The area of a square field is 8 hectares.
Question: The area of a square field is 8 hectares. How long would a man take to cross it diagonally by walking at the rate of 4 km per hour? Solution: We know that 1 hectare = 10000 m2.So, Area of square field = 8 hectares = 80000 m2. $\Rightarrow \frac{1}{2}(\text { diagonal })^{2}=80000$ $\Rightarrow(\text { diagonal })^{2}=160000$ $\Rightarrow$ diagonal $=\sqrt{160000}=400 \mathrm{~m} .$ Now, distance to be travelled = 400 m speed is given to be $4 \mathrm{~km} / \mathrm{h}=4 \times \frac{10...
Read More →Prove that two lines that are respectively
Question: Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other. Solution: Given Let lines l and m are two intersecting lines. Again, let n and p be another two lines which are perpendicular to the intersecting lines meet at point D. To prove Two lines n and p intersecting at a point. Proof Suppose we consider lines n and p are not intersecting, then it means they are parallel to each other i.e., n || p (i) Since, lines n and pare perpendicular t...
Read More →A salesman has the following record of sales during
Question: A salesman has the following record of sales during three months for three itemsA,BandCwhich have different rates of commission Find out the rates of commission on itemsA,BandCby using determinant method. Solution: Letx,yandzbe the rates of commission on items A, B and C respectively. Based on the given data, we get $90 x+100 y+20 z=800$ $130 x+50 y+40 z=900$ $60 x+100 y+30 z=850$ $D=\left|\begin{array}{ccc}9 10 2 \\ 13 5 4 \\ 6 10 3\end{array}\right|$ [Expressing the equation as a det...
Read More →Find the length of the diagonal of a square whose area is 128 cm2.
Question: Find the length of the diagonal of a square whose area is 128 cm2. Also, find its perimeter. Solution: Area of the square = 128 cm2 Area $=\frac{1}{2} d^{2} \quad$ (where $d$ is a diagonal of the square) $\Rightarrow 128=\frac{1}{2} d^{2}$ $\Rightarrow d^{2}=256$ $\Rightarrow d=16 \mathrm{~cm}$ Now, Area = Side $^{2}$ $\Rightarrow 128=$ Side $^{2}$ $\Rightarrow$ Side $=11.31 \mathrm{~cm}$ Perimeter = 4(Side) $=4(11.31)$ $=45.24 \mathrm{~cm}$ Thus, the perimeter of the square is 45.24 c...
Read More →Prove that through a given point,
Question: Prove that through a given point, we can draw only one perpendicular to a given line. Solution: Given Consider a line l and a point P. Construction Draw two intersecting lines passing through the point $P$ and which is perpendicular to $l$. To prove Only one perpendicular line can be drawn through a given point i.e., to prove $\angle P=0^{\circ}$ Proof In $\triangle A P B, \quad \angle A+\angle P+\angle B=180^{\circ}$ [by angle sum property of a triangle is $180^{\circ}$ ] $\Rightarrow...
Read More →Find the length of the diagonal of a square whose area is 128 cm2.
Question: Find the length of the diagonal of a square whose area is 128 cm2. Also, find its perimeter. Solution: Area of the square = 128 cm2 Area $=\frac{1}{2} d^{2} \quad$ (where $d$ is a diagonal of the square) $\Rightarrow 128=\frac{1}{2} d^{2}$ $\Rightarrow d^{2}=256$ $\Rightarrow d=16 \mathrm{~cm}$ Now, Area = Side $^{2}$ $\Rightarrow 128=$ Side $^{2}$ $\Rightarrow$ Side $=11.31 \mathrm{~cm}$ Perimeter = 4(Side) $=4(11.31)$ $=45.24 \mathrm{~cm}$ Thus, the perimeter of the square is 45.24 c...
Read More →A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission
Question: A salesman has the following record of sales during three months for three itemsA,BandCwhich have different rates of commission Find out the rates of commission on itemsA,BandCby using determinant method. Solution: Letx,yandzbe the rates of commission on items A, B and C respectively. Based on the given data, we get $90 x+100 y+20 z=800$ $130 x+50 y+40 z=900$ $60 x+100 y+30 z=850$ $D=\left|\begin{array}{ccc}9 10 2 \\ 13 5 4 \\ 6 10 3\end{array}\right|$ [Expressing the equation as a det...
Read More →What must be added to each of the following expressions to make it a whole square?
