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Class 7 R S Aggarwal Maths Solutions

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20. Areas of Triangles 20D, 20E, 20F

 

EXERCISE -20D. AREAS OF TRIANGLES.

1.      Area Of triangle  =   

2.      Area of right angle =  

3.      area of an equilateral triangle with side a =

4.      Hero’s formula : If a, b, b, be the sides of a triangle ABC and

s =   (a + b + c), then,

area of triangle ABC =  

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Q1.         I]  b = 42 cm, h = 25 cm. Area Of triangle  =   

                =  

                Ii]     Area Of triangle  =    

Iii]    b = 8dm = 80 cm,  h = 35 cm.   Area Of triangle  =  

Q2.         Area Of triangle  =     => 72 =    => h =        

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Q3.         Area Of triangle  =     =>  224 =    = 16 m.

 

Q4.         Area Of triangle  =  90 = 12.   => b =   

 

Q5.         Let the height be y. Then b = 3y.                 we know, Area Of triangle  =   

                Area =   Cost of cultivation = area x rate(per hectare)

π  14580 = A  x 1080.      A =  

Since, 1hectare = 10000 m2.

                13.5 = 135000 m2 = area of triangular region.

                135000 =   =>   y2 =     = 300.

            height = 300m & base = 300 x 3 = 900 m.

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Q6.         Area Of triangle  =  

                129.5 = 7.4 b.     b =  

 

Q7.         Given,    BC = 1.2 m,  AC = 3.7 m, AB = ?,  area =?

                Now,     AB2 + BC2 = AC2 .   =>   AB2 =  AC2 - BC2 = (3.7)2 – (1.2)2

                AB2 = 13.69 – 1.44 = 12.25 m.     => AB =  = 3.5 m.

                  Area Of triangle  = 3.5 = 2.1 m2.

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Q8.         Area Of triangle  =    =>   1014 =  

π  1014  x 2 = 12y2.             y2 =          y =   = 13.

   AB = 3y = 3 x 13 = 39 cm. & BC = 4y = 4 x 13 = 52 cm.        

 

Q9.         AC2 = AB2 + BC2.       =>  1002 = y2 + 80cm.   =>  10000 – 6400 = 3600 = y2

                Y = 60 cm.    => Area Of triangle  = 2400cm2.

                           [since, it is in square. 100 x 100]

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Q10.       I]  Area of equilateral triangle with side 18 cm.

                =       =  

                Ii]    =     =  = 1.73 x 100  = 173 cm2.

 

Q11.       Area of equilateral triangle =.             => (16 x   )cm2  =   .             

π  a2 =     => a =     =  8 cm.

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Q12.       Area of equilateral triangle =.     =   . ----(1)

                Area of triangle =       -------(2)

                From (1) and (2)

                               => h = 12  12 x 1.73  = 20.76 cm.

 

Q13.       I]  area of triangle =       .

                S =  

                   =  

                  = 7 x 3 x 2 x 2 = 84 m2.     

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Q14.       S =     area =   

                =    =        

                =   =  11 x 11 x 3 x 2 = 726 cm2.

                Also  area =             h =  

 

Q15.       Sides of triangle are in ratio 13:14:15.   Perimeter = 84 cm.

                    13y + 14y + 15y = 84.   => 42y = 84.    => y = 2.

                     sides are 13y = 13 x 2 = 26 cm.    14y = 14 x 2 = 28 cm.

                  15y = 15 x 2 = 30 cm.      s =   

                Area of triangle =   

 = 

                    =   

= 14 x 3 x 4 x 2 = 336 cm2.  www.rsmaths99.com

 

Q16.       Sides are 42 cm, 34 cm, and20 cm.

                S =     Area of triangle =    

                =      =

                =         = 6 x 2 x 2 x 7 x 2   = 336 cm2.

                Also, area of triangle =   .       21h = 336      h =  16   req. height = 16cm.

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Q17.       S =  .     Area of triangle =    

                =        =  

                 =           = 6 x 3 x 4 x 6  = 43 cm2.

 

Q18.       Let the equal sides be y cm each. ATQ, perimeter = 32 cm

                  y + y + 12 = 32   => 2y = 20   => y = 10 cm.

                  S =     = 16.     www.rsmaths99.com

      Area of triangle =    

                 =       =  

                =        =    = 4 x 3 x 4 = 48 cm2.

 

Q19.       Triangle ABC =    =  =  145.6.

                Triangle ADC =      .

                Then, area of quadrilateral ABCD =   145.6 +    = 312 cm2.

 

Q20.       Triangle ACD ,  S =        =  = 

                Area ACD =      =  

                    =     = 6 x 100 = 600cm2.

                Triangle ABC  S=      =      www.rsmaths99.com

                        =   

                =     = 336.   Area of ABCD = 600 + 336 = 936 cm2.

 

Q21.       Area of ABCD = 36 X 24 = 864 m2.     Area of AED =  . 

                Area of shaded region = 864 – 180 = 684 m2.

