Class 10th Maths - Triangles Case Study Questions and Answers 2022 - 2023
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Triangles Case Study Questions With Answer Key
10th Standard CBSE
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Reg.No. :
Maths
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In a classroom, students were playing with some pieces of cardboard as shown below.
All of a sudden, teacher entered into classroom. She told students to arrange all pieces. On seeing this beautiful image, she observed that \(\Delta\) ADH is right angled triangle, which contains.
(i) right triangles ABJ and IGH.
(ii) quadrilateral GFJI
(iii) squares JKLM and LCBK
(iv) rectangles MLEF and LCDE.
After observation, she ask certain questions to students. Help them to answer these questions.
(i) If an insect (small ant) walks 24 m from H to F, then walks 6 m to reach at M, then walks 4 m to reach at L and finally crossing K, reached at J. Find the distance between initial and final position of insect.(a) 25m (b) 26m (c) 27m (d) 28m (ii) If m, n and r are the sides of right triangle ABJ, then which of the following can be correct?
(a) m2+n2= r2 (b) m2+n2+r2-=0 (c) m2 + n2 = 2r2 (d) none of these (iii) If \(\Delta\)ABJ ~ \(\Delta\)ADH, then which similarity criterion is used here?
(a) AA (b) SAS (c) AAS (d) SSS (iv) If \(\angle\)ABJ = 90° and B, J are mid points of sides AD and AH respectively and BJ || DH, then which of the following option is false?
\((a) \triangle A B J \sim \triangle A D H\) \((b) 2 B J=D H\) \((c) A J^{2}=J B^{2}+A B^{2}\) \((d) \frac{A B}{B D}=\frac{A J}{A H}\) (v) If \(\Delta\)PQR is right triangle with QM \(\perp\) PR, then which of the following is not correct?
\((a) \Delta P M Q \sim \Delta P Q R\) \((b) Q R^{2}=P R^{2}-P Q^{2}\) \((c) P R^{2}=P Q+Q R\) \((d) \Delta P M Q \sim \Delta Q M R\) (a) -
An aeroplane leaves an airport and flies due north at a speed of 1200km /hr. At the same time, another aeroplane leaves the same station and flies due west at the speed of 1500 km/hr as shown below. After \(1 \frac{1}{2}\) hr both the aeroplanes reaches at point P and Q respectively.
(i) Distance travelled by aeroplane towards north after \(1 \frac{1}{2}\) hr is(a) 1800 km (b) 1500 km (c) 1400km (d) 1350 km (ii) Distance travelled by aeroplane towards west after \(1 \frac{1}{2}\) hr is
(a) 1600 km (b) 1800 km (c) 2250km (d) 2400 km (iii) In the given figure,\(\angle\)POQ is
(a) 70° (b) 90° (c) 80° (d) 100° (iv) Distance between aeroplanes after \(1 \frac{1}{2}\) hr is
\((a) 450 \sqrt{41} \mathrm{~km}\) \((b) 350 \sqrt{31} \mathrm{~km}\) \((c) 125 \sqrt{12} \mathrm{~km}\) \((d) 472 \sqrt{41} \mathrm{~km}\) (v) Area of \(\Delta\)POQ is
(a) 185000km2 (b) 179000km2 (c) 186000km2 (d) 2025000 km2 (a) -
Rohit's father is a mathematician. One day he gave Rohit an activity to measure the height of building. Rohit accepted the challenge and placed a mirror on ground level to determine the height of building. He is standing at a certain distance so that he can see the top of the building reflected from mirror. Rohit eye level is at 1.8m above ground. The distance of Rohit from mirror and that of building from mirror are 1.5 m and 2.5 m respectively.
Based on the above information, answer the following questions.
(i) Two similar triangles formed in the above figure is\((a) \Delta A B M and \Delta C M D\) \((b) \Delta A M B and \Delta C D M\) \((c) \Delta A B M and \Delta C D M\) (d) None of these (ii) Which criterion of similarity is applied here?
(a) AA similarity criterion (b) SSS similarity criterion (c) SAS similarity criterion (d) ASA similarity criterion (iii) Height of the building is
(a) 1m (b) 2m (c) 3m (d) 4m (iv) In \(\Delta\)ABM, if LBAM = 30°, then LMCD is equal to
(a) 40° (b) 30° (c) 65° (d) 90° (v) If \(\Delta\)ABM and \(\Delta\)CDMare similar where CD = 6 ern, MD = 8 cm and BM = 24 ern, then AB is equal to
(a) 16cm (b) 18cm (c) 12cm (d) 14cm (a) -
Meenal was trying to find the height of tower near his house. She is using the properties of similar triangles. The height of Meenal's house is 20 m. When Meenal's house casts a shadow of 10m long on the ground, at the same time, tower casts a shadow of 50 m long and Arun's house casts a shadow of 20 m long on the ground as shown below.
