Two chords AB and AC of a circle with
Question: Two chords AB and AC of a circle with centre O are on the opposite sides of OA. Then, OAB = OAC. Solution: False In figure, AB and AC are two chords of a circle. Join OB and OC. In ΔOAB and ΔOAC, In $\triangle O A B$ and $\triangle O A C_{1}$ $O A=O A$ [common side] $O B=O C$ [both are the radius of circle] Here, we are not able to show that either the any angle or third side is equal and $\triangle O A B$ is not congruent to $\triangle O A C$. $\therefore \quad \angle O A B \neq \angl...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{3 x+5}{4 x+2}=\frac{3 x+4}{4 x+7}$ Solution: $\frac{3 x+5}{4 x+2}=\frac{3 x+4}{4 x+7}$ or, $12 x^{2}+20 x+21 x+35=12 x^{2}+16 x+6 x+8 \quad$ [Cross multiply $]$ or, $12 \mathrm{x}^{2}-12 \mathrm{x}^{2}+41 \mathrm{x}-22 \mathrm{x}=8-35$ or, $19 \mathrm{x}=-27$ or, $\mathrm{x}=\frac{-27}{19}$ Thus, $\mathrm{x}=\frac{-27}{19}$ is the solution of the given equation Check : Substituting $\mathrm{x}=\frac{-27}{19}$ in the given equa...
Read More →Two chords AB and CD of a circle
Question: Two chords AB and CD of a circle are each at distances 4 cm from the centre. Then, AB = CD. Solution: TrueBecause, the chords equidistant from the centre of circle are equal in length....
Read More →In figure, if ∠AOB = 90°
Question: In figure, if AOB = 90 and ABC = 30, then CAO is equal to (a)30 (b)45 (c)90 (d)60 Solution: (d) In AOAB, OAB + ABO + BOA = 180 OAB + OAB + 90 = 180 = 2OAB = 180- 90 [angles opposite to equal sides are equal] [angle sum property of a triangle] [from Eq. (i)] = OAB = 90/2 = 45 (i) In ΔACB, ACB + CBA + CAB = 180 45+ 30+ CAB = 180 = CAB = 180 75 = 105 CAO+ OAB = 105 CAO + 45 = 105 CAO = 105 45 = 60...
Read More →In the given figure, PQRS represents a flower bed.
Question: In the given figure,PQRSrepresents a flower bed. IfOP= 21 m andOR= 14 m, find the area of the flower bed. Solution: Area of the flower bed is the difference between the areas of sectors OPQ and ORS. Area of the flower bed $=\frac{\theta}{360} \times \pi\left(\mathrm{PO}^{2}-\mathrm{OR}^{2}\right)$ $=\frac{90}{360} \times \frac{22}{7}\left(21^{2}-14^{2}\right)$ $=\frac{1}{4} \times \frac{22}{7} \times 35 \times 7$ $=192.5 \mathrm{~m}^{2}$...
Read More →In the given figure, PQRS represents a flower bed.
Question: In the given figure,PQRSrepresents a flower bed. IfOP= 21 m andOR= 14 m, find the area of the flower bed. Solution: Area of the flower bed is the difference between the areas of sectors OPQ and ORS. Area of the flower bed $=\frac{\theta}{360} \times \pi\left(\mathrm{PO}^{2}-\mathrm{OR}^{2}\right)$ $=\frac{90}{360} \times \frac{22}{7}\left(21^{2}-14^{2}\right)$ $=\frac{1}{4} \times \frac{22}{7} \times 35 \times 7$ $=192.5 \mathrm{~m}^{2}$...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{2}{3 x}-\frac{3}{2 x}=\frac{1}{12}$ Solution: $\frac{2}{3 x}-\frac{3}{2 x}=\frac{1}{12}$ or, $\frac{4-9}{6 x}=\frac{1}{12}$ or, $\frac{-5}{x}=\frac{1}{2}$ or, $\mathrm{x}=-10[$ After $c$ ross multiplication $]$ Thus, $x=-10$ is the solution of the given equation. Check : Substituting $\mathrm{x}=-10$ in the given equation, we get: L.H.S. $=\frac{2}{3(-10)}-\frac{3}{2(-10)}=\frac{2}{-30}-\frac{3}{-20}=\frac{4-9}{-60}=\frac{-5}{...
