Sketch the graphs of the following curves on the same scale and the same axes:
Question: Sketch the graphs of the following curves on the same scale and the same axes: (i) $y=\cos x$ and $y=\cos \left(x-\frac{\pi}{4}\right)$ (ii) $y=\cos 2 x$ and $y=\cos 2\left(x-\frac{\pi}{4}\right)$ (iii) $y=\cos x$ and $y=\cos \frac{x}{2}$ (iv) $y=\cos ^{2} x$ and $y=\cos x$ Solution: (i) First, we draw the graph ofy= cosx. Let us now draw the graph of $y=\cos \left(x-\frac{\pi}{4}\right)$. $y=\cos \left(\mathrm{x}-\frac{\pi}{4}\right)$ $\Rightarrow y-0=\cos \left(x-\frac{\pi}{4}\right)...
Read More →Write the truth value (T/F) of each of the following statements:
Question: Write the truth value (T/F) of each of the following statements: (i) Any two similar figures are congruent.(ii) Any two congruent figures are similar.(iii) Two polygons are similar, if their corresponding sides are proportional.(iv) Two polygons are similar, if their corresponding angles are proportional.(v) Two triangles are similar if their corresponding sides are proportional.(vi) Two triangles are similar if their corresponding angles are proportional. Solution: (i) False (ii) True...
Read More →If D and E are points on sides AB and AC respectively
Question: If D and E are points on sides AB and AC respectively of a ∆ABC such that DE || BC and BD = CE. Prove that ∆ABC is isosceles. Solution: It is given that in $\triangle A B C, D E \| B C$ and $B D=C E$. We have to prove that∆ABC is isosceles.By Thales theorem we have $\frac{A D}{B D}=\frac{A E}{E C}$ $\Rightarrow A D=A E$ Now $B D=C E$ and $A D=A E$ So $A D+B D=A E+C E$ Hence $A B=A C$ So, $\triangle \mathrm{ABC}$ is isosceles...
Read More →A circus tent is cylindrical to a height of 3 m and conical above it.
Question: A circus tent is cylindrical to a height of 3 m and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent. Solution: Given diameter = 105 m, Radius = 105/2 m = 52.5 m Therefore curved surface area of circus tent $=\pi r l+2 \pi r h$ $=(22 / 7 * 52.5 * 53)+(2 * 22 / 7 * 52.5 * 3)$ $=8745+990=9735 \mathrm{~m}^{2}$ Therefore length of the canvas required for tent $=\frac{\text {...
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Question: $\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y$ Solution: The given differential equation is: $\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y$ $\frac{d y}{d x}=\frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\...
Read More →In three line segments OA, OB and OC points L, M, N
Question: In three line segments OA, OB and OC points L, M, N respectively are so chosen that LM || AB and MN || BC but neither of L, M, N nor of A, B, C are collinear. Show that LN || AC. Solution: In∆OAB,sinceLM∥AB,thenOLLA=OMMBByBPT........1In∆OBC,sinceMN∥BC,thenOMMB=ONNCByBPT⇒ONNC=OMMB.........2from1and2,wegetOLLA=ONNC...
Read More →Sketch the graphs of the following trigonometric functions:
Question: Sketch the graphs of the following trigonometric functions: (i) $f(x)=\cos \left(x-\frac{\pi}{4}\right)$ (ii) $g(x)=\cos \left(x+\frac{\pi}{4}\right)$ (iii) $h(x)=\cos ^{2} 2 x$ (iv) $\phi(x)=2 \cos \left(x-\frac{\pi}{6}\right)$ (v) $\Psi(x)=\cos 3 x$ (vi) $u(x)=\cos ^{2} \frac{x}{2}$ (vii) $f(x)=\cos \pi x$ (viii) $g(x)=\cos 2 \pi x$ Solution: (i) $y=\cos \left(\mathrm{x}-\frac{\pi}{4}\right)$ $\Rightarrow y-0=\cos \left(x-\frac{\pi}{4}\right)$ ...(i) On shifting the origin at $\left(...
