then find the value of 
so that 
may be a unit vector.Solution: $\vec{a}=2 \hat{\imath}-4 \hat{\jmath}+5 \hat{k}$ Given: $(\lambda \vec{a})$ is a unit vector Since $(\lambda \vec{a})$ is a unit vector, so I $\lambda \vec{a} \mathrm{I}=1$...
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}} =...
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}} =...
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????????????????...
Answer: ABCD is a rhombus. The diagonals of a rhombus perpendicularly bisect each other. ∠???????????? = 900 ???????? = 12 ???????? ???????? = 5 ???????? In right triangle AOB,...
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 ∠???????????? = ∠???????????? (???????????????????????????????? ???????????????????????? ???????? ???????? ∥ ????????) ∠???????????? = ∠????????????...
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...
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 +...
Answer: In ∆ADE and ∆ABC, ∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ???????????????????????? ???????? ???????? ∥ ????????)...
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 ~...
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.
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.
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...
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.
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.
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.
Answer: The two triangles are similar if and only if The corresponding sides are in proportion. The corresponding angles are equal.
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 3x – 4y + 1 = 0 …(i) 5x + y – 1 = 0 …(ii) Now, we find the point of...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 2x + 3y – 2 = 0 …(i) x – 2y + 1 = 0 …(ii) Now, we find the point of...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 2x – 3y + 1 = 0 …(i) x + y – 2 = 0 …(ii) Now, we find the point of...
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 : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x – 7y + 5 = 0 …(i) 3x + y – 7 = 0 …(ii) Now, we find the point of...
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 : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 2x – 3y = 0 …(i) 4x – 5y = 2 …(ii) Now, we find the point of intersection...
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 : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 5x – 3y = 1 …(i) 2x + 3y = 23 …(ii) Now, we find the point of intersection...
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 : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x – y = 1 …(i) 2x – 3y + 1 = 0 …(ii) Now, we find the point of intersection...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x + y = 9 …(i) 2x – 3y + 7 = 0 …(ii) Now, we find the point of intersection...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x – y = 7 …(i) 2x + y = 2 …(ii) Now, we find the point of intersection of...
Answer: Let the smallest term be x, and the largest term be 2x Then AP formed= x, ?, ?, 2x so, 360°= 4/2 [x+ 2x]....[We know that → a+(n-1) d= last term= 2x] ⇒ 180°= 3x ⇒ x= 60° Now, 60° is least...
Answer: θ will be in degrees. Arc-length can be given by the formula : θ / 360° × 2πr Hence it is given that 121 cm is the arclength. ⇒ 121 = θ / 360° × 2πr = 121 = θ / 360° × 2 × 22 / 7 × 180 = 121...
Answer: Radius = 1500 m. Train speed at rate of 66km/hr = 18.33 m/s Therefore, Distance covered in 1 second = 18.33 m Distance covered in 10 second = 18.33 × 10 = 183.33m We know that θ = Distance /...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x – 2y + 3 = 0 …(i) 2x – 3y + 4 = 0 …(ii) Now, we find the point of...
Answer: Given that Number of revolutions per minute = 180 Then per second, it will be = 180/60 =3 We know that In one complete revolution, the wheel turns at an angle of 2π rad. Then for 3 complete...
Answer : Let the new origin be (h, k) = (1, -2) Then, the transformation formula become: x = X + 1 and y = Y + (-2) = Y – 2 Substituting the value of x and y in the given equation, we get 2x2 + y2 –...
Answer : Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we get xy – x – y + 1 = 0...
x2 – y2 – 2x + 2y = 0 Answer : Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we get...
Answer : Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we get xy – y2 – x + y = 0...
Answer: For 20 minutes = θ = 4 x 30° = 120° We know that l = r × θ $ l=42\times 120\times \frac{\pi }{180} \\ & l=88}$ Therefore, the length is equal to 88 cm.
Answer: We know that l = r × θ Here l = length of arc = 11 cm R = radius = length of pendulum = 45 cm We need to find θ 11 = 45 x θ
Answer: Diameter = 30 cm Length of chord = 15 cm Radius = 15 cm [ r = 0.5 x diameter ] The created triangle in the circle is an equilateral triangle because the radius is equal to the length of the...