Read More →In figure, BC is a diameter of the circle
Question: In figure, BC is a diameter of the circle and BAO = 60. Then, ADC is equal to OA = OB [both are the radius of circle] (a) $30^{\circ}$ (d) $45^{\circ}$ (d) $60^{\circ}$ (d) $120^{\circ}$ Solution: (c)In ΔAOB, OBA = BAO [angles opposite to equal sides are equal] OBA = 60 [ BAO = 60, given] ABC=ADC [angles in the same segment AC are equal] ADC = 60...
Read More →A park is in the form of a rectangle 120 m by 90 m.
Question: A park is in the form of a rectangle 120 m by 90 m. At the centre of the park there is a circular lawn as shown in the figure. The area of the park excluding the lawn is 2950 m2. Find the radius of the circular lawn. Solution: Area of the rectangle $=l \times b$ $=120 \times 90$ $=10800$ sq. $\mathrm{m}$ Area of the park excluding the lawn = 2950 m2Area of the circular lawn = Area of the parkArea of the park excluding the lawn $=10800-2950$ $=7850 \mathrm{~m}^{2}$ Area of the circular ...
Read More →A park is in the form of a rectangle 120 m by 90 m.
Question: A park is in the form of a rectangle 120 m by 90 m. At the centre of the park there is a circular lawn as shown in the figure. The area of the park excluding the lawn is 2950 m2. Find the radius of the circular lawn. Solution: Area of the rectangle $=l \times b$ $=120 \times 90$ $=10800$ sq. $\mathrm{m}$ Area of the park excluding the lawn = 2950 m2Area of the circular lawn = Area of the parkArea of the park excluding the lawn $=10800-2950$ $=7850 \mathrm{~m}^{2}$ Area of the circular ...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{6}{2 x-(3-4 x)}=\frac{2}{3}$ Solution: $\frac{6}{2 \mathrm{x}-(3-4 \mathrm{x})}=\frac{2}{3}$ or $\frac{6}{6 \mathrm{x}-3}=\frac{2}{3}$ or $12 \mathrm{x}-6=18 \quad$ [After c ross multiplication] or $12 \mathrm{x}=18+6$ or $\mathrm{x}=\frac{24}{12}$ or $\mathrm{x}=2$ Thus, $\mathrm{x}=2$ is the solution of the given equation. Check: Substituting $x=2$ in the given equation, we get: L.H.S. $=\frac{6}{2 \times 2-(3-4 \times 2)}=\...
Read More →ABCD is a cyclic quadrilateral
Question: ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ADC = 140, then BAC is equal to (a)80 (b)50 (c)40 (d)30 Solution: (b)Given, ABCD is a cyclic quadrilateral and ADC = 140. We know that, sum of the opposite angles in a cyclic quadrilateral is 180. We know that, sum of the opposite angles in a cyclic quadrilateral is $180^{\circ}$. $\angle A D C+\angle A B C=180^{\circ}$ $\Rightarrow \quad 140^{\circ}+\angle A B C=180^{\circ}$ $\Rightarrow \qua...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{y-(7-8 y)}{9 y-(3+4 y)}=\frac{2}{3}$ Solution: $\frac{\mathrm{y}-(7-8 \mathrm{y})}{9 \mathrm{y}-(3+4 \mathrm{y})}=\frac{2}{3}$ or $\frac{9 \mathrm{y}-7}{5 \mathrm{y}-3}=\frac{2}{3}$ or $27 \mathrm{y}-21=10 \mathrm{y}-6$ [After c ross multipl ication] or $27 \mathrm{y}-10 \mathrm{y}=-6+21$ or $17 \mathrm{y}=15$ or $\mathrm{y}=\frac{15}{17}$ Thus, $\mathrm{y}=\frac{10}{17}$ is the solution of the given equation. Check : Substitu...
