Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Question: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. Solution: It is given that: Diameter of the cone (d) = 24 m Radius of the cone(r) = 24/2= 12 m Slant height of the cone (l) = 21 m T. S. A = Curved Surface Area of Cone + Area of Circular Base $=\pi * r * l+\pi * r^{2}$ $=(22 / 7 * 12 * 21)+(22 / 7 * 12 * 12)=22 / 7 * 12(12+21)=1244.57 \mathrm{~m}^{2}$ Therefore the total surface area of the cone is $1244.57 \mathrm{~m}^{2}$....
Read More →If x = sin
Question: Ifx= sin14x+ cos20x, then write the smallest interval in which the value ofxlie. Solution: If $x=0^{\circ}, 90^{\circ}, 180^{\circ}, 270^{\circ}, 360^{\circ}$, then $\sin ^{14} x+\cos ^{20} x$ will always be $1 .$ The smallest interval in which the value of $x$ lie is $(0,1]$....
Read More →In the given figure, RS || DB || PQ. If CP = PD = 11 cm and DR = RA = 3 cm.
Question: In the given figure, RS || DB || PQ. If CP = PD = 11 cm and DR = RA = 3 cm. Then the values ofxandyare respectively. (a) 12, 10(b) 14, 6(c) 10, 7(d) 16, 8 Solution: Given: RS || DB || PQ. CP = PD = 11cm and DR = RA = 3cm To find: the value ofxand y respectively. In∆ASRand∆ABD,ASR=ABQCorrespondinganglesA=ACommon∆ASR~∆ABDAASimilarity $\frac{\mathrm{AR}}{\mathrm{AD}}=\frac{\mathrm{AS}}{\mathrm{AB}}=\frac{\mathrm{RS}}{\mathrm{DB}}$ $\frac{3}{6}=\frac{\text { RS }}{\text { DB }}$ $\frac{1}{...
Read More →Write the least value of cos
Question: Write the least value of cos2x+ sec2x. Solution: We know: cosxcan take the minimum value of1-1. $\cos ^{2} x+\sec ^{2} x$ $=\frac{\cos ^{4} x+1}{\cos ^{2} x}$ $=\frac{(-1)^{4}+1}{(-1)^{2}}$ = 2...
Read More →Write the least value of cos
Question: Write the least value of cos2x+ sec2x. Solution: We know: cosxcan take the minimum value of1-1. $\cos ^{2} x+\sec ^{2} x$$=\frac{\cos ^{4} x+1}{\cos ^{2} x}$ $=\frac{(-1)^{4}+1}{(-1)^{2}}$ = 2...
Read More →Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.
Question: Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm. Solution: It is given that: Base radius of the cone (r) = 5.25 cm Slant height of the cone (l) = 10 cm Curved Surface Area (C. S. A) = rl $=22 / 7^{*} 5.25 * 10=165 \mathrm{~cm}^{2} .$ Therefore the curved surface area of the cone is $165 \mathrm{~cm}^{2}$...
Read More →If tan A + cot A = 4,
Question: If tanA+ cotA= 4, then write the value of tan4A+ cot4A. Solution: $\tan A+\cot A=4$ Squaring both the sides : $\tan ^{2} A+\cot ^{2} A+2=16$ $\Rightarrow \tan ^{2} A+\cot ^{2} A=14$ Squaring both the sides again: $\tan ^{4} A+\cot ^{4} A+2=196$ $\Rightarrow \tan ^{4} A+\cot ^{4} A=194$...
Read More →The general solution of the differential equation
Question: The general solution of the differential equation $\frac{d y}{d x}=e^{x+y}$ is A. $e^{x}+e^{-y}=\mathrm{C}$ B. $e^{x}+e^{y}=\mathrm{C}$ C. $e^{-x}+e^{y}=\mathrm{C}$ D. $e^{-x}+e^{-y}=\mathrm{C}$ Solution: $\frac{d y}{d x}=e^{x+y}=e^{x} \cdot e^{y}$ $\Rightarrow \frac{d y}{e^{y}}=e^{x} d x$ $\Rightarrow e^{-y} d y=e^{x} d x$ Integrating both sides, we get: $\int e^{-y} d y=\int e^{x} d x$ $\Rightarrow-e^{-y}=e^{x}+k$ $\Rightarrow e^{x}+e^{-y}=-k$ $\Rightarrow e^{x}+e^{-y}=c$ $(c=-k)$ He...
