• Home
  • Courses
  • Online Test
  • Contact
    Have any question?
    +91-8287971571
    contact@dronstudy.com
    Login
    DronStudy
    • Home
    • Courses
    • Online Test
    • Contact

      Class 9 Maths

      • Home
      • All courses
      • Class 09
      • Class 9 Maths
      CoursesClass 09MathsClass 9 Maths
      • 1.Number System
        14
        • Lecture1.1
          Numbers and its Classification, Rational Number and its Representation on the Number Line 50 min
        • Lecture1.2
          Decimal Representation of Rational Number and Conversion of Decimal Number into Rational Number 01 hour
        • Lecture1.3
          Representation of Decimal Number on Number Line, Finding Irrational Numbers Between Given Two Numbers 46 min
        • Lecture1.4
          Representing Irrational Number on the Number Line 41 min
        • Lecture1.5
          Rationalisation 54 min
        • Lecture1.6
          Introduction to Exponents, Laws of Exponents, BODMAS Rule 41 min
        • Lecture1.7
          Integral Exponents of Real Numbers 57 min
        • Lecture1.8
          Integral Exponents of Real Numbers and Solving Equation for x 01 hour
        • Lecture1.9
          Rational Exponents of Real Numbers 52 min
        • Lecture1.10
          Examples Based on Rational Exponents of Real Numbers 30 min
        • Lecture1.11
          Chapter Notes – Number System
        • Lecture1.12
          NCERT Solutions – Number System Exercise 1.1 – 1.6
        • Lecture1.13
          R D Sharma Solutions Number System
        • Lecture1.14
          Revision Notes – Number System
      • 2.Polynomials
        10
        • Lecture2.1
          Introduction to Polynomials and Some Basics Questions 58 min
        • Lecture2.2
          Value of Polynomial, Zeroes of Polynomial and Relationship between Zeroes and Coefficient of Linear Equation 01 hour
        • Lecture2.3
          Division of Polynomials and Division Algorithm and Remainder Theorem 51 min
        • Lecture2.4
          Factorising by Spliting Method 01 hour
        • Lecture2.5
          Factorisation By Different Methods and Factor Theorem 27 min
        • Lecture2.6
          Factorisation of Polynomial by Using Algebric Identities 01 hour
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.5
        • Lecture2.9
          R S Aggarwal Solutions Polynomials
        • Lecture2.10
          Revision Notes Polynomials
      • 3.Coordinate Geometry
        8
        • Lecture3.1
          Representation of Position on Number Line, Finding the location by Using the Axis 01 hour
        • Lecture3.2
          Quadrant, Coordinates of Point on Axis, Distance from Axis 01 hour
        • Lecture3.3
          Finding the Area of Triangle and Rectangle and Mirror Image of Point On Axis 57 min
        • Lecture3.4
          Chapter Notes – Coordinate Geometry
        • Lecture3.5
          NCERT Solutions – Coordinate Geometry Exercise 3.1 – 3.3
        • Lecture3.6
          R D Sharma Solutions Coordinate Geometry
        • Lecture3.7
          R S Aggarwal Solutions Coordinate Geometry
        • Lecture3.8
          Revision Notes Coordinate Geometry
      • 4.Linear Equations
        8
        • Lecture4.1
          Introduction of Linear Equation in Two Variables and Graphical Representation 01 hour
        • Lecture4.2
          Example to find the value of Constant and Graph of Linear Equation in Two Variables 57 min
        • Lecture4.3
          Equations of lines Parallel to the x-axis and y-axis 46 min
        • Lecture4.4
          Chapter Notes – Linear Equations
        • Lecture4.5
          NCERT Solutions – Linear Equations Exercise 4.1 – 4.4
        • Lecture4.6
          R D Sharma Solutions Linear Equations
        • Lecture4.7
          R S Aggarwal Solutions Linear Equations
        • Lecture4.8
          Revision Notes Linear Equations
      • 5.Euclid's Geometry
        7
        • Lecture5.1
          Euclid’s Definitions and Euclid’s Axioms 32 min
        • Lecture5.2
          Eulcid’s Postulates and Playfair’s Axiom 37 min
        • Lecture5.3
          Chapter Notes – Euclid’s Geometry
        • Lecture5.4
          NCERT Solutions – Euclid’s Geometry Exercise 5.1 5.2
        • Lecture5.5
          R D Sharma Solutions Euclid’s Geometry
        • Lecture5.6
          R S Aggarwal Solutions Euclid’s Geometry
        • Lecture5.7
          Revision Notes Euclid’s Geometry
      • 6.Lines and Angles
        10
        • Lecture6.1
          Introduction to Lines and Angles, Types of Angle and Axioms 01 hour
        • Lecture6.2
          Vertically opposite Angles and Angle bisector 46 min
        • Lecture6.3
          Angle made by a transversal with two Lines and Parallel Lines 01 hour
        • Lecture6.4
