eSaral - JEE | NEET | Class 8-10 Exam Prep App

ABCD is a rhombus.

Question: ABCDis a rhombus. If ACB= 40, find ADB. Solution: In a rhombus, the diagonals are perpendicular. $\therefore \angle \mathrm{BPC}=90^{\circ}$ From $\Delta$ BPC, the sum of angles is $180^{\circ}$. $\therefore \angle \mathrm{CB} P+\angle \mathrm{BP} C+\angle P B C=180^{\circ}$ $\angle \mathrm{CB} P=180^{\circ}-\angle \mathrm{BP} C-\angle P B C$ $\angle \mathrm{CBP}=180^{\circ}-40^{\circ}-90^{\circ}=50^{\circ}$ $\angle A D B=\angle \mathrm{CBP}=50^{\circ}$ (alternate angle)...

The diagonals of a quadrilateral are perpendicular to each other.

Question: The diagonals of a quadrilateral are perpendicular to each other. Is such a quadrilateral always a rhombus? If your answer is 'No', draw a figure to justify your answer. Solution: No, it is not so. Diagonals of a rhombus are perpendicular and bisect each other. Along with this, all of its sides are equal. In the figure given below, the diagonals are perpendicular to each other, but do not bisect each other....

The diagonals of a parallelogram are not perpendicular. Is it a rhombus?

Question: The diagonals of a parallelogram are not perpendicular. Is it a rhombus? Why or why not? Solution: No, it is not a rhombus. This is because diagonals of a rhombus must be perpendicular....

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) A rhombus is a parallelogram in which ...... (ii) A square is a rhombus in which ...... (iii) A rhombus has all its sides of ...... length. (iv) The diagonals of a rhombus ...... each other at ...... angles. (v) If the diagonals of a parallelogram bisect each other at right angles, then it is a ...... Solution: (i) A rhombus is a parallelogram in whichadjacent sides are equal. (ii) A square is a rhombus...

Which of the following statements are true for a rhombus?

Question: Which of the following statements are true for a rhombus? (i) It has two pairs of parallel sides. (ii) It has two pairs of equal sides. (iii) It has only two pairs of equal sides. (iv) Two of its angles are at right angles. (v) Its diagonals bisect each other at right angles. (vi) Its diagonals are equal and perpendicular. (vii) It has all its sides of equal lengths. (viii) It is a parallelogram. (ix) It is a quadrilateral. (x) It can be a square. (xi) It is a square. Solution: (i) Tru...

In a parallelogram ABCD, AB = 10 cm, AD = 6 cm.

Question: In a parallelogramABCD,AB= 10 cm,AD= 6 cm. The bisector of AmeetsDCinE,AEandBCproduced meet atF. Find te lengthCF. Solution: $\mathrm{AE}$ is the bisector of $\angle \mathrm{A}$. $\therefore \angle \mathrm{DAE}=\angle \mathrm{BAE}=x$ $\angle \mathrm{BAE}=\angle \mathrm{AED}=x$ (alternate angles) Since opposite angles in $\Delta \mathrm{ADE}$ are equal, $\Delta \mathrm{ADE}$ is an isosceles triangle. $\therefore \mathrm{AD}=\mathrm{DE}=6 \mathrm{~cm}$ (sides opposite to equal angles) $\...

Solve the following system of equations by matrix method:

Question: Solve the following system of equations by matrix method: (i) $x+y-z=3$ $2 x+3 y+z=10$ $3 x-y-7 z=1$ (ii) $x+y+z=3$ $2 x-y+z=-1$ $2 x+y-3 z=-9$ (iii) $6 x-12 y+25 z=4$ $4 x+15 y-20 z=3$ $2 x+18 y+15 z=10$ (iv) $3 x+4 y+7 z=14$ $2 x-y+3 z=4$ $x+2 y-3 z=0$ (v) $\frac{2}{x}-\frac{3}{y}+\frac{3}{z}=10$ $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=10$ $\frac{3}{x}-\frac{1}{y}+\frac{2}{z}=13$ (vi) $5 x+3 y+z=16$ $2 x+y+3 z=19$ $x+2 y+4 z=25$ (vii) $3 x+4 y+2 z=8$ $2 y-3 z=3$ $x-2 y+6 z=-2$ (viii) $2...

Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF.

Question: PointsEandFlie on diagonalACof a parallelogram ABCD such thatAE=CF. What type of quadrilateral isBFDE? Solution: In the IIgm $\mathrm{ABCD}:$ $\mathrm{AO}=\mathrm{OC} \ldots \ldots$ (i) (diagonals of a parallelogram bisect each other) $\mathrm{AE}=\mathrm{CF} \ldots \ldots$ (ii) (given) $\mathrm{Subtracting}$ (ii) from (i): $\mathrm{AO}-\mathrm{AE}=\mathrm{OC}-\mathrm{CF}$ $\mathrm{EO}=\mathrm{OF} \ldots \ldots$ (iii) $\mathrm{In} \Delta \mathrm{DOE}$ and $\Delta \mathrm{BOF}:$ $\mathr...

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, the total area of canvas required is(a) 1760 m2(b) 2640 m2(c) 3960 m2(d) 7920 m2 Solution: (d) 7920 m2Total area of the canvas required = (Curved surface area of the cylinder) + (Curved surface area of the cone) $=(2 \pi r h+\pi r l)$ $=\left(2 \times \frac{22}{7} \times \frac{105}{2} \times 4\right)+\left(\frac{22}{7} \times \frac{105}{2} \times 40\right)\left[d...

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, the total area of canvas required is(a) 1760 m2(b) 2640 m2(c) 3960 m2(d) 7920 m2 Solution: (d) 7920 m2Total area of the canvas required = (Curved surface area of the cylinder) + (Curved surface area of the cone) $=(2 \pi r h+\pi r l)$ $=\left(2 \times \frac{22}{7} \times \frac{105}{2} \times 4\right)+\left(\frac{22}{7} \times \frac{105}{2} \times 40\right)\left[d...

If the radii of the ends of a bucket are 5 cm and 15 cm and it is 24 cm high, then its surface area is

Question: If the radii of the ends of a bucket are 5 cm and 15 cm and it is 24 cm high, then its surface area is(a) 1815.3 cm2(b) 1711.3 cm2(c) 2025.3cm2(d) 2360 cm2 Solution: (b) 1711.3 cm2LetRandrbe the radii of the top and base of the bucket, respectively, and lethandlbe its height and slant height.Then, $R=15 \mathrm{~cm}, r=5 \mathrm{~cm}, h=24 \mathrm{~cm}$ $l=\sqrt{h^{2}+(R-r)^{2}}$ $=\sqrt{(24)^{2}+(15-5)^{2}}$ $=\sqrt{576+100}$ $=\sqrt{676}$ $=26 \mathrm{~cm}$ Surface area of the bucket...

If the radii of the ends of a bucket are 5 cm and 15 cm and it is 24 cm high, then its surface area is

Question: If the radii of the ends of a bucket are 5 cm and 15 cm and it is 24 cm high, then its surface area is(a) 1815.3 cm2(b) 1711.3 cm2(c) 2025.3cm2(d) 2360 cm2 Solution: (b) 1711.3 cm2LetRandrbe the radii of the top and base of the bucket, respectively, and lethandlbe its height and slant height.Then, $R=15 \mathrm{~cm}, r=5 \mathrm{~cm}, h=24 \mathrm{~cm}$ $l=\sqrt{h^{2}+(R-r)^{2}}$ $=\sqrt{(24)^{2}+(15-5)^{2}}$ $=\sqrt{576+100}$ $=\sqrt{676}$ $=26 \mathrm{~cm}$ Surface area of the bucket...

Diagonals of a parallelogram ABCD intersect at O.

Question: Diagonals of a parallelogramABCDintersect atO.ALandCMare drawn perpendiculars toBDsuch thatLandMlie onBD. IsAL=CM? Why or why not? Solution: In $\Delta \mathrm{AOL}$ and $\Delta \mathrm{CMO}:$ $\angle \mathrm{AOL}=\angle \mathrm{COM}($ vertically opposite angle $) \ldots(i)$ $\angle \mathrm{ALO}=\angle \mathrm{CMO}=90^{\circ}(e$ ach right angle $) \ldots($ ii $)$ Usin $g$ angle sum property $:$ $\angle \mathrm{AOL}+\angle \mathrm{ALO}+\angle L A O=180^{\circ} \ldots \ldots \ldots($ iii...

