The following figures are parallelograms.
Question: The following figures are parallelograms. Find the degree values of the unknownsx,y,z. Solution: (i) $O$ pposite angles of a parallelogram are same. $\therefore \mathrm{x}=\mathrm{z}$ and $\mathrm{y}=100^{\circ}$ Also, $\mathrm{y}+\mathrm{z}=180^{\circ} \quad\left(s\right.$ um of adjacent angles of a quadrilateral is $\left.180^{\circ}\right)$ $\mathrm{z}+100^{\circ}=180^{\circ}$ $\mathrm{x}=180^{\circ}-100^{\circ}$ $\mathrm{x}=80^{\circ}$ $\therefore \mathrm{x}=80^{\circ}, \mathrm{y}=...
Read More →A drinking glass is in the shape of the frustum of a cone of height 21 cm
Question: A drinking glass is in the shape of the frustum of a cone of height 21 cm with 6 cm and 4 cm as the diameters of its two circular ends. Find the capacity of the glass. Solution: Let R and r be the radii of the top and base, respectively, of the drinking glass and let its height be h. Then, $R \Rightarrow \frac{6}{2} \mathrm{~cm}=3 \mathrm{~cm}, r \Rightarrow \frac{4}{2} \mathrm{~cm}=2 \mathrm{~cm}, h=21 \mathrm{~cm}$ Capacity of the glass = Capacity of the frustum of the cone $=\frac{\...
Read More →A drinking glass is in the shape of the frustum of a cone of height 21 cm
Question: A drinking glass is in the shape of the frustum of a cone of height 21 cm with 6 cm and 4 cm as the diameters of its two circular ends. Find the capacity of the glass. Solution: Let R and r be the radii of the top and base, respectively, of the drinking glass and let its height be h. Then, $R \Rightarrow \frac{6}{2} \mathrm{~cm}=3 \mathrm{~cm}, r \Rightarrow \frac{4}{2} \mathrm{~cm}=2 \mathrm{~cm}, h=21 \mathrm{~cm}$ Capacity of the glass = Capacity of the frustum of the cone $=\frac{\...
Read More →The diameter of a sphere is 42 cm.
Question: The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire. Solution: Radius of the sphere $=\frac{42}{2}=21 \mathrm{~cm}$ Volume of the sphere $=\frac{4}{3} \pi r^{3}$ $=\left(\frac{4}{3} \pi \times 21 \times 21 \times 21\right) \mathrm{cm}^{3}$ Radius of the wire $=\frac{2.8}{2}=1.4 \mathrm{~cm}$ Let the length of the wire be $\mathrm{h} \mathrm{cm}$. Then, Volume of the wire $=\pi r^{2} h$ $=\left(\pi \times \fra...
Read More →The diameter of a sphere is 42 cm.
Question: The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire. Solution: Radius of the sphere $=\frac{42}{2}=21 \mathrm{~cm}$ Volume of the sphere $=\frac{4}{3} \pi r^{3}$ $=\left(\frac{4}{3} \pi \times 21 \times 21 \times 21\right) \mathrm{cm}^{3}$ Radius of the wire $=\frac{2.8}{2}=1.4 \mathrm{~cm}$ Let the length of the wire be $\mathrm{h} \mathrm{cm}$. Then, Volume of the wire $=\pi r^{2} h$ $=\left(\pi \times \fra...
Read More →Given below is a parallelogram ABCD.
Question: Given below is a parallelogramABCD. Complete each statement along with the definition or property used. (i)AD= (ii) DCB= (iii)OC= (iv) DAB+ CDA= Solution: The correct figure is (i) $\mathrm{AD}=\mathrm{BC}$ (opposite sides of a parallelogram are equal) (ii) $\angle \mathrm{DCB}=\angle \mathrm{BAD}$ (opposite angles are equal) (iii) $\mathrm{OC}=\mathrm{OA}$ (diagonals of a prallelogram bisect each other) (iv)$\angle \mathrm{DAB}+\angle \mathrm{CDA}=180^{\circ}$ (the sum of two adjacent...
Read More →A solid metallic sphere of diameter 21 cm is melted and recast into small cones of diameter 3.5 cm and height 3 cm each.
Question: A solid metallic sphere of diameter 21 cm is melted and recast into small cones of diameter 3.5 cm and height 3 cm each. Find the number of cones so formed. Solution: Radius of sphere $=\frac{21}{2} \mathrm{~cm}$ Volume of the metallic sphere $=\frac{4}{3} \pi \mathrm{r}^{3}$ $=\left(\frac{4}{3} \pi \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}\right) \mathrm{cm}^{3}$ Radius of cone $=\frac{3.5}{2} \mathrm{~cm}$ Height of cone $=3 \mathrm{~cm}$ Volume of each small cone $...
Read More →A mason has made a concrete slab.
Question: A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular? Solution: (i) By measuring each angle - Each angle of a rectangle is $90^{\circ} .$(ii) By measuring the length of the diagonals - Diagonals of a rectangle are equal.(iii) By measuring the sides of rectangle - Each pair of opposite sides are equal....