          Lines Parallel to same Line; Exterior angle of a triangle; Angle Sum property of a Triangle 52 min
        • Lecture6.5
          Angle Sum Property of Triangle Including Interior and Exterior Angles 01 hour
        • Lecture6.6
          Chapter Notes – Lines and Angles
        • Lecture6.7
          NCERT Solutions – Lines and Angles Exercise 6.1 – 6.3
        • Lecture6.8
          R D Sharma Solutions Lines and Angles
        • Lecture6.9
          R S Aggarwal Solutions Lines and Angles
        • Lecture6.10
          Revision Notes Lines and Angles
      • 7.Triangles
        11
        • Lecture7.1
          Introduction, Types of Triangle, Altitude & Median of Triangle, Congruences, SAS criterion of Congruence 42 min
        • Lecture7.2
          Questions Based on SAS Criterion 01 hour
        • Lecture7.3
          ASA Criterion of Congruence and Its based Questions 42 min
        • Lecture7.4
          Questions Based on AAS and SSS Criterion 29 min
        • Lecture7.5
          RHS Criterion of Congruence and Its based Questions, Questions Based on Multiple Criterion 01 hour
        • Lecture7.6
          Inequalities of Triangle 42 min
        • Lecture7.7
          Inequalities of Triangle cont., Distance Between a line and The Point, Miscellaneous Questions 48 min
        • Lecture7.8
          Chapter Notes – Triangles
        • Lecture7.9
          NCERT Solutions – Triangles Exercise 7.1 – 7.5
        • Lecture7.10
          R D Sharma Solutions Triangles
        • Lecture7.11
          Revision Notes Triangles
      • 8.Quadrilaterals
        13
        • Lecture8.1
          Congruence of Triangles- SAS Congruence Criteria, ASA Congruence Criteria, SSS Congruence Criteria, RHS Congruence Criteria; Types of Quadrilateral- Parallelogram, Rectangle, Rhombus, Square, Trapezium, Kite 01 hour
        • Lecture8.2
          Theorem 1; Properties of parallelogram- Theorem 2, Theorem 3; 01 hour
        • Lecture8.3
          Theorem 4; Theorem 5; 24 min
        • Lecture8.4
          Conditions for a Quadrilateral to be a parallelogram- Theorem 6 01 hour
        • Lecture8.5
          Theorem 8 01 hour
        • Lecture8.6
          Theorem 9 51 min
        • Lecture8.7
          Theorem 9(Mid-point Theorem); 01 hour
        • Lecture8.8
          Theorem 10 27 min
        • Lecture8.9
          Chapter Notes – Quadrilaterals
        • Lecture8.10
          NCERT Solutions – Quadrilaterals Exercise 8.1 8.2
        • Lecture8.11
          R D Sharma Solutions Quadrilaterals
        • Lecture8.12
          R S Aggarwal Solutions Quadrilaterals
        • Lecture8.13
          Revision Notes Quadrilaterals
      • 9.Area of Parallelogram
        11
        • Lecture9.1
          Parallelograms on the Same Base and between the Same Parallels; A Diagonal of a parallelogram divides it into two triangles of Equal Area 57 min
        • Lecture9.2
          Questions based on parallelogram and Triangle are on same base & parallel lines and Area of Triangle 01 hour
        • Lecture9.3
          Triangles on the same Base and between the same Parallels 56 min
        • Lecture9.4
          Miscellaneous Questions Based on Triangles on the same Base and between the same Parallels 54 min
        • Lecture9.5
          Sums Based on Median of a Triangle 52 min
        • Lecture9.6
          Two Triangles on the same Base and equal Area lie b/w the same parallel 01 hour
        • Lecture9.7
          Summary of All Concepts and Miscellaneous Questions 56 min
        • Lecture9.8
          Chapter Notes – Area of Parallelogram
        • Lecture9.9
          NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4
        • Lecture9.10
          R D Sharma Solutions Area of Parallelogram
        • Lecture9.11
          Revision Notes Area of Parallelogram
      • 10.Constructions
        7
        • Lecture10.1
          Construction of Perpendicular Bisector of a Line Segment, Different Angles and Angle bisector 42 min
        • Lecture10.2
          Construction of Different types of Triangle 43 min
        • Lecture10.3
          Chapter Notes – Constructions
        • Lecture10.4
          NCERT Solutions – Constructions Exercise
        • Lecture10.5
          R D Sharma Solutions Constructions
        • Lecture10.6
          R S Aggarwal Solutions Constructions
        • Lecture10.7
          Revision Notes Constructions
      • 11.Circles
        11
        • Lecture11.1
          Introduction, Important Terms related to circle, Theorem-1, 2, 3 and 4 01 hour
        • Lecture11.2
          Problems Solving 48 min
        • Lecture11.3
          Problems Based on Parallel Chords, Common Chord of Circles 01 hour
        • Lecture11.4
          Problems Based on Concentric Circles, Triangle in Circle 41 min
        • Lecture11.5
