Answer: Given, AM : MB = 1 : 2 $\frac{M B}{A M}=\frac{2}{1}$ Adding 1 to both sides, $\frac{M B}{A M}+1=\frac{2}{1}+1$ $\frac{M_{B}+A M}{A M}=\frac{2+1}{1}$ $\frac{A B}{A...
and 
. If ∆DEF ~ ∆GHK then find the measures of ∠F.Answer: If two triangles are similar then the corresponding angles of the two triangles are equal. ∆DEF ~ ∆GHK ∴ ∠???? = ∠???? = 570 In ∆ DEF Using the a????????????????...
is 
Solution: $\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}^{2}$ Comparing with $\frac{d y}{d x}+P y=Q$ Integrating factor $=\mathrm{x}$ General solution is $\mathrm{yx}=\int...
Answer: ABCD is a rhombus. The diagonals of a rhombus perpendicularly bisect each other. ∠???????????? = 900 ???????? = 12 ???????? ???????? = 5 ???????? In right triangle AOB,...
is 
Solution: $\frac{d y}{d x}+y \operatorname{Cot} x=2 \operatorname{Cos} x$ Comparing with $\frac{d y}{d x}+P y=Q$ Integrating factor is $e^{\int \cot x d x}=\operatorname{Sin} x$ General solution is...
Solution: $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \tan \mathrm{x}=\mathrm{Secx}$ Comparing with $\frac{d y}{d x}+P y=Q$ Integrating factor $e^{\int \tan x d x}=\operatorname{Sec} x$ General...
is 
Solution: $\frac{d y}{d x}=\frac{y}{x}+\sin \frac{y}{x}$ Let $\mathrm{y}=\mathrm{vx}$ $\begin{array}{l} \mathrm{dy} / \mathrm{dx}=\mathrm{v}+\mathrm{xdv} / \mathrm{d} \mathrm{x} \\...
Answer: We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides. ABC is an...
Answer: If two triangles are similar, then the ratio of their areas is equal to the squares of their corresponding sides. $\frac{\text { area of } smaller triangle}{\text { area of } larger...
Answer: In ∆ AOB and COD ∠???????????? = ∠???????????? (???????????????????????????????? ???????????????????????? ???????? ???????? ∥ ????????) ∠???????????? = ∠????????????...
is 
Solution: $\begin{array}{r} (x-y) d y+(x+y) d x=0 \\ \qquad \frac{d y}{d x}=\frac{x+y}{y-x} \end{array}$ Let $\mathrm{y}=\mathrm{vx}$ $\begin{array}{l} \mathrm{dy} /...
Answer: Given, ∆ ABC ~ ∆ DEF If two triangles are similar then the ratio of their areas is equal to the ratio of the squares of their corresponding sides. $\frac{\text { area }(\triangle A B...
is 
Solution: $\begin{array}{r} 2 \mathrm{xydy}+\left(\mathrm{x}^{2} \quad \mathrm{y}^{2}\right) \mathrm{d} \mathrm{x}=0 \\ \frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y} \end{array}$ Let...
Answer: The altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides. ABC is an equilateral...
Answer: Let AB be A ladder and B is the window at 8 m above the ground C. In right triangle ABC By using Pythagoras theorem, ????????2 = ????????2 + ????????2 102 = 82 +...
is 
Solution: $\mathrm{x} \frac{d y}{d x}=y+x \tan \frac{y}{x}$ Dividing both sides by $x$, we get, $\frac{d y}{d x}=\frac{y}{x}+\tan \frac{y}{x}$ Let $\mathrm{y}=\mathrm{vx}$ Differentiating both...
Answer: In ∆ADE and ∆ABC, ∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ???????????????????????? ???????? ???????? ∥ ????????)...
is. 
Solution: $\begin{array}{l} \mathrm{x}^{2} \frac{d y}{d x}=x^{2}+x y+y^{2} \\ \frac{d y}{d x}=1+\frac{y}{x}+\frac{y^{2}}{x^{2}} \end{array}$ Let $\mathrm{y}=\mathrm{vx}$ $\begin{array}{l}...