Read More →The cost of fencing a circular field at the rate of Rs 25 per metre is Rs 5500.
Question: The cost of fencing a circular field at the rate of Rs 25 per metre is Rs 5500. The field is to be ploughed at the rate of 50 paise per m2. Find the cost of ploughing the field.$\left[\right.$ Take $\left.\pi=\frac{22}{7}\right]$ Solution: Circumference $=\frac{\text { Total cost of fencing }}{\text { Rate of fencing }}=\frac{5500}{25}=220$ Let the radius of the circle ber.Now, Circumference $=220$ $\Rightarrow 2 \pi r=220$ $\Rightarrow r=35 \mathrm{~cm}$ Now, Area of field $=\pi r^{2...
Read More →In figure, if ∠DAB = 60° ,
Question: In figure, if $\angle D A B=60^{\circ}, \angle A B D=50^{\circ}$, then $\angle A C B$ is equal to (a)60 (b)50 (c)70 (d)80 Thinking Process Use the theorem that angles in the same segment of a circle are equal and further simplify it. Solution: (c)Given, DAB = 60, ABD = 50 Since, ADB = ACB (i) [angles in same segment of a circle are equal] In ΔABD, ABD + ADB + DAB = 180 [by angle sum property of a triangle] 50+ ADB + 60 = 180 = ADB = 180 110 = 70 $\Rightarrow \angle \mathrm{ADB}=180^{\c...
Read More →A square ABCD is inscribed in a circle of radius r.
Question: A square ABCD is inscribed in a circle of radiusr. Find the area of the square. Solution: Let the diameter of the square bedand having circumscribed circle of radiusr.We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.d= 2rNow, Area of square $=\frac{1}{2} d^{2}=\frac{1}{2}(2 r)^{2}=2 r^{2}$ sq units Hence, the area of the square ABCD is 2r2sq units....
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{2 x}{3 x+1}=-3$ Solution: $\frac{2 x}{3 x+1}=-3$ or $2 \mathrm{x}=-9 \mathrm{x}-3$ [ After c ross multipl ication ] or $2 \mathrm{x}+9 \mathrm{x}=-3$ or $11 \mathrm{x}=-3$ or $\mathrm{x}=\frac{-3}{11}$ Thus, $\mathrm{x}=\frac{-3}{11}$ is the solution of the given equation. Check : Substituting $\mathrm{x}=\frac{-3}{11}$ in the given equation, we get: L.H.S. $=\frac{2\left(\frac{-3}{11}\right)}{3\left(\frac{-3}{11}\right)+1}=\f...
Read More →A wire is bent to form a square enclosing an area of 484 cm2.
Question: A wire is bent to form a square enclosing an area of 484 cm2.Using the same wire, a circle is formed. Find the area of the circle. Solution: Area of the circle = 484 cm2 Area of the square $=\operatorname{Side}^{2}$ $\Rightarrow 484=$ Side $^{2}$ $\Rightarrow 22^{2}=$ Side $^{2}$ $\Rightarrow$ Side $=22 \mathrm{~cm}$ Perimeter of the square $=4 \times$ Side Perimeter of the square $=4 \times 22$ $=88 \mathrm{~cm}$ Length of the wire = 88 cmCircumference of the circle = Length of the wi...
Read More →A wire is bent to form a square enclosing an area of 484 cm2.
Question: A wire is bent to form a square enclosing an area of 484 cm2.Using the same wire, a circle is formed. Find the area of the circle. Solution: Area of the circle = 484 cm2 Area of the square $=\operatorname{Side}^{2}$ $\Rightarrow 484=$ Side $^{2}$ $\Rightarrow 22^{2}=$ Side $^{2}$ $\Rightarrow$ Side $=22 \mathrm{~cm}$ Perimeter of the square $=4 \times$ Side Perimeter of the square $=4 \times 22$ $=88 \mathrm{~cm}$ Length of the wire = 88 cmCircumference of the circle = Length of the wi...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{1-9 y}{19-3 y}=\frac{5}{8}$ Solution: $\frac{1-9 y}{19-3 y}=\frac{5}{8}$ or $8-72 \mathrm{y}=95-15 \mathrm{y}[$ After c ross multiplication $]$ or $95-15 \mathrm{y}=8-72 \mathrm{y}$ or $72 \mathrm{y}-15 \mathrm{y}=8-95$ or $57 \mathrm{y}=-87$ or $\mathrm{y}=\frac{-87}{57}$ or $\mathrm{y}=\frac{-29}{19}$ Thus $\mathrm{y}=\frac{-29}{19}$ is the solution of the given equation. Check : Substituting $\mathrm{y}=\frac{-29}{19}$ in t...