Question: What must be added to each of the following expressions to make it a whole square? (i) 4x2 12x+ 7 (ii) 4x2 20x+ 20 Solution: (i) Let us consider the following expression: $4 x^{2}-12 x+7$ The above expression can be written as: $4 x^{2}-12 x+7=(2 x)^{2}-2 \times 2 x \times 3+7$ It is evident that if 2xis considered as the first term and 3 is considered as the second term, 2 is required to be added to the above expression to make it a perfect square. Therefore, 7 must become 9. Therefor...
Read More →Find the area and perimeter of a square plot of land whose diagonal is 24 m long.
Question: Find the area and perimeter of a square plot of land whose diagonal is $24 \mathrm{~m}$ long. [Take $\sqrt{2}=1.41$ ] Solution: Area of the square $=\frac{1}{2} \times$ Diagonal $^{2}$ $=\frac{1}{2} \times 24 \times 24$ $=288 \mathrm{~m}^{2}$ Now, let the side of the square bexm.Thus, we have: Area $=$ Side $^{2}$ $\Rightarrow 288=x^{2}$ $\Rightarrow x=12 \sqrt{2}$ $\Rightarrow x=16.92$ Perimeter $=4 \times$ Side $=4 \times 16.92$ $=67.68 \mathrm{~m}$ Thus, the perimeter of the square ...
Read More →Find the area and perimeter of a square plot of land whose diagonal is 24 m long.
Question: Find the area and perimeter of a square plot of land whose diagonal is $24 \mathrm{~m}$ long. [Take $\sqrt{2}=1.41$ ] Solution: Area of the square $=\frac{1}{2} \times$ Diagonal $^{2}$ $=\frac{1}{2} \times 24 \times 24$ $=288 \mathrm{~m}^{2}$ Now, let the side of the square bexm.Thus, we have: Area $=$ Side $^{2}$ $\Rightarrow 288=x^{2}$ $\Rightarrow x=12 \sqrt{2}$ $\Rightarrow x=16.92$ Perimeter $=4 \times$ Side $=4 \times 16.92$ $=67.68 \mathrm{~m}$ Thus, the perimeter of the square ...
Read More →A transversal intersects two parallel lines.
Question: A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel. Solution: Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles APG and CQP, respectively. To prove $E P \| F Q$ Proof Given, $A B \| C D$ $\Rightarrow \quad \angle A P G=\angle C Q P$[corresponding angles] $\Rightarrow$ $\frac{1}{2} \angle A P G=\frac{1}{2} \angle C Q P$...
Read More →Solve the following
Question: Ifx2+y2= 29 andxy= 2, find the value of (i)x + y (ii)x y (iii)x4+y4 Solution: (i) We have: $(x+y)^{2}=x^{2}+2 x y+y^{2}$ $\Rightarrow(x+y)=\pm \sqrt{x^{2}+2 x y+y^{2}}$ $\Rightarrow(x+y)=\pm \sqrt{29+2 \times 2} \quad\left(\because x^{2}+y^{2}=29\right.$ and $\left.x y=2\right)$ $\Rightarrow(x+y)=\pm \sqrt{29+4}$ $\Rightarrow(x+y)=\pm \sqrt{29+4}$ $\Rightarrow(x+y)=\pm \sqrt{33}$ (ii) We have: $(x-y)^{2}=x^{2}-2 x y+y^{2}$ $\Rightarrow(x-y)=\pm \sqrt{x^{2}-2 x y+y^{2}}$ $\Rightarrow(x+...
Read More →The cost of painting the four walls of a room 12 m long at ₹30 per m2 is ₹7560 and the cost of covering the floor with mat at ₹25 per m2 is ₹2700.
Question: The cost of painting the four walls of a room 12 m long at₹30 per m2is₹7560 and the cost of covering the floor with mat at₹25 per m2is₹2700. Find the dimensions of the room. Solution: As, the rate of covering the floor $=₹ 25$ per $\mathrm{m}^{2}$ And, the cost of covering the floor $=₹ 2700$ So, the area of the floor $=\frac{2700}{25}$ $\Rightarrow$ length $\times$ breadth $=108$ $\Rightarrow 12 \times$ breadth $=108$ $\Rightarrow$ breadth $=\frac{108}{12}$ $\therefore$ breadth $=9 \m...