Answer: When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal. ∆ABC ~ ∆DEF $\therefore \frac{A B}{D E}=\frac{B C}{E F}$ $\frac{A B}{2 A B}=\frac{6}{E...
, PR = 9cm ∠???? = and PQ = 4.5 cm. Show that ∆ ABC ~ ∆ PQR and state that similarity criterion.Answer: In ∆ABC and ∆PQR ∠???? = ∠???? = 700 $\frac{A B}{P Q}=\frac{A C}{P R}$ $\frac{3}{4.5}=\frac{6}{9}$ $\frac{1}{1.5}=\frac{1}{1.5}$ By SAS similarity criterion, ∆ABC ~...
is 
Solution: $\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}$ Let $\mathrm{y}=\mathrm{vx}$ $\begin{array}{l} \mathrm{dy} / \mathrm{dx}=\mathrm{v}+\mathrm{xdv} / \mathrm{dx} \\ \frac{x^{2}...
Answer: Using midpoint theorem, The segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of...
Converse of Pythagoras theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
Solution: $\begin{array}{r} \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{1-\mathrm{x}^{2}} \sqrt{1-\mathrm{y}^{2}} \\ \frac{d y}{\sqrt{1-y^{2}}}=\sqrt{1-x^{2}} d x \end{array}$ $\operatorname{Let}...
Pythagoras theorem: The square of the hypotenuse is equal to the sum of the squares of the other two sides. Here, the hypotenuse is the longest side and it’s always opposite the right angle.
SAS-similarity criterion: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.
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Solution: $\log \left(\frac{d y}{d x}\right)=(a x+b y)$ $\begin{array}{l} \frac{d y}{d x}=e^{a x+b y} \\ \frac{d y}{e^{b y}}=e^{a x} d x \end{array}$ On integrating on both sides we get...
SSS-similarity criterion for similarity of triangles: If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are...
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Solution: $\begin{array}{r} x \sqrt{1+y^{2}} \mathrm{dx}+\mathrm{y} \sqrt{1+x^{2}} \mathrm{dy}=0 \\ \frac{y d y}{\sqrt{1+y^{2}}}=\frac{-x d x}{\sqrt{1+x^{2}}} \end{array}$ Let $1+y^{2}=t$ and...
AA-similarity criterion: If two angles are correspondingly equal to the two angles of another triangle, then the two triangles are similar.
AAA-similarity criterion: If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and hence the triangles are similar.
Midpoint theorem: The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side.
is 
Solution: $\left(1+x^{2}\right) d y-x y d x=0$ $\frac{d y}{y}=\frac{x}{1+x^{2}} d x$ Let $1+\mathrm{x}^{2}=\mathrm{t}$ $\begin{array}{r} 2 \mathrm{x} \mathrm{dx}=\mathrm{dt} \\ \frac{d y}{y}=\frac{d...
Thale’s 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.
is 
Solution: $\frac{d y}{d x}+y \log y \operatorname{Cot} x=0$ Let $\log \mathrm{y}=\mathrm{t}$ On differentiating we get $\begin{array}{l} \frac{1}{y} d y=d t \\ \frac{d t}{t}=-\operatorname{Cot} x d...
Basic proportionality theorem: If a line is draw parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio.
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Solution: $\begin{array}{l} x \operatorname{Cosydy}=\left(x e^{x} \log x+e^{x}\right) d x \\ \operatorname{Cosydy}=\frac{x \operatorname{exlogx}+e x}{x} d x \end{array}$ On integrating on both sides...
Answer: The two triangles are similar if and only if The corresponding sides are in proportion. The corresponding angles are equal.
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Solution: $\cos x(1+\cos y) d x-\sin y(1+\sin x) d y=0$ Let $1+\cos y=t$ and $1+\sin x=u$ On differentiating both equations, we get $-\sin y d y=d t$ and $\cos x d x=d u$ Substitute this in the...