 

Q22.       Area of ABCD =40 x 25 = 1000 cm2.  www.rsmaths99.com

                Area of triangle ASP =    Area of triangle PBQ =    

                ar of triangle QCR =   and ar of triangle RDS =

                Total unshaded region area = 125 x 4 = 500 cm2. 

Total shaded region = 1000 – 500 = 500 cm2.

Q23.       I]  Area ABCD = 18 x 10 = 180 cm2.    www.rsmaths99.com

   In triangle BCE :   AREA =  

                In triangle EDE:  area  of triangle EDF =   

                Total unshaded region = 40 + 30 = 70 cm2. Area of unshaded region = 180 -70 = 110 cm2.

               

                Ii]  Area ABCD = 20 X 20 = 400 cm2.  Area of triangle MQR =

                  Area of triangle RSL =   area LPM =

                Total unshaded region = 250 cm2. Area of shaded region = 400 – 250 = 150 cm2.    

Q24.       In triangle ABD:  Area ABD =   = 60 cm2. www.rsmaths99.com

                In triangle BDC:  Area BCD = .

                Total area = 156 cm2.

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EXERCISE- 20E,CIRCUMFERENCE AND AREA OF A CIRCLE

Circumference of circle = 2r,  

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Q1.         Given, r – 15 cm.   Circumference of circle = 2r = 2 x 3.14 x 15 = 94.2 cm.

Q2.         I] Circumference of circle = 2r = 2 x

                Ii]  Circumference of circle = 2r = 2 x

Q3.         I]  diameter = 35 cm.  r = 

                 Circumference of circle = 2r =  2 x

                Ii]  d = 4.9 m.  r =

                   Circumference of circle = 2r = 2 x   

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Q4.         Circumference of circle = 2r = 2 x  r =

 

Q5.         D = ?,     Circumference of circle = 2r = 2 x  

                   d = 10.15 x 2 = 20.3 m.

 

Q6.         Let the radius of the circle be r cm. Then, its circumference = 2r cm.

                Now, circumference. – diameter = 30 cm.   [given]

                (2r – 2r) = 30.  => 2r()=30.   => 2r 

π   2r x      =>  r =      www.rsmaths99.com           

   Radius of the given circle is

Q7.         Let the radii of the given circle be 5y & 3y respectively. And let there circumference

be C1 & C2 respectively.. then, C1 = 2 x  x 5y = 10y &   C2 = 2 x  x 3y = 6y.

                       =        => C1  : C2 =5 : 3. Hence, the ratio of the circumference of the given circle is  5 : 3.

 

Q8.         Circumference of circle = 2r = 2 x     

                Since, he cycling  8 km in   1 hr.

                                                1 km  in      www.rsmaths99.com

                0.132 km  in        = 59.4 sec.  

 

Q9.         Let the inner radius be r m and outer radius be R m . Then,        

                2r  = 528            and        2R = 616 .  

                2 x      2 x 616. 

=>           r =         =>(R – r ) = 98 – 84 = 14 m.

Hence, the width of the track is 14 m.  www.rsmaths99.com

 

Q10.         Let the inner and outer radii of the track be r m and (r + 10.5)m respectively.

2r = 330 m .      2 x        r =    

                Then, outer radius = 52.5 m + 10.5 = 63 m.

                Circumference of the outer circle = 2 x

                Rate of fencing = Rs. 20 per meter.   Cost of fencing = 396 x 20 = Rs. 7920.

 

Q11.       Given,     r = 98 cm.,    R =  1m 26cm = 126 cm.  www.rsmaths99.com

                In 1st circle:  Circumference of circle = 2 x   98 = 616 cm.

                In 2nd circle : Circumference of circle = 2 x   126 = 792 cm.

                (R – r ) = 792 – 616 = 176 cm.  Hence, 2nd circle is  176 cm longer than 1st  circle.

 

Q12.       Perimeter = 8.8 x 3 = 26.4 cm.   Circumference of circle = 2r .

ATQ, [perim of triangle = coc]    www.rsmaths99.com

π  26.4  = 2 x   r = 4.2 m.  Hence, diameter = 4.2 x 2= 8.4 m.

 

Q13.       Perimeter pf rhombus = 4 x 33 = 132 cm. 

ATQ,  Perimeter of rhombus = Circumference of circle . then,

Circumference of circle =  2r.

π     132  =  2 x   r =      

Q14.       Perimeter of rectangle = 2 x (l + b ) = 2 x (18.7 + 14.3 )= 66 cm.

                Circumference of circle =  2r.    => 66 = 2 x    www.rsmaths99.com

                  r =  

 

Q15.       ATQ, radius of circle = 35 cm.    

Circumference of circle =  2r =  2 x =220 cm.

                 perimeter of square = 220 cm.  => Each side =  

 

Q16.       Length of the hour hand = R = 4.2 cm.   www.rsmaths99.com

                Distance covered by hour hand in 1 day = 2 x

                Length of min hand = r = 7 cm.

                    its Circumference = 2 x

                   Distance covered by min hand in 1 hr = 44 cm.

