The surface areas of a sphere and a cube are equal.
Question: The surface areas of a sphere and a cube are equal. Find the ratio of their volumes. Solution: Surface area of the sphere $=4 \pi r^{2}$ Surface area of the cube $=6 a^{2}$ Therefore, $4 \pi r^{2}=6 a^{2}$ $\Rightarrow 2 \pi r^{2}=3 a^{2}$ $\Rightarrow r^{2}=\frac{3 a^{2}}{2 \pi}$ $\Rightarrow r=\sqrt{\frac{3}{2 \pi}} a$ Ratio of their volumes $=\frac{4 / 3 \pi r^{3}}{a^{3}}=\frac{4 \pi r^{3}}{3 a^{3}}$ $=\frac{4 \pi}{3 a^{3}} \times \frac{3 \sqrt{3} a^{3}}{2 \pi \sqrt{2 \pi}} \quad\le...
Read More →The surface areas of a sphere and a cube are equal.
Question: The surface areas of a sphere and a cube are equal. Find the ratio of their volumes. Solution: Surface area of the sphere $=4 \pi r^{2}$ Surface area of the cube $=6 a^{2}$ Therefore, $4 \pi r^{2}=6 a^{2}$ $\Rightarrow 2 \pi r^{2}=3 a^{2}$ $\Rightarrow r^{2}=\frac{3 a^{2}}{2 \pi}$ $\Rightarrow r=\sqrt{\frac{3}{2 \pi}} a$ Ratio of their volumes $=\frac{4 / 3 \pi r^{3}}{a^{3}}=\frac{4 \pi r^{3}}{3 a^{3}}$ $=\frac{4 \pi}{3 a^{3}} \times \frac{3 \sqrt{3} a^{3}}{2 \pi \sqrt{2 \pi}} \quad\le...
Read More →onstruct a quadrilateral ABCD given that AB = 4 cm,
Question: onstruct a quadrilateralABCDgiven thatAB= 4 cm,BC= 3 cm, A= 75, B= 80 and C= 120. Solution: Steps of construction: Step I: Draw $\mathrm{AB}=4 \mathrm{~cm} .$ Step II : Construct $\angle \mathrm{XAB}=75^{\circ}$ at $\mathrm{A}$ and $\angle \mathrm{ABY}=80^{\circ}$ at B. Step III : With B as the centre and radius $3 \mathrm{~cm}$, cut off BC $=3 \mathrm{~cm} .$ Step IV : At C, draw $\angle \mathrm{BCZ}=120^{\circ}$ such that it meets AX at D. The quadrilateral so obtained is the require...
Read More →If the angles of a triangle are x,
Question: If the angles of a triangle are x, y and 40 and the difference between the two angles x and y is 30. Then, find the value of x and y, Solution: Given that, x, y and 40 are the angles of a triangle. x + y + 40 = 180 [since, the sum of all the angles of a triangle is 180] $\Rightarrow \quad x+y=140^{\circ}$$\ldots$ (i) Also, $x-y=30^{\circ}$....(ii) On adding Eqs. (i) and (ii), we get $2 x=170^{\circ}$ $\Rightarrow \quad x=85^{\circ}$ On putting $x=85^{\circ}$ in Eq. (i), we get $85^{\ci...
Read More →Construct a quadrilateral BDEF, where DE = 4.5 cm,
Question: Construct a quadrilateralBDEF, whereDE= 4.5 cm,EF= 3.5 cm,FB= 6.5 cm, F= 50 and E= 100. Solution: Steps of construction: Step I : Draw $\mathrm{EF}=3.5 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{DEF}=100^{\circ}$ at $\mathrm{E}$. Step III : With $\mathrm{E}$ as the centre and radius $4.5 \mathrm{~cm}$, cut off $\mathrm{DE}=4.5 \mathrm{~cm}$. Step IV : Construct $\angle \mathrm{EFB}=50^{\circ}$ at F. Step V : With F as the centre and radius $6.5 \mathrm{~cm}$, cut off $\mathrm{F...
Read More →A hemispherical bowl of internal radius 9 cm is full of water.
Question: A hemispherical bowl of internal radius 9 cm is full of water. This water is to be filled in cylindrical bottles of diameter 3 cm and height 4 cm. Find the number of bottles needed in which the water can be filled. Solution: Radius of hemisphere = 9 cm. Volume of hemisphere $=\frac{2}{3} \pi \mathrm{r}^{3}$ $=\left(\frac{2}{3} \pi \times 9 \times 9 \times 9\right) \mathrm{cm}^{3}$ Radius of each bottle $=\frac{3}{2} \mathrm{~cm}$ Height of each bottle $=4 \mathrm{~cm}$ Volume of each b...