Read More →A hemispherical bowl of internal diameter 30 cm is full of a liquid.
Question: A hemispherical bowl of internal diameter 30 cm is full of a liquid. This liquid is poured into cylindrical bottles of diameter 5 cm and height 6 cm each. How many bottles are required? Solution: Radius of hemispherical ball $=\frac{30}{2}=15 \mathrm{~cm}$ Volume of hemispherical bowl $=\frac{2}{3} \pi r^{3}$ $=\left(\frac{2}{3} \pi \times 15 \times 15 \times 15\right) \mathrm{cm}^{3}$ Radius of each bottle $=\frac{5}{2} \mathrm{~cm}$ Height of each bottle $=6 \mathrm{~cm}$ Volume of e...
Read More →ABC is a right-angled trianle and O is the mid-point of the side opposite to the right angle.
Question: ABCis a right-angled trianle andOis the mid-point of the side opposite to the right angle. Explain whyOis equidistant fromA,BandC. Solution: (i) Construct a triangle $\mathrm{ABC}$ right angle at $\mathrm{B}$. (ii) Suppose $\mathrm{O}$ is the mid point of $\mathrm{AC}$. (iii) Complete the rectangle $\mathrm{ABCD}$ having $\mathrm{AC}$ as its diagonal. Since diagonals of a rectangle are equal and they bisect each other, $\mathrm{O}$ is the midpoint of both $\mathrm{AC}$ and $\mathrm{BD}...
Read More →ABC is a right-angled trianle and O is the mid-point of the side opposite to the right angle.
Question: ABCis a right-angled trianle andOis the mid-point of the side opposite to the right angle. Explain whyOis equidistant fromA,BandC. Solution: (i) Construct a triangle $\mathrm{ABC}$ right angle at $\mathrm{B}$. (ii) Suppose $\mathrm{O}$ is the mid point of $\mathrm{AC}$. (iii) Complete the rectangle $\mathrm{ABCD}$ having $\mathrm{AC}$ as its diagonal. Since diagonals of a rectangle are equal and they bisect each other, $\mathrm{O}$ is the midpoint of both $\mathrm{AC}$ and $\mathrm{BD}...
Read More →Name the quadrilaterals whose diagonals:
Question: Name the quadrilaterals whose diagonals: (i) bisect each other (ii) are perpendicular bisector of each other (iii) are equal. Solution: (i) Rhombus, parallelogram, rectangle and square (ii) Rhombus and square (iii) Rectangle and square...
Read More →A solid metal cone with base radius 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each.
Question: A solid metal cone with base radius 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls formed. Solution: Radius of cone = 12 cmHeight of cone = 24 cm Volume of the metallic cone $=\frac{1}{3} \pi r^{2} h$ $=\frac{1}{3} \pi \times(12)^{2} \times 24$ Radius of spherical ball $=\frac{6}{2} \mathrm{~cm}=3 \mathrm{~cm}$ Volume of each spherical ball $=\frac{4}{3} \pi \mathrm{r}^{3}$ $=\frac{4}{3} \pi \times(3)^{3}$ Number of balls ...
Read More →Explain how a square is
Question: Explain how a square is (i) a quadrilateral? (ii) a parallelogram? (iii) a rhombus? (iv) a rectangle? Solution: (i) Since a square has four sides, it is a quadrilateral. (ii) Since the opposite sides are parallel and equal, it is a parallelogram. (iii) Since the diagonals bisect each other and all the sides are equal, it is a rhombus. (iv) Since the opposite sides are equal and all the angles are right angles, it is a rectangle....
Read More →Identify all the quadrilaterals that have:
Question: Identify all the quadrilaterals that have: (i) Four sides of equal length (ii) Four right angles Solution: (i) If all four sides are equal, then it can be either a square or a rhombus. (ii) All four right angles, make it either a rectangle or a square....
Read More →The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm,
Question: The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. Find the curved surface area of the bucket. Solution: Let R and r be the radii of the top and base of the bucket, respectively, and letlbe its slant height.Then, curved surface area of the bucket= Curved surface area of the frustum of the cone $=\pi l(R+r)$ $=\frac{22}{7} \times 45 \times(28+7) \mathrm{cm}^{2}$ $=\left(\frac{22}{7} \times 45 \times 35\right) \mathrm{cm}^{2}$ $=4950 \math...
Read More →Draw a square whose each side measures 4.8 cm.
Question: Draw a square whose each side measures 4.8 cm. Solution: (i) Draw side $\mathrm{AB}=4.8 \mathrm{~cm}$. (ii) From A, make an angle of $90^{\circ}$ and cut it at $4.8 \mathrm{~cm}$ and mark it point D. (iii) From B, make an angle of $90^{\circ}$ and cut it at $4.8 \mathrm{~cm}$ and mark it point C. (iv) Join C and D. Thus, $\mathrm{ABCD}$ is the required square....