Read More →In the given figure, if ∠ADE = ∠ABC, then CE =
Question: In the given figure, if ADE = ABC, then CE = (a) 2(b) 5(c) 9/2(d) 3 Solution: Given: $\angle \mathrm{ADE}=\angle \mathrm{ABC}$ To find: The value of CE Since $\angle \mathrm{ADE}=\angle \mathrm{ABC}$ $\therefore \mathrm{DE} \| \mathrm{BC}$ (Two lines are parallel if the corresponding angles formed are equal) According to basic proportionality theorem if a line is parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. In ∆AB...
Read More →Find the total surface area of a right circular cone with radius 6 cm and height 8 cm.
Question: Find the total surface area of a right circular cone with radius 6 cm and height 8 cm. Solution: It is given that: Radius of the cone (r) = 6 cm Height of the cone (h) = 8 cm Total Surface Area of the Cone (T. S. A) =? $l^{2}=r^{2}+h^{2}$ $=6^{2}+8^{2}$ = 36 + 64 = 100 l = 10 cm T. S. A = Curved Surface Area of Cone + Area of Circular Base $=\pi * r * 1+\pi * r^{2}$ $=(22 / 7 * 6 * 10)+(22 / 7 * 6 * 6)$ $=\frac{1320+792}{7}$ $=301.171 \mathrm{~cm}^{2}$ Therefore the area of the base is...
Read More →If cot (α + β) = 0,
Question: If cot ( + ) = 0, then write the value of sin ( + 2). Solution: $\cot (\alpha+\beta)=0$ $\Rightarrow \alpha+\beta=\frac{\pi}{2}$ (1) $\beta=\frac{\pi}{2}-\alpha$ (2) $\alpha=\frac{\pi}{2}-\beta$ (3) Now, $\sin (\alpha+2 \beta)=\sin (\alpha+\beta+\beta)$ $=\sin \left(\frac{\pi}{2}+\frac{\pi}{2}-\alpha\right)$ $=\sin (\pi-\alpha)$ $=\sin \alpha$ Now, $\sin (\alpha+2 \beta)=\sin (\alpha+2 \beta)$ $=\sin \left(\frac{\pi}{2}-\beta+2 \beta\right)$ $=\sin \left(\frac{\pi}{2}+\beta\right)$ $=\...
Read More →In the given figure, the value of x for which DE || AB is
Question: In the given figure, the value ofxfor which DE || AB is (a) 4(b) 1(c) 3(d) 2 Solution: Given: In∆ABC, DE || AB. To find: the value ofx According to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. In ∆ABC, DE || AB $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ $\frac{x+3}{3 x+19}=\frac{x}{3 x+4}$ $(x+3)(3 x+4)=(x)(3 x+19)$ $3 x^{2}+4 x+9 x+12=3 x^{2}...
Read More →Write the value of cos
Question: Write the value of cos 1 + cos 2 + cos 3 + ... + cos 180. Solution: $\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots+\cos 180^{\circ}$ $=\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots+\cos 88^{\circ}+\cos 89^{\circ}+\cos 90^{\circ}+\cos (180-89)^{\circ}+\cos (180-88)^{\circ}+\ldots$$+\cos (180-1)^{\circ}+\cos 180^{\circ} \quad\left[\cos \left(180^{\circ}-\theta\right)=-\cos \theta\right]$ $=\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots+\cos 88^{\circ}+\cos 89^{\circ}+\co...
Read More →In a culture, the bacteria count is 1,00,000.
