The angles of a cyclic quadrilateral ABCD
Question: The angles of a cyclic quadrilateral ABCD are A (6x +10), B = (5x), C = (x+ y) and D = (3y 10).Find x and y and hence the values of the four angles. Solution: We know that, by property of cyclic quadrilateral, Sum of opposite angles = 180 A + C = (6x + 10) + (x + y) = 180 $\left[\because \angle A=(6 x+10)^{\circ}, \angle C=(x+y)^{\circ}\right.$, given $]$ $\Rightarrow \quad 7 x+y=170$...(i) and $\angle B+\angle D=(5 x)^{\circ}+(3 y-10)^{\circ}=180^{\circ}$ $\left[\because \angle B=(5 x...
Read More →Construct a rhombus with side 6 cm and one diagonal 8 cm.
Question: Construct a rhombus with side 6 cm and one diagonal 8 cm. Measure the other diagonal. Solution: Steps of construction : Step 1: Draw $\mathrm{AC}=8 \mathrm{~cm} .$ Step 2: With $\mathrm{A}$ as the centre and radius $=6 \mathrm{~cm}$, draw arcs on both sides. Step 3: With $\mathrm{C}$ as the centre and radius $=6 \mathrm{~cm}$, draw arcs on both sides, intersecting the previous arcs at points B and D. Step 4: Join $\mathrm{BD}=8.9 \mathrm{~cm} .$ Thus, $\mathrm{ABCD}$ is the required rh...
Read More →Construct a parallelogram PQRS such that PQ = 5.2 cm,
Question: Construct a parallelogramPQRSsuch thatPQ= 5.2 cm,PR= 6.8 cm andQS= 8.2 cm. Solution: In a parallelogram opposite sides are equal. Thus, we have to construct a quadrilateral PQRS in which $\mathrm{PQ}=5.2 \mathrm{~cm}, \mathrm{PR}=6.8 \mathrm{~cm}$ and $\mathrm{QS}=8.2 \mathrm{~cm} .$ Steps of construction: Step I: Draw QS $=8.2 \mathrm{~cm}$ Step II : With Q as the centre and radius $5.2 \mathrm{~cm}$, draw an arc. Step III : With S as the centre and radius $5.2 \mathrm{~cm}$, draw an ...
Read More →An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet.
Question: An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket. Also, find the volume of water the bucket can hold, in litres. Solution: We have, Radius of the upper end of the frustum, $R=\frac{45...
Read More →Show that each one of the following systems of linear equation is inconsistent:
Question: Show that each one of the following systems of linear equation is inconsistent: (i) $2 x+5 y=7$ $6 x+15 y=13$ (ii) $2 x+3 y=5$ $6 x+9 y=10$ (iii) $4 x-2 y=3$ $6 x-3 y=5$ (iv) $4 x-5 y-2 z=2$ $5 x-4 y+2 z=-2$ $2 x+2 y+8 z=-1$ (v) $3 x-y-2 z=2$ $2 y-z=-1$ $3 x-5 y=3$ (vi) $x+y-2 z=5$ $x-2 y+z=-2$ $-2 x+y+z=4$ Solution: (i) The given system of equations can be expressed as follows: $A X=B$ Here, $A=\left[\begin{array}{cc}2 5 \\ 6 15\end{array}\right], X=\left[\begin{array}{c}x \\ y\end{ar...
Read More →Construct a quadrilateral XYZW in which XY = 5 cm,
Question: Construct a quadrilateralXYZWin whichXY= 5 cm,YZ= 6 cm,ZW= 7 cm,WX= 3 cm andXZ= 9 cm. Solution: Steps of construction: Step I : Draw XZ $=9 \mathrm{~cm}$ Step II : With X as the centre and radius $5 \mathrm{~cm}$, draw an arc above XZ. Step III : With Z as the centre and radius $6 \mathrm{~cm}$, draw an arc to intersect the arc drawn in Step II at Y above XZ. Step IV : With Z as the centre and radius $7 \mathrm{~cm}$, draw an arc below XZ. Step V: With X as the centre and radius $3 \ma...