Amswer; $ Angle\text{ }in\text{ radians}=Angle\text{ }in\text{ }\deg \times \frac{\pi }{180} $ $ \theta =\frac{l}{r} $ $ where\text{ }\theta \text{ }is\text{ }central\text{ }angle,\text{...
Answer: $ Angle\text{ }in\text{ radians}=Angle\text{ }in\text{ }\deg \times \frac{\pi }{180} $ $ \theta =\frac{l}{r} $ $ where\text{ }\theta \text{ }is\text{ }central\text{ }angle,\text{...
x2 + xy – 3x – y + 2 = 0 Answer : Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we...
Answer : Let (h, k) be the point to which the origin is shifted. Then, x = -4, y = 2, X = 3 and Y = -2 ∴ x = X + h and y = Y + k ⇒ -4 = 3 + h and 2 = -2 + k ⇒ h = -7 and k = 4 Hence, the origin must...
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 : Let the new origin be (h, k) = (2, -1) and (x, y) = (-3, 5) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: x = X + h and y = Y + k ⇒ -3 = X + 2 and...
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 : Let the new origin be (h, k) = (0, -2) and (x, y) = (3, 2) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: x = X + h and y = Y + k ⇒ 3 = X + 0 and 2...
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: $ Angle\text{ }in\text{ radians}=Angle\text{ }in\text{ }\deg \times \frac{\pi }{180} $ $ \theta =\frac{l}{r} $ where θ is central angle, l=length of arc, r=radius $ \therefore...
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: $ Angle\text{ }in\text{ radians}=Angle\text{ }in\text{ }\deg \times \frac{\pi }{180} $ $ \theta =\frac{1}{r} $ where θ is central angle, l=length of arc, r=radius $ \therefore...
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...
Answer: In ∆BED and ∆ACB, ∠???????????? = ∠???????????? = 90o ∵ ∠???? + ∠???? = 180o ∴ BD || AC ∠???????????? = ∠???????????? By AA similarity theorem, ∆ BED ~ ∆ ACB $\begin{array}{l}...
Answer: Given, ∠???????????? = ∠???????????? ∵ DA || BC ∴ ∠???????????? = ∠???????????? ∆ DAF ~ ∆ BEF $\begin{array}{l} \frac{{AF}}{{EF}} = \frac{FD}{FB}\\ \end{array}$...
Answer: Given, $\begin{array}{l} \frac{{AP}}{{AB}} = \frac{2}{6} = \frac{1}{3}\\ \frac{{AQ}}{{AC}} = \frac{3}{9} = \frac{1}{3}\\ \\ \frac{{AP}}{{AB}} = \frac{{AQ}}{{AC}} \end{array}$ By AA...
Answer: $\therefore \text{ a}ngle\text{ }in\text{ }degrees=\frac{\pi }{5}\times \frac{180}{\pi }={{36}^{\circ }}$ Let, two acute angles are x and y so, ATQ, x - y= 36°......(1) x+ y= 90°......(2)...
Answer: Let a - d, a, a + d be the three angles of the triangle that form AP. Given that the greatest angle is double the least. Now, a + d = 2(a - d) 2a - 2d = a + d a = 3d ……(1) Now by angle sum...
$ iii)\,\,\frac{5}{6} $ $ iv)\,\,-4 $ Answer: (iii) Formula : Angle in degrees = $Angle\,in\,radians\times \frac{180}{\pi }$ The angle in minutes is equal to the decimal of the angle in radians...
Answer : Let the new origin be (h, k) = (-3, -2) and (x, y) = (3, -5) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: x = X + h and y = Y + k ⇒ 3 = X – 3 and...
Answer: Given, ABC is a right-angled triangle BD is the altitude drawn from the right angle to the hypotenuse. In ∆ DBA and ∆ DCB, ∠???????????? = ∠????????????...
Answer : Let the new origin be (h, k) = (1, 2) and (x, y) = (3, -4) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: x = X + h and y = Y + k ⇒ 3 = X + 1 and...