Question: In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present? Solution: Letybe the number of bacteria at any instantt. It is given that the rate of growth of the bacteria is proportional to the number present. $\therefore \frac{d y}{d t} \propto y$ $\Rightarrow \frac{d y}{d t}=k y$ (where $k$ is a constant) $\Rightarrow \frac{d y}{y}=k d t...
Read More →Write the value of 2
Question: Write the value of 2 (sin6x+ cos6x) 3 (sin4x+ cos4x) + 1. Solution: $2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$ $=2\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cdot \cos ^{2} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$ $=2 \cdot 1\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cdot \cos ^{2} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$ $=2\left(\sin ^{4} x+\cos ^{4} x\right)-2 \sin ^{2} x \cdot \cos ^{2} x-3\le...
Read More →The height of a cone 21 cm.
Question: The height of a cone 21 cm. Find the area of the base if the slant height is 28 cm. Solution: It is given that: Height of the traffic cone (h) = 21 cm Slant height of the traffic cone (l) = 28 cm Now we know that, $\mathrm{I}^{2}=\mathrm{r}^{2}+\mathrm{h}^{2}$ $28^{2}=r^{2}+21^{2}$ $r^{2}=28^{2}-21^{2}$ $r=\sqrt{77} \mathrm{~cm}$ Area of the circular base $=\pi r^{2}$ $=\frac{22}{7} *(7 \sqrt{7})^{2}=1078 \mathrm{~cm}^{2}$ Therefore the area of the base is $1078 \mathrm{~cm}^{2}$...
Read More →Write the value of 2
Question: Write the value of 2 (sin6x+ cos6x) 3 (sin4x+ cos4x) + 1. Solution: $2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$ $=2\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cdot \cos ^{2} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$ $=2 \cdot 1\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cdot \cos ^{2} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$ $=2\left(\sin ^{4} x+\cos ^{4} x\right)-2 \sin ^{2} x \cdot \cos ^{2} x-3\le...
Read More →In the given figure the measure of ∠D and ∠F are respectively
Question: In the given figure the measure of $\angle \mathrm{D}$ and $\angle \mathrm{F}$ are respectively (a) $50^{\circ}, 40^{\circ}$ (b) $20^{\circ}, 30^{\circ}$ (c) $40^{\circ}, 50^{\circ}$ (d) $30^{\circ}, 20^{\circ}$ Solution: $\mathrm{ABAC}=\mathrm{EFED} \angle \mathrm{A}=\angle \mathrm{E}=130^{\circ}$ $\triangle \mathrm{ABC} \sim \triangle \mathrm{EFD} \quad$ (SAS Similarity) $\triangle \mathrm{ABC} \sim \triangle \mathrm{EFD}$ $\therefore \angle \mathrm{F}=\angle \mathrm{B}=30^{\circ} \a...
Read More →A circular wire of radius 15 cm is cut and bent so as to lie along the circumference of a loop of radius 120 cm.
Question: A circular wire of radius 15 cm is cut and bent so as to lie along the circumference of a loop of radius 120 cm. Write the measure of the angle subtended by it at the centre of the loop. Solution: Circumference of the circle of radius 15 cm: $2 \pi \mathrm{r}$ $=2 \times 3.14 \times 15 \mathrm{~cm}$ $=94.2 \mathrm{~cm}$ Now, 94.2 cm will be the length of arc(l)lfor the circle with radius 120 cm. We know: $l=r \theta$ Here, $\theta$ is measured in radians. $\therefore 94.2=120 \times \t...
Read More →The radius of a cone is 7 cm and area of curved surface is 176 cm2
Question: The radius of a cone is $7 \mathrm{~cm}$ and area of curved surface is $176 \mathrm{~cm}^{2}$. Find the slant height. Solution: It is given that: Radius of cone(r) = 7 cm Curved Surface Area (C. S. A) $=176 \mathrm{~cm}^{2}$ Slant Height of tent (l) = ? Now we know, C.S.A = rl ⟹ rl= 176 ⟹ 22/7 7 l = 176 $\Rightarrow \mathrm{l}=\frac{176 * 7}{22 * 7}=8 \mathrm{~cm}$ Therefore the slant height of the cone is 8 cm....
