The ratio between the radius of the base and the height of a cylinder is 2 : 3.
Question: The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm3, the total surface area of the cylinder is(a) 308 cm2(b) 462 cm2(c) 540 cm2(d) 770 cm2 Solution: (d) 770 cm2Let the common multiple be x.Let the radius of the cylinder be 2xcm and its height be 3xcm. Then, volume of the cylinder $=\pi r^{2} h$ $=\frac{22}{7} \times(2 x)^{2} \times 3 x$ Therefore, $\frac{22}{7} \times(2 x)^{2} \times 3 x=1617$ $\Rightarrow \frac{22}{7} \times 4 x^{2...
Read More →The sum of the interior angles of a polygon is three times the sum of its exterior angles.
Question: The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sided of the polygon. Solution: $\left\{(2 \mathrm{n}-4) \times 90^{\circ}\right\}=3 \times\left(\frac{360^{\circ}}{\mathrm{n}} \times \mathrm{n}\right)$ $\Rightarrow(\mathrm{n}-2) \times 180=3 \times 360$ $\Rightarrow \mathrm{n}-2=6 $\therefore \mathrm{n}=8$...
Read More →For the pair of equations λx + 3y + 7 = 0
Question: For the pair of equations x + 3y + 7 = 0 and 2x + 6y -14 = 0. To have infinitely many solutions, the value of should be 1. Is the statement true? Give reasons. Solution: No, the given pair of linear equations x + 3y+7 = 0 and 2x + 6y-14 = 0 Here, a1= , b1= 3 c1=7; a2=2, b2= 6,c2= -14 If $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$, then system has infinitely many solutions. $\Rightarrow$$\frac{\lambda}{2}=\frac{3}{6}=-\frac{7}{14}$ $\because$ $\frac{\lambda}{2}=\frac{3...
Read More →In a convex hexagon, prove that the sum of all interior angle is equal to twice the sum of
Question: In a convex hexagon, prove that the sum of all interior angle is equal to twice the sum of its exterior angles formed by producing the sides in the same order. Solution: For a convex hexagon, interior angle $=\left(\frac{2 n-4}{n} \times 90^{\circ}\right)$ For a hexagon, $n=6$ $\therefore$ Interior angle $=\left(\frac{12-4}{6} \times 90^{\circ}\right)$ $=\left(\frac{8}{6} \times 90^{\circ}\right)$ $=120^{\circ}$ So, the sum of all the interior angles $=120^{\circ}+120^{\circ}+120^{\cir...
Read More →The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3.
Question: The curved surface area of a cylindrical pillar is 264 m2and its volume is 924 m3. The height of the pillar is(a) 4 m(b) 5 m(c) 6 m(d) 7 m Solution: (c) 6 mThe curved surface area of a cylindrical pillar $=2 \pi r h$ Therefore, $2 \pi r h=264$ Volume of a cylinder $=\pi r^{2} h$ Therefore, $\pi r^{2} h=924$ Hence, $\frac{\pi r^{2} h}{2 \pi r h}=\frac{924}{264}$ $\Rightarrow \frac{r}{2}=\frac{924}{264}$ $\Rightarrow r=\left(\frac{924 \times 2}{264}\right)$ $\Rightarrow r=7 \mathrm{~m}$ ...
Read More →The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3.
Question: The curved surface area of a cylindrical pillar is 264 m2and its volume is 924 m3. The height of the pillar is(a) 4 m(b) 5 m(c) 6 m(d) 7 m Solution: (c) 6 mThe curved surface area of a cylindrical pillar $=2 \pi r h$ Therefore, $2 \pi r h=264$ Volume of a cylinder $=\pi r^{2} h$ Therefore, $\pi r^{2} h=924$ Hence, $\frac{\pi r^{2} h}{2 \pi r h}=\frac{924}{264}$ $\Rightarrow \frac{r}{2}=\frac{924}{264}$ $\Rightarrow r=\left(\frac{924 \times 2}{264}\right)$ $\Rightarrow r=7 \mathrm{~m}$ ...
Read More →The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3.