Read More →Show that each one of the following systems of linear equation is inconsistent:
Question: Show that each one of the following systems of linear equation is inconsistent: (i) $2 x+5 y=7$ $6 x+15 y=13$ (ii) $2 x+3 y=5$ $6 x+9 y=10$ (iii) $4 x-2 y=3$ $6 x-3 y=5$ (iv) $4 x-5 y-2 z=2$ $5 x-4 y+2 z=-2$ $2 x+2 y+8 z=-1$ (v) $3 x-y-2 z=2$ $2 y-z=-1$ $3 x-5 y=3$ (vi) $x+y-2 z=5$ $x-2 y+z=-2$ $-2 x+y+z=4$ Solution: (i) The given system of equations can be expressed as follows: $A X=B$ Here, $A=\left[\begin{array}{cc}2 5 \\ 6 15\end{array}\right], X=\left[\begin{array}{c}x \\ y\end{ar...
Read More →In a competitive examination,
Question: In a competitive examination, 1 mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly? Solution: Let x be the number of correct answers of the questions in a competitive examination, then (120 x) be the number of wrong answers of the questions. Then, by given condition, $x \times 1-(120-x) \times \frac{1}{2}=90$ $\Rightarrow$$x-60+\frac{x}{2}=90$ $\Rightarro...
Read More →Construct a quadrilateral ABCD such that AB = BC = 5.5 cm,
Question: Construct a quadrilateralABCDsuch thatAB=BC= 5.5 cm,CD= 4 cm,DA= 6.3 cm andAC= 9.4 cm. MeasureBD. Solution: Steps of construction: Step I: Draw $\mathrm{AB}=5.5 \mathrm{~cm}$ Step II : With $\mathrm{B}$ as the centre and radius $\mathrm{BC}=5.5 \mathrm{~cm}$, draw an arc. Step III : With $\mathrm{A}$ as the centre and radius $\mathrm{AC}=9.4 \mathrm{~cm}$, draw an arc to intersect the arc drawn in Step II at C. Step IV: With $\mathrm{C}$ as the centre and radius $\mathrm{CD}=4 \mathrm{...
Read More →A shopkeeper gives books on rent for reading.
Question: A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days and an additional charge for each day thereafter. Latika paid ₹ 22 for a book kept for six days, while Anand paid ₹ 16 for the book kept for four days. Find the fixed charges and the charge for each extra day. Solution: Let Latika takes a fixed charge for the first two day is ₹ x and additional charge for each day thereafter is ₹ y. Now by first condition. Latika paid ₹ 22 for a book kept for ...
Read More →Construct a quadrilateral ABCD in which AB = 4.4 cm,
Question: Construct a quadrilateralABCDin whichAB= 4.4 cm,BC= 4 cm,CD= 6.4 cm,DA= 3.8 cm andBD= 6.6 cm. Solution: First, we draw a rough sketch of the quadrilateral $\mathrm{ABCD}$ and write down its dimensions along the sides. We may divide the quadrilateral into two constructible triangles $\mathrm{ABD}$ and $\mathrm{BCD}$. Step I : Draw $\mathrm{BD}=6.6 \mathrm{~cm}$ Step II : With $\mathrm{B}$ as the centre and radius $\mathrm{BC}=4 \mathrm{~cm}$, draw an arc. Step III : With $\mathrm{D}$ as...
Read More →A solid rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm
Question: A solid rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast intoa hollow cylindrical pipe of internal radius 30 cm and thickness5 cm. Find the length of the pipe. Solution: We have, Length of the rectangular block, $l=4.4 \mathrm{~m}$, Breadth of the rectangular block, $\mathrm{b}=2.6 \mathrm{~m}$, Height of the rectangular block, $h=1 \mathrm{~m}$, Internal radius of the cylindrical pipe, $r=30 \mathrm{~cm}=0.3 \mathrm{~m}$ and Thickness of the pipe $=5 \mathrm{~cm}=0.05 \ma...