Based on the above information, answer the following questions.
(i) What is the height of tower?(a) 100 m (b) 50 m (c) 15 m (d) 45 m (ii) What will be the length of shadow of tower when Meenal's house casts a shadow of 15 m?
(a) 45 m (b) 70 m (c) 75 m (d) 72 m (iii) Height of Aruns house is
(a) 80 m (b) 75 m (c) 60 m (d) 40 m (iv) If tower casts a shadow of 40 rn, then find the length of shadow of Arun's house
(a) 18 m (b) 17 m (c) 16 m (d) 14 m (v) If tower casts a shadow of 40 m, then what will be the length of shadow of Meenal's house?
(a) 7 m (b) 9 m (c) 4 m (d) 8 m (a) -
In the backyard of house, Shikha has some empty space in the shape of a \(\Delta\)PQR. She decided to make it a garden. She divided the whole space into three parts by making boundaries AB and CD using bricks to grow flowers and
vegetables where ABIICDIIQR as shown in figure.
Based on the above information, answer the following questions.
(i) The length of AB is(a) 3m (b) 4m (c) 5m (d) 6m (ii) The length of CD is
(a) 4m (b) 5m (c) 6m (d) 7m (iii) Area of whole empty land is
(a) 90 m2 (b) 60m2 (c) 32m2 (d) 72m2 (iv) Area of \(\Delta\)PAB is
\((a) \frac{45}{4} \mathrm{~m}^{2}\) \((b) \frac{45}{8} \mathrm{~m}^{2}\) \((c) \frac{8}{45} \mathrm{~m}^{2}\) \((d) \frac{4}{45} \mathrm{~m}^{2}\) (v) Area of \(\Delta\)PCD is
\((a) \frac{12}{245} \mathrm{~m}^{2}\) \((b) \frac{245}{12} \mathrm{~m}^{2}\) \((c) \frac{243}{8} \mathrm{~m}^{2}\) \((d) \frac{245}{8} \mathrm{~m}^{2}\) (a) -
Minister of a state went to city Q from city P. There is a route via city R such that PR \(\perp\)RQ. PR = 2x km and RQ = 2(x + 7) km. He noticed that there is a proposal to construct a 26 km highway which directly connects the two cities P and Q.
Based on the above information, answer the following questions.
(i) Which concept can be used to get the value of x?(a) Thales theorem (b) Pythagoras theorem (c) Converse ofthales theorem (d) Converse of Pythagoras theorem (ii) The value of x is
(a) 4 (b) 6 (c) 5 (d) 8 (iii) The value of PR is
(a) 10 km (b) 20 km (c) 15 km (d) 25 km (iv) The value of RQ is
(a) 12 km (b) 24 km (c) 16 km (d) 20 km (v) How much distance will be saved in reaching city Q after the construction of highway?
(a) 10 km (b) 9 km (c) 4 km (d) 8 km (a) -
Class teacher draw the shape of quadrilateral on board. Ankit observed the shape and explored on his notebook in different ways as shown below.
Based on the above information, answer the following questions.
(i) In if ABCD is a trapezium with AB || CD, E and F are points on non-parallel sides AD and BC respectively such that EF || AB, then \(\frac{A E}{E D}=\)(a) \(\frac{B E}{C D}\) (b) \(\frac{A B}{C D}\) (c) \(\frac{B F}{F C}\) (d) None of these (ii) In if AB || CD, and DO = 3x - 19, OB = x - 5, OC = x - 3 and AO = 3, then the value of x can be
(a) 5 or 8 (b) 8 or 9 (c) 10 or 12 (d) 13 or 14 (iii) In if OD = 3x - 1, OB = 5x - 3, OC = 2x + 1 and AO = 6x - 5, then the value of x is
(a) 0 (b) 1 (c) 2 (d) 3 (iv) In \(\Delta\) ABC, if PQ || BC and AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, then AB + PQ is equal to
(a) 7.2 cm (b) 5.9 cm (c) 2.6 cm (d) 8.4 cm (v) In \(\Delta\)DEF, if RS || EF, DR = 4x - 3, DS = 8x - 7, ER = 3x - 1 and FS = 5x - 3, then the value of x is
(a) 1 (b) 5.9 cm (c) 2.6 cm (d) 8.4 cm (a) -
Ankita wants to make a toran for Diwali using some pieces of cardboard. She cut some cardboard pieces as shown below. If perimeter of \(\Delta\)ADE and \(\Delta\)BCE are in the ratio 2: 3, then answer the following questions.