Read More →Sketch the graphs of the following trigonometric functions:
Question: Sketch the graphs of the following trigonometric functions: (i) $f(x)=\cos \left(x-\frac{\pi}{4}\right)$ (ii) $g(x)=\cos \left(x+\frac{\pi}{4}\right)$ (iii) $h(x)=\cos ^{2} 2 x$ (iv) $\phi(x)=2 \cos \left(x-\frac{\pi}{6}\right)$ (v) $\Psi(x)=\cos 3 x$ (vi) $u(x)=\cos ^{2} \frac{x}{2}$ (vii) $f(x)=\cos \pi x$ (viii) $g(x)=\cos 2 \pi x$ Solution: (i) $y=\cos \left(\mathrm{x}-\frac{\pi}{4}\right)$ $\Rightarrow y-0=\cos \left(x-\frac{\pi}{4}\right)$ ...(i) On shifting the origin at $\left(...
Read More →A tent is in the form of a right circular cylinder surmounted by a cone.
Question: A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent. Solution: It is given that Diameter of cylinder = 24 m there fore radius = daimeter/2 = 24/2= 12 cm Also radius of cone = 12 m Height of cylinder = 11 m Height of cone = 16 - 11 = 5 m Slant height of cone $=\sqrt{5^{2}+12^{2}}$ = 13...
Read More →Fill in the blanks using the correct word given in brackets :
Question: Fill in the blanks using the correct word given in brackets : (i) All circles are ......... (congruent, similar).(ii) All squares are ........(similar, congruent).(iii) All .......... triangles are similar (isosceles, equilateral):(iv) Two triangles are similar, if their corresponding angles are .......... (proportional, equal)(v) Two triangles are similar, if their corresponding sides are ........... (proportional, equal)(vi) Two polygons of the same number of sides are similar, if (a...
Read More →A conical tent is 10 m high and the radius of it base is 24 m.
Question: A conical tent is $10 \mathrm{~m}$ high and the radius of it base is $24 \mathrm{~m}$. Find the slant height of the tent. If the cost of 1 $\mathrm{m}$ canvas is Rs 70 , find the cost of canvas required for the tent. Solution: It is given that Height of the conical tent (h) = 10 m Radius of conical tent (r) = 24 m Let slant height of conical tent be l $\mathrm{I}^{2}=\mathrm{h}^{2}+\mathrm{r}^{2}$ $=10^{2}+24^{2}=100+576$ $=676 \mathrm{~m}^{2}$ ⟹ l = 26 m Thus, the slant height of the ...
Read More →M and N are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether MN || QR.
Question: M and N are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether MN || QR. (i) PM = 4 cm, QM = 4.5 cm, PN = 4 cm, NR = 4.5 cm(ii) PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, PN = 0.32 cm Solution: (1)lt is given that $P M=4 \mathrm{~cm}, Q M=4.5 \mathrm{~cm}, P N=4 \mathrm{~cm}$ and $N R=4.5 \mathrm{~cm}$. We have to check that $M N \| O R$ or not. According to Thales theorem we have $\frac{P M}{Q M}=\frac{P N}{N R}$ $\Rightarrow \frac{4}{4...
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Question: $x d y-y d x=\sqrt{x^{2}+y^{2}} d x$ Solution: $x d y-y d x=\sqrt{x^{2}+y^{2}} d x$\ $\Rightarrow x d y=\left[y+\sqrt{x^{2}+y^{2}}\right] d x$ $\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}$ ...(1) Let $F(x, y)=\frac{y+\sqrt{x^{2}+y^{2}}}{x}$ $\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\sqrt{(\lambda x)^{2}+(\lambda y)^{2}}}{\lambda x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}=\lambda^{0} \cdot F(x, y)$ Therefore, the given differential equation is a homogeneous equation. To solve it, we...
Read More →In Fig. 4.35, state if PQ || EF.