Question: The curved surface area of a cylindrical pillar is 264 m2and its volume is 924 m3. The height of the pillar is(a) 4 m(b) 5 m(c) 6 m(d) 7 m Solution: (c) 6 mThe curved surface area of a cylindrical pillar $=2 \pi r h$ Therefore, $2 \pi r h=264$ Volume of a cylinder $=\pi r^{2} h$ Therefore, $\pi r^{2} h=924$ Hence, $\frac{\pi r^{2} h}{2 \pi r h}=\frac{924}{264}$ $\Rightarrow \frac{r}{2}=\frac{924}{264}$ $\Rightarrow r=\left(\frac{924 \times 2}{264}\right)$ $\Rightarrow r=7 \mathrm{~m}$ ...
Read More →Are the following pair of linear
Question: Are the following pair of linear equations consistent? Justify your answer, (i) $-3 x-4 y=12$ and $4 y+3 x=12$ (ii) $\frac{3}{5} x-y=\frac{1}{2}$ and $\frac{1}{5} x-3 y=\frac{1}{6}$ (iii) $2 a x+b y=a$ and $4 a x+2 b y-2 a=0 ; a, b \neq 0$ (iv) $x+3 y=11$ and $2(2 x+6 y)=22$ Solution: Conditions for pair of linear equations are consistent $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ [unique solution] and $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ [infinitely many s...
Read More →The ratio of the total surface area to the lateral surface area of a cylinder with base radius 80 cm and height 20 cm is
Question: The ratio of the total surface area to the lateral surface area of a cylinder with base radius 80 cm and height 20 cm is(a) 2 : 1(b) 3 : 1(c) 4 : 1(d) 5 : 1 Solution: (d) 5 : 1 Ratio of the total surface area to the lateral surface area $=\frac{\text { Total surface area }}{\text { Lateral surface area }}$ $=\frac{2 \pi r(h+r)}{2 \pi r h}$ $=\frac{h+r}{h}$ $=\frac{(20+80)}{20}$ $=\frac{100}{20}$ $=\frac{5}{1}$ $=5: 1$ Hence, the required ratio is 5:1....
Read More →The measure of angles of a hexagon are x°,
Question: The measure of angles of a hexagon arex, (x 5), (x 5), (2x 5), (2x 5), (2x+ 20). Find the value ofx. Solution: Since the sum of all the angles of a hexagon is $720^{\circ}$, we get: $x^{\circ}+(x-5)^{\circ}+(x-5)^{\circ}+(2 x-5)^{\circ}+(2 x-5)^{\circ}+(2 x+20)^{\circ}=720^{\circ}$ $\Rightarrow x^{\circ}+x^{\circ}-5^{\circ}+x^{\circ}-5^{\circ}+2 x-5^{\circ}+2 x-5^{\circ}+2 x+20^{\circ}=720^{\circ}$ $\Rightarrow 9 x-20^{\circ}+20^{\circ}=720^{\circ}$ $\Rightarrow 9 x=720^{\circ}$ $\ther...
Read More →Find the number of degrees in each exterior
Question: Find the number of degrees in each exterior angle of a regular pentagon. Solution: Each exterior angle $=\left(\frac{360}{\mathrm{n}}\right)^{\circ}$ For a regular pentagon, $n=5$. $\therefore$ Exterior angle $=\left(\frac{360}{5}\right)^{\circ}=72^{\circ}$...
Read More →The curved surface are of a cylinder is 1760 cm2
Question: The curved surface are of a cylinder is 1760 cm2and its base radius is 14 cm. The height of the cylinder is(a) 10 cm(b) 15 cm(c) 20 cm(d) 40 cm Solution: (c) 20 cm Curved surface area of the cylinder $=2 \pi r h$ $=2 \times \frac{22}{7} \times 14 \times h$ Therefore, $2 \times \frac{22}{7} \times 14 \times h=1760$ $\Rightarrow h=\frac{1760}{88} \mathrm{~cm}$ $\Rightarrow h=20 \mathrm{~cm}$ Hence, the height of the cylinder is 20 cm....
Read More →Find the number of sides of a regular polygon, when each of its angles has a measure of
Question: Find the number of sides of a regular polygon, when each of its angles has a measure of (i) 160 (ii) 135 (iii) 175 (iv) 162 (v) 150 Solution: (i) Each interior angle $=\left(\frac{2 n-4}{n} \times 90\right)^{\circ}$ So, $\left(\frac{2 n-4}{n} \times 90\right)^{\circ}=160^{\circ}$ $\Rightarrow \frac{2 n-4}{n}=\frac{160^{\circ}}{90^{\circ}}$ $\Rightarrow \frac{2 n-4}{n}=\frac{16}{9}$ $\Rightarrow 18 n-36=16 n$ $\Rightarrow 2 n=36$ $\therefore n=18$ (ii) Each interior angle $=\left(\frac{...