(i) If the two triangles here are similar by SAS similarity rule, then their corresponding proportional sides are\((a) \frac{A E}{C E}=\frac{D E}{B E}\) \((b) \frac{B E}{A E}=\frac{C E}{D E}\) \(\text { (c) } \frac{A D}{C E}=\frac{B E}{D E}\) (d) None of these (ii) Length of BC =
(a) 2 cm (b) 4 cm (c) 5 cm (d) None of these (iii) Length of AD =
(a) 10/3 cm (b) 9/4 cm (c) 5/3 cm (d) 4/3 cm (iv) Length of ED =
(a) 4/3 cm (b) 8/3 cm (c) 7/3 cm (d) None of these (v) Length of AE =
\((a) \frac{2}{3} \times B E\) \((b) \sqrt{A D^{2}-D E^{2}}\) \((c) \frac{2}{3} \times \sqrt{B C^{2}-C E^{2}}\) (d) All of these (a) -
Aruna visited to her uncle's house. From a point A, where Aruna was standing, a bus and building come in a straight line as shown in the figure.
Based on the above information, answer the following questions.
(i) Which similarity criteria can be seen in this case, if bus and building are considered in a straight line?(a) AA (b) SAS (c) SSS (d) ASA (ii) If the distance between Aruna and the bus is twice as much as the height of the bus, then the height of the bus is
(a) 40 m (b) 12.5 m (c) 15 m (d) 20 m (iii) If the distance of Aruna from the building is twelve times the height of the bus, then the ratio of the heights of bus and building is
(a) 3:1 (b) 1:4 (c) 1:6 (d) 2:3 (iv) What is the ratio of the distance between Aruna and top of bus to the distance between the tops of bus and building?
(a) 1:5 (b) 1:6 (c) 2:5 (d) Can't be determined (v). What is the height of the building?
(a) 50 m (b) 75 m (c) 120 m (d) 30 m (a) -
Two hotels are at the ground level on either side of a mountain. On moving a certain distance towards the top of the mountain two huts are situated as shown in the figure. The ratio between the distance from hotel B to hut-2 and that ofhut-2 to mountain top is 3: 7.
Based on the above information, answer the following questions.
(i) What is the ratio of the perimeters of the triangle formed by both hotels and mountain top to the triangle formed by both huts and mountain top?(a) 5: 2 (b) 10: 7 (c) 7: 3 (d) 3: 10 (ii) The distance between the hotel A and hut-I is
(a) 2.5 miles (b) 29 miles (c) 4.29 miles (d) 1.5 miles (iii) If the horizontal distance between the hut -1 and hut -2 is 8 miles, then the distance between the two hotels is
(a) 2.4 miles (b) 11.43 miles (c) 9 miles (d) 7 miles (iv) If the distance from mountain top to hut-1 is 5 miles more than that of distance from hotel B to mountain top, then what is the distance between hut-2 and mountain top?
(a) 3.5 miles (b) 6 miles (c) 5.5 miles (d) 4 miles (v) What is the ratio of areas of two parts formed in the complete figure?
(a) 53: 21 (b) 10: 41 (c) 51: 33 (d) 49:51 (a) -
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. She is pulling the string at the rate of 5 cm per second. Nazima’s friend observe her position and draw a rough sketch by using A, B, C and D positions of tip, point directly under the tip of the rod, fish and Nazima’s position (see the below figure). Assuming that her string (from the tip of her rod to the fly) is taut, answer the following questions:
(a) What is the length AC?(a) 2 m (b) 3 m (c) 4 m (d) 5 m (b) What is the length of string pulled in 12 seconds?
(a) 6 m (b) 0.3 m (c) 0.6 m (d) 3 m (c) What is the length of string after 12 seconds?
(a) 2.4 m (b) 2.7 m (c) 2 m (d) 2.2 m (d) What will be the horizontal distance of the fly from her after 12 seconds?
(a) 2.7 m (b) 2.78 m (c) 2.58 m (d) 2.2 m (e) The given problem is based on which concept?
(a) Triangles (b) Co-ordinate geometry (c) Height and Distance (d) None of these (a) -
Points A and B are on the opposite edges of a pond as shown in below figure. To find the distance between the two points, Ram makes a right-angled triangle using rope connecting B with another point C at a distance of 12 m, connecting C to point D at a distance of 40 m from point C and then connecting D to the point A which is at a distance of 30 m from D such that ∠ADC = 900.
(i) Which property of geometry will be used to find the distance AC?(a) Similarity of Triangles (b) Thales Theorem (c) Pythagoras Theorem (d) Quadratic Equation (ii) What is the distance AC?
(a) 50m (b) 12m (c) 100m (d) 70m (iii) Which is the following does not form a Pythagoras triplet?
