A room 4.9 m long and 3.5 m broad is covered with carpet, leaving an uncovered margin of 25 cm all around the room.
Question: A room 4.9 m long and 3.5 m broad is covered with carpet, leaving an uncovered margin of 25 cm all around the room. If the breadth of the carpet is 80 cm, find its cost at Rs 80 per meter. Solution: Width of the room left uncovered = 0.25 mNow, Length of the room to be carpeted $=4.9-(0.25+0.25)=4.9-0.5=4.4 \mathrm{~m}$ Breadth of the room be carpeted $=3.5-(0.25+0.25)=3.5-0.5=3 \mathrm{~m}$ Area to be carpeted $=4.4 \times 3=13.2 \mathrm{~m}^{2}$ Breadth of the carpet = 80 cm = 0.8 mW...
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Question: $x-y+z=3$ $2 x+y-z=2$ $-x-2 y+2 z=1$ Solution: Using the equations we get $D=\left|\begin{array}{rrr}1 -1 1 \\ 2 1 -1 \\ -1 -2 2\end{array}\right|$ $\Rightarrow 1(2-2)+1(4-1)+1(-4+1)=0$ $D_{1}=\left|\begin{array}{rrr}3 -1 1 \\ 2 1 -1 \\ 1 -2 2\end{array}\right|$ $\Rightarrow 3(2-2)+1(4+1)+1(-4-1)=0$ $D_{2}=\left|\begin{array}{rrr}1 3 1 \\ 2 2 -1 \\ -1 1 2\end{array}\right|$ $\Rightarrow 1(4+1)-3(4-1)+1(2+2)=0$ $D_{3}=\left|\begin{array}{rrr}1 -1 3 \\ 2 1 2 \\ -1 -2 1\end{array}\right|$...
Read More →Find the values of the following expressions:
Question: Find the values of the following expressions: (i) $16 x^{2}+24 x+9$, when $x=\frac{7}{4}$ (ii) $64 x^{2}+81 y^{2}+144 x y$, when $x=11$ and $y=\frac{4}{3}$ (iii) $81 x^{2}+16 y^{2}-72 x y$, when $x=\frac{2}{3}$ and $y=\frac{3}{4}$ Solution: (i) Let us consider the following expression: $16 x^{2}+24 x+9$ now, $16 x^{2}+24 x+9=(4 x+3)^{2} \quad\left(\right.$ Using identity $\left.(a+b)^{2}=a^{2}+2 a b+b^{2}\right)$ $\Rightarrow 16 x^{2}+24 x+9=\left(4 \times \frac{7}{4}+3\right)^{2} \qua...
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Question: $x-y+z=3$ $2 x+y-z=2$ $-x-2 y+2 z=1$ Solution: Using the equations we get $D=\left|\begin{array}{rrr}1 -1 1 \\ 2 1 -1 \\ -1 -2 2\end{array}\right|$ $\Rightarrow 1(2-2)+1(4-1)+1(-4+1)=0$ $D_{1}=\left|\begin{array}{rrr}3 -1 1 \\ 2 1 -1 \\ 1 -2 2\end{array}\right|$ $\Rightarrow 3(2-2)+1(4+1)+1(-4-1)=0$ $D_{2}=\left|\begin{array}{rrr}1 3 1 \\ 2 2 -1 \\ -1 1 2\end{array}\right|$ $\Rightarrow 1(4+1)-3(4-1)+1(2+2)=0$ $D_{3}=\left|\begin{array}{rrr}1 -1 3 \\ 2 1 2 \\ -1 -2 1\end{array}\right|$...
Read More →Two lines are respectively perpendicular
Question: Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. Solution: Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p m, p n, q m, q n To prove p||g Proof Since, m || n and p is perpendicular to m and n. $\therefore$$\angle 1=\angle 10=90^{\circ}$[corresponding angles] Similarly, $\angle 2=\angle 9=90^{\circ}$ [corresponding angles] $[\because p \perp m$ and $p \perp n]$...