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Solution: $\begin{array}{l} \frac{d y}{d x}=\frac{-2 x y}{x^{2}+1} \\ \frac{d y}{y}=\frac{-2 x d x}{x^{2}+1} \end{array}$ Let $\mathrm{x}^{2}+1=\mathrm{t}$ On differentiating on both sides we get $2...
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Solution: $\begin{array}{l} \frac{d y}{d x}=\frac{1-\operatorname{Cos} x}{1+\operatorname{Cos} x} \\ \frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}, \\ \frac{d y}{d x}=\tan...
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Solution: $\frac{\mathrm{dy}}{\mathrm{dx}}+\sqrt{\frac{1-\mathrm{y}^{2}}{1-\mathrm{x}^{2}}}=0$ $\frac{-d y}{\sqrt{1-y^{2}}}=\frac{d x}{\sqrt{1-x^{2}}}$ On integrating on both sides, we get $-\sin...
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Solution: $\begin{array}{l} \frac{d y}{d x}=e^{x+y}+x^{2} e^{y} \\ \left(e^{-y}\right) d y=\left(e^{x}+e^{2}\right) d x \end{array}$ On integrating on both sides, we get $\begin{array}{l}...
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Solution: $\begin{array}{l} \frac{d y}{d x}=1-x+y-x y \\ \frac{d y}{d x}=1-x+y(1-x) \\ \frac{d y}{d x}=(1+y)(1-x) \\ \frac{d y}{1+y}=(1-x) d x \end{array}$ On integrating on both sides, we get $\log...
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Solution: $\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$ On integrating on both sides, we get $\begin{array}{l} \tan ^{-1} y=\tan ^{-1} x+c \\ \frac{y-x}{1+y x}=\mathrm{c} \\...
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Solution: $x \frac{d y}{d x}=\cot y$ Separating the variables, we get, $\begin{array}{c} \frac{d y}{\cot y}=\frac{d x}{x} \\ \tan y d y=\frac{d x}{x} \end{array}$ Integrating both sides, we get,...
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Solution: $\begin{array}{l} \text { Given } x d y+y d x=0 \\ x d y=-y d x \end{array}$ $-\frac{d y}{y}=\frac{d x}{d x}$ On integrating on both sides we get, $\begin{array}{l} -\log y=\log x+c \\...
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Solution: $\begin{array}{l} \left(e^{x}+1\right) y d y=(y+1) e^{x} d x \\ \frac{y d y}{y+1}=\frac{e^{x} d x}{\left(e^{x}+1\right)} \end{array}$ On differentiating on both sides we get $e^{x} d x=d...
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Solution: $\begin{array}{c} \frac{d y}{d x}=e^{x+y} \\ \frac{d y}{d x}=e^{x} e^{y} \\ e^{-y} d y=e^{x} d x \end{array}$ On integrating on both sides, we get $\begin{array}{c}...
Answer: Naman pulls in the string at the rate of 5 cm per second. Hence, after 12 seconds the length of the string he will pulled is given by: 12 × 5 = 60 cm or 0.6 m In ∆BMC By...
Answer: Adding $\Rightarrow A D^{2}+B C \cdot D L+\left(\frac{B C}{2}\right)^{2}$ and $A B^{2}=A D^{2}-B C-D L+\left(\frac{B C}{2}\right)^{2}$, $\begin{aligned} AC^{2}+A...
(ii) 
Answers: (i) In right triangle ALD, Using Pythagoras theorem, $A C^{2}=A L^{2}+L C^{2}$ $\Rightarrow A D^{2}-D L^{2}+(D L+D C)^{2}$ $\Rightarrow A D^{2}-D L^{2}+\left(D...
hours?Answer: Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr Distance...
Answer: ABC as an isosceles triangle, right angled at B. AB = BC Applying Pythagoras theorem in right-angled triangle ABC, ????????2 = ????????2 + ????????2 = 2????????2...
Answer: Draw AE⊥BC, meeting BC at D. Applying Pythagoras theorem in right-angled triangle AED, ABC is an isosceles triangle and AE is the altitude and we know that the altitude...