(a) (7, 24, 25) (b) (15, 8, 17) (c) (5, 12, 13) (d) (21, 20, 28) (iv) Find the length AB.
(a) 12m (b) 38m (c) 50m (d) none of these (v) Find the length of the rope used.
(a) 120m (b) 70m (c) 82m (d) none of these (a) -
Rahul is studying in X Standard. He is making a kite to fly it on a Sunday. Few questions came to his mind while making the kite. Give answers to his questions by looking at the figure.
(i) Rahul tied the sticks at what angles to each other?(a) 30° (b) 60° (c) 90° (d) 60° (ii) Which is the correct similarity criteria applicable for smaller triangles at the upper part of this kite?
(a) RHS (b) SAS (c) SSA (d) AAS (iii) Sides of two similar triangles are in the ratio 4:9. Corresponding medians of these triangles are in the ratio,
(a) 2:3 (b) 4:9 (c) 81:16 (d) 16:81 (iv) In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. This theorem is called as,
(a) Pythagoras theorem (b) Thales theorem (c) Converse of Thales theorem (d) Converse of Pythagoras theorem (v) What is the area of the kite, formed by two perpendicular sticks of length 6 cm and 8 cm?
(a) 48 cm2 (b) 14 cm2 (c) 24 cm2 (d) 96 cm2 (a) -
On one day, a poor girl of height 90 cm is looking for a lamp-post for completing her homework as in her area power is not there and she finds the same at some distance away from her home. After completing the homework, she is walking away from the base of a lamp-post at a speed of 1.2 m/s. The lamp is 3.6 m above the ground (see below figure).
(i) Find her distance from the base of the lamp post.(a) 1.2 m (b) 3.6 m (c) 4.8 m (d) none of these (ii) Find the correct similarity criteria applicable for triangles ABE and CDE.
(a) AA (b) SAS (c) SSS (d) AAS (iii) Find the length of her shadow after 4 seconds.
(a) 1.2 m (b) 3.6 m (c) 4.8 m (d) none of these (iv) Sides of two similar triangles are in the ratio 9:16. Find the ratio of Corresponding areas of these triangles.
(a) 9:16 (b) 3:4 (c) 81:256 (d) 18:32 (v) Find the ratio AC:CE.
(a) 1: 3 (b) 3 : 1 (c) 1 : 4 (d) 4 : 1 (a) -
Anjali places a mirror on level ground to determine the height of a tree (see the diagram). She stands at a certain distance so that she can see the top of the tree reflected from the mirror. Anjali's eye level is 2 m above the ground. The distance of Anjali and the tree from the mirror are 1.4 m and 2.8 m respectively.
(i) What are the two \(\triangle\)s formed in the above diagram, which are used to calculate the height of the tree?(a) \(\triangle\)s QMM' and PQM (b) \(\triangle\)s PQM and RSM (c) \(\triangle\)s RM'M and MRS (d) \(\triangle\)s PM'M and RM'M (ii) State the criterion of similarity, that will be used in the above found triangles.
(a) RHS (b) SSS (c) SAS (d) AA (iii) What is the height of the tree?
(a) 4 m (b) 5 cm (c) 6 m (d) 7 cm (iv) What is the distance between Rashmi and Gulmohar tree?
(a) 3.2 m (b) 5.2 m (c) 4.2 m (d) 2.2 m (a)
Case Study
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Answers
Triangles Case Study Questions With Answer Key Answer Keys
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(i) (b): As JKLM is a square.
\(\therefore\) ML=JM=4m
So, JF = 6 + 4 = 10m
Required distance between initial and final position of insect = HJ
\(\begin{array}{l} =\sqrt{(H F)^{2}+(J F)^{2}} \\ =\sqrt{(24)^{2}+(10)^{2}} \\ =\sqrt{676}=26 \mathrm{~m} \end{array}\)
(ii) (a): By Pythagoras, n2 + m2 = r2
(iii) (a) : In \(\Delta\)ABJ and \(\Delta\)ADH
<B = <D= 90°
<A = <A (common)
\(\therefore\) By AA similarity criterion, \(\Delta\)ABJ \(\sim\) \(\Delta\)ADH
(iv) (d): Since,\(\Delta A B J \sim \Delta A D H\) [ By AA similarity criterion]
\(\therefore \quad \frac{A B}{A D}=\frac{A J}{A H}\)
(v) (c): Since, PR2 = PQ2 + QR2 [By Pythagoras theorem] -
(i) (a): Speed = 1200 km/hr
\(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\)
\(\therefore\) Required distance = Speed x Time
\(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\)
(ii) (c): Speed = 1500 km/hr
Time = \(\frac{3}{2}\) hr.