Read More →Bisectors of interior ∠B and exterior
Question: Bisectors of interior B and exterior ACD of a ΔABC intersect at the point T. Prove that BTC = BAC. Thinking Process For obtaining the interior required result use the property that the exterior angle of a triangle is equal to the sum of the two opposite angles of triangle. Solution: Given In AABC, produce SC to D and the bisectors of ABC and ACD meet at point T. To prove BTC = BAC Proof $\ln \triangle A B C, \angle C$ is an exterior angle. $\therefore$$\angle A C D=\angle A B C+\angle ...
Read More →Solve this
Question: $x-y+3 z=6$ $x+3 y-3 z=-4$ $5 x+3 y+3 z=10$ Solution: Using the equations, we get $D=\left|\begin{array}{ccc}1 -1 3 \\ 1 3 -3 \\ 5 3 3\end{array}\right|=1(9+9)+1(3+15)+3(3-15)$ $=18+18-36=0$ $D_{1}=\left|\begin{array}{ccc}6 -1 3 \\ -4 3 -3 \\ 10 3 3\end{array}\right|=6(9+9)+1(-12+30)+3(-12-30)$ $=108+18-126=0$ $D_{2}=\left|\begin{array}{ccc}1 6 3 \\ 1 -4 -3 \\ 5 10 3\end{array}\right|=1(-12+30)-6(3+15)+3(10+20)$ $=18-108+90=0$ $D_{3}=\left|\begin{array}{ccc}1 -1 6 \\ 1 3 -4 \\ 5 3 10\e...
Read More →If 2x + 3y = 14 and 2x − 3y = 2, find the value of xy.
Question: If 2x+ 3y= 14 and 2x 3y= 2, find the value ofxy. [Hint: Use (2x+ 3y)2 (2x 3y)2= 24xy] Solution: We will use the identity $(a+b)(a-b)=a^{2}-b^{2}$ to obtain the value of $x y$. Squaring $(2 x+3 y)$ and $(2 x-3 y)$ both and then subtracting them, we get: $(2 x+3 y)^{2}-(2 x-3 y)^{2}=\{(2 x+3 y)+(2 x-3 y)\}\{(2 x+3 y)-(2 x-3 y)\}=4 x \times 6 y=24 x y$ $\Rightarrow(2 x+3 y)^{2}-(2 x-3 y)^{2}=24 x y$ $\Rightarrow 24 x y=(2 x+3 y)^{2}-(2 x-3 y)^{2}$ $\Rightarrow 24 x y=(14)^{2}-(2)^{2}$ $\R...
Read More →The dimensions of a room are 14 m × 10 m × 6.5 m. There are two doors and 4 windows in the room.
Question: The dimensions of a room are 14 m 10 m 6.5 m. There are two doors and 4 windows in the room. Each door measures 2.5 m 1.2 m and each window measures 1.5 1 m. Find the cost of painting the four walls of the room at Rs 35 per m2. Solution: The room has four walls to be painted. Area of these walls $=2(l \times h)+2(b \times h)$ $=(2 \times 14 \times 6.5)+(2 \times 10 \times 6.5)$ $=312 \mathrm{~m}^{2}$ Now, Area of the two doors $=(2 \times 2.5 \times 1.2)=6 \mathrm{~m}^{2}$ Area of the ...
Read More →Solve this
Question: $2 x+y-2 z=4$ $x-2 y+z=-2$ $5 x-5 y+z=-2$ Solution: Using the equations we get $D=\left|\begin{array}{rrr}2 1 -2 \\ 1 -2 1 \\ 5 -5 1\end{array}\right|$ $\Rightarrow 2(-2+5)-1(1-5)-2(-5+10)=0$ $D_{1}=\left|\begin{array}{ccc}4 1 -2 \\ -2 -2 1 \\ -2 -5 1\end{array}\right|$ $\Rightarrow 4(-2+5)-1(-2+2)-2(10-4)=0$ $D_{2}=\left|\begin{array}{ccc}2 4 -2 \\ 1 -2 1 \\ 5 -2 1\end{array}\right|$ $\Rightarrow 2(-2+2)-4(1-5)-2(-2+10)=0$ $D_{3}=\left|\begin{array}{ccc}2 1 4 \\ 1 -2 -2 \\ 5 -5 -2\end...
