\displaystyle \text{SCALE FACTOR}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{A scale drawing of an object has the same shape as the actual object but may have a different}
\displaystyle \text{size. The scale of a drawing compares a length in the drawing with the corresponding length}
\displaystyle \text{of the actual object. The scale is generally written as a ratio. The ratio of corresponding}
\displaystyle \text{lengths in two similar figures is called the scale factor.}
\displaystyle \text{Scale factor}=\frac{\text{Length in image}}{\text{Corresponding length in actual object}}
\displaystyle \text{Two figures are similar if they have the same shape, though their sizes may be different.}
\displaystyle \text{Resizing, reflecting, translating or rotating a figure does not change its shape. In the}
\displaystyle \text{photograph shown above, the side view of a train engine has a scale factor of }1:200.
\displaystyle \text{Thus, a length of }1\text{ cm in the photograph represents }200\text{ cm, or }2\text{ m, of the actual engine.}

\displaystyle \text{This means that a length of }1\text{ cm on the photograph corresponds to a length of }200\text{ cm, or}
\displaystyle 2\text{ m, on the actual engine. The scale can also be written as the ratio of two corresponding lengths.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{If the length of the model is }11\text{ cm, then the overall length of the engine,}
\displaystyle \text{including the couplings, is:}
\displaystyle \text{(a) }22\text{ cm}\qquad\text{(b) }220\text{ cm}\qquad\text{(c) }220\text{ m}\qquad\text{(d) }22\text{ m}
\displaystyle \text{Answer:}
\displaystyle \text{Scale factor}=1:200
\displaystyle \text{Length of the model}=11\text{ cm}
\displaystyle \text{Actual length}=11\times200=2200\text{ cm}
\displaystyle =\frac{2200}{100}=22\text{ m}
\displaystyle \therefore \text{The overall length of the engine is }22\text{ m.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Which of the following transformations produces a similar figure?}
\displaystyle \text{(a) The figure is flipped horizontally}\qquad\text{(b) The figure is dilated by a scale factor}
\displaystyle \text{(c) The figure is translated down}\qquad\text{(d) The figure is not a mirror image of another}
\displaystyle \text{Answer:}
\displaystyle \text{A dilation changes the size of a figure while preserving its shape.}
\displaystyle \text{The corresponding sides remain proportional and the corresponding angles remain equal.}
\displaystyle \therefore \text{A figure dilated by a scale factor is similar to the original figure.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the actual width of the door if its width in the photograph is }0.35\text{ cm?}
\displaystyle \text{(a) }0.7\text{ m}\qquad\text{(b) }0.7\text{ cm}\qquad\text{(c) }0.07\text{ cm}\qquad\text{(d) }0.07\text{ m}
\displaystyle \text{Answer:}
\displaystyle \text{Scale factor}=1:200
\displaystyle \text{Width of the door in the photograph}=0.35\text{ cm}
\displaystyle \text{Actual width}=0.35\times200=70\text{ cm}
\displaystyle =0.7\text{ m}
\displaystyle \therefore \text{The actual width of the door is }0.7\text{ m.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If two similar triangles have a scale factor }5:3,\text{ which statement regarding}
\displaystyle \text{the two triangles is true?}
\displaystyle \text{(a) The ratio of their perimeters is }15:1
\displaystyle \text{(b) Their altitudes have a ratio }25:15
\displaystyle \text{(c) Their medians have a ratio }10:4
\displaystyle \text{(d) Their angle bisectors have a ratio }11:5
\displaystyle \text{Answer:}
\displaystyle \text{In similar triangles, corresponding linear measurements are in the same ratio as the}
\displaystyle \text{corresponding sides.}
\displaystyle \text{Given scale factor}=5:3
\displaystyle 25:15=5:3
\displaystyle \therefore \text{The corresponding altitudes can be in the ratio }25:15.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the length of }AB\text{ in the given figure.}
\displaystyle \text{(a) }8\text{ cm}\qquad\text{(b) }6\text{ cm}\qquad\text{(c) }4\text{ cm}\qquad\text{(d) }10\text{ cm} \displaystyle \text{Answer:}
\displaystyle \text{From the figure, }\triangle ABC\sim\triangle ADE.
\displaystyle AB=x\text{ cm},\qquad BD=4\text{ cm}
\displaystyle \therefore AD=(x+4)\text{ cm}
\displaystyle BC=3\text{ cm},\qquad DE=6\text{ cm}
\displaystyle \text{Since corresponding sides of similar triangles are proportional,}
\displaystyle \frac{AB}{AD}=\frac{BC}{DE}
\displaystyle \frac{x}{x+4}=\frac{3}{6}=\frac{1}{2}
\displaystyle 2x=x+4
\displaystyle x=4
\displaystyle \therefore AB=4\text{ cm}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