\(\therefore\) Required distance = Speed x Time
\(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\)
(iii) (b): Clearly, directions are always perpendicular to each other.
\(\therefore \quad \angle P O Q=90^{\circ}\)
(iv) (a): Distance between aeroplanes after \(1\frac{1}{2}\) hour
\(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\)
(v) (d): Area of \(\Delta\)POQ= \(\frac{1}{2}\)x base x height
\(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\) -
(i) (c): Since,\(\angle\)B = \(\angle\)D = 90°, \(\angle\)AMB = \(\angle\)CMD
(\(\because\) Angle of incident = Angle of reflection)
\(\therefore\) By similarity criterion, \(\Delta\)ABM \(\sim\) \(\Delta\)CDM
(ii) (a)
(iii) (c) : \(\because\) \(\Delta\)ABM \(\sim\) \(\Delta\)CDM
\(\begin{array}{l} \therefore \quad \frac{A B}{C D}=\frac{B M}{D M} \Rightarrow \frac{A B}{1.8}=\frac{2.5}{1.5} \\ \Rightarrow \quad A B=\frac{2.5 \times 1.8}{1.5}=3 \mathrm{~m} \end{array}\)
(iv) (b): Since \(\Delta\)ABM \(\sim\) \(\Delta\)CDM
\(\because\) \(\angle\)A=\(\angle\)C=30° [\(\therefore\) Corresponding angles of similar triangles are also equal]
(v) (b): Since \(\Delta\)ABM \(\sim\) \(\Delta\)CDM
\(\therefore \frac{A B}{C D}=\frac{B M}{M D} \Rightarrow \quad \frac{A B}{6}=\frac{24}{8} \Rightarrow A B=18 \mathrm{~cm}\) -
(i) (a): Since \(\Delta\)ABC - \(\Delta\)PQR
\(\therefore \quad \frac{A B}{P Q}=\frac{B C}{Q R} \Rightarrow \frac{x}{20}=\frac{50}{10} \Rightarrow x=100\)
Thus, height of tower is 100 m.
(ii) (c): Since \(\Delta\)ABC - \(\Delta\)PQR
\(\therefore \quad \frac{100}{20}=\frac{x}{15} \Rightarrow x=\frac{1500}{20}=75 \mathrm{~m}\)
(iii) (d): Since, the shapes are similar
\(\therefore \quad \frac{x}{20}=\frac{20}{10} \)
\(\Rightarrow \quad x=\frac{20 \times 20}{10}=40 \mathrm{~m}\)
(iv) (b): Since, the shapes are similar, so, \(\frac{40}{100}=\frac{x}{40}\)
\(\Rightarrow\) x = 16m
(v) (d): Since, the shapes are similar, so \(\frac{20}{100}=\frac{x}{40}\)
\(\Rightarrow \quad x=\frac{20 \times 40}{100}=8 \mathrm{~m}\)
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(i) (a): In \(\Delta\)PAB and \(\Delta\)PQR,
\(\angle\)P = \(\angle\)P (Common)
\(\angle\)A = \(\angle\)Q (Corresponding angles)
By AA similarity criterion, \(\Delta\)PAB \(\sim\) \(\Delta\)PQR
\(\therefore \frac{A B}{Q R}=\frac{P A}{P Q} \Rightarrow \frac{A B}{12}=\frac{6}{24} \Rightarrow A B-3 \mathrm{~m}\)
(ii) (d): Similarly, \(\Delta\)PCD and \(\Delta\)PQR are similar
\(\therefore \frac{P C}{P Q}=\frac{C D}{Q R} \Rightarrow \frac{14}{24}=\frac{C D}{12} \Rightarrow C D=7 \mathrm{~m}\)
(iii) (a): Area of whole empty land
= \(\frac{1}{2}\)x base x height = \(\frac{1}{2}\) x 1 x15= 90 m2.