                   Distance covered by min hand in 1 day = 44 x 24 = 1056 cm.

                    sum of distance = 26.4 + 1056 = 1082.4 cm.

 

Q17.       Given, diameter = 140 cm, then, r = 70 cm.

                Let R be the radius of outer edge of parapet. Then,  2r = 616.

                R =          width of parapet = R – r = 98 – 70 = 28 cm.

 

Q18.       No. of rotations = 2000.  www.rsmaths99.com

                Diameter of wheal = 98 cm.

                  radius of wheal = 49 cm.

                  distance covered in 1 revolution =   2r.  =   2 x

                  Distance covered in 2000 revolutions = 2000 x 298 = 596000 cm.

 

Q19.       Diameter of wheel = 70 cm.          r = 35 cm.

                Distance covered in 250 revolutions  = 250 x 2r = 250 x 2 x

                 = 250 x 220 = 55000 cm.  www.rsmaths99.com

 

Q20.       Diameter of wheel = 77 cm.           r =  cm.

                Distance covered in 1 revolution =   2r =  2 x   = 242 cm.

                No. of revolution covering 242 cm = 1

                  No. of revolution in covering 1 cm =

                                “              “              “   121 km  = 121 x 1000 x 100 =    = 50000.

Q21.       Let radius of wheel = r cm.  Distance covered in 1 revolution =  2r.   Now,

                Dist. covered in 5000 revolution = 11000 m.   www.rsmaths99.com

                  “         “              1     “                      =         2r =     .

π    r =          = 35 cm.

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EXERCISE – 20F. AREA OF A CIRCLE   

 

Area of a circle = r2.

Q1.         I]    Area of a circle = r2 =    

                Ii]   Area of a circle = r2 = 

Q2.         I]  D = 28 cm, then r = 14 cm.   area = r2    = 

                Ii]   d = 1.4 m,  then r =   www.rsmaths99.com

                        area =   r2 =        x     x      =      

 

Q3.         Circumference = 264 cm.   =>  2r = 264.    => r =     

                  area =  r2 =    x 42 x 42 =  22 x 252 =   5544 cm2.

 

Q4.         Circumference = 35.2 m.     => 2r =        =>   r =       

                =   

                  area = r2  =   x    =   =

 

Q5.         Area of circle = 616 cm2.    =>  r2 = 616     => r2 =  

π  r =     =     = 7 x 2 = 14 cm.

   circumference = 2     2 x

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Q6.         Area of circle = 1386 m2.       r2 = 1386.  =>  r2 =

π  r2  = (21)2      => r = 21. 

π     circumference = 2     2 x  = 22 x 6 = 132 m2.

 

Q7.         Let r and R be the radii of two circles.  Given,      

                Then, ratio of area =        = .

 

Q8.         Area over which the horse can graze = area of circle of radius 21 cm.

                =   r2  =  x 21 x 21   =  =  22 x 63 = 1386 m2.

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Q9.         Area of square = 121 cm2.          (side)2 = 121 = (11)2  => side = 11cm.

                  circumference of circle = perimeter of square.

π  =   2r = 4 x 11  => .     => r = 7 cm.

π  Area of circle = r2 =

 

 Q10.      Let the side of square be = a cm 

π  perimeter of square = circumference of circle

π  4 x a =  2r   =>  4a =    2 x 28       => a = 44 cm.

                  area of square = (side)2   = (44)2  = 44 x 44 = 1936 cm2.

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Q11.       Area of remaining sheet = ar. Of acrylic – 64 x ar. Of 1 button.

                = 34 x 24 – 64 x r2    =  816 – 64 x     x   = 816 – 88 = 728 cm2.

 

Q12.       Area of remaining portion = ar. Of rectangle – ar. Of circle

                = 90 x 32 – x 14 x 14    = 2880 - 2880 – 22 x 28.

                = 2880 – 616 = 2264 m2.

                  cost of turfing = area x rate    = 2264 x 50 = Rs. 113200.

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Q13.       Area of the square = (14 x 14)cm2= 196 cm2.

                Sum of the area of 4 quadrants =

                   area of the shaded region = (196 - 154)cm2 = 42 cm2.

 

Q14.       Area to be grazed =  r2    =  = 154 cm2.

 

Q15.       Let R , r1 , r2  be the radius of the circles respectively

                R =  cm ,   d = 21 x      r1 =    d =21 x   r2 =     

                    area of the shaded region = area of bigger circle -   area of two smaller circles

=   R2 -  r12 -  r2 2   www.rsmaths99.com

            =   cm2.

               =    =    2

                =     2 .         

 

Q16.      Remaining area = area of rectangle – (4 x area of 1 quadrant + area of circle)

                = 8 x 6 – (4 x  x r2 + r2)

            =48 - 2r2 = 48 – 2 x  x 2 x 2 = 48 -  =      

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