(ii) 
Answers: (i) Adding $b^{2}=p^{2}+a x+\frac{a^{2}}{x}$ and $c^{2}=p^{2}-a x+\frac{a^{2}}{x}$, $b^{2}+c^{2}=p^{2}+a x+\frac{a^{2}}{4}+p^{2}-a x+\frac{a^{2}}{4}$ $b^{2}+c^{2}=2...
(ii) 
Answers: (i) In right-angled triangle AEC, Applying Pythagoras theorem, ????????2 = ????????2 + ????C2 $b^{2}=h^{2}+\left(x+\frac{a}{2}\right)^{2}$ …(1) In right – angled...
CD ⊥ AB Prove that 
Answer: Given, ∠???????????? = 90o ???????? ⊥ ????B In ∆ ACB and ∆ CDB ∠???????????? = ∠???????????? = 90o ∠???????????? = ∠???????????? By AA similarity-criterion, ∆ ACB ~ ∆CDB...
Answer: In right-angled triangle AED, applying Pythagoras theorem, ????????2 = ????????2 + ????????2 … (????) In right-angled triangle AED, applying Pythagoras theorem,...
Answer: Let ABCD be the rhombus with diagonals meeting at O. AC = 24 cm BD = 10 cm We know that the diagonals of a rhombus bisect each other at angles. Applying Pythagoras...
Answer: Let ABCD be the rectangle with diagonals AC and BD meeting at O. AB = CD = 30 cm BC = AD = 16 cm Applying Pythagoras theorem in right-angled triangle ABC, we get:...
Answer: Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. D will be the midpoint of BC. Applying Pythagoras theorem in right-angled triangle...
Answer: Let AD, BE and CF be the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, D, E and F are the midpoint of BC, AC and AB, In right-angled ∆ABD, AB = 2a BD = a...
Answer: In isosceles ∆ ABC, AB = AC = 2a units BC = a unit Let AD be the altitude drawn from A that meets BC at D. D is the midpoint of BC. BD = BC = ????2...
Answer: Given, ∆ ABC is an isosceles triangle. AB = AC = 13 cm The altitude from A on BC meets BC at D. D is the midpoint of BC. AD = 5 cm ∆ ????????????...
Answer: Applying Pythagoras theorem in right-angled triangle POR, ????????2 = ????????2 + ????????2 ????????2 = 62 + 82 =>36 + 64 =>100 ???????? = √100 =>10 ???????? In...
Answer: Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A. In right triangle ABC By using Pythagoras theorem,...
Answer: Let the two poles be DE and AB and the distance between their bases be BE. DE = 9 m AB = 14 m BE = 12 m Draw a line parallel to BE from D, meeting AB at C. DC = 12 m AC...
Answer: Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, AB = 20 m BC = 15 m Applying Pythagoras theorem in...
Answer: Let AB and AC be the ladder and height of the building. Given, AB = 13 m AC = 12 m In right-angled triangle ABC, $A B^{2}=A C^{2}+B C^{2}$ $B...
Answer: Let the man starts from point D and goes 10 m due south at E. He then goes 24 m due west at F. In right ∆DEF, DE = 10 m EF = 24 m $D F^{2}=E F^{2}+D E^{2}$ $D...
Answer: Let the man starts from point A and goes 80 m due east to B. From B, he goes 150 m due north to c. In right- angled triangle ABC, $A C^{2}=A B^{2}+B C^{2}$...
Answer: Given, The sum of the two sides must be equal to the square of the third side. The three sides of the triangle - a, b and c. P = (a-1) cm, q = 2 √???? ???????? ???????????? ???? = (???? + 1)...
Answers: Given, The sum of the two sides must be equal to the square of the third side. The three sides of the triangle - a, b and c. (i) a = 1.4 cm b= 4.8 cm c= 5 cm ????2 + ????2 = (1.4) 2 + (4.8)...
Answers: Given, The sum of the two sides must be equal to the square of the third side. The three sides of the triangle - a, b and c. (i) a = 9 cm b = 16 cm c = 18 cm ????2 + ????2 = 92 + 162...