(iv) (b): Since, \(\Delta\)PAB \(\sim\) \(\Delta\)PQR
\(\therefore \frac{\operatorname{ar}(\Delta P A B)}{a r(\Delta P Q R)}=\left(\frac{P A}{P Q}\right)^{2}=\left(\frac{6}{24}\right)^{2}=\frac{1}{16}\)
\(\Rightarrow \quad \operatorname{ar}(\Delta P A B)=\frac{1}{16} \times 90=\frac{45}{8} \mathrm{~m}^{2}\) \(\left[\because \operatorname{ar}(\Delta P Q R)=90 \mathrm{~m}^{2}\right]\)
(v) (d): Since, \(\Delta\)PCD \(\sim\) \(\Delta\)PQR
\(\therefore \frac{\operatorname{ar}(\Delta P C D)}{\operatorname{ar}(\Delta P Q R)}=\left(\frac{P C}{P Q}\right)^{2}=\left(\frac{14}{24}\right)^{2}=\left(\frac{7}{12}\right)^{2}\)
\(\Rightarrow \operatorname{ar}(\Delta P C D)=\frac{90 \times 49}{144}=\frac{245}{8} \mathrm{~m}^{2}\) -
(i) (b)
(ii) (c): Using Pythagoras theorem, we have PQ2 = PR2 + RQ2
\(\Rightarrow(26)^{2}=(2 x)^{2}+(2(x+7))^{2} \Rightarrow 676=4 x^{2}+4(x+7)^{2} \)
\(\Rightarrow 169=x^{2}+x^{2}+49+14 x \Rightarrow x^{2}+7 x-60=0\)
\(\Rightarrow x^{2}+12 x-5 x-60=0 \)
\(\Rightarrow x(x+12)-5(x+12)=0 \Rightarrow(x-5)(x+12)=0 \)
\(\Rightarrow x=5, x=-12\)
\(\therefore \quad x=5\) [Since length can't be negative]
(iii) (a) : PR = 2x = 2 x 5 = 10 km
(iv) (b): RQ= 2(x + 7) = 2(5 + 7) = 24 km
(v) (d): Since, PR + RQ = 10 + 24 = 34 km Saved distance = 34 - 26 = 8 km -
(i) (c)
(ii) (b): Since \(\Delta\)AOB ~ \(\Delta\)COD [ByAA similarity criterion]
\(\therefore \frac{A O}{O C}=\frac{B O}{O D} \Rightarrow \frac{3}{x-3}=\frac{x-5}{3 x-19}\)
\(\Rightarrow 3(3 x-19)=(x-5)(x-3) \)
\(\Rightarrow 9 x-57=x^{2}-3 x-5 x+15 \Rightarrow x^{2}-17 x+72=0 \)
\(\Rightarrow(x-8)(x-9)=0 \Rightarrow x=8 \text { or } 9\)
(iii) (c) : Since, \(\Delta\)AOB ~ \(\Delta\)COD [ByAA similarity criterion]
\(\therefore \frac{A O}{O C}=\frac{B O}{O D} \Rightarrow \frac{6 x-5}{2 x+1}=\frac{5 x-3}{3 x-1}\)
\(\Rightarrow(6 x-5)(3 x=1)=(5 x-3)(2 x+1) \)
\(\Rightarrow \quad 18 x^{2}-6 x-15 x+5=10 x^{2}+5 x-6 x-3 \)
\(\Rightarrow \quad 8 x^{2}-20 x+8=0 \Rightarrow 2 x^{2}-5 x+2=0\)
From options, x = 2 is the only value that satisfies this equation.
(iv) (d): Since, \(\Delta\)APQ ~ \(\Delta\)ABC [ByAA similarity criterion]
\(\therefore \quad \frac{A P^{\circ}}{A B}=\frac{A Q}{A C}=\frac{P Q}{B C} \Rightarrow \frac{2.4}{A B}=\frac{2}{5}=\frac{P Q}{6} \)
\(\therefore \quad A B=\frac{2.4 \times 5}{2}=6 \mathrm{~cm} \text { and } P Q=\frac{2 \times 6}{5}=2.4 \mathrm{~cm} \)
\(\therefore \quad A B+P Q=6+2.4=8.4 \mathrm{~cm}\)
(v) (a): Since, \(\Delta\)DRS ~ \(\Delta\)DEF [ByAA similarity criterion]
\(\therefore \quad \frac{D E}{D R}=\frac{D F}{D S} \Rightarrow \frac{D E}{D R}-1=\frac{D F}{D S}-1 \)
\(\Rightarrow \frac{D E-D R}{D R}=\frac{D F-D S}{D S} \Rightarrow \frac{E R}{D R}=\frac{F S}{D S} \)
\(\Rightarrow \quad \frac{D R}{E R}=\frac{D S}{F S} \Rightarrow \frac{4 x-3}{3 x-1}=\frac{8 x-7}{5 x-3} \)
\(\Rightarrow \quad 20 x^{2}-12 x-15 x+9=24 x^{2}-8 x-21 x+7 \)
\(\Rightarrow \quad 4 x^{2}-2 x-2=0 \Rightarrow 2 x^{2}-x-1=0\)
Only option (a) i.e., x = 1 satisfies this equation. -
(i) (b): If \(\Delta\)AED and \(\Delta\)BEC, are similar by SAS similarity rule, then their corresponding proportional sides are \(\frac{B E}{A E}=\frac{C E}{D E}\)
(ii) (c): By Pythagoras theorem, we have
\(\begin{array}{l} B C=\sqrt{C E^{2}+E B^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16+9} \\ =\sqrt{25}=5 \mathrm{~cm} \end{array}\)
(iii) (a): Since \(\Delta\)ADE and \(\Delta\)BCE are similar.