Read More →A hemispherical bowl of internal radius 9 cm is full of water.
Question: A hemispherical bowl of internal radius 9 cm is full of water. This water is to be filled in cylindrical bottles of diameter 3 cm and height 4 cm. Find the number of bottles needed in which the water can be filled. Solution: Radius of hemisphere = 9 cm. Volume of hemisphere $=\frac{2}{3} \pi \mathrm{r}^{3}$ $=\left(\frac{2}{3} \pi \times 9 \times 9 \times 9\right) \mathrm{cm}^{3}$ Radius of each bottle $=\frac{3}{2} \mathrm{~cm}$ Height of each bottle $=4 \mathrm{~cm}$ Volume of each b...
Read More →Construct a quadrilateral ABCD, in which AB = BC = 3 cm,
Question: Construct a quadrilateralABCD, in whichAB=BC= 3 cm,AD= 5 cm, A= 90 and B= 105. Solution: Steps of construction: Step I : Draw $\mathrm{AB}=3 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{DAB}=90^{\circ}$ at A. Step III : With A as the centre and radius $5 \mathrm{~cm}$, cut off $\mathrm{AD}=5 \mathrm{~cm}$. Step IV : Construct $\angle \mathrm{ABC}=105^{\circ}$ at B. Step V : With B as the centre and radius $3 \mathrm{~cm}$, cut off BC $=3 \mathrm{~cm}$. Step VI : Join CD. The quad...
Read More →If (x + 1) is a factor of 2x3 + ax2 + 2bx+l,
Question: If (x + 1) is a factor of 2x3+ ax2+ 2bx+l, then find the value of a and b given that 2a 3b = 4. Solution: Given that, (x + 1) is a factor of f(x) = 2xs+ ax2+ 2bx + 1, then f(-1) = 0. [if (x + a) is a factor of f(x) = ax2+ bx + c, then f(-) = 0] $\Rightarrow \quad 2(-1)^{3}+a(-1)^{2}+2 b(-1)+1=0$ $\Rightarrow \quad-2+a-2 b+1=0$ $\Rightarrow \quad a-2 b-1=0$$\ldots($ i) Also, $2 a-3 b=4$ $\Rightarrow$ $3 b=2 a-4$ $\Rightarrow$ $b=\left(\frac{2 a-4}{3}\right)$ Now, put the value of $b$ in...
Read More →Construct a quadrilateral PQRS, in which ∠Q = 45°,
Question: Construct a quadrilateralPQRS, in which Q = 45, R= 90,QR= 5 cm,PQ= 9 cm and Rs = 7 cm. Solution: Steps of construction: Step I : Draw QR $=5 \mathrm{~cm} .$ Step II : Construct $\angle \mathrm{PQR}=45^{\circ}$ at Q. Step III : With Q as the centre and radius $9 \mathrm{~cm}$, cut off QP $=9 \mathrm{~cm} .$ Step IV : Construct $\angle \mathrm{QRS}=90^{\circ}$ at R. Step V : With R as the centre and radius $7 \mathrm{~cm}$, cut off RS $=7 \mathrm{~cm} .$ Since, the line segment PQ and RS...
Read More →A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm.
Question: A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy. Solution: Radius of the hemisphere = Radius of the cone = 7 cm Height of the cone $=(31-7) \mathrm{cm}=24 \mathrm{~cm}$ Slant height of the cone, $l=\sqrt{r^{2}+h^{2}}$ $=\sqrt{(7)^{2}+(24)^{2}}$ $=\sqrt{49+576}$ $=\sqrt{625}$ $=25 \mathrm{~cm}$ Total surface area of the toy = (Curved surface area of the hemisphere) + (Curved s...
Read More →A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm.
Question: A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy. Solution: Radius of the hemisphere = Radius of the cone = 7 cm Height of the cone $=(31-7) \mathrm{cm}=24 \mathrm{~cm}$ Slant height of the cone, $l=\sqrt{r^{2}+h^{2}}$ $=\sqrt{(7)^{2}+(24)^{2}}$ $=\sqrt{49+576}$ $=\sqrt{625}$ $=25 \mathrm{~cm}$ Total surface area of the toy = (Curved surface area of the hemisphere) + (Curved s...