Read More →In an isosceles triangle ABC,
Question: In an isosceles triangle ABC, if AB = AC = 25 cm and BC = 14 cm, the measure of altitude from A on BC is (a) 20 cm(b) 22 cm(c) 18 cm(d) 24 cm Solution: Given: In an isosceles triangle ABC, AB = AC = 25 cm and BC = 14cm. To find: Measure of altitude from A on BC Since AD BC, so BD = CD =BC2=14cm2=7cmIn right∆ABD,AB2=AD2+BD2⇒AD2=AB2-BD2⇒AD2=252-72=625-49=576⇒AD=24cm We got the result as $(d)$...
Read More →Write the value of sin
Question: Write the value of sin 10 + sin 20 + sin 30 + ... + sin 360. Solution: $\sin 10^{\circ}+\sin 20^{\circ}+\ldots+\sin 170^{\circ}+\sin 180^{\circ}+\sin \left(360^{\circ}-170^{\circ}\right)+\sin \left(360^{\circ}-160^{\circ}\right)+\ldots$$+\sin \left(360^{\circ}-20^{\circ}\right)+\sin \left(360^{\circ}-10^{\circ}\right)+\sin 360^{\circ}$ $=\sin 10^{\circ}+\sin 20^{\circ}+\ldots+\sin 180^{\circ}-\sin 170^{\circ}-\sin 160^{\circ}-\ldots-\sin 20^{\circ}-\sin 10^{\circ}+\sin 360^{\circ}$ $\l...
Read More →In a bank, principal increases continuously at the rate of $5 %$ per year.
Question: In a bank, principal increases continuously at the rate of $5 \%$ per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years $\left(e^{0.5}=1.648\right)$. Solution: Letpandtbe the principal and time respectively. It is given that the principal increases continuously at the rate of 5% per year. $\Rightarrow \frac{d p}{d t}=\left(\frac{5}{100}\right) p$ $\Rightarrow \frac{d p}{d t}=\frac{p}{20}$ $\Rightarrow \frac{d p}{p}=\frac{d t}{20}$ Integrating...
Read More →If in two triangle ABC and DEF, ∠A = ∠E, ∠B = ∠F, then which of the following is not true?
Question: If in two triangle $\mathrm{ABC}$ and $\mathrm{DEF}, \angle \mathrm{A}=\angle \mathrm{E}, \angle \mathrm{B}=\angle \mathrm{F}$, then which of the following is not true? (a)BCDF=ACDE(b)ABDE=BCDF(c)ABEF=ACDE(d)BCDF=ABEF Solution: In ΔABC and ΔDEF $\angle \mathrm{A}=\angle \mathrm{E}$ $\angle \mathrm{B}=\angle \mathrm{F}$ $\therefore \triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are similar triangles. Hence $\frac{\mathrm{AB}}{\mathrm{EF}}=\frac{\mathrm{BC}}{\mathrm{FD}}=\frac{\mat...
Read More →The radius of a cone is 5cm and vertical height is 12 cm.
Question: The radius of a cone is 5cm and vertical height is 12 cm. Find the area of the curved surface. Solution: It is given that: Radius of cone (r) = 5 cm Height of the tent (h) = 12 cm Slant Height of tent (l) = ? Curved Surface Area (C. S. A) = ? Now we know we that, $I=r^{2}+h^{2}$ $\mathrm{l}^{2}=\mathrm{r}^{2}+\mathrm{h}^{2}$ $I^{2}=5^{2}+12^{2}$ $1^{2}=25+144$ ⟹ l = 13 cm Now, C.S.A =rl = 3.14 * 5 * 12 $=204.28 \mathrm{~cm}^{2}$ Therefore the curved surface area of the cone is $204.28 ...
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