Read More →Solve the following
Question: If $x+\frac{1}{x}=12$, find the value of $x-\frac{1}{x}$ Solution: Let us consider the following equation: $x+\frac{1}{x}=12$ Squaring both sides, we get: $\left(x+\frac{1}{x}\right)^{2}=(12)^{2}=144$ $\Rightarrow\left(x+\frac{1}{x}\right)^{2}=144$ $\Rightarrow x^{2}+2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=144 \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$ $\Rightarrow x^{2}+2+\frac{1}{x^{2}}=144$ $\Rightarrow x^{2}+\frac{1}{x^{2}}=142$ (Subtracting 2 from both sides...
Read More →A 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running through its
Question: A 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running through its middle, one parallel to its length and the other parallel to its breadth. Find the cost of gravelling the roads at Rs 40 per m2. Solution: The length and breadth of the lawn are80 m and 64 m, respectively.The layout of the roads is shown in the figure below: Area of the road $A B C D=80 \times 5=400 \mathrm{~m}^{2}$ Area of the road $P Q R S=64 \times 5=320 \mathrm{~m}^{2}$ Clearly, the areaEFGH is common...
Read More →Solve the following equations
Question: $x+y-z=0$ $x-2 y+z=0$ $3 x+6 y-5 z=0$ Solution: Using the equations we get $D=\left|\begin{array}{ccr}1 1 -1 \\ 1 -2 1 \\ 3 6 -5\end{array}\right|$ $\Rightarrow 1(10-6)-1(-5-3)-1(6+6)=0$ $D_{1}=\left|\begin{array}{ccr}0 1 -1 \\ 0 -2 1 \\ 0 6 -5\end{array}\right|$ $\Rightarrow 0(10-6)-1(0-0)-1(0+0)=0$ $D_{2}=\left|\begin{array}{rrr}1 0 -1 \\ 1 0 1 \\ 3 0 -5\end{array}\right|$ $\Rightarrow 1(0-0)-0(-5-3)-1(0-0)=0$ $D_{3}=\left|\begin{array}{ccc}1 1 0 \\ 1 -2 0 \\ 3 6 0\end{array}\right|$...
Read More →If two lines intersect prove that the vertically
Question: If two lines intersect prove that the vertically opposite angles are equal Solution: Given Two lines AB and CD intersect at point O. To prove (i) $\angle A O C=\angle B O D$ (ii)$\angle A O D=\angle B O C$ Proof (i) Since, ray $O A$ stands on line $C D$. $\therefore$ $\angle A O C+\angle A O D=180^{\circ}$ [linear pair axiom]...(1) Since, ray $O D$ stands on line $A B$. $\therefore$ $\angle A O D+\angle B O D=180^{\circ}$ [linear pair axiom] ...(ii) From Eqs. (i) and (ii). $\angle A O ...
Read More →Solve the following
Question: If $x+\frac{1}{x}=9$, find the value of $x^{4}+\frac{1}{x^{4}}$. Solution: Let us consider the following equation: $x+\frac{1}{x}=9$ Squaring both sides, we get: $\left(x+\frac{1}{x}\right)^{2}=(9)^{2}=81$ $\Rightarrow\left(x+\frac{1}{x}\right)^{2}=81$ $\Rightarrow x^{2}+2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=81$ $\Rightarrow x^{2}+2+\frac{1}{x^{2}}=81$ $\Rightarrow x^{2}+\frac{1}{x^{2}}=79$ (Subtracting 2 from both sides) Now, squaring both sides again, we get: $\l...
Read More →Solve this
Question: $x+2 y=5$ $3 x+6 y=15$ Solution: Using the equations, we get $D=\left|\begin{array}{cc}1 2 \\ 3 6\end{array}\right|=6-6=0$ $D_{1}=\left|\begin{array}{cc}5 2 \\ 15 6\end{array}\right|=30-30=0$ $D_{2}=\left|\begin{array}{cc}1 5 \\ 3 15\end{array}\right|=15-15=0$ $\therefore D=D_{1}=D_{2}$ Hence, the system of linear equation has infinitely many solutions....
Read More →In a rectangular park of dimensions 50 m × 40 m, a rectangular pond is constructed
Question: In a rectangular park of dimensions 50 m 40 m, a rectangular pond is constructed so that the area of grass strip of uniform width surrounding the pond would be 1184 m2. Find the length and breadth of the pond. Solution: It is given that the dimensions of rectangular park is 50 m 40 m. Area of the rectangular park = 50 40 = 2000 m2Area of the grass surrounding the pond = 1184 m2Now,Area of the rectangular pond= Area of the rectangular park Area of the grass surrounding the rectangular p...
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