Read More →Construct a quadrilateral ABCD, in which AD = 3.5 cm,
Question: Construct a quadrilateralABCD, in whichAD= 3.5 cm,AB= 4.4 cm,BC= 4.7 cm, A= 125 and B= 120. Solution: Steps of construction: Step I : Draw $\mathrm{AB}=4.4 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{BAD}=125^{\circ}$ at A. Step III : With A as the centre and radius $3.5 \mathrm{~cm}$, cut off $\mathrm{AD}=3.5 \mathrm{~cm}$. Step IV : Construct $\angle \mathrm{ABC}=120^{\circ}$ at B. Step V : With B as the centre and radius $4.7 \mathrm{~cm}$, cut off $\mathrm{BC}=4.7 \mathrm{~c...
Read More →Construct a quadrilateral ABCD given BC = 6.6 cm,
Question: Construct a quadrilateralABCDgivenBC= 6.6 cm,CD= 4.4 cm,AD= 5.6 cm and D= 100 and C= 95. Solution: Steps of construction : Step I : Draw DC $=4.4 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{ADC}=100^{\circ}$ at D. Step III : With D as the centre and radius $5.6 \mathrm{~cm}$, cut off DA $=5.6 \mathrm{~cm}$. Step IV : Construct $\angle \mathrm{BCD}=95^{\circ}$ at C. Step V : With C as the centre and radius $6.6 \mathrm{~cm}$, cut off CB $=6.6 \mathrm{~cm}$. Step VI : Join AB. The...
Read More →The radius of the base and the height of a solid right circular cylinder are in the ratio 2 : 3 and its volume is 1617 cm3.
Question: The radius of the base and the height of a solid right circular cylinder are in the ratio 2 : 3 and its volume is 1617 cm3. Find the total surface area of the cylinder. Solution: Let the radius of the cylinder be2xcm and its height be3x cm. Then, volume of the cylinder $=\pi r^{2} h$ $=\frac{22}{7} \times(2 x)^{2} \times 3 x$ Therefore, $\frac{22}{7} \times(2 x)^{2} \times 3 x=1617$ $\Rightarrow \frac{22}{7} \times 4 x^{2} \times 3 x=1617$ $\Rightarrow \frac{22}{7} \times 12 x^{3}=1617...
Read More →Construct a quadrilateral PQRS, in which PQ = 3.5 cm,
Question: Construct a quadrilateralPQRS, in whichPQ= 3.5 cm,QR= 2.5 cm,RS= 4.1 cm, Q= 75 and R= 120. Solution: Steps of construction: Step I : Draw QR $=2.5 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{PQR}=75^{\circ}$ at Q. Step III : With Q as the centre and radius $3.5 \mathrm{~cm}$, cut off QP $=3.5 \mathrm{~cm}$. Step IV : Construct $\angle \mathrm{QRS}=120^{\circ}$ at R. Step V : With R as the centre and radius $4.1 \mathrm{~cm}$, cut off RS $=4.1 \mathrm{~cm}$. Step VI : Join PS The...
Read More →The radius of the base and the height of a solid right circular cylinder are in the ratio 2 : 3 and its volume is 1617 cm3.
Question: The radius of the base and the height of a solid right circular cylinder are in the ratio 2 : 3 and its volume is 1617 cm3. Find the total surface area of the cylinder. Solution: Let the radius of the cylinder be2xcm and its height be3x cm. Then, volume of the cylinder $=\pi r^{2} h$ $=\frac{22}{7} \times(2 x)^{2} \times 3 x$ Therefore, $\frac{22}{7} \times(2 x)^{2} \times 3 x=1617$ $\Rightarrow \frac{22}{7} \times 4 x^{2} \times 3 x=1617$ $\Rightarrow \frac{22}{7} \times 12 x^{3}=1617...
Read More →The radius of the base and the height of a solid right circular cylinder are in the ratio 2 : 3 and its volume is 1617 cm3.
Question: The radius of the base and the height of a solid right circular cylinder are in the ratio 2 : 3 and its volume is 1617 cm3. Find the total surface area of the cylinder. Solution: Let the radius of the cylinder be2xcm and its height be3x cm. Then, volume of the cylinder $=\pi r^{2} h$ $=\frac{22}{7} \times(2 x)^{2} \times 3 x$ Therefore, $\frac{22}{7} \times(2 x)^{2} \times 3 x=1617$ $\Rightarrow \frac{22}{7} \times 4 x^{2} \times 3 x=1617$ $\Rightarrow \frac{22}{7} \times 12 x^{3}=1617...
Read More →Write an equation of a Line passing through
Question: Write an equation of a Line passing through the point representing solution of the pair of Linear equations x + y = 2 and 2x y = 1, How many such lines can we find? Solution: Plotting the points A (2,0) and B (0, 2), we get the straight line AB. Plotting the points C (0, 1)andD (1/2, 0), we get the straight line CD. The lines AB and CD intersect at E(1,1), Hence, infinite lines can pass through the intersection point of linear equations x + y = 2 and 2x y = 1 i.e., E(1,1) like as y = x...