          Equal Chords and Their Distance, Theorem-5, 6, 7, 8 01 hour
        • Lecture11.6
          Circle Through Points, Theorem-9, Arc of a Circle, Theorem-10 01 hour
        • Lecture11.7
          Chapter Notes – Circles
        • Lecture11.8
          NCERT Solutions – Circles Exercise
        • Lecture11.9
          R D Sharma Solutions Circles
        • Lecture11.10
          R S Aggarwal Solutions Circles
        • Lecture11.11
          Revision Notes Circles
      • 12.Heron's Formula
        8
        • Lecture12.1
          Introduction of Heron’s Formula; Area of Different Triangles 01 hour
        • Lecture12.2
          Area of Triangles Using Heron’s Formula (Type-1 level-1 and 2) 53 min
        • Lecture12.3
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-1) 32 min
        • Lecture12.4
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-2) 38 min
        • Lecture12.5
          Chapter Notes – Heron’s Formula
        • Lecture12.6
          NCERT Solutions – Heron’s Formula Exercise 12.1 12.2
        • Lecture12.7
          R D Sharma Solutions Heron’s Formula
        • Lecture12.8
          Revision Notes Heron’s Formula
      • 13.Surface Area and Volume
        16
        • Lecture13.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere 40 min
        • Lecture13.2
          Type-1: Surface Area and Volume of Cube and Cuboid 27 min
        • Lecture13.3
          Level-2, Question based on Cube and Cuboid, Level-1 Questions based on Cylinder 28 min
        • Lecture13.4
          Level-2 Questions based on Cylinder, Type-3 Surface Area and Volume of Right Circular Cone (Level-1) 25 min
        • Lecture13.5
          Level-2 Questions based on Right Circular Cone, Level-1& 2 Questions Based on Sphere & Hemisphere 27 min
        • Lecture13.6
          Type-5 Short Answer Types Questions, 12 min
        • Lecture13.7
          Chapter Notes – Surface Area and Volume of a Cuboid and Cube
        • Lecture13.8
          Chapter Notes – Surface Area and Volume of A Right Circular Cylinder
        • Lecture13.9
          Chapter Notes – Surface Area ad Volume of A Right Circular Cone
        • Lecture13.10
          Chapter Notes – Surface Area ad Volume of A Sphere
        • Lecture13.11
          NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9
        • Lecture13.12
          R D Sharma Solutions Surface Areas And Volumes of A cuboid and A cube
        • Lecture13.13
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cylinder
        • Lecture13.14
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cone
        • Lecture13.15
          R D Sharma Solutions Surface Areas And Volumes of A Sphere
        • Lecture13.16
          Revision Notes Surface Areas And Volumes
      • 14.Statistics
        15
        • Lecture14.1
          Data; Presentation of Data; Understanding Basic terms; 01 hour
        • Lecture14.2
          Questions; 52 min
        • Lecture14.3
          Bar Graph, Histogram, Frequency Polygon; 01 hour
        • Lecture14.4
          Histogram for Grouped Frequency Distribution (Class interval in Inclusive form), Frequency Polygon, Measures of Central Tendency, 01 hour
        • Lecture14.5
          Examples Based on Mean of Ungrouped Data 56 min
        • Lecture14.6
          Direct Method of Finding Mean of Ungrouped Data, Median of Cont. Freq. Distribution (Grouped Data) 01 hour
        • Lecture14.7
          Mean; Mean for an ungrouped freq. Dist. Table 25 min
        • Lecture14.8
          Chapter Notes – Statistics – Tabular Representation of Statistical Data
        • Lecture14.9
          Chapter Notes – Statistics – Graphical Representation of Statistical Data
        • Lecture14.10
          Chapter Notes – Statistics – Measures of Central Tendency
        • Lecture14.11
          NCERT Solutions – Statistics Exercise 14.1 – 14.4
        • Lecture14.12
          R D Sharma Solutions Tabular Representation of Statistical Data
        • Lecture14.13
          R D Sharma Solutions Graphical Representation of Statistical Data
        • Lecture14.14
          R S Aggarwal Solutions Statistics
        • Lecture14.15
          Revision Notes Statistics
      • 15.Probability
        8
        • Lecture15.1
          Activities; Understanding term Probability; Experiments; 44 min
        • Lecture15.2
          Event; Probability; Empirical Probability- Tossing a coin; 31 min
        • Lecture15.3
          Rolling a dice; Miscellaneous Example; 27 min
        • Lecture15.4
          Chapter Notes – Probability
        • Lecture15.5
          NCERT Solutions – Probability Exercise 15.1
        • Lecture15.6
          R D Sharma Solutions Probability
        • Lecture15.7
          R S Aggarwal Solutions Probability
        • Lecture15.8
          Revision Notes Probability