Read More →The height of a cylinder is 14 cm and its curved surface area is 264 cm2.
Question: The height of a cylinder is 14 cm and its curved surface area is 264 cm2. The volume of the cylinder is(a) 308 cm3(b) 396 cm3(c) 1232 cm3(d) 1848 cm3 Solution: (b) 396 cm3 Curved surface area of the cylinder $=2 \pi r h=2 \times \frac{22}{7} \times r \times 14$ Therefore, $2 \times \frac{22}{7} \times r \times 14=264$ $\Rightarrow r=\frac{264}{88}$ $\Rightarrow r=3 \mathrm{~cm}$ Hence, the volume of the cylinder $=\pi r^{2} h$ $=\left(\frac{22}{7} \times 3 \times 3 \times 14\right) \ma...
Read More →In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively.
Question: In a quadrilateral $A B C D, C O$ and $D O$ are the bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle C O D=\frac{1}{2}(\angle A+\angle B)$. Solution: $\angle \mathrm{COD}=180^{\circ}-(\angle \mathrm{OCD}+\angle \mathrm{ODC})$ $=180^{\circ}-\frac{1}{2}(\angle \mathrm{C}+\angle \mathrm{D})$ $=180^{\circ}-\frac{1}{2}\left[360^{\circ}-(\angle \mathrm{A}+\angle \mathrm{B})\right]$ $=180^{\circ}-180^{\circ}+\frac{1}{2}(\angle \mathrm{A}+\angle \mathrm{B})$ $=\frac{1}{2...
Read More →The diameter of a cylinder is 28 cm and its height is 20 cm.
Question: The diameter of a cylinder is 28 cm and its height is 20 cm. The total surface area of the cylinder is(a) 2993 cm2(b) 2992 cm2(c) 2292 cm2(d) 2229 cm2 Solution: (b) 2992 cm2 The total surface area of the cylinder $=2 \pi r+2 r^{2}$ $=2 \pi r(h+r)$ $=\left[2 \times \frac{22}{7} \times 14 \times(20+14)\right] \mathrm{cm}^{2} \quad\left[d=28 \mathrm{~cm} \Rightarrow r=\frac{28}{2} \mathrm{~cm}=14 \mathrm{~cm}\right]$ $=\frac{20944}{7} \mathrm{~cm}^{2}$ $=2992 \mathrm{~cm}^{2}$...
Read More →In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5.
Question: In a quadrilateralABCD, the anglesA, B, CandDare in the ratio 1 : 2 : 4 : 5. Find the measure of each angle of the quadrilateral. Solution: Let $x$ be the measure of each angle. Then the ratio becomes $x: 2 x: 4 x: 5 x$. Since, the sum of all angles in a quadrilateral is $360^{\circ}$, we have: $x+2 x+4 x+5 x=360^{\circ}$ $\Rightarrow 12 x=360^{\circ}$ $\Rightarrow x=\frac{360^{\circ}}{12}$ $\Rightarrow x=30^{\circ}$ Thus, the angles are : $x=30^{\circ}$ $2 x=60^{\circ}$ $4 x=120^{\cir...
Read More →The diameter of the base of a cylinder is 4 cm and its height is 14 cm.
Question: The diameter of the base of a cylinder is 4 cm and its height is 14 cm. The volume of the cylinder is(a) 176 cm3(b) 196 cm3(c) 276 cm3(d) 352 cm3 Solution: (a) 176 cm3 Volume of the cylinder $=\pi r^{2} h$ $=\left(\frac{22}{7} \times 2 \times 2 \times 14\right) \quad\left[d=4 \mathrm{~cm} \Rightarrow r=\frac{4}{2} \mathrm{~cm}=2 \mathrm{~cm}\right]$ $=176 \mathrm{~cm}^{3}$...
Read More →Do the following equations represent
Question: Do the following equations represent a pair of coincident lines? Justify your answer. (i) $3 x+\frac{1}{7} y=3$ and $7 x+3 y=7$ (ii) $-2 x-3 y=1$ and $6 y+4 x=-2$ (iii) $\frac{x}{2}+y+\frac{2}{5}=0$ and $4 x+8 y+\frac{5}{16}=0$ Solution: Condition for coincident lines, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ (i) No, given pair of linear equations $3 x+\frac{y}{7}-3=0$ and $\quad 7 x+3 y-7=0$, where, $\quad a_{1}=3, b_{1}=\frac{1}{7}, c_{1}=-3$; $a_{2}=7, b_{2}=3, ...