Answer: Given, ABC is a right-angled triangle and BD is the altitude drawn from the right angle to the hypotenuse. In ∆ BDC and ∆ ABC, ∠???????????? = ∠???????????? = 90o ∠???? = ∠????...
Answer: In ∆ BDA and ∆ BAC, ∠???????????? = ∠???????????? = 90o ∠???????????? = ∠???????????? By AA similarity theorem, ∆ BDA - ∆ BAC $\begin{array}{l} \frac{{AD}}{{AC}} =...
Answer: Given, ∆ ABC - ∆ DEF Their corresponding sides will be proportional. Ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides. $\begin{array}{l}...
Answer: Given, Triangles ABC and PQR are similar. $\begin{array}{l} \frac{{Perimeter( ABC)}}{{ Perimeter( PQR)}} = \frac{{AB}}{{PQ}}\\ \end{array}$ $\begin{array}{l} \frac{{32}}{{24}} =...
Answer: Given, ∠???????????? = ∠???????????? ???????????? ∠???? = ∠???? Let DE be X cm By AA similarity theorem, ∆ ADE - ∆ ABC $\begin{array}{l} \frac{{AD}}{{AB}} =...
Answers: (i) Let OA be X cm. ∵ ∆ OAB - ∆ OCD $\begin{array}{l} \frac{{OA}}{{OC}} = \frac{{AB}}{{CD}}\\ \end{array}$ $\begin{array}{l} \frac{{x}}{{3.5}} =...
Answers: (i) Given, ∆ ODC - ∆ OBA ∠???????????? = ∠???????????? = 45 o (ii) Given, ∆ ODC- ∆ OBA ∠???????????? = ∠???????????? = 70...
Answers: (i) Given, DB is a straight line. ∠???????????? + ∠???????????? = 180o ∠???????????? = 180o − 115o = 65o (ii) In ∆ DOC, ∠???????????? +...
Answer: In ∆ ABC ∠A + ∠???? + ∠???? = 1800 (Angle Sum Property) 800 + ∠???? + 700 = 180o ∠???? = 30o ∠???? = ∠???? ???????????? ∠???? = ∠???? By AA similarity, ∆ ABC - ∆ MNR...
Answers: (iii) Given, $\begin{array}{l} \frac{{CA}}{{QR}} = \frac{{8}}{{6}}\\ \end{array}$ $\begin{array}{l} \frac{{CA}}{{QR}} =...
Answers: (i) Given, ∠BAC = ∠PQR = 50o ∠ABC = ∠QPR = 60 o ∠ACB = ∠PRQ = 70 o By AAA similarity theorem, ∆ ABC – QPR (ii) Given, $\begin{array}{l}...
Let the equation of line AB be x – 4y + 7 = 0 and point C be (-2, 3) CD is perpendicular to the line AB, and we need to find: Equation of Perpendicular drawn from point C Coordinates of D Let the...
Answer: In triangle BQO, BR bisects angle B. Applying angle bisector theorem, $\begin{array}{l} \frac{{QR}}{{PR}} = \frac{{BQ}}{{BP}}\\ \end{array}$ Hence,...
Answer: Given, AD = AE …(i) AB = AC …(ii) Subtracting AD from both sides, AB – AD = AC – AD AB – AD = AC - AE (Since, AD = AE) BD = EC …(iii) Dividing equation (i) by...
AC. If PQ produced meets BC at R, prove that R is the midpoint of BC.Answer: We know that the diagonals of a parallelogram bisect each other. Therefore, CS = $\begin{array}{l} \frac{{1}}{{2}} \end{array}$ AC …(i) Also, it is given that CQ =...
Let the equation of line AB be 4x – 3y + 5 = 0 and point C be (0, 0) CD is perpendicular to the line AB, and we need to find: Equation of Perpendicular drawn from point C Coordinates of D Let the...
Answer: Given, BC is bisected at D. ∴ BD = DC OD =OX The diagonals OX and BC of quadrilateral BOCX bisect each other. BOCX is a parallelogram. BO || CX and BX || CO BX ||...