        NCERT Solutions – Heron’s Formula Exercise 12.1 12.2

        Exercise 12.1

        Q.1     A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
        Sol.

        To find the area of an equilateral triangle using Heron’s formula.|
        If a, b, c be the lengths of sides BC, CA and AB of Δ ABC and if s=12(a+b+c), then
        area of Δ ABC = s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        Since Δ ABC is given as equilateral of side = a
        Therefore here a = b = c (= a) and s=12(a+a+a)=3a2
        Now , s−a=3a2−a=a2
        s−b=3a2−a=a2
        and s−c=3a2−a=a2
        Therefore Area=3a2×a2×a2×a2−−−−−−−−−−−−−√
        =a2×a23–√=3√a24 … (1)
        To find the area when its perimeter = 180 cm
        Here a + a + a = 180 cm
        ⇒ 3a = 180 cm
        ⇒ a = 60 cm
        Therefore Required area =3√4×(60)2 cm2
        =3√4×3600 cm2
        [On putting a = 60 in (1)]
        =9003–√cm2


        Q.2      The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of Rs 5000 per m2peryear. A company hired one of its walls for 3 months. How much rent did it pay?

        1
        Sol.

        First we find the semi perimeter of triangular side measuring 122 m, 22m and 120 m.
        Let a = 122 m, b = 22 m and c = 120 m
        Therefore s=12(a+b+c)=12(122+22+120)m
        =(12×264)m=132m
        Now, s – a = (132 – 122) m = 10 m
        s – b = (132 – 22) m = 110 m
        and s – c = (132 – 120) m = 12 m
        Therefore Area of triangular wall =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√ [By Heron’s formula]
        =132×10×110×12−−−−−−−−−−−−−−−−√m2
        =11×12×10×10×11×12−−−−−−−−−−−−−−−−−−−−−−√m2
        =10×10×11×11×12×12−−−−−−−−−−−−−−−−−−−−−−√m2
        =(10×11×12)m2=1320m2
        Rent charges = Rs 5000 per m2 per year
        Therefore Rent charged from a company for 3 month
        =Rs(5000×1320×312)
        =Rs16,50,000


        Q.3      There is a slide in a park. One of its side walls has been painted in blue colour with a message “KEEP THE PARK GREEN AND CLEAN” (See figure). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

        2
        Sol.

        First find the semi perimeter of side wall measuring 15 m, 11 m and 6 m.
        Let a = 15 m, b = 11 m, c = 6 m
        and s=12(a+b+c)=12(15+11+6)m
        =(12×32)m=16m
        Now, s – a = (16 – 15)m = 1 m
        s – b = (16 – 11) m = 5 m
        and s – c = (16 – 6) m = 10 m
        Area of side wall =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        =16×1×5×10−−−−−−−−−−−−−√m2
        =4×4×5×5×2−−−−−−−−−−−−−−√m2
        =4×52–√m2
        =202–√m2
        Therefore Area painted in blue colour = Area of side wall
        =202–√m2


        Q.4      Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
        Sol.