The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm.

Question: The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm. Its volume is(a) 60060 cm3(b) 80080 cm3(c) 70040 cm3(d) 80160 cm3 Solution: (b) 80080 cm3LetRandrbe the radii of the top and base of the bucket, respectively, and lethbe its height.Then, $R=35 \mathrm{~cm}, r=14 \mathrm{~cm}, h=40 \mathrm{~cm}$ Volume of the bucket = Volume of the frustum of the cone $=\frac{1}{3} \pi h\left[R^{2}+r^{2}+R r\right] \mathrm{cm}^{3}$ $=\frac{1}{3} \times \fra...

In Fig. 17.31, ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A.

Question: In Fig. 17.31,ABCDis a parallelogram,CEbisects CandAFbisects A. In each of the following, if the statement is true, give a reason for the same: (i) A= C (ii) $\angle F A B=\frac{1}{2} \angle A$ (iii) $\angle D C E=\frac{1}{2} \angle C$ (iv) $\angle C E B=\angle F A B$ (v)CE||AF Solution: (i) True, since opposite angles of a parallelogram are equal. (ii) True, as AF is the bisector ofA. (iii) True, as CE is the bisector ofC. (iv) True CEB =DCE........(i) (alternate angles) DCE=FAB.........

The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm.

Question: The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm. Its volume is(a) 60060 cm3(b) 80080 cm3(c) 70040 cm3(d) 80160 cm3 Solution: (b) 80080 cm3LetRandrbe the radii of the top and base of the bucket, respectively, and lethbe its height.Then, $R=35 \mathrm{~cm}, r=14 \mathrm{~cm}, h=40 \mathrm{~cm}$ Volume of the bucket = Volume of the frustum of the cone $=\frac{1}{3} \pi h\left[R^{2}+r^{2}+R r\right] \mathrm{cm}^{3}$ $=\frac{1}{3} \times \fra...

The total surface area of a hemisphere of radius 7 cm is

Question: The total surface area of a hemisphere of radius 7 cm is(a) (588) cm2(b) (392) cm2(c) (147) cm2(d) (98) cm2 Solution: (c) (147) cm2Radius of the hemisphere = 7 cm Total surface area of the hemisphere $=$ Curved surface area of hemisphere $+$ Area of the circle $=3 \pi r^{2}$ $=(3 \pi \times 7 \times 7) \mathrm{cm}^{2}$ $=(147 \pi) \mathrm{cm}^{2}$...

Diagonals of parallelogram ABCD intersect at O as shown in Fig. 17.30. XY contains O,

Question: Diagonals of parallelogramABCDintersect atOas shown in Fig. 17.30.XYcontainsO, andX,Yare points on opposite sides of the parallelogram. Give reasons for each of the following: (i)OB=OD (ii) OBY= ODX (iii) BOY= DOX (iv) ∆BOY ∆DOX Now, state ifXYis bisected atO. Solution: (i) Diagonals of a parallelogram bisect each other. (ii) Alternate angles (iii) Vertically opposite angles (iv)In $\Delta$ BOY and $\Delta$ DOX: $\mathrm{OB}=\mathrm{OD} \quad$ (diagonals of a parallelogram bisect each ...

In Fig. 17.29, suppose it is known that DE = DF.

Question: In Fig. 17.29, suppose it is known thatDE=DF. Then, is ΔABCisosceles? Why or why not? Solution: In $\Delta$ FDE : DE $=$ DF $\therefore \angle \mathrm{FED}=\angle \mathrm{DFE} \ldots \ldots \ldots \ldots$ (i) (angles opposite to equal sides) In the IIg BDEF : $\angle \mathrm{FBD}=\angle \mathrm{FED} \ldots \ldots$ (ii) (opposite angles of a parallelogram are equal) In the II $^{\mathrm{gm}}$ DCEF: $\angle \mathrm{DCE}=\angle \mathrm{DFE} \ldots$... (iii) (opposite angles of a parallelo...