ABC and DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ||AB and PR||BD are drawn, meeting AC at Q and CD at R respectively. Prove that QR||AD.Answer: In ∆ CAB, PQ || AB. Applying Thales' theorem, $\begin{array}{l} \frac{{CP}}{{PB}} = \frac{{CQ}}{{QA}}\\ \end{array}$ Applying Thales theorem in DBDC, PR||DM...
ABC, M and N are points on the sides AB and AC respectively such that BM= CN. If MN = B C then show that MN||BCAnswer: In ∆ABC, ∠B = ∠ ???? ∴ AB = AC (Sides opposite to equal angle are equal) Subtracting BM from both sides, AB – BM = AC – BM AB – BM = AC – CN (∵BM =CN) AM =AN...
Answer: In trapezium ABCD, AB ‖ CD and the diagonals AC and BD intersect at O $\begin{array}{l} \frac{{AO}}{{OC}} = \frac{{BO}}{{OD}}\\ \frac{{5x-7}}{{2x+1}} =...
Answer: Let the trapezium be ABCD with E and F as the mid Points of AD and BC, AD and BC to meets at P. In ∆ PAB, DC || AB. Applying Thales’ theorem, $\begin{array}{l}...
Answer: Given, ABCD is a parallelogram In ∆ DMC and ∆ NMB ∠DMC = ∠NMB (Vertically opposite angle) ∠DCM = ∠NBM (Alternate angles) By AAA- Similarity ∆DMC ~ ∆NMB $\begin{array}{l}...
Answer : The given equations are y = x …(i) y = 2x …(ii) and y – 3x = 4 …(iii) Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC From eq. (i) and (ii), we get x = 0...
Answer : The given equations are x = 0 …(i) y = 1 …(ii) and 2x + y = 2 …(iii) Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC From eq. (i) and (ii), we get x = 0...
The given equations are x + y = 6 …(i) x – 3y = 2 …(ii) and 5x – 3y + 2 = 0 or 5x – 3y = -2 …(iii) Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC Firstly, we...
$ i)\,\,\frac{5\pi }{12} $ $ ii)\,\,-\frac{18\pi }{5} $ Answer:
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Answer : Given that 3x – y – 2 = 0, 5x + ky – 3 = 0 and 2x + y – 3 = 0 are concurrent We know that, The lines a1x + b1y + c1 = 0, a1x + b1y + c1 = 0 and a1x + b1y + c1 = 0 are concurrent if It is...
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Answer : Given: 3x – 4y + 5 = 0, 7x – 8y + 5 = 0 and 4x + 5y = 45 or 4x + 5y – 45 = 0 To show: Given lines are concurrent The lines a1x + b1y + c1 = 0, a1x + b1y + c1 = 0 and a1x + b1y + c1 = 0 are...
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Answer : Suppose the given two lines intersect at a point P(2, 3). Then, (2, 3) satisfies each of the given equations. So, taking equation x + 7y = 23 Substituting x = 2 and y = 3 Lhs = x + 7y = 2 +...
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Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. ∴ 4x + 3y = 5 or 4x + 3y – 5 = 0 …(i) and x = 2y – 7 or x – 2y + 7 = 0...
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Answer: The angle specified is negative. Adding or removing 360 degrees from an angle has no effect on its position. As a result, Angle can be expressed as=-400° + 360° = -40°. Because the angle is...
Answer: The angle specified is negative. Adding or removing 360 degrees from an angle has no effect on its position. Therefore, Angle can also be written as=-310° + 360° = 50° • Draw a straight line...
Answer: Given angle can be completely written in degree as = -220° -220° = 360° - 220° = 140°
Answer: The given angle is negative Adding or removing 360 degrees from an angle has no effect on its position. Therefore, Angle can also be written as=-40° + 360° = 320 • Draw a straight line from...
Answer: The given angle is greater than 360° Adding or subtracting 360° from a particular angle doesn't change its position. Therefore, Angle can also be written at as = 430° – 360° = 70° • Draw a...
Answer: • Draw a straight line from point A to point B. • Make a mark at B. The vertex of the angle is represented by this dot. • Place the protractor's centre at B and the protractor's baseline...