        Let a, b and c be the sides of a triangle such that
        a = 18 cm, b = 10 cm and a + b + c = 42 cm
        Therefore c = 42 – a – b
        ⇒ c = (42 – 18 – 10) cm = 14 cm
        Now, s=12(a+b+c)=12×42cm=21cm
        s−a=(21−18)cm=3cm
        s−b=(21−10)cm=11cm
        and s−c=(21−14)cm=7cm
        Therefore Area of the triangle =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        =21×3×11×7−−−−−−−−−−−−−√cm2
        =3×7×3×11×7−−−−−−−−−−−−−−−√cm2
        =3×3×7×7×11−−−−−−−−−−−−−−−√cm2
        =3×711−−√cm2=2111−−√cm2


        Q.5      Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
        Sol.

        Let the sides be a, b and c of the triangle.
        Therefore a : b : c = 12 : 17 : 25
        ⇒ a12=b17=c25=k(say)
        ⇒ a = 12 k, b = 17 k and c = 25 k
        Also, perimeter = 540 cm
        ⇒ a + b + c = 540
        ⇒ 12k + 17k + 25 k = 540
        ⇒ 54 k = 540
        k=54054 i.e., k = 10
        Thus a = 12 × 10 = 120 cm, b = 17 × 10 cm = 170 and c = 25 × 10 = 250 cm
        Now, s=12×540cm=270cm
        s – a = (270 – 120) cm = 150 cm
        s – b = (270 – 170) cm = 100 cm
        and s – c = (270 – 250) cm = 20 cm
        Area of the triangle
        =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        =270×150×100×20−−−−−−−−−−−−−−−−−√cm2
        =10027×15×10×2−−−−−−−−−−−−−√cm2
        =1003×3×3×3×5×5×2×2−−−−−−−−−−−−−−−−−−−−−−−√cm2
        =100×3×3×5×2cm2=9000cm2


        Q.6      An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
        Sol.

        Here a = b = 12 cm and a + b + c = perimeter = 30 cm
        ⇒ c = (30 – a – b) cm
        ⇒ c = (30 –12 – 12) cm
        =(30 – 24) cm
        = 6 cm
        3
        Now , s=12×30cm=15cm
        s−a=(15−12)cm=3cm
        s−b=(15−12)cm=3cm
        and s−c=(15−6)cm=9cm
        Therefore, Area of the triangle =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        =15×3×3×9−−−−−−−−−−−−√cm2
        =5×3×3×3×3×3−−−−−−−−−−−−−−−−−√cm2
        =3×35×3−−−−√cm2=915−−√cm2
        Hence, the area of triangle is = 915−−√ cm2.

         

        Exercise 12.2

        Q.1      A park, in the shape of a quadrilateral ABCD, has ∠C= 90º, AB =  9 m , BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
        Sol.

        Area of ΔBCD=12×BC×CD
        =(12×12×5)m2
        =30m2

        4Using Pythagoras theorem, we have
        BD2=BC2+CD2 ⇒ BD2=122+52
        ⇒ BD2=144+25 ⇒ BD2=169
        ⇒ BD=169−−−√=13m
        For :ΔABD:
        Let a = 13 m, b = 8 m and c = 9 m
        Now, s=12(a+b+c)=12(13+8+9)m
        =12×30m=15m
        s−a=(15−13)m=2m
        s−b=(15−8)m=7m
        and s−c=(15−9)m=6m
        Therefore AreaofΔABD=15×2×7×6−−−−−−−−−−−−√m2
        =3×5×2×7×2×3−−−−−−−−−−−−−−−−−√m2
        =2×2×3×3×5×7−−−−−−−−−−−−−−−−−√m2
        =2×335−−√m2
        divide
        =6×5.9m2(approx)
        =35.4m2(approx)
        Therefore Required area = ΔABD+ΔBCD
        =35.4m2+30m2=65.4m2


        Q.2      Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
        Sol.