Find the values of x and y in

Question: Find the values of x and y in the following rectangle Solution: By property of rectangle, Lengths are equal, i.e., $C D=A B$ $\Rightarrow$ $x+3 y=13$ $\ldots$ (i) Breadth are equal, i.e., $A D=B C$ $\Rightarrow$ $3 x+y=7$ ...(ii) On multiplying Eq. (ii) by 3 and then subtracting Eq. (i), we get On putting $x=1$ in Eq. (i), we get $3 y=12 \Rightarrow y=4$ Hence, the required values of $x$ and $y$ are 1 and 4 , respectively....

In Fig. 17.29, BDEF and DCEF are each a parallelogram.

Question: In Fig. 17.29,BDEFandDCEFare each a parallelogram. Is it true thatBD=DC? Why or why not? Solution: In parallelogram BDEF $\therefore \mathrm{BD}=\mathrm{EF} \quad \ldots$ (i) (opposite sides of a parallelogram are equal) In parallelogram DCEF $\mathrm{CD}=\mathrm{EF} \quad \ldots$ (ii) $\quad$ (opposite sides of a parallelogram are equal) From equations (i) and (ii) $\mathrm{BD}=\mathrm{CD}$...

The surface area of a sphere is 154 cm2.

Question: The surface area of a sphere is 154 cm2. The volume of the sphere is (a) $179 \frac{2}{3} \mathrm{~cm}^{3}$ (b) $359 \frac{1}{3} \mathrm{~cm}^{3}$ (c) $1437 \frac{1}{3} \mathrm{~cm}^{3}$ (d) None of these Solution: (a) $179 \frac{2}{3} \mathrm{~cm}^{3}$ Surface area of a sphere $=4 \pi r^{2}$ Therefore, $4 \pi r^{2}=154$ $\Rightarrow 4 \times \frac{22}{7} \times r^{2}=154$ $\Rightarrow r^{2}=\left(154 \times \frac{7}{88}\right)$ $\Rightarrow r^{2}=\frac{49}{4}$ $\Rightarrow r^{2}=\left...

The surface area of a sphere is 154 cm2.

Question: The surface area of a sphere is 154 cm2. The volume of the sphere is (a) $179 \frac{2}{3} \mathrm{~cm}^{3}$ (b) $359 \frac{1}{3} \mathrm{~cm}^{3}$ (c) $1437 \frac{1}{3} \mathrm{~cm}^{3}$ (d) None of these Solution: (a) $179 \frac{2}{3} \mathrm{~cm}^{3}$ Surface area of a sphere $=4 \pi r^{2}$ Therefore, $4 \pi r^{2}=154$ $\Rightarrow 4 \times \frac{22}{7} \times r^{2}=154$ $\Rightarrow r^{2}=\left(154 \times \frac{7}{88}\right)$ $\Rightarrow r^{2}=\frac{49}{4}$ $\Rightarrow r^{2}=\left...

Write a pair of linear equations

Question: Write a pair of linear equations which has the unique solution x = 1 and y = 3. How many such pairs can you write? Solution: Condition for the pair of system to have unique solution $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ Let the equations are, $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ Since, $x=-1$ and $y=3$ is the unique solution of these two equations, then $a_{1}(-1)+b_{1}(3)+c_{1}=0$ $\Rightarrow \quad-a_{1}+3 b_{1}+c_{1}=0$$\ldots$ (i) and $\quad a_{2}(-1)+b_...

In Fig. 17.28, ABCD and AEFG are parallelograms.

Question: In Fig. 17.28,ABCDandAEFGare parallelograms. If C= 55, what is the measure of F? Solution: Both the parallelograms $\mathrm{ABCD}$ and $\mathrm{AEFG}$ are similar. $\therefore \angle \mathrm{C}=\angle \mathrm{A}=55^{\circ}$ (opposite angles of a parallelogram are equal) $\therefore \angle \mathrm{A}=\angle F=55^{\circ} \quad$ (opposite angles of a parallelogram are equal)...

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