Read More →Construct a quadrilateral ABCD, where AB = 4.2 cm,
Question: Construct a quadrilateralABCD, whereAB= 4.2 cm,BC= 3.6 cm,CD= 4.8 cm, B= 30 and C= 150. Solution: Steps of construction : Step I : Draw $\mathrm{BC}=3.6 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{ABC}=30^{\circ}$ at B. Step III : With B as the centre and radius $4.2 \mathrm{~cm}$, cut off BA $=4.2 \mathrm{~cm} .$ Step IV : Construct $\angle \mathrm{BCD}=150^{\circ}$ at C. Step V : With C as the centre and radius $4.8 \mathrm{~cm}$, cut off $\mathrm{CD}=4.8 \mathrm{~cm} .$ Step ...
Read More →Construct a quadrilateral ABCD, in which AB = 6 cm,
Question: Construct a quadrilateralABCD, in whichAB= 6 cm,BC= 4 cm,CD= 4 cm, B= 95 and C= 90. Solution: Steps of construction: Step I: Draw BC $=4 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{ABC}=95^{\circ}$ at B. Step III : With B as the centre and radius $6 \mathrm{~cm}$, cut off BA $=6 \mathrm{~cm}$. Step IV : Construct $\angle \mathrm{BCD}=90^{\circ}$ at C. Step V : With C as the centre and radius $4 \mathrm{~cm}$, cut off BA $=4 \mathrm{~cm}$. Step VI : Join CD. The quadrilateral so ...
Read More →Draw the graph of the pair of equations 2x + y= 4
Question: Draw the graph of the pair of equations 2x + y= 4 and 2x y = 4. Write the vertices of the triangle formed by these lines and the y-axis, find the area of this triangle? Solution: The given pair of llnear equations Table for line $2 x+y=4$, and table for line $2 x-y=4$, Graphical representation of both lines. Here, both lines and $Y$-axis form a $\triangle A B C$. Hence, the vertices of a $\triangle A B C$ are $A(0,4) B(2,0)$ and $C(0,-4)$. $\therefore \quad$ Required area of $\triangle...
Read More →Two cubes, each of volume 64 cm3, are joined end to end.
Question: Two cubes, each of volume 64 cm3, are joined end to end. Find the total surface area of the resulting cuboid. Solution: Volume of the cube =a3 Therefore, $a^{3}=64$ $\Rightarrow a^{3}=(4)^{3}$ $\Rightarrow a=4 \mathrm{~cm}$ Each side of the cube = 4 cmThen, Length of the cuboid $\Rightarrow(2 \times 4) \mathrm{cm}=8 \mathrm{~cm}$ Breadth of the cuboid =4 cmHeight of the cuboid = 4 cm Total surface area of the cuboid $=2(l b+b h+l h)$ $=2[(8 \times 4)+(4 \times 4)+(8 \times 4)] \mathrm{...
Read More →Two cubes, each of volume 64 cm3, are joined end to end.
Question: Two cubes, each of volume 64 cm3, are joined end to end. Find the total surface area of the resulting cuboid. Solution: Volume of the cube =a3 Therefore, $a^{3}=64$ $\Rightarrow a^{3}=(4)^{3}$ $\Rightarrow a=4 \mathrm{~cm}$ Each side of the cube = 4 cmThen, Length of the cuboid $\Rightarrow(2 \times 4) \mathrm{cm}=8 \mathrm{~cm}$ Breadth of the cuboid =4 cmHeight of the cuboid = 4 cm Total surface area of the cuboid $=2(l b+b h+l h)$ $=2[(8 \times 4)+(4 \times 4)+(8 \times 4)] \mathrm{...
Read More →By the graphical method,
Question: By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them. (i) 3x + y + 4 = 0,6x- 2y + 4 = 0 (ii) x 2y 6, 3x 6y = 0 (iii) x + y = 3, 3x + 3y =9 Solution: (I) Given pair of equations is $3 x+y+4=0$ $\ldots$ (i) and $6 x-2 y+4=0$ ...(ii) On comparing with $a x+b y+c=0$, we get $a_{1}=3, b_{1}=1$ and $c_{1}=4$ [from Eq. (i)] $a_{2}=6, b_{2}=-2$ and $c_{2}=4$ [from Eq. (ii)] Here, $\frac{a_{1}}{a_{2}}=\frac{3}{6}=\frac{1}{2} ; \f...
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