Read More →If 3x + 5y = 11 and xy = 2,
Question: If 3x+ 5y= 11 andxy= 2, find the value of 9x2+ 25y2 Solution: We have: $(3 x+5 y)^{2}=(3 x)^{2}+2(3 x)(5 y)+(5 y)^{2}$ $\Rightarrow(3 x+5 y)^{2}=9 x^{2}+30 x y+25 y^{2}$ $\Rightarrow 9 x^{2}+25 y^{2}=(3 x+5 y)^{2}-30 x y$ $\Rightarrow 9 x^{2}+25 y^{2}=11^{2}-30 \times 2 \quad(\because 3 x+5 y=11$ and $x y=2)$ $\Rightarrow 9 x^{2}+25 y^{2}=121-60$ $\Rightarrow 9 x^{2}+25 y^{2}=61$...
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Question: $3 x-y+2 z=6$ $2 x-y+z=2$ $3 x+6 y+5 z=20$ Solution: Given: $3 x-y+2 z=6$ $D=\left|\begin{array}{ccc}3 -1 2 \\ 2 -1 1 \\ 3 6 5\end{array}\right|$ $3(-5-6)+1(10-3)+2(12+3)=4$ Since $D$ is non-zero, the system of linear equations is consistent and has a unique solution. $D_{1}=\left|\begin{array}{ccc}6 -1 2 \\ 2 -1 1 \\ 20 6 5\end{array}\right|$ $=6(-5-6)+1(10-20)+2(12+20)$ $=-66-10+64$ $=-12$ $D_{2}=\left|\begin{array}{ccc}3 6 2 \\ 2 2 1 \\ 3 20 5\end{array}\right|$ $=3(10-20)-6(10-3)+2...
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Question: $3 x-y+2 z=6$ $2 x-y+z=2$ $3 x+6 y+5 z=20$ Solution: Given: $3 x-y+2 z=6$ $D=\left|\begin{array}{ccc}3 -1 2 \\ 2 -1 1 \\ 3 6 5\end{array}\right|$ $3(-5-6)+1(10-3)+2(12+3)=4$ Since $D$ is non-zero, the system of linear equations is consistent and has a unique solution. $D_{1}=\left|\begin{array}{ccc}6 -1 2 \\ 2 -1 1 \\ 20 6 5\end{array}\right|$ $=6(-5-6)+1(10-20)+2(12+20)$ $=-66-10+64$ $=-12$ $D_{2}=\left|\begin{array}{ccc}3 6 2 \\ 2 2 1 \\ 3 20 5\end{array}\right|$ $=3(10-20)-6(10-3)+2...
Read More →If x − y = 7 and xy = 9,
Question: Ifxy= 7 andxy= 9, find the value ofx2+y2 Solution: We have: $(x-y)^{2}=x^{2}-2 x y+y^{2}$ $\Rightarrow x^{2}+y^{2}=(x-y)^{2}+2 x y$ $\Rightarrow x^{2}+y^{2}=7^{2}+2 \times 9 \quad(\because x-y=7$ and $x y=9)$ $\Rightarrow x^{2}+y^{2}=7^{2}+2 \times 9$ $\Rightarrow x^{2}+y^{2}=49+18$ $\Rightarrow x^{2}+y^{2}=67$...
Read More →A ΔABC is right angled at A.
Question: A ΔABC is right angled at A. L is a point on BC such that AL BC. Prove that BAL = ACB. Solution: Given In ΔABC, A = 90 and AL BC To prove BAL = ACB Proof In ΔABC and ΔLAC, BAC = ALC [each 90] (i) and ABC = ABL [common angle] (ii) On adding Eqs. (i) and (ii), we get BAC + ABC = ALC + ABL (iii) Again, in ΔABC, BAC + ACB + ABC = 180 [sum of all angles of a triangle is 180] =BAC+ABC = 1 80-ACB (iv) In ΔABL, ABL + ALB + BAL = 180 [sum of all angles of a triangle is 180] = ABL+ ALC = 180 BAL...
Read More →If x + y = 4 and xy = 2,
Question: Ifx+y= 4 andxy= 2, find the value ofx2+y2 Solution: We have: $(x+y)^{2}=x^{2}+2 x y+y^{2}$ $\Rightarrow x^{2}+y^{2}=(x+y)^{2}-2 x y$ $\Rightarrow x^{2}+y^{2}=4^{2}-2 \times 2 \quad(\because x+y=4$ and $x y=2)$ $\Rightarrow x^{2}+y^{2}=16-4$ $\Rightarrow x^{2}+y^{2}=12$...
