Two equal chords AB and CD of a circle
Question: Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. Solution: Given Two equal chords $A B$ and $C D$ of a circle intersecting at a point $P$. To prove $P B=P D$ Construction Join $O P$, draw $O L \perp A B$ and $O M \perp C D$ Proof We have, $A B=C D$ $\Rightarrow \quad O L=O M$ [equal chords are equidistant from the centre] In $\triangle O L P$ and $\triangle O M P$, $O L=O M$ [proved above] $\angle O L P=\angle O M P$ [each 90^{\circ] } $\...
Read More →If the cost of 93 m of a certain kind of plastic sheet is Rs 1395,
Question: If the cost of 93 m of a certain kind of plastic sheet is Rs 1395, then what would it cost to buy 105 m of such plastic sheet? Solution: Let the cost of the plastic sheet per metre be Rs $x$. If more sheets are bought, the cost will also be more. Therefore, it is a direct variation. We get: $93: 105=1395: x$ $\Rightarrow \frac{93}{105}=\frac{1395}{x}$ Applying cross muliplication, we get: $x=\frac{105 \times 1395}{93}$ $=1575$ Thus, the required cost will be Rs 1,575 ....
Read More →The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km2.
Question: The slant height of a conical mountain is 2.5 km and the area of its baseis 1.54 km2. Find the height of the mountain. Solution: Letr,handlbe the base radius, the height and the slant height of the conical mountain, respectively. As, the area of the base $=1.54 \mathrm{~km}^{2}$ $\Rightarrow \pi r^{2}=1.54$ $\Rightarrow \frac{22}{7} \times r^{2}=1.54$ $\Rightarrow r^{2}=\frac{1.54 \times 7}{22}$ $\Rightarrow r^{2}=0.49$ $\Rightarrow r=\sqrt{0.49}$ $\Rightarrow r=0.7 \mathrm{~km}$ Now, ...
Read More →The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km2.
Question: The slant height of a conical mountain is 2.5 km and the area of its baseis 1.54 km2. Find the height of the mountain. Solution: Letr,handlbe the base radius, the height and the slant height of the conical mountain, respectively. As, the area of the base $=1.54 \mathrm{~km}^{2}$ $\Rightarrow \pi r^{2}=1.54$ $\Rightarrow \frac{22}{7} \times r^{2}=1.54$ $\Rightarrow r^{2}=\frac{1.54 \times 7}{22}$ $\Rightarrow r^{2}=0.49$ $\Rightarrow r=\sqrt{0.49}$ $\Rightarrow r=0.7 \mathrm{~km}$ Now, ...
Read More →Solve the matrix equation
Question: Solve the matrix equation $\left[\begin{array}{ll}5 4 \\ 1 1\end{array}\right] X=\left[\begin{array}{cc}1 -2 \\ 1 3\end{array}\right]$, where $X$ is a $2 \times 2$ matrix. Solution: Let: $A=\left[\begin{array}{ll}5 4\end{array}\right.$ $\left.\begin{array}{ll}1 1\end{array}\right]$ $B=\left[\begin{array}{ll}1 -2\end{array}\right.$ $\begin{array}{ll}1 3]\end{array}$ Now, $|\mathrm{A}|=\mid \begin{array}{ll}5 4\end{array}$ $1 \quad 1 \mid=5-4=1 \neq 0$ Hence, $A$ is invertible. If $C_{i ...
Read More →Anupama takes 125 minutes in walking a distance of 100 metre.
Question: Anupama takes 125 minutes in walking a distance of 100 metre. What distance would she cover in 315 minutes? Solution: Let the distance travelled in 315 minutes bexkm. If the distance travelled is more, the time needed to cover it will also be more. Therefore, it is a direct variation. We get: $125: 315=100: x$ $\Rightarrow \frac{125}{315}=\frac{100}{x}$ Applying cross muliplication, we get: $x=\frac{100 \times 315}{125}$ $=252$ Thus, Anupama would cover 252 metre in 315 minutes....
Read More →If the volumes of two cones are in the ratio of 1:4
Question: If the volumes of two cones are in the ratio of 1:4 and their diameters are in the ratio of 4:5, then find the ratio of their heights. Solution: Let $r$ and $R$ be the base radii, $h$ and $H$ be the heights, $v$ and $V$ be the volumes of the two given cones. We have, $\frac{2 r}{2 R}=\frac{4}{5}$ or $\frac{r}{R}=\frac{4}{5} \quad \cdots$ (i) and $\frac{v}{V}=\frac{1}{4}$ $\Rightarrow \frac{\left(\frac{1}{3} \pi r^{2} h\right)}{\left(\frac{1}{3} \pi R^{2} H\right)}=\frac{1}{4}$ $\Righta...
Read More →Rohit bought 12 registers for Rs 156, find the cost of 7 such registers.
Question: Rohit bought 12 registers for Rs 156, find the cost of 7 such registers. Solution: Let the cost of 7 registers beRsx. If he buys less number of registers, the cost will also be less. Therefore, it is a direct variation. We get: $12: 7=156: x$ $\Rightarrow \frac{12}{7}=\frac{156}{x}$ Applying cross muliplication, we get: $x=\frac{156 \times 7}{12}$ $=91$ Thus, the cost of 7 such registers will be Rs 91 ....