Read More →There are some students in the two examination halls A and B.
Question: There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B but, if 20 students are sent from B to A, the number of students in A becomes double the number of students in B, then find the number of students in the both halls. Solution: Let the number of students in halls A and 8 are x and y, respectively. Now, by given condition, x-10=y+10 ⇒ x y = 20 (i) and (x + 20) = 2 (y-20) ⇒ x-2y=-60 (ii) On subt...
Read More →Construct a quadrilateral PQRS, in which PQ = 4 cm,
Question: Construct a quadrilateralPQRS, in whichPQ= 4 cm,QR= 5 cm, P= 50, Q= 110 and R= 70. Solution: Steps of construction: Step I: Draw $\mathrm{PQ}=4 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{XPQ}=50^{\circ}$ at $\mathrm{P}$ and $\angle \mathrm{PQY}=110^{\circ}$ at $\mathrm{Q}$. Step III : With $Q$ as the centre and radius $5 \mathrm{~cm}$, cut off $Q R=5 \mathrm{~cm}$. Step IV: At R, draw $\angle \mathrm{QRZ}=70^{\circ}$ such that it meets PX at S. The quadrilateral so obtained is ...
Read More →Two numbers are in the ratio 5 : 6.
Question: Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5, then find the numbers. . Solution: Let the two numbers be x and y. Then, by first Condition, ratio of these two numbers = 5:6 $x: y=5: 6$ $\Rightarrow$$\frac{x}{y}=\frac{5}{6} \Rightarrow y=\frac{6 x}{5}$$\ldots$ (i) and by second condition, then, 8 is subtracted from each of the numbers, then ratio becomes $4: 5$. $\frac{x-8}{y-8}=\frac{4}{5}$ $\Rightarrow \quad 5 x-40=4 y-32$ $\R...
Read More →From a solid cylinder whose height is 15 cm and diameter 16 cm, a conical cavity of the same height and same diameter is hollowed out.
Question: From a solid cylinder whose height is 15 cm and diameter 16 cm, aconical cavity of the same height and same diameter is hollowed out.Find the total surface area of the remaining solid. (Use= 3.14) Solution: We have, Height of the cylinder $=$ Height of the cone $=h=15 \mathrm{~cm}$ and Radius of the cylinder $=$ Radius of the cone $=r=\frac{16}{2}=8 \mathrm{~cm}$ Also, the slant height of the cone, $l=\sqrt{h^{2}+r^{2}}$ $=\sqrt{15^{2}+8^{2}}$ $=\sqrt{225+64}$ $=\sqrt{289}$ $=17 \mathr...
Read More →Construct a quadrilateral ABCD, where ∠A = 65°,
Question: Construct a quadrilateralABCD, where A= 65, B= 105, C= 75BC= 5.7 cm andCD= 6.8 cm. Solution: We know that the sum of all the angles in a quadrilateral is 360 . i. e., $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ}$ $\Rightarrow \angle \mathrm{D}=115^{\circ}$ Steps of Construction: Step I : Draw BC $=5.7 \mathrm{~cm} .$ Step II : Construct $\angle \mathrm{XBC}=105^{\circ}$ at B and $\angle \mathrm{BCY}=105^{\circ}$ at C. Step III : With C as the cen...
Read More →A solid is composed of a cylinder with hemispherical ends.
Question: A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each hemispherical end is 7 cm, find the surface area of the solid. Solution: Radius of the hemispherical end = 7 cmHeight of the hemispherical end = 7 cm Height of the cylindrical part $=(104-2 \times 7) \mathrm{cm}=90 \mathrm{~cm}$ Surface area of the solid = 2(curved surface area of the hemisphere) + (curved surface area of the cylinder) $=\left[2\left(2 \pi r^{2}\...
Read More →A solid is composed of a cylinder with hemispherical ends.