Question: In Fig. 4.35, state if PQ || EF. Solution: It is given thatEP =3 cm,PG= 3.9 cm,FQ =3.6 cm andFQ= 2.4 cm. We have to check that $P Q \| E F$ or not. According to Thales theorem we have PGGE=GQFQ Now, $3.93 \neq 3.62 .4$ Hence, it is not proportional. So, $P Q$ H EF....
Read More →The slant height and base diameter of a conical tomb are 25 m and 14 m respectively.
Question: The slant height and base diameter of a conical tomb are $25 \mathrm{~m}$ and $14 \mathrm{~m}$ respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per $100 \mathrm{~m}^{2}$. Solution: It is given that Slant height of conical tomb (l) = 25 m Base radius (r) of tomb = 14/2 = 7m Curved surface area of conical length tomb = rl = 22/7 7 25 $=550 \mathrm{~m}^{2}$ Cost of white washing $100 \mathrm{~m}^{2}$ area $=$ Rs 210 Cost of white washing $550 \mat...
Read More →Sketch the graphs of the following pairs of functions on the same axes:
Question: Sketch the graphs of the following pairs of functions on the same axes: (i) $f(x)=\sin x, g(x)=\sin \left(x+\frac{\pi}{4}\right)$ (ii) $f(x)=\sin x, g(x)=\sin 2 x$ (iii) $f(x)=\sin 2 x, g(x)=2 \sin x$ (iv) $f(x)=\sin \frac{x}{2}, g(x)=\sin x$ Solution: (i) $f(x)=\sin x, g(x)=\sin \left(x+\frac{\pi}{4}\right)$ Clearly, $\sin x$ and $\sin \left(x+\frac{\pi}{4}\right)$ is a periodic function with period $2 \pi$. The graphs of $f(x)=\sin x$ and $g(x)=\sin \left(x+\frac{\pi}{4}\right)$ on d...
Read More →Sketch the graphs of the following pairs of functions on the same axes:
Question: Sketch the graphs of the following pairs of functions on the same axes: (i) $f(x)=\sin x, g(x)=\sin \left(x+\frac{\pi}{4}\right)$ (ii) $f(x)=\sin x, g(x)=\sin 2 x$ (iii) $f(x)=\sin 2 x, g(x)=2 \sin x$ (iv) $f(x)=\sin \frac{x}{2}, g(x)=\sin x$ Solution: (i) $f(x)=\sin x, g(x)=\sin \left(x+\frac{\pi}{4}\right)$ Clearly, $\sin x$ and $\sin \left(x+\frac{\pi}{4}\right)$ is a periodic function with period $2 \pi$. The graphs of $f(x)=\sin x$ and $g(x)=\sin \left(x+\frac{\pi}{4}\right)$ on d...
Read More →In a ∆ABC, D and E are points on AB and AC respectively such that DE || BC.
Question: In a ∆ABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE. Solution: It is given that $A D=2.4 \mathrm{~cm}, A E=3.2 \mathrm{~cm}, D E=2 \mathrm{~cm}$ and $B C=5 \mathrm{~cm}$. We have to find BD and CE. Since DE∥BC, AB is transversal, thenADE = ABC (corresponding angles)Since DE∥BC, AC is a transversal, thenAED = ACB (corresponding angles)In ∆ADE and ∆ABC,ADE = ABC (proved above)AED = ACB (proved ab...
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Question: $x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$ Solution: The given differential equation is: $x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$ $\frac{d y}{d x}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}$ ...(1) Let $F(x, y)=\frac{x^{2}-2 y^{2}+x y}{x^{2}}$ $\therefore F(\lambda x, \lambda y)=\frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda x)(\lambda y)}{(\lambda x)^{2}}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}=\lambda^{0} \cdot F(x, y)$ Therefore, the given differential equation is a homogeneous equation. To solve it,...
Read More →Curved surface area of a cone is 308 cm2
Question: Curved surface area of a cone is $308 \mathrm{~cm}^{2}$ and its slant height is $14 \mathrm{~cm}$. Find the radius of the base and total surface area of the cone. Solution: (1)It is given that Slant height of cone = 14 cm Let radius of circular end of cone = r Curved surface area of cone = rl $\Rightarrow 308 \mathrm{~cm}^{2}=22 / 7 * \mathrm{r} * 14$ ⟹ r = 308/44 = 7 cm Thus radius of circular end of cone = 7 cm. (ii) It is given that C.S.A $=308 \mathrm{~cm}^{2}$ We know that total s...