 

\displaystyle \text{COORDINATE GEOMETRY}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{To conduct Sports Day activities in a school, lines have been drawn with chalk powder at a}
\displaystyle \text{distance of }1\text{ m from each other on a rectangular ground }ABCD.\text{ Along }AD,\ 100\text{ flowerpots}
\displaystyle \text{have been placed at a distance of }1\text{ m from each other, as shown in the figure. Niharika}
\displaystyle \text{runs }\frac14\text{ of the distance }AD\text{ on the second line and posts a green flag. Preet runs }\frac15
\displaystyle \text{of the distance }AD\text{ on the eighth line and posts a red flag.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find the position of the green flag.}
\displaystyle \text{(a) }(2,25)\qquad\text{(b) }(2,0.25)\qquad\text{(c) }(25,2)\qquad\text{(d) }(0,-25)
\displaystyle \text{Answer:}
\displaystyle AD=100\text{ m}
\displaystyle \text{Distance covered by Niharika}=\frac14\times100=25\text{ m}
\displaystyle \text{She runs along the second line, so the }x\text{-coordinate is }2.
\displaystyle \text{Her distance along }AD\text{ gives the }y\text{-coordinate as }25.
\displaystyle \therefore \text{The position of the green flag is }(2,25).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the position of the red flag.}
\displaystyle \text{(a) }(8,0)\qquad\text{(b) }(20,8)\qquad\text{(c) }(8,20)\qquad\text{(d) }(8,0.2)
\displaystyle \text{Answer:}
\displaystyle AD=100\text{ m}
\displaystyle \text{Distance covered by Preet}=\frac15\times100=20\text{ m}
\displaystyle \text{He runs along the eighth line, so the }x\text{-coordinate is }8.
\displaystyle \text{His distance along }AD\text{ gives the }y\text{-coordinate as }20.
\displaystyle \therefore \text{The position of the red flag is }(8,20).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the distance between the two flags?}
\displaystyle \text{(a) }\sqrt{41}\qquad\text{(b) }\sqrt{11}\qquad\text{(c) }\sqrt{61}\qquad\text{(d) }\sqrt{51}
\displaystyle \text{Answer:}
\displaystyle \text{Green flag}=(2,25),\qquad\text{Red flag}=(8,20)
\displaystyle \text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
\displaystyle =\sqrt{(8-2)^2+(20-25)^2}
\displaystyle =\sqrt{6^2+(-5)^2}
\displaystyle =\sqrt{36+25}=\sqrt{61}
\displaystyle \therefore \text{The distance between the two flags is }\sqrt{61}\text{ m.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Rashmi has to post a blue flag exactly halfway between the two flags. Where should}
\displaystyle \text{she post the flag?}
\displaystyle \text{(a) }(5,22.5)\qquad\text{(b) }(10,22)\qquad\text{(c) }(2,8.5)\qquad\text{(d) }(2.5,20)
\displaystyle \text{Answer:}
\displaystyle \text{The blue flag is at the midpoint of }(2,25)\text{ and }(8,20).
\displaystyle \text{Midpoint}=\left(\frac{2+8}{2},\frac{25+20}{2}\right)
\displaystyle =\left(\frac{10}{2},\frac{45}{2}\right)
\displaystyle =(5,22.5)
\displaystyle \therefore \text{Rashmi should post the blue flag at }(5,22.5).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Joy has to post a flag at one-fourth of the distance from the green flag along the}
\displaystyle \text{line segment joining the green and red flags. Where should he post the flag?}
\displaystyle \text{(a) }(3.5,23.75)\qquad\text{(b) }(0.5,12.5)\qquad\text{(c) }(2.25,8.5)\qquad\text{(d) }(25,20)
\displaystyle \text{Answer:}
\displaystyle \text{Let the green flag be }G(2,25)\text{ and the red flag be }R(8,20).
\displaystyle \text{The required point }P\text{ is one-fourth of the distance from }G\text{ to }R.
\displaystyle \therefore GP:PR=1:3
\displaystyle \text{By the section formula,}
\displaystyle P=\left(\frac{1(8)+3(2)}{1+3},\frac{1(20)+3(25)}{1+3}\right)
\displaystyle =\left(\frac{14}{4},\frac{95}{4}\right)
\displaystyle =(3.5,23.75)
\displaystyle \therefore \text{Joy should post the flag at }(3.5,23.75).
\displaystyle \therefore \text{The corrected option (a) is correct.}
\displaystyle \\