Read More →In the figure, DE || QR and AP and BP
Question: In the figure, DE || QR and AP and BP are bisectors of EAB and RBA, respectively. Find APB. Solution: Given, $D E \| Q R$ and $A P$ and $P B$ are the bisectors of $\angle E A B$ and $\angle R B A$, respectively. We know that, the interior angles on the same side of transversal are supplementary. $\therefore$ $\angle E A B+\angle R B A=180^{\circ}$ $\Rightarrow \quad \frac{1}{2} \angle E A B+\frac{1}{2} \angle R B A=\frac{180^{\circ}}{2} \quad$ [dividing both sides by 2] $\Rightarrow \q...
Read More →Solve the following
Question: If $x^{2}+\frac{1}{x^{2}}=18$, find the values of $x+\frac{1}{x}$ and $x-\frac{1}{x}$. Solution: Let us consider the following expression: $x+\frac{1}{x}$ Squaring the above expression, we get: $\left(x+\frac{1}{x}\right)^{2}=x^{2}+2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=x^{2}-2+\frac{1}{x^{2}} \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$ $\Rightarrow\left(x+\frac{1}{x}\right)^{2}=x^{2}+2+\frac{1}{x^{2}}$ $\Rightarrow\left(x+\frac{1}{x}\right)^{2}=20 \quad\left(\be...
Read More →The length and the breadth of a rectangular garden are in the ratio 9 : 5.
Question: The length and the breadth of a rectangular garden are in the ratio 9 : 5. A path 3.5 m wide, running all around inside it has an area of 1911 m2. Find the dimensions of the garden. Solution: Let the length and breadth of the garden be 9xm and 5xm, respectively,Now, Area of the garden $=(9 x \times 5 x)=45 x^{2}$ Length of the garden excluding the path $=(9 x-7)$ Breadth of the garden excluding the path $=(5 x-7)$ Area of the path $=45 x^{2}-[(9 x-7)(5 x-7)]$ $\Rightarrow 1911=45 x^{2}...
Read More →The length and the breadth of a rectangular garden are in the ratio 9 : 5.
Question: The length and the breadth of a rectangular garden are in the ratio 9 : 5. A path 3.5 m wide, running all around inside it has an area of 1911 m2. Find the dimensions of the garden. Solution: Let the length and breadth of the garden be 9xm and 5xm, respectively,Now, Area of the garden $=(9 x \times 5 x)=45 x^{2}$ Length of the garden excluding the path $=(9 x-7)$ Breadth of the garden excluding the path $=(5 x-7)$ Area of the path $=45 x^{2}-[(9 x-7)(5 x-7)]$ $\Rightarrow 1911=45 x^{2}...
Read More →Solve the following
Question: If $x-\frac{1}{x}=3$, find the values of $x^{2}+\frac{1}{x^{2}}$ and $x^{4}+\frac{1}{x^{4}}$. Solution: Let us consider the following equation: $x-\frac{1}{x}=3$ Squaring both sides, we get: $\left(x-\frac{1}{x}\right)^{2}=(3)^{2}=9$ $\Rightarrow\left(x-\frac{1}{x}\right)^{2}=9$ $\Rightarrow x^{2}-2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=9$ $\Rightarrow x^{2}-2+\frac{1}{x^{2}}=9$ $\Rightarrow x^{2}+\frac{1}{x^{2}}=11$ (Adding 2 to both sides) Squaring both sides again...
Read More →In the figure, BA || ED and BC || EF.
Question: In the figure, BA || ED and BC || EF. Show that ABC + DEF = 180. Solution: Given BA || ED and BC || EF To show, ABC + DEF = 180 Construction Draw a ray PE opposite to ray EF. Proof in the figure, $B C \| E F$ $\therefore \angle E P B+\angle P B C=180^{\circ}$[sum of cointerior angles is $180^{\circ}$ ] ...(i) Now, $A B \| E D$ and $P E$ is a transversal line. $\therefore \quad \angle E P B=\angle D E F \quad$ [corresponding angles] ...(ii) From Egs. (i) and (ii), $\quad \angle D E F+\a...
Read More →A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide.