Read More →A circle has radius √2 cm.
Question: A circle has radius 2 cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45. Solution: Draw a circle having centre $O$. Let $A B=2 \mathrm{~cm}$ be a chord of a circle. A chord $A B$ is divided by the line OM in two equal segments. To prove $\angle A P B=45^{\circ}$ Here, $\quad A N=N B=1 \mathrm{~cm}$ and $O B=\sqrt{2} \mathrm{~cm}$ In $\triangle O N B$, $O B^{2}=O N^{2}+N B^{2}$ [use Pythagoras the...
Read More →Rohit bought 12 registers for Rs 156, find the cost of 7 such registers.
Question: Rohit bought 12 registers for Rs 156, find the cost of 7 such registers. Solution: Let the cost of 7 registers beRsx. If he buys less number of registers, the cost will also be less. Therefore, it is a direct variation. We get: $12: 7=156: x$ $\Rightarrow \frac{12}{7}=\frac{156}{x}$ Applying cross muliplication, we get: $x=\frac{156 \times 7}{12}$ $=91$ Thus, the cost of 7 such registers will be Rs 91 ....
Read More →Complite the following tables given that x varies directly as y.
Question: Complite the following tables given thatxvaries directly asy. (i) (ii) (iii) (iv) (v) Solution: Here, $x$ and $y$ vary directly. $\therefore x=k y$ (i) $x=2.5$ and $y=5$ i.e., $2.5=k \times 5$ $\Rightarrow k=\frac{2.5}{5}=0.5$ For $y=8$ and $k=0.5$, we have : $x=k y$ $\Rightarrow x=8 \times 0.5=4$ For $y=12$ and $k=0.5$, we have : $x=k y$ $\Rightarrow x=12 \times 0.5=6$ For $x=15$ and $k=0.5$, we have : $x=k y$ $\Rightarrow 15=0.5 \times y$ $\Rightarrow y=\frac{15}{0.5}=30$ (ii) $x=5$ ...
Read More →A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m.
Question: (i) A 5-m-wide cloth is used to make a conical tent of base diameter 14 mand height 24 m. Find the cost of cloth used at the rate of₹25 per metre. (ii) The radius and height of a solid right-circular cone are in the ratio of 5 : 12. If its volume is 314 cm3, find the total surface area. [Take = 3.14.] Solution: (i) We have, the height of the cone, $h=24 \mathrm{~m}$, the base diameter of the cone, $d=14 \mathrm{~m}$ Also, the base radius of the cone, $r=\frac{d}{2}=\frac{14}{2}=7 \math...
Read More →A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m.
Question: (i) A 5-m-wide cloth is used to make a conical tent of base diameter 14 mand height 24 m. Find the cost of cloth used at the rate of₹25 per metre. (ii) The radius and height of a solid right-circular cone are in the ratio of 5 : 12. If its volume is 314 cm3, find the total surface area. [Take = 3.14.] Solution: (i) We have, the height of the cone, $h=24 \mathrm{~m}$, the base diameter of the cone, $d=14 \mathrm{~m}$ Also, the base radius of the cone, $r=\frac{d}{2}=\frac{14}{2}=7 \math...
Read More →If bisectors of opposite angles of a cyclic
Question: If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. Thinking Process Use the property of cyclic quardrilateral, the sum of opposite angles of cyclic quadrilateral is supplementary. Further, simplify it to prove the required result. Solution: Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of D and B, respectively. To prove PQ is the diameter of a circl...
Read More →If the total surface area of a solid hemisphere is 462 cm2,
Question: If the total surface area of a solid hemisphere is 462 cm2, then find itsvolume. Solution: As, the total surface area of the solid hemisphere $=462 \mathrm{~cm}^{2}$ $\Rightarrow 3 \pi r^{2}=462$ $\Rightarrow 3 \times \frac{22}{7} \times r^{2}=462$ $\Rightarrow r^{2}=\frac{462 \times 7}{3 \times 22}$ $\Rightarrow r^{2}=49$ $\Rightarrow r^{2}=\sqrt{49}$ $\Rightarrow r=7 \mathrm{~cm}$ Now, the volume of the solid hemisphere $=\frac{2}{3} \pi r^{3}$ $=\frac{2}{3} \times \frac{22}{7} \time...
Read More →In figure, AB and CD are two chords of a circle
Question: In figure, AB and CD are two chords of a circle intersecting each other at point E. Prove that AEC = (angle subtended by arc C x A at centre + angle subtended by arc DYB at the centre). Solution: Given In a figure, two chords AB and CD intersecting each other at point E. To prove $\quad \angle A E C=\frac{1}{2}$ [angle subtended by arc $C \times A$ at centre $+$ angle subtended by $\operatorname{arc} D Y B$ at the centre] Construction Extend the line $D O$ and $B O$ at the points $l$ a...