Answer: Given, D and E are midpoints of AB and AC. Applying midpoint theorem, DE ‖ BC. By B.P.T., $\frac{A D}{A B}=\frac{A E}{A C}$ ∠???? = ∠???? Applying SAS similarity...
Answer: Given, DE || BC. ∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ????????????????????????) ∠???????????? = ∠????????????...
Answer: In ∆ABC and ∆ADC, ∠???????????? = ∠???????????? = 900 ∠???????????? = ∠???????????? (????????????????????????) By AA similarity, ∆ BAC~ ∆ ADC. The ratio of the areas of...
Answer: Given, DE || BC ∴ ∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ????????????????????????) ∠???????????? = ∠????????????...
Answer: Given, $\frac{A P}{A B}=\frac{1}{1+3}=\frac{1}{4}$ $\frac{A Q}{A C}=\frac{1.5}{1.5+4.5}=\frac{1.5}{6}=\frac{1}{4}$ $\frac{A P}{A B}=\frac{A Q}{A C}$ ∠???? = ∠???? By SAS...
Answer: Let the two triangles be ABC and PQR with medians AM and PN, The ratio of areas of two similar triangles will be equal to the ratio of squares of their corresponding...
Answer: Given, Triangles are similar The areas of these triangles will be equal to the ratio of squares of their corresponding sides. The ratio of areas of two similar triangles...
Answer: Let the two triangles be ABC and DEF with altitudes AP and DQ, Given, ∆ ABC ~ ∆ DEF. The ratio of areas of two similar triangles is equal to the ratio of squares of...
Answer: Given, ∆ABC ~ ∆DEF. The ration of the areas of these triangles will be equal to the ratio of squares of their corresponding sides. The ratio of areas of two similar...
Answer: Given, Triangles are similar. The ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides. Let the longest side of smaller triangle be X cm....
Answer: Given, ???????? ( ∆ ???????????? ) = 4???????? (∆ ???????????? ) \begin{equation} \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{4}{1} \end{equation} ∵ ∆ABC...
Answer: Given, ∆ ABC ~ ∆ PQR The ratio of the areas of triangles will be equal to the ratio of squares of their corresponding sides. \begin{equation} \frac{\operatorname{ar}(\triangle A B...
Answer: Given, ∆ ABC ~ ∆ DEF. The ratio of the areas of these triangles will be equal to the ration of squares of their corresponding sides. $\eqalign{ & \frac{ar(\vartriangle...
Answer: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the...
Answer: Given, AB and CD are two chords ∠???????????? + ∠???????????? = 180o …(1) ∠???????????? + ∠???????????? = 180p …(2) Using (1) and (2), ∠???????????? = ∠???????????? ∠???? = ∠A...
Answer: Given, AB and CD are two chords In ∆ PAC and ∆ PDB ∠???????????? = ∠???????????? ∠???????????? = ∠???????????? By ???????? ????????????????????????????????????????...
Answer: In ∆ ABC, P and Q are mid points of AB and AC respectively. So, PQ || BC, and PQ = $\begin{array}{l} \frac{{1}}{{2}} \end{array}$ ???????? …(1) Similarly, in...
. Prove that ∆ ACB ~ ∆ DCE.Answer: Given, $\begin{array}{l} \frac{{AC}}{{BD}} = \frac{CB}{CE}\\ \frac{{AC}}{{CB}} = \frac{BD}{CE}\\ \frac{{AC}}{{CB}} = \frac{CD}{CE}\\ \end{array}$ ∠1 = ∠2 ∠???????????? =...
Answer: Given. ∆ABC is an isosceles triangle. CA = CB ∠???????????? = ∠???????????? 180o − ∠???????????? = 180o − ∠???????????? ∠???????????? = ∠???????????? AP × BQ = ????????2...
Answer: Let AB be the vertical stick and BC be its shadow. AB = 7.5 m, BC = 5 m Let PQ be the tower and QR be its shadow. QR = 24 m Let the length of PQ be x m. In ∆ ABC...