Exercise17A Word Problem Mensuration Part 2 Class 9

 Mensuration  

1) In △ABC, AB = 4 cm, altitude CF = 6 cm and AC = 3 cm. Find the length of altitude BE.

Solution : 

Given :

In △ABC, 

AB = 4 cm, 

CF = 6 cm 

AC = 3 cm.

Area of △ABC = 1/2 * base * altitude

                          = 1/2 * AB * CF

                          = 1/2 * 4 * 6

                          = 12 cm2 

Base  AC  = 3 cm

Length of altitude BE = area * 2/ base = 12 * 2/ 3 = 8cm

2) In △PQR , medians QS and RT intersect at right angles at G. If QS = 9cm and RT = 12cm.

Find  i) length of QG

         ii) area of △TQR

        iii) area of △PQR

Solution : 

In △PQR, median QS and RT interest at right angles at G.

QS =9 cm, RT = 12cm

In △ GQR, ∠G =  90° 

QG = 9 * 2/3 = 6cm

RG = 12 * 2/3 = 8cm

         △ GQR = 1/2 * GQ * GR

                       = 1/2 * 6 *8

                      

                       = 24 cm2 

But ratio is TG : GR = 1: 2

Area △TQR = 24 * 3/2 = 36 cm2 

But T is the mid point of PQ

Area △PQR = 2 * Area △TQR

                     = 2 * 36 = 72 cm2 

3) The base of isosceles triangle ia 16 cm. If the height on the base is 4 cm shorter than the equal sides, find the sides, height and area of the triangle.

Solution :

 In isosceles △ABC, AB = AC, AD ⊥ BC

Base BC = 16 cm

Let each equal side = x cm

Area of △ABC = 1/2 * base * altitude

                         = 1/2 * 16 * ( x - 4)

                         = 8( x - 4) cm2

AD ⊥ BC

 so, D is the mid point of BC

BD = DC = 16/2 = 8cm

Now, in right △ABD, we have

AB2 = AD2 + BD2

x2 = (x - 4)2 + 82

x2 = x2 + 16 - 8x + 64

8x = 80 

x = 80 / 8

x = 10

Length of each equal side = 10 cm

height of the triangle = x - 4

                                  = 10 - 4 = 6 cm

area of triangle ABC = 8( 10 - 4) = 8 * 6 = 48 cm2 

Q 4 ) The base and height of a triangle are in the ratio 7:5. If the area of the triangle is 280 cm2, find its  base and height.

Solution :

Area of triangle = 280 cm2 

Ratio of the base and height of the triangle = 7 : 5

Let the length of base of triangle be 7x , then 

Height of triangle = 5x

Area = 1/2 * base * height

         = 1/2 * 2x  *5x

         =  35/2 * x2 

280   =  35/2 * x2 

 

x2  = 280 * 2 /  35 = 16 = 42 

x = 4

     

   Base = 7x = 7 * 4 = 28 cm

Height = 5x = 5 * 4 = 20 cm

Q 5 ) In the figure given alongside,

∠ ADB = 90° , AD = 9 cm , BD = 12 cm, BC = 25 cm and AC = 20 cm.

Find the i) length of AB ii) area of shaded part.

Solution :

 In right angled ADB ,

i) By pythagoras theorem, we have 

  AB2 = AD2  + BD2 

 

          = 92 + 122 

        

          = 81 + 144 

      

          = 225

AB    = 15 cm 

ii) Let a = BC = 25 cm ; b = AC = 20 cm;

  C = AB = 15 cm

  s =  a + b + c / 2 = 25 + 20 + 15 / 2 = 30 cm

Thus, area  △ABC , 

= √s( s - a) ( s - b) ( s = c)

= √30 (30 - 25 ) ( 30 - 20)(30 - 15)

= √  30 * 5 * 10 * 15

=√  2 * 3 * 5 * 5 * 2 * 5 * 3 * 5

= 2 * 3 * 5 * 5

= 150 cm2  

area  of △ABD = 1/2 * AB *  BD

                          = 1/2 * 9 * 12

                          = 54  cm2  

Area of shaded part =

area  of △ABC - area  of △ABD

= 150 - 54 = 96 cm2

Q 6) The perimeter of an isosceles triangle is 42cm. The base is 1 1/2 times the equal side. Find the length of each side.

Solution:

Given, perimeter of an isosceles triangle = 42 cm

Base= 1 1/2 * on of equal side

let each equal side =x

Perimeter of triangle = x + x + 3/2 x = 7/2 x

7/2 * x = 42 

 x = 42 * 2 /7 = 12

Each equal side  = 12cm

and base = 12 * 3/2 = 18 cm

lengths of sides are 12 cm, 12 cm, 18 cm.  

Q 7) In △PQR, ∠Q = 90° and QR is 4 cm more than PQ. If area of  △PQR = 96  cm2. Find the sides of triangle.

Solution:

In △PQR, 

Given,

 ∠Q = 90° and QR = 4 cm more than PQ.

And  area of  △PQR = 96  cm2.

Let PQ = x cm

then QR = ( x+ 4) cm

 area of  △PQR = 96  cm2.

                 96     = 1/2 * PQ * QR

                 96     = 1/2 * x * (x + 4)

       x(x + 4 )     =  96 * 2 

         x2 + 4x      = 192

        x2 + 4x - 192 = 0

        x2 + 16x -12x -192 = 0

          x (x+16) -12(x + 16) = 0

              ( x +16) ( x - 12) = 0

             x+16 = 0 or x -12 =0

x = -16 or x = 12

x = -16  is not possible being negative

PQ = 12 cm, QR = 12 + 4 = 16 cm

Then by pythagoras theorem, we have

     

PR = √PQ 2 +  QR2    

      = √12 2 +  16

      = √144 + 256  

     

      = √400

      

      = 20 cm

Required sides are 12 cm, 16cm, 20cm.

Q 8) Find the area of equilateral triangle with side 8cm. ( Take  √3 = 1.732 )

Solution :

Length of each side of an equilateral triangle = 8 cm

Area = √3/4 * side2

        =  1.732 /4 * 8 *8

        =  27.712cm2

Q 9) Find the cost of levelling a triangular piece of land measuring 80 m, 41 m and 41m, if the rate is ₹ 12 per m2

 Solution :

Sides of triangular piece of land arc 80 m, 41 m, 41 m.

Let a = 80, b = 41, c = 41;

s = a + b + c / 2 = 80 + 41 + 41 / 2 = 162/2 = 81

    Using Heron's formula

    Area of triangle = √ s( s- a) (s - b ) ( s - c)

                           = √ 81( 81- 80) ( 81 - 41) ( 81 - 41)

                           = √ 81 * 1 * 40 * 40

                         

                              = √ 9 * 9 * 2 *20 * 2 * 20 

                              =  9 * 2 * 20

                              = 360 m2

Given rate of levelling = ₹ 12 per m2

Total cost of levelling = 360 * 12  = ₹ 4320