 

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{The Class X students of a school in Krishnagar have been allotted a rectangular plot of land}
\displaystyle \text{for a gardening activity. Gulmohar saplings are planted along the boundary at a distance of}
\displaystyle 1\text{ m from each other. A triangular grassy lawn }PQR\text{ is present in the plot, as shown in the}
\displaystyle \text{figure. The students are to sow seeds of flowering plants in the remaining area of the plot.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{Taking }A\text{ as the origin, find the coordinates of }P\text{.}
\displaystyle \text{(a) }(4,6)\qquad\text{(b) }(6,4)\qquad\text{(c) }(0,6)\qquad\text{(d) }(4,0)
\displaystyle \text{Answer:}
\displaystyle \text{Taking }A\text{ as the origin, the horizontal distance of }P\text{ from }A\text{ is }4\text{ m.}
\displaystyle \text{The vertical distance of }P\text{ from }AD\text{ is }6\text{ m.}
\displaystyle \therefore P=(4,6).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What will be the coordinates of }R\text{ if }C\text{ is taken as the origin?}
\displaystyle \text{(a) }(8,6)\qquad\text{(b) }(3,10)\qquad\text{(c) }(10,3)\qquad\text{(d) }(0,6)
\displaystyle \text{Answer:}
\displaystyle \text{When }A\text{ is the origin, }R=(6,5).
\displaystyle \text{The dimensions of the rectangular plot are }16\text{ m}\times8\text{ m.}
\displaystyle \text{Taking }C\text{ as the origin, the horizontal distance of }R\text{ from }C=16-6=10\text{ m.}
\displaystyle \text{Its vertical distance from }BC=8-5=3\text{ m.}
\displaystyle \therefore R=(10,3).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What will be the coordinates of }Q\text{ if }C\text{ is taken as the origin?}
\displaystyle \text{(a) }(6,13)\qquad\text{(b) }(-6,13)\qquad\text{(c) }(-13,6)\qquad\text{(d) }(13,6)
\displaystyle \text{Answer:}
\displaystyle \text{When }A\text{ is the origin, }Q=(3,2).
\displaystyle \text{Taking }C\text{ as the origin, the horizontal distance of }Q\text{ from }C=16-3=13\text{ m.}
\displaystyle \text{Its vertical distance from }BC=8-2=6\text{ m.}
\displaystyle \therefore Q=(13,6).
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Calculate the area of the triangle if }A\text{ is taken as the origin.}
\displaystyle \text{(a) }4.5\text{ m}^2\qquad\text{(b) }6\text{ m}^2\qquad\text{(c) }8\text{ m}^2\qquad\text{(d) }6.25\text{ m}^2
\displaystyle \text{Answer:}
\displaystyle P=(4,6),\qquad Q=(3,2),\qquad R=(6,5)
\displaystyle \text{Area of }\triangle PQR
\displaystyle =\frac12\left|4(2-5)+3(5-6)+6(6-2)\right|
\displaystyle =\frac12\left|-12-3+24\right|
\displaystyle =\frac12\times9=4.5\text{ m}^2
\displaystyle \therefore \text{The area of the triangular grassy lawn is }4.5\text{ m}^2.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Calculate the area of the triangle if }C\text{ is taken as the origin.}
\displaystyle \text{(a) }8\text{ m}^2\qquad\text{(b) }5\text{ m}^2\qquad\text{(c) }6.25\text{ m}^2\qquad\text{(d) }4.5\text{ m}^2
\displaystyle \text{Answer:}
\displaystyle \text{Taking }C\text{ as the origin, }P=(12,2),\ Q=(13,6),\ R=(10,3).
\displaystyle \text{Area of }\triangle PQR
\displaystyle =\frac12\left|12(6-3)+13(3-2)+10(2-6)\right|
\displaystyle =\frac12\left|36+13-40\right|
\displaystyle =\frac12\times9=4.5\text{ m}^2
\displaystyle \therefore \text{The area of the triangular grassy lawn is }4.5\text{ m}^2.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