Read More →In Fig. 16.21, the bisectors of ∠A and ∠B meet at a point P.
Question: In Fig. 16.21, the bisectors ofAandBmeet at a pointP. IfC= 100 andD= 50, find the measure ofAPB. Solution: $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ}$ $\Rightarrow \angle \mathrm{A}+\angle \mathrm{B}+100^{\circ}+50^{\circ}=360^{\circ}$ $\Rightarrow \angle \mathrm{A}+\angle \mathrm{B}=210^{\circ} \quad \ldots(\mathrm{i})$ In $\triangle \mathrm{APB}$, we have $:$ $\frac{1}{2} \angle \mathrm{A}+\frac{1}{2} \angle \mathrm{B}+\angle \mathrm{APB}=180...
Read More →The volumes of two cubes are in the ratio 1 : 27.
Question: The volumes of two cubes are in the ratio 1 : 27. The ratio of their surface area is(a) 1 : 3(b) 1 : 8(c) 1 : 9(d) 1 : 18 Solution: (c) 1 : 9Let the edges of the two cubes beaandb. Then, ratio of their volumes $=\frac{a^{3}}{b^{3}}$ Therefore, $\frac{a^{3}}{b^{3}}=\frac{1}{27}$ $\Rightarrow\left(\frac{a}{b}\right)^{3}=\left(\frac{1}{3}\right)^{3}$ $\Rightarrow \frac{a}{b}=\frac{1}{3}$ The ratio of their surface areas $=\frac{6 a^{2}}{6 b^{2}}$ Therefore, $\frac{6 a^{2}}{6 b^{2}}=\frac{...
Read More →The sides of a quadrilateral are produced in order.
Question: The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles? Solution: The sides of the quadrilateral $\mathrm{ABCD}$ are produced in order (according to figure). Now, we need to find the sum of the exterior angles. Since the angles made on the same side of straight line are $180^{\circ}$, i.e., linear pair, we have: $\mathrm{a}+\mathrm{x}+\mathrm{b}+\mathrm{y}+\mathrm{c}+\mathrm{z}+\mathrm{w}+\mathrm{d}=180^{\circ}+180^{\circ}+180^{\circ}+180^{\circ...
Read More →Rainfall in an area is 5 cm. The volume of the water that falls on 2 hectares of land is
Question: Rainfall in an area is 5 cm. The volume of the water that falls on 2 hectares of land is(a) 100 m3(b) 10 m3(c) 1000 m3(d) 10000 m3 Solution: (c) 1000 m3 Volume of water that falls on 2 hectares of land $=($ Area of the ground $\times$ Amount of rain in $\mathrm{m})$ $=\left(2 \times 1000 \times \frac{5}{100}\right) \mathrm{m}^{3} \quad\left(\because 5 \mathrm{~cm}=\frac{5}{100} \mathrm{~m}, 2\right.$ heactares $\left.=2 \times 1000 \mathrm{~m}^{2}\right)$ $=1000 \mathrm{~m}^{3}$...
Read More →In Fig. 16.20, find the measure of ∠MPN.
Question: In Fig. 16.20, find the measure ofMPN. Solution: Since the sum of all the angles of a quadrilateral is $360^{\circ}$, we have: $45^{\circ}+90^{\circ}+90^{\circ}+\angle \mathrm{MPN}=360^{\circ}$ $\Rightarrow 225^{\circ}+\angle \mathrm{MPN}=360^{\circ}$ $\therefore \angle \mathrm{MPN}=135^{\circ}$...
Read More →If the sum of the two angles of a quadrilateral is 180°.
Question: If the sum of the two angles of a quadrilateral is 180. What is the sum of the remaining two angles? Solution: Let $(x+y)$ be the sum of the remaining two angles. Since, the sum of all the angles of a quadrilateral is $360^{\circ}$, we have : $180^{\circ}+(x+y)^{\circ}=360^{\circ}$ $\Rightarrow(x+y)^{\circ}=180^{\circ}$ $\therefore$ The sum of the remaining two angles is $180^{\circ}$....
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