Read More →The volume of a hemisphere is
Question: The volume of a hemisphere is $2425 \frac{1}{2} \mathrm{~cm}^{3}$. Find its curved surface area. Solution: As, volume of hemisphere $=2425 \frac{1}{2} \mathrm{~cm}^{3}$ $\Rightarrow \frac{2}{3} \pi r^{3}=2425 \frac{1}{2}$ $\Rightarrow \frac{2}{3} \times \frac{22}{7} \times r^{3}=\frac{4851}{2}$ $\Rightarrow r^{3}=\frac{4851 \times 3 \times 7}{2 \times 2 \times 22}$ $\Rightarrow r^{3}=\frac{441 \times 3 \times 7}{2 \times 2 \times 2}$ $\Rightarrow r^{3}=\frac{21^{3}}{2^{3}}$ $\Rightarro...
Read More →Two cubes each of volume 27 cm3 are joined end to end to form a solid.
Question: Two cubes each of volume 27 cm3are joined end to end to form a solid. Find the surface area of the resulting cuboid. Solution: As, volume of a cube $=27 \mathrm{~cm}^{3}$ $\Rightarrow$ (edge) $^{3}=27$ $\Rightarrow$ edge $=\sqrt[3]{27}$ $\Rightarrow$ edge $=3 \mathrm{~cm}$ The length of the resulting cuboid, $l=3+3=6 \mathrm{~cm}$, its breadth, $b=3 \mathrm{~cm}$ and its height, $h=3 \mathrm{~cm}$ Now, the surface area of the resulting cuboid $=2(l b+b h+h l)$ $=2(6 \times 3+3 \times 3...
Read More →Fill in the blanks in each of the following so as to make the statement true:
Question: Fill in the blanks in each of the following so as to make the statement true: (i) Two quantities are said to vary.... with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same. (ii)xandyare said to vary directly with each other if for some positive numberk,...... =k. (iii) Ifu= 3v, thenuandvvary .... with each other. Solution: (i) directly (ii) $x a$ nd $y a$ re said to vary directly with each other if $\frac{x}{y}=k$, wh...
Read More →In which of the following tables x and y vary directly?
Question: In which of the following tablesxandyvary directly? (i) (ii) (iii) (iv) Solution: If $x$ and $y$ vary directly, the ratio of the corresponding values of $x$ and $y$ remain $s$ constant. (i) $\frac{x}{y}=\frac{7}{21}=\frac{1}{3}$ $\frac{x}{y}=\frac{9}{27}=\frac{1}{3}$ $\frac{x}{y}=\frac{13}{39}=\frac{1}{3}$ $\frac{x}{y}=\frac{21}{63}=\frac{1}{3}$ $\frac{x}{y}=\frac{25}{75}=\frac{1}{3}$ In all the cases, the ratio is the same. Therefore, $x$ and $y$ vary directly. (ii) $\frac{x}{y}=\frac...
Read More →If ABC is an equilateral triangle inscribed
Question: If ABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C, then prove that PA is angle bisector of BPC. Solution: Given ΔABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C. To prove PA is an angle bisector of BPC. Construction Join PB and PC. Proof Since, $\triangle A B C$ is an equilateral triangle. $\angle 3=\angle 4=60^{\circ}$ Now,$\a...
Read More →Which of the following quantities vary directly with each other?
Question: Which of the following quantities vary directly with each other? (i) Number of articles (x) and their price (y). (ii) Weight of articles (x) and their cost (y). (iii) Distancexand timey, speed remaining the same. (iv) Wages (y) and number of hours (x) of work. (v) Speed (x) and time (y) (distance covered remaining the same). (vi) Area of a land (x) and its cost (y). Solution: (i) The number of articles is directly related to the price. Therefore, they will vary directly with each other...
Read More →Explain the concept of direct variation.
Question: Explain the concept of direct variation. Solution: When two variables are connected to each other in such a way that if we increase the value of one variable, the value of other variable also increases and vice - versa. Similarly, if we decrease the value of one variable, the value of other variable also decreases and vice - versa. Therefore, if the ratio between two variables remains constant, it is said to be in direct variation....
Read More →Solve this
Question: If $A=\left[\begin{array}{ccc}-1 2 0 \\ -1 1 1 \\ 0 1 0\end{array}\right]$, show that $A^{2}=A^{-1}$ Solution: We have, $A=\left[\begin{array}{lll}-1 2 0\end{array}\right.$ $\begin{array}{lll}-1 1 1\end{array}$ $\left.\begin{array}{lll}0 1 0\end{array}\right]$ Now, $A^{2}=\left[\begin{array}{lll}-1 2 0\end{array}\right.$ $\begin{array}{lll}-1 1 1\end{array}$ $\left.\begin{array}{lll}0 1 0\end{array}\right]\left[\begin{array}{lll}-1 2 0\end{array}\right.$ $-1 \quad 1 \quad 1$ $0 \quad 1...
Read More →If two chords AB and CD of a circle
Question: If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. Solution: Given In a circle AYDZBWCX, two chords AB and CD intersect at right angles. To prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle. Construction Draw a diameter EF parallel to CD having centre M. Proof Since, CD||EF arc EC = arc PD (i) arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF (ii) We know that,$\o...
Read More →