 

\displaystyle \text{CIRCLES}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{A Ferris wheel, also known as a big wheel, is an amusement ride consisting of a large rotating}
\displaystyle \text{upright wheel with passenger-carrying cabins attached to its rim. As the wheel rotates, the}
\displaystyle \text{cabins remain upright, usually due to gravity.}
\displaystyle \text{After taking a ride on a Ferris wheel, Aarti observes her friends enjoying the ride. She becomes}
\displaystyle \text{curious about the different angles formed by the wheel and represents the situation using the}
\displaystyle \text{figure shown below. Here, }PR\text{ and }PQ\text{ are tangents to the circle at }R\text{ and }Q,
\displaystyle \text{respectively, }O\text{ is the centre of the circle and }\angle RPQ=30^\circ. \displaystyle \\

\displaystyle \textbf{Question 1: }\text{In the given figure, find }\angle ROQ\text{.}
\displaystyle \text{(a) }60^\circ\qquad\text{(b) }100^\circ\qquad\text{(c) }150^\circ\qquad\text{(d) }90^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Since the radius is perpendicular to the tangent at the point of contact,}
\displaystyle OR\perp PR\qquad\text{and}\qquad OQ\perp PQ
\displaystyle \therefore \angle ORP=\angle OQP=90^\circ
\displaystyle \text{In quadrilateral }ORPQ,
\displaystyle \angle ROQ+\angle ORP+\angle RPQ+\angle OQP=360^\circ
\displaystyle \angle ROQ+90^\circ+30^\circ+90^\circ=360^\circ
\displaystyle \angle ROQ=150^\circ
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find }\angle RQP\text{.}
\displaystyle \text{(a) }75^\circ\qquad\text{(b) }60^\circ\qquad\text{(c) }30^\circ\qquad\text{(d) }90^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle \therefore PR=PQ
\displaystyle \therefore \triangle PRQ\text{ is an isosceles triangle.}
\displaystyle \angle PRQ=\angle RQP
\displaystyle \text{Let }\angle PRQ=\angle RQP=x.
\displaystyle x+x+30^\circ=180^\circ
\displaystyle 2x=150^\circ
\displaystyle x=75^\circ
\displaystyle \therefore \angle RQP=75^\circ.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find }\angle RSQ\text{.}
\displaystyle \text{(a) }60^\circ\qquad\text{(b) }75^\circ\qquad\text{(c) }100^\circ\qquad\text{(d) }30^\circ
\displaystyle \text{Answer:}
\displaystyle \text{By the tangent-chord theorem, the angle between tangent }QP\text{ and chord }QR
\displaystyle \text{is equal to the angle in the alternate segment subtended by chord }QR.
\displaystyle \therefore \angle RQP=\angle RSQ
\displaystyle \text{From Question 2, }\angle RQP=75^\circ.
\displaystyle \therefore \angle RSQ=75^\circ.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find }\angle ORP\text{.}
\displaystyle \text{(a) }90^\circ\qquad\text{(b) }70^\circ\qquad\text{(c) }100^\circ\qquad\text{(d) }60^\circ
\displaystyle \text{Answer:}
\displaystyle \text{A tangent to a circle is perpendicular to the radius through the point of contact.}
\displaystyle \text{Since }PR\text{ is tangent to the circle at }R,\ OR\perp PR.
\displaystyle \therefore \angle ORP=90^\circ.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

 