Question: (i) A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide. If the area of the path is 420 m2., find the width of the path.(ii) A carpet is laid on the floor of a room 8 m by 5 m. There is a border of constant width all around the carpet. If the area of the border is 12 m2, find its width. Solution: (i) Area of the rectangular field $=54 \times 35=1890 \mathrm{~m}^{2}$ Let the width of the path bexm. The path is shown in the following diag...
Read More →Solve the following
Question: If $x+\frac{1}{x}=20$, find the value of $x^{2}+\frac{1}{x^{2}}$ Solution: Let us consider the following equation: $x+\frac{1}{x}=20$ Squaring both sides, we get: $\left(x+\frac{1}{x}\right)^{2}=(20)^{2}=400$ $\Rightarrow\left(x+\frac{1}{x}\right)^{2}=400$ $\Rightarrow x^{2}+2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=400 \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$ $\Rightarrow x^{2}+2+\frac{1}{x^{2}}=400$ $\Rightarrow x^{2}+\frac{1}{x^{2}}=398$ (Subtracting 2 from bo...
Read More →A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide.
Question: (i) A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide. If the area of the path is 420 m2., find the width of the path.(ii) A carpet is laid on the floor of a room 8 m by 5 m. There is a border of constant width all around the carpet. If the area of the border is 12 m2, find its width. Solution: (i) Area of the rectangular field $=54 \times 35=1890 \mathrm{~m}^{2}$ Let the width of the path bexm. The path is shown in the following diag...
Read More →In the figure, BA || ED and BC || EF.
Question: In the figure, BA || ED and BC || EF. Show that ABC = DEF. Solution: Given BA || ED and BC || EF. To show ABC = DEF. Construction Draw a ray EP opposite to ray ED. Proof In the figure, $B A \| E D$ or $B A \| D P$ $\therefore \quad \angle A B P=\angle E P C$[corresponding angles] $\Rightarrow \quad \angle A B C=\angle E P C$ $\ldots($ i) Again, $B C \| E F$ or $P C \| E F$ $\therefore \quad \angle D E F=\angle E P C \quad$ [corresponding angles] ...(ii) From Eqs. (i) and (ii), $\angle ...
Read More →Find the value of x, if:
Question: Find the value ofx, if: (i) 4x= (52)2 (48)2 (ii) 14x= (47)2 (33)2 (iii) 5x= (50)2 (40)2 Solution: (i) Let us consider the following equation: $4 x=(52)^{2}-(48)^{2}$ Using the identity $(a+b)(a-b)=a^{2}-b^{2}$, we get: $4 x=(52)^{2}-(48)^{2}$ $4 x=(52+48)(52-48)$ $4 x=100 \times 4=400$ $\Rightarrow 4 x=400$ $\Rightarrow x=100$ (Dividing both sides by 4) (ii) Let us consider the following equation: $14 x=(47)^{2}-(33)^{2}$ Using the identity $(a+b)(a-b)=a^{2}-b^{2}$, we get: $14 x=(47)^...
Read More →In the given figure, bisectors AP and BQ
Question: In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l ||m. Solution: Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles CAB and ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore PAB = ABQ [alternate interior angles] = 2 PAB = 2 ABQ [multiplying both sides by 2] So, alternate interior angles are equal. We know that, if two alternate interior angles are equal, then lines are para...
Read More →AP and BQ are the bisectors of the two alternate
Question: AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. Solution: Given In the figure l || m, AP and BQ are the bisectors of EAB and ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. EAB = ABH [dividing both sides by 2] PAB =ABQ [AP and BQ are the bisectors of EAB and ABH] Since, PAB and ABQ are alternate interior angles with two line...
Read More →Simplify the following using the identities:
Question: Simplify the following using the identities: (i) $\frac{58^{2}-42^{2}}{16}$ (ii) $178 \times 178-22 \times 22$ (iii) $\frac{198 \times 198-102 \times 102}{96}$ (iv) $1.73 \times 1.73-0.27 \times 0.27$ (v) $\frac{8.63 \times 8.63-1.37 \times 1.37}{0.726}$ Solution: (i) Let us consider the following expression: $\frac{58^{2}-42^{2}}{16}$ Using the identity $(a+b)(a-b)=a^{2}-b^{2}$, we get: $\frac{58^{2}-42^{2}}{16}=\frac{(58+42)(58-42)}{16}$ $\Rightarrow \frac{58^{2}-42^{2}}{16}=\frac{10...
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