\(\therefore \quad \frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{A D}{B C} \)
\(\Rightarrow \frac{2}{3}=\frac{A D}{5} \Rightarrow A D=\frac{5 \times 2}{3}=\frac{10}{3} \mathrm{~cm}\)
(iv) (b):\(\frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{E D}{C E} \)
\(\Rightarrow \frac{2}{3}=\frac{E D}{4} \Rightarrow E D=\frac{4 \times 2}{3}=\frac{8}{3} \mathrm{~cm}\)
(v) (d) : \(\frac{\text { Perimeter of } \Delta A D E}{\text { Perimeter of } \Delta B C E}=\frac{A E}{B E} \Rightarrow \frac{2}{3} B E=A E\)
\(\Rightarrow A E=\frac{2}{3} \sqrt{B C^{2}-C E^{2}} \)
\(\text { Also, in } \triangle A E D, A E=\sqrt{A D^{2}-D E^{2}}\) -
Let BC represents the height of bus and DE represents the height of building.
(i) (a) : In \(\Delta\)ABC and \(\Delta\)ADE,
\(\angle\)A = \(\angle\)A (Common)
\(\angle\)B = \(\angle\)D (Corresponding angles)
\(\therefore\) \(\Delta\)ABC \(\sim\) \(\Delta\)ADE, (By AA similarity criteria)
(ii) (b): We have, AB = 2BC
\(\Rightarrow B C=\frac{25}{2}=12.5 \mathrm{~m}\)
So, height of bus = 12.5 m
(iii) (c) : We have, AD = 12 BC
\(\Rightarrow A D=12 \times 12.5=150 \mathrm{~m} \)
\(\because \Delta A B C \sim \Delta A D E \)
\(\therefore \frac{A B}{A D}=\frac{B C}{D E} \Rightarrow \frac{B C}{D E}=\frac{25}{150}=\frac{1}{6}\)
So, ratio of heights of bus and building is 1 : 6.
(iv) (a): Since, \(\Delta\)ABC \(\sim\) \(\Delta\)ADE
\(\Rightarrow \quad \frac{A B}{A D}=\frac{A C}{A E} \Rightarrow \frac{A C}{A E}=\frac{1}{6}\)
\(\Rightarrow \quad \frac{A C}{A E-A C}=\frac{1}{6-1} \Rightarrow \frac{A C}{E C}=\frac{1}{5}\)
\(\therefore\) Required ratio = 1 : 5
(v) (b): Height of the building = DE
\(\text { Now, } \frac{B C}{D E}=\frac{1}{6} \)
\(\Rightarrow \quad D E=6 B C=6 \times 12.5=75 \mathrm{~m}\) -
(i) (b): Let \(\Delta\)ABC is the triangle formed by both hotels and mountain top. \(\Delta\)CDE is the triangle formed by both huts and mountain top. Clearly, DE || AB and so
\(\triangle A B C \sim \triangle D E C\) [By AA-similarity criterion]
Now, required ratio = Ratio of their corresponding sides \(=\frac{B C}{E C}=\frac{10}{7}\) i.e., 10:7.
(ii) (c): Since, DE || AB, therefore
\(\frac{C D}{A D}=\frac{C E}{E B} \Rightarrow \frac{10}{A D}=\frac{7}{3} \Rightarrow A D=\frac{10 \times 3}{7}=4.29 \text { miles }\)
(iii) (b) : Since, \(\triangle A B C \sim \triangle D E C\)
\(\therefore \quad \frac{B C}{E C}=\frac{A B}{D E}\) [ \(\because\) Corresponding sides of similar triangles are proportional]
\(\Rightarrow \frac{10}{7}=\frac{A B}{8} \Rightarrow A B=\frac{80}{7}=11.43 \text { miles }\)
(iv) (a) Given, DC= 5+ BC.
Clearly, BC = 10-5 = 5 miles
Now, CE = \(\frac{7}{10}\)x BC = \(\frac{7}{10}\) x 5 = 3.5 miles
(v) (d) :Clearly the radio of areas of two angles (i,e \(\triangle A B C \sim \triangle D E C\))
\(\begin{array}{l} =\left(\frac{B C}{E C}\right)^{2}=\left(\frac{10}{7}\right)^{2}=\frac{100}{49} \\ \therefore \quad \text { Required ratio }=\frac{\operatorname{ar}(\Delta C D E)}{a r(E B A D)}=\frac{49}{100-49}=\frac{49}{51} \end{array}\) -
(i) (b): AC2 = AB2 + BC2 [ By Pythagoras theorem ]
⇒ AC2 = (1.8)2 + (2.4)2
⇒ AC2 = 3.24 + 5.76
⇒ AC2 = 9
⇒ AC = 3m
(ii) (c): She pulls the string at the rate of 5cm/s
∴ String pulled in 12 second = 12 x 5 = 60cm
= 0.6
(iii) (a): Length of string out after 12 second is AP.