Read More →Sketch the graphs of the following functions:
Question: Sketch the graphs of the following functions:(i)f(x) = 2 sinx, 0 x (ii) $g(x)=3 \sin \left(x-\frac{\pi}{4}\right), 0 \leq x \leq \frac{5 \pi}{4}$ (iii) $h(x)=2 \sin 3 x, 0 \leq x \leq \frac{2 \pi}{3}$ (iv) $\phi(x)=2 \sin \left(2 x-\frac{\pi}{3}\right), 0 \leq x \leq \frac{7 \pi}{5}$ (v) $\psi(x)=4 \sin 3\left(x-\frac{\pi}{4}\right), 0 \leq x \leq 2 \pi$ (vi) $\theta(x)=\sin \left(\frac{x}{2}-\frac{\pi}{4}\right), 0 \leq x \leq 4 \pi$ (vii) $u(x)=\sin ^{2} x, 0 \leq x \leq 2 \pi v(x)=|...
Read More →Sketch the graphs of the following functions:
Question: Sketch the graphs of the following functions:(i)f(x) = 2 sinx, 0 x (ii) $g(x)=3 \sin \left(x-\frac{\pi}{4}\right), 0 \leq x \leq \frac{5 \pi}{4}$ (iii) $h(x)=2 \sin 3 x, 0 \leq x \leq \frac{2 \pi}{3}$ (iv) $\phi(x)=2 \sin \left(2 x-\frac{\pi}{3}\right), 0 \leq x \leq \frac{7 \pi}{5}$ (v) $\psi(x)=4 \sin 3\left(x-\frac{\pi}{4}\right), 0 \leq x \leq 2 \pi$ (vi) $\theta(x)=\sin \left(\frac{x}{2}-\frac{\pi}{4}\right), 0 \leq x \leq 4 \pi$ (vii) $u(x)=\sin ^{2} x, 0 \leq x \leq 2 \pi v(x)=|...
Read More →In a ∆ABC, P and Q are points on sides AB and AC respectively,
Question: In a ∆ABC, P and Q are points on sides AB and AC respectively, such that PQ || BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find the AB and PQ Solution: It is given that $A P=2.4 \mathrm{~cm}, A Q=2 \mathrm{~cm}, Q C=3 \mathrm{~cm}$ and $B C=6 \mathrm{~cm}$. We have to find $A B$ and $P Q$. So $\frac{A P}{P B}=\frac{A Q}{Q C}$ (by Thales theorem) Then $\frac{2.4}{P B}=\frac{2}{3}$ $\Rightarrow 2 P B=2.4 \times 3 \mathrm{~cm}$ $\Rightarrow P B=\frac{2.4 \times 3}{2} \mathrm{~...
Read More →The diameters of two cones are equal.
Question: The diameters of two cones are equal. If their slant height are in the ratio 5:4, find the ratio of their curved surfaces. Solution: It is given that Diameters of two cones are equal Therefore their radius are also equal i.e r1= r2 Let the ratio of slant height be x Thereforel1= 5x l2= 4x Therefore Ratio of curved surface area = c1/c2 $\Rightarrow c_{1} / c_{2}=\pi r_{1} l_{1} / \pi r_{2} l_{2}$ $=\frac{\pi r_{1} * 5 x}{\pi r_{2} * 4 x}$ = 5/4 Ratio of curved surface area is 5: 4....
Read More →In a ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC
Question: In a $\triangle A B C, D$ and $E$ are points on the sides $A B$ and $A C$ respectively. For each of the following cases show that $D E \| B C$ : (i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 and AE = 1.8 cm.(iii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm Solution: (i) It is given that $D$ and $E$ are point on sides $A B$ and $A C$. We have to prove thatDE || BC. Accord...
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