Question: A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each hemispherical end is 7 cm, find the surface area of the solid. Solution: Radius of the hemispherical end = 7 cmHeight of the hemispherical end = 7 cm Height of the cylindrical part $=(104-2 \times 7) \mathrm{cm}=90 \mathrm{~cm}$ Surface area of the solid = 2(curved surface area of the hemisphere) + (curved surface area of the cylinder) $=\left[2\left(2 \pi r^{2}\...
Read More →The age of the father is twice the sum
Question: The age of the father is twice the sum of the ages of his two children. After 20 yr, his age will be equal to the sum of the ages of his children. Find the age of the father. Solution: Let the present age (in year) of father and his two children be x, y and z yr, respectively. Now by given condition, x=2(y+z) (i) and after 20 yr, (x + 20) = (y + 20) + (z + 20) ⇒ y+z +40 = x +20 ⇒ y + z = x 20 On putting the value of (y + z) in Eq. (i) and get the present age of father x =2(x -20) x = 2...
Read More →Construct a quadrilateral ABCD when BC = 5.5 cm,
Question: Construct a quadrilateralABCDwhenBC= 5.5 cm,CD= 4.1 cm, A= 70, B= 110 and D= 85. Solution: We know that the sum of all the angles in a quadrilateral is 360 . i. e., $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ}$ $\Rightarrow \angle \mathrm{C}=95^{\circ}$ Steps of construction: Step I : Draw BC $=5.5 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{XBC}=110^{\circ}$ at A and $\angle \mathrm{BCY}=95^{\circ}$. Step III : With C as the centre and ra...
Read More →Two years ago, Salim was thrice as old as
Question: Two years ago, Salim was thrice as old as his daughter and six years later, he will be four year older than twice her age. How old are they now? Solution: Let Salim and his daughters age be x and y yr respectively. Now, by first condition Two years ago, Salim was thrice as old as his daughter. i.e. $\quad x-2=3(y-2) \Rightarrow x-2=3 y-6$ $\Rightarrow \quad x-3 y=-4$ ....(i) and by second condition, six years later. Salim will be four years older than twice her age. $x+6=2(y+6)+4$ $\Ri...
Read More →Construct a quadrilateral PQRS, where PQ = 3.5 cm,
Question: Construct a quadrilateralPQRS, wherePQ= 3.5 cm,QR= 6.5 cm, P= R= 105 and S= 75. Solution: We know that the sum of all the angles in a quadrilateral is 360 . i. e., $\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}+\angle \mathrm{S}=360^{\circ}$ $\Rightarrow \angle \mathrm{Q}=75^{\circ}$ Steps of construction: Step I : Draw PQ $=3.5 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{XPQ}=105^{\circ}$ at P and $\angle \mathrm{PQY}=75^{\circ}$ at Q. Step III : With Q as the centre an...
Read More →The slant height of the frustum of a cone is 4 cm and the perimeters (i.e. circumferences) of its circular ends are 18 cm and 6 cm.
Question: The slant height of the frustum of a cone is 4 cm and the perimeters (i.e. circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. Solution: Let R and r be the radii of the top and base of the frustum of the cone, respectively, and its slant height bel. Then, $2 \pi R=18 \Rightarrow \pi R=9 \quad \ldots(i)$ $2 \pi \mathrm{r}=6 \Rightarrow \pi \mathrm{r}=3 \quad \ldots(i i)$ Curved surface area of the frustum $=\pi l(R+r)$ $=l \times(\pi R+\...
Read More →Construct a quadrilateral ABCD, where AB = 5.5 cm,
Question: Construct a quadrilateralABCD, whereAB= 5.5 cm,BC= 3.7 cm, A= 60, B= 105 and D= 90. Solution: We know that the sum of all the angles in a quadrilateral is 360 . i.e., $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ}$ $\Rightarrow \angle \mathrm{C}=105^{\circ}$ Steps of construction: Step I : Draw $\mathrm{AB}=5.5 \mathrm{~cm}$. Step II : Construct $\angle \mathrm{XAB}=60^{\circ}$ at A and $\angle \mathrm{ABY}=105^{\circ}$. Step III: With B as the cen...
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