        Since AC2=AB2+BC2
        (as52=32+42i.e.,25=9+16)
        Therefore ∠ABC= 90º

        5Area of rt ∠dΔABC=12×AB×BC
        =(12×3×4)cm2=6cm2
        For ΔACD :
        Let a = 5 cm, b = 4cm and c = 5 cm. Then ,
        s=12(a+b+c)=12(5+4+5)cm
        =12×14cm=7cm
        Now, s−a=(7−5)cm=2cm
        s−b=(7−4)cm=3cm
        and s−c=(7−5)cm=2cm
        Area of ΔACD=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        =7×2×3×2−−−−−−−−−−−√cm2
        =221−−√cm2
        divide1
        =2×4.6cm2(approx.)
        =9.2cm2(approx.)
        Area of quadrilateral ABCD
        AreaofΔABC+AreaofΔACD
        =(6+9.2)cm2=15.2cm2approx


        Q.3      Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

        6Sol.       Area of region I:

        Region I is Enclosed by a triangle of sides a = 5 cm, b = 5 cm and c = 1 cm

        Let s be the perimeter of the triangle. Then,
        Now, s=12(a+b+c)
        =12(5+5+1)m
        ⇒s=112cm

         AreaofregionI=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√cm2

        ⇒AreaofregionI=112×(112−5)×(112−5)×(112−1)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√cm2

        ⇒AreaofregionI=112×12×12×92−−−−−−−−−−−−−√cm2=311√4cm2

        ⇒AreaofregionI=34×3.32=2.49cm2

        Area of region II:

        Region II is a rectangle of length 6.5 cm and breadth 1 cm.

         AreaofregionII=6.5×1=6.5cm2

        Area of region III:

        Region III is an isosceles trapezium.

        In ΔABE, We have

        AB2  = AE2 + BE2

        ⇒                       1    =   0.25 +  BE2

        ⇒                       BE  = 0.75−−−−√�=34−−√

        AreaofregionIII=12(AD+BC)×BE=12(2+1)×34−−√cm2

        ⇒AreaofregionIII=33√4cm2=1.3cm2

        Area of region IV:

        Region IV forms a right triangle whose two sides are of lengths 6 cm and 1.5 cm.

        AreaofregionIV=12×6×1.5cm2=4.5cm2

        Area of region V:

        Region IV and V are congruent.

        Area of region V = 4.5 cm2

        Hence, total area of the paper used = (2.49 + 6.5 + 1.3 + 4.5 + 4.5) cm2

        = 19.29 cm2


        Q.4    A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
        Sol.

        For the triangle :
        Let its sides be a, b and c such that
        a = 26 cm, b = 28 cm and c = 30. Then,
        s=12(26+28+30)cm
        =12×84cm=42cm
        Now, s−a=(42−26)cm=16cm
        s−b=(42−28)cm=14cm
        and s−c=(42−30)cm=12cm
        Area of the triangle
        =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        =42×16×14×12−−−−−−−−−−−−−−√cm2
        =2×3×7×4×4×2×7×2×2×3−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√cm2
        =2×2×2×2×3×3×4×4×7×7−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√cm2
        =(2×2×3×4×7)cm2=336cm2
        For the parallelogram :
        Area = base × height
        Therefore Height=AreaBase
        [Since Area of parallelogram = Area of Δ (given)
        Therefore Area of parallelogram = 336 cm2 and its base = 28 cm]
        =(33628)cm
        =12cm


        Q.5     A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
        Sol.

        We know that the diagonals of the rhombus bisect each other at right angles. Using Pythagoras theorem, we have OD=AD2−AO2−−−−−−−−−−√=302−242−−−−−−−−√m
        =(30+24)(30−24)−−−−−−−−−−−−−−−√m
        =54×6−−−−−√m
        =9×6×6−−−−−−−−√m
        =(3×6)m=18m
        7Area of one ΔAOD=(12×24×18)m2=216m2
        Therefore Area of rhombus =4×ΔAOD
        =(4×216)m2=864m2
        Grass area for 18 cows =864m2
        Grass area for 1 cow =(86418)m2=48m2


        Q.6      An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

        8Sol.

        In one triangular piece, let a = 20 cm, b = 50 cm and c = 50 cm
        Now, s=12(a+b+c)=12(20+50+50)cm
        =12×120cm=60cm
        s−a=(60−20)cm=40cm
        s−b=(60−50)cm=10cm
        and s−c=(60−50)cm=10cm
        Therefore Area of one triangular piece
        =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        =60×40×10×10−−−−−−−−−−−−−−√cm2=2006–√cm2
        Therefore Area of 5 red triangles =10006–√cm2
        and , area of 5 green triangles =10006–√cm2


        Q.7     A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it ?
        9Sol.