Read More →The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm,
Question: The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. Find the curved surface area of the bucket. Solution: Let R and r be the radii of the top and base of the bucket, respectively, and letlbe its slant height.Then, curved surface area of the bucket= Curved surface area of the frustum of the cone $=\pi l(R+r)$ $=\frac{22}{7} \times 45 \times(28+7) \mathrm{cm}^{2}$ $=\left(\frac{22}{7} \times 45 \times 35\right) \mathrm{cm}^{2}$ $=4950 \math...
Read More →Find the solution of the pair of equations
Question: Find the solution of the pair of equations$\frac{x}{10}+\frac{y}{5}-1=0$ $\frac{x}{8}+\frac{y}{6}=15$ and find $\mathrm{A}$, if $\mathrm{y}=\lambda \mathrm{x}+5$ Solution: Given pair of equations is $\frac{x}{10}+\frac{y}{5}-1=0$ $\ldots($ i) and $\frac{x}{8}+\frac{y}{6}=15$ .....(ii) Now, multiplying both sides of Eq. (i) by LCM $(10,5)=10$, we get $x+2 y-10=0$ $\Rightarrow$ $x+2 y=10$ ....(iii) Again, multiplying both sides of Eq. (iv) by LCM $(8,6)=24$, we get $3 x+4 y=360$ ...(iv) ...
Read More →Draw a rectangle whose one side measures 8 cm
Question: Draw a rectangle whose one side measures 8 cm and the length of each of whose diagonals is 10 cm. Solution: (i) Draw a side AB, equal to 8 cm. (ii) With A as the centre, draw an arc of length 10 cm. (iii) Draw $\angle \mathrm{ABX}=90^{\circ}$, which intersects the arc at $\mathrm{C}$. (iv) Draw $\angle B A Y=90^{\circ}$. (v) With C as the centre, draw an arc of length 8 cm. (vi) Join CD. Thus, ABCD is the required rectangle....
Read More →A circus tent is cylindrical to a height of 4 m and conical above it.
Question: A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, then find the total area of the canvas required. Solution: We have, Height of the cylindrical part, $H=4 \mathrm{~m}$, Radius of the base, $r=\frac{105}{2} \mathrm{~m}$ and Slant height of the conical part, $l=40 \mathrm{~m}$ Now, The total area of canvas required = CSA of conical part + CSA of cylindrical part $=\pi r l+2 \pi r H$ $=\pi r(l+2 H)$ $=\frac{22}{7} ...
Read More →Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2.
Question: Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii? Solution: Let the radii of the two cylinders be $r$ and $R$; and the heights be $h$ and $H$. We have, $\frac{h}{H}=\frac{1}{2} \quad \ldots . .(\mathrm{i})$ Now, Volume of the first cylinder $=$ Volume of the second sphere $\Rightarrow \pi r^{2} h=\pi R^{2} H$ $\Rightarrow \frac{h}{H}=\frac{R^{2}}{r^{2}}$ $\Rightarrow \frac{1}{2}=\frac{R^{2}}{r^{2}}$ $\Rightarrow \frac...
Read More →Find the length of the diagonal of a rectangle whose sides are 12 cm
Question: Find the length of the diagonal of a rectangle whose sides are 12 cm and 5 cm. Solution: Using Pythagoras theorem : $A D^{2}+D C^{2}=A C^{2}$ $5^{2}+12^{2}=A C^{2}$ $25+14=A C^{2}$ $169=A C^{2}$ $A C=\sqrt{169}$ $=13 \mathrm{~cm}$ Thus, length of the diagonal is $13 \mathrm{~cm}$....
Read More →The sides of a rectangle are in the ratio 4 : 5.
Question: The sides of a rectangle are in the ratio 4 : 5. Find its sides if the perimeter is 90 cm. Solution: Let the side be $x \mathrm{~cm}$ and $y \mathrm{~cm}$. So, we have : $2(x+y)=90$ Sides are in the ratio $4: 5$. $\therefore \quad y=\frac{5 x}{4}$ Putting the value of $y$ : $2\left(x+\frac{5 x}{4}\right)=90$ $\frac{4 x+5 x}{4}=45$ $9 x=180$ $x=20$ $\therefore y=\frac{5 \times 20}{4}=25$ Thus, the sides of the rectangle will be $20 \mathrm{~cm}$ and $25 \mathrm{~cm}$....
Read More →Find the number of solid spheres, each of diameter 6 cm,
Question: Find the number of solid spheres, each of diameter 6 cm, that could be moulded to form a solid metallic cylinder of height 45 cm and diameter 4 cm. Solution: We have, Radius of the sphere, $R=\frac{6}{2}=3 \mathrm{~cm}$, Radius of the cylinder, $r=\frac{4}{2}=2 \mathrm{~cm}$ and Height of the cylinder, $h=45 \mathrm{~cm}$ Now, The number of solid spheres $=\frac{\text { Volume of the Cylinder }}{\text { Volume of the sphere }}$ $=\frac{\pi r^{2} h}{\left(\frac{4}{3} \pi R^{3}\right)}$ ...
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