⇒ AP = AC – String pulled by Nazima in 12 seconds.
⇒ AP = (3 − 0.6)m
= 2.4m
(iv) (b): In â–³ADB, AB2 + BP2 = AP2 ⇒ (1.8)2 + BP2 = (2.4)2
⇒ BP2 = 5.76 − 3.24 ⇒ BP2 = 5.76 − 3.24
⇒ BP2 = 2.52 ⇒ BP = 1.58 m
Horizontal distance of fly = BP + 1.2m Horizontal distance of fly =1.58m + 1.2m
∴ Horizontal distance of fly = 2.78m
(v) (a): Triangles -
(i) (c): Pythagoras Theorem
(ii) (a): ADC is a right-angled triangle. By Pythagoras theorem we get AC2 = (30)2 + (40)2
⇒ AC2 = 900 + 1600
⇒ AC2 = 2500
⇒ AC = 50m
(iii) (d): (21, 20, 28)
since,\((28)^{2} \neq(20)^{2}+(21)^{2}\)
(iv) (b): AC = 50m and BC = 12m
⇒ AB = AC - BC
⇒ AB = 50 - 12 = 38m
(v) (c): Length of the rope used = 30 + 40 + 12
= 82m -
(i) (c): In this kite shape, diagonals bisect each other at 90o.
(ii) (b): SAS
(iii) (b): Since ratio of sides of similar triangles = Ratio of their corresponding medians Therefore, the required ratio is 4 : 9.
(iv) (d): Converse of Pythagoras theorem
(v) (c): Since ,length sticks are 6 cm and 8 cm or length of diagonals
Then, Area of kite \(=\frac{1}{2} \times d_{1} \times d_{2}\)
\(=\frac{6 \times 8}{2}\)
= 24 cm2 -
(i) (c): Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post.
From the figure, DE is the shadow of the girl. Let DE be x metres.
Now, her distance from the base of the lamp = BD = 1.2 m x 4 = 4.8 m.
(ii) (a): In \(\triangle\) ABE and \(\triangle\)CDE, \(\angle\)B = \(\angle\)D ( Each is of 90o)
and \(\angle\)E = \(\angle\)E (same angle)
So, \(\Delta \mathrm{ABE} \sim \Delta \mathrm{CDE}\) (AA similarity criterion)
(iii) (d): \(\Delta \mathrm{ABE} \sim \Delta \mathrm{CDE}\)
\(\Rightarrow \frac{\mathrm{BE}}{\mathrm{DE}}=\frac{\mathrm{AB}}{\mathrm{CD}}\)
\(\Rightarrow \frac{4.8+x}{x}=\frac{3.6}{0.9} \quad \Rightarrow 4.8+x=4 x\)
\(\Rightarrow 3 x=4.8 \ \Rightarrow \ x=1.6\)
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
(iv) (b): Since ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides,
Ratio of areas of similar triangles = \(\sqrt{9}\) : \(\sqrt{16}\)
= 3 : 4
(v) (b): \(\frac{A E}{C E}=\frac{B E}{D E}=\frac{4.8+1.6}{1.6}=\frac{6.4}{1.6}=4\)
\(\Rightarrow\) AE = 4 CE
\(\Rightarrow\) AC + CE = 4 CE
\(\Rightarrow A C=3 C E \Rightarrow \frac{A C}{C E}=\frac{3}{1}\) -
(i) (b): (b) \(\triangle\)s PQM and RSM
(ii) (d): AA similarity
As \(\angle \mathrm{PQM}=\angle \mathrm{RSM}=90^{\circ}\)
and angle of incidence is equal to angle of reflection.
Then, \(\angle \mathrm{PMQ}=\angle \mathrm{RMS}\)
(iii) (a): Since, \(\Delta \mathrm{PMQ} \sim \Delta \mathrm{RMS}\)
Then, \(\frac{\mathrm{PQ}}{\mathrm{RS}}=\frac{\mathrm{QM}}{\mathrm{MS}} \Rightarrow \mathrm{PQ}=\frac{2.8 \times 2}{1.4}\)
= 4m
(iv) (c): Distance between Rashmi and Gulmohar tree
QS = QM + MS = 2.8 + 1.4 = 4.2 m