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{Varun has been selected by his school to design a logo for Sports Day T-shirts for students}
\displaystyle \text{and staff. The logo design is shown in the figure, and he is working on different fonts and}
\displaystyle \text{colours according to the theme. In the given figure, a circle with centre }O\text{ is inscribed}
\displaystyle \text{in }\triangle ABC\text{ and touches the sides }AB,\ BC\text{ and }CA\text{ at }D,\ E\text{ and }F,\text{ respectively.} \displaystyle \text{The lengths of }AB,\ BC\text{ and }CA\text{ are }12\text{ cm},\ 8\text{ cm and }10\text{ cm, respectively.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find the length of }AD\text{.}
\displaystyle \text{(a) }7\text{ cm}\qquad\text{(b) }8\text{ cm}\qquad\text{(c) }5\text{ cm}\qquad\text{(d) }9\text{ cm}
\displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from the same external point to a circle are equal.}
\displaystyle \therefore AD=AF,\quad BD=BE,\quad CE=CF
\displaystyle \text{Let }AD=AF=x,\quad BD=BE=y,\quad CE=CF=z.
\displaystyle x+y=12,\qquad y+z=8,\qquad z+x=10
\displaystyle (x+y)+(x+z)-(y+z)=12+10-8
\displaystyle 2x=14
\displaystyle x=7
\displaystyle \therefore AD=7\text{ cm.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the length of }BE\text{.}
\displaystyle \text{(a) }8\text{ cm}\qquad\text{(b) }5\text{ cm}\qquad\text{(c) }2\text{ cm}\qquad\text{(d) }9\text{ cm}
\displaystyle \text{Answer:}
\displaystyle AD=7\text{ cm}
\displaystyle AB=AD+DB
\displaystyle 12=7+DB
\displaystyle DB=5\text{ cm}
\displaystyle \text{Since tangents from }B\text{ are equal, }BD=BE.
\displaystyle \therefore BE=5\text{ cm.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the length of }CF\text{.}
\displaystyle \text{(a) }9\text{ cm}\qquad\text{(b) }5\text{ cm}\qquad\text{(c) }2\text{ cm}\qquad\text{(d) }3\text{ cm}
\displaystyle \text{Answer:}
\displaystyle BE=5\text{ cm}
\displaystyle BC=BE+EC
\displaystyle 8=5+EC
\displaystyle EC=3\text{ cm}
\displaystyle \text{Since tangents from }C\text{ are equal, }CE=CF.
\displaystyle \therefore CF=3\text{ cm.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If the radius of the circle is }4\text{ cm, find the area of }\triangle OAB\text{.}
\displaystyle \text{(a) }20\text{ cm}^2\qquad\text{(b) }36\text{ cm}^2\qquad\text{(c) }24\text{ cm}^2\qquad\text{(d) }48\text{ cm}^2
\displaystyle \text{Answer:}
\displaystyle \text{Since }OD\text{ is the radius through the point of contact, }OD\perp AB.
\displaystyle AB=12\text{ cm},\qquad OD=4\text{ cm}
\displaystyle \text{Area of }\triangle OAB=\frac12\times AB\times OD
\displaystyle =\frac12\times12\times4
\displaystyle =24\text{ cm}^2
\displaystyle \therefore \text{The area of }\triangle OAB\text{ is }24\text{ cm}^2.
\displaystyle \therefore \text{Option (c) is correct, based on the stated radius of }4\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the area of }\triangle ABC\text{.}
\displaystyle \text{(a) }50\text{ cm}^2\qquad\text{(b) }60\text{ cm}^2\qquad\text{(c) }100\text{ cm}^2\qquad\text{(d) }90\text{ cm}^2
\displaystyle \text{Answer:}
\displaystyle AB=12\text{ cm},\qquad BC=8\text{ cm},\qquad CA=10\text{ cm}
\displaystyle s=\frac{12+8+10}{2}=15\text{ cm}
\displaystyle \text{By Heron's formula,}
\displaystyle \text{Area of }\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{15(15-12)(15-8)(15-10)}
\displaystyle =\sqrt{15\times3\times7\times5}
\displaystyle =\sqrt{1575}=15\sqrt7\text{ cm}^2
\displaystyle \therefore \text{The area of }\triangle ABC\text{ is }15\sqrt7\text{ cm}^2\text{, approximately }39.69\text{ cm}^2.
\displaystyle \therefore \text{None of the given options is correct.}
\displaystyle \\

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