        ABCD is a square such that AC = BC = 32 cm and AEF is an isosceles triangle in which AE = AF = 6 cm
        For the area of shades I and II
        Clearly from the figure
        Area of shade I = Area of shade II = Area of Δ CDB
        =12×DB×CM [Since ABCD is in square shapes]
        =12×32×16 sq cm
        =256sqcm
        For the area of shade III (i.e., Δ AEF)
        EL=LF=12EF
        =12×8=4cm
        11and AE=6cm(given)
        Therefore AL=AE2−EL2−−−−−−−−−−√
        =36−16−−−−−−√=20−−√=25–√cm
        Area of shade III =12×EF×AL
        =12×8×25–√sqcm
        =85–√sqcm
        =8×2.24sqcm(approx)
        =17.92sqcm(approx)


        Q.8     A floral design on floor is made up of 16 tiles which are triangular , the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 p per cm2.

        10Sol.

        For one triangular tile :
        Let a = 9 cm, b = 28 cm and c = 35 cm
        Now , s=12(a+b+c)=12(9+28+35)cm
        =12×72cm=36cm
        s−a=(36−9)cm=27cm
        s−b=(36−28)cm=8cm
        and s−c=(36−35)cm=1cm
        Area of one tile =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        =36×27×8×1−−−−−−−−−−−−−√cm2
        =9×4×9×3×4×2−−−−−−−−−−−−−−−−−√cm2
        =2×3×4×4×9×9−−−−−−−−−−−−−−−−−√cm2=4×96–√cm2
        =366–√cm2=36×2.45cm2approx.
        =88.2cm2approx.
        Therefore Area of 16 such tiles =(16×88.2)cm2
        =1411.2cm2approx.
        Cost of polishing @ 50 P per cm2=Rs(1411.2×50100)
        =Rs705.60approx.


        Q.9     A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non- parallel sides are 14 m and 13 m. Find the area of the field.
        Sol.          Various length are as marked in the figure.
        12

        Clearly DM = CN
        ⇒ DM2=CN2
        [Distance between parallel sides are always equal]
        ⇒ (13)2−x2=142−(15−x)2
        ⇒ 169−x2=196−(225−30x+x2)
        ⇒ 50x=169−196+625=598
        ⇒ x=59850=11.96
        Now DM=AD2−AM2−−−−−−−−−−√=132−(11.96)2−−−−−−−−−−−√
        =(13+11.96)(13−11.96)−−−−−−−−−−−−−−−−−−−−√
        =24.96×1.04−−−−−−−−−−√=25.9584−−−−−−√=5.09
        Area of the trapezium =12×(AB+CD)×DM
        =12×(25+10)×5.09sq.m
        =12×35×5.09sq.m
        =89.075sq.m

        Prev Chapter Notes – Heron’s Formula
        Next R D Sharma Solutions Heron’s Formula

        Leave A Reply Cancel reply

        Your email address will not be published. Required fields are marked *

        All Courses

        • Backend
        • Chemistry
        • Chemistry
        • Chemistry
        • Class 08
          • Maths
          • Science
        • Class 09
          • Maths
          • Science
          • Social Studies
        • Class 10
          • Maths
          • Science
          • Social Studies
        • Class 11
          • Chemistry
          • English
          • Maths
          • Physics
        • Class 12
          • Chemistry
          • English
          • Maths
          • Physics
        • CSS
        • English
        • English
        • Frontend
        • General
        • IT & Software
        • JEE Foundation (Class 9 & 10)
          • Chemistry
          • Physics
        • Maths
        • Maths
        • Maths
        • Maths
        • Maths
        • Photography
        • Physics
        • Physics
        • Physics
        • Programming Language
        • Science
        • Science
        • Science
        • Social Studies
        • Social Studies
        • Technology

        Latest Courses

        Class 8 Science

        Class 8 Science

        ₹8,000.00
        Class 8 Maths

        Class 8 Maths

        ₹8,000.00
        Class 9 Science

        Class 9 Science

        ₹10,000.00

        Contact Us

        +91-8287971571

        contact@dronstudy.com

        Company

        • About Us
        • Contact
        • Privacy Policy

        Links

        • Courses
        • Test Series

        Copyright © 2021 DronStudy Pvt. Ltd.

        Login with your site account

        Lost your password?

        Modal title

        Message modal