Read More →Find the inverse of each of the following matrices
Question: Find the inverse of each of the following matrices and verify that $A^{-1} A=I_{3}$. (i) $\left[\begin{array}{lll}1 3 3 \\ 1 4 3 \\ 1 3 4\end{array}\right]$ (ii) $\left[\begin{array}{lll}2 3 1 \\ 3 4 1 \\ 3 7 2\end{array}\right]$ Solution: (i) $A=\left[\begin{array}{lll}1 3 3 \\ 1 4 3 \\ 1 3 4\end{array}\right]$ Now, $C_{11}=\left|\begin{array}{ll}4 3 \\ 3 4\end{array}\right|=7, C_{12}=-\left|\begin{array}{ll}1 3 \\ 1 4\end{array}\right|=-1$ and $C_{13}=\left|\begin{array}{cc}1 4 \\ 1 ...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{2 x+1}{3 x-2}=\frac{5}{9}$ Solution: $\frac{2 x+1}{3 x-2}=\frac{5}{9}$ or $18 \mathrm{x}+9=15 \mathrm{x}-10[$ After c ross multiplication $]$ or $18 \mathrm{x}-15 \mathrm{x}=-10-9$ or $3 \mathrm{x}=-19$ or $\mathrm{x}=\frac{-19}{3}$ Thus, $\mathrm{x}=\frac{-19}{3}$ is the solution of the given equation. Check: Substituting $\mathrm{x}=\frac{-19}{3}$ in the given equation, we get: L. H.S. $=\frac{2\left(\frac{-19}{3}\right)+1}{...
Read More →In figure, if ∠DAB = 60° ,
Question: In figure, if $\angle D A B=60^{\circ}, \angle A B D=50^{\circ}$, then $\angle A C B$ is equal to (a)60 (b)50 (c)70 (d)80 Thinking Process Use the theorem that angles in the same segment of a circle are equal and further simplify it. Solution: (c)Given, DAB = 60, ABD = 50 Since, ADB = ACB (i) [angles in same segment of a circle are equal] In ΔABD, ABD + ADB + DAB = 180 [by angle sum property of a triangle] 50+ ADB + 60 = 180 = ADB = 180 110 = 70...
Read More →Find the area of the shaded region in the given figure,
Question: Find the area of the shaded region in the given figure, if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of the circle. Solution: In right triangle ABCAC2= AB2+ BC2= 82+62= 64 + 36= 100 AC2= 100⇒ AC = 10 cm Now, Radius of circle $(\mathrm{OA})=\frac{1}{2} \mathrm{AC}=5 \mathrm{~cm}$ Area of the shaded region = Area of circle Area of rectangle OABC $=\pi(\mathrm{OA})^{2}-\mathrm{AB} \times \mathrm{BC}$ $=\frac{22}{7} \times(5)^{2}-8 \times 6$ $=78.57-48$ $=30.57 \math...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{2 y+5}{y+4}=1$ Solution: $\frac{2 y+5}{y+4}=1$ or $2 y+5=y+4$ or $2 y-y=4-5$ or $y=-1$ Thus, $y=-1$ is the solution of the given equation. Check : Substituting $\mathrm{y}=-1$ in the given equation, we get: L. H.S. $=\frac{2(-1)+5}{-1+4}=\frac{-2+5}{3}=\frac{3}{3}=1$ R.H.S. $=1$ $\therefore$ L. H.S. $=$ R. H.S. for $\mathrm{y}=-1$....
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