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Question-1

Find the area of a triangle in which
(i)  a = b = c = 4 cm
(ii) In right angled triangle, hypotenuse is 13 cm and one side is 5 cm.

Solution:
(i) Area of a equilateral triangle =  =  ´ (4)2 = 6.928 cm2                                                                     
(ii) Area of right angled triangle = ´ base × height
Hypotenuse = 13cm, one side = 5cm
(Hypotenuse)2= (one side)2 + (other side)2
    132 = x2 + 52
    ⇒ x = 12 cm
Area of the
right angled triangle ´ 12 ´ 5 = 30 cm2.

Question-2

The cost of levelling a park at the rate of 2/km2 is 2,700. If the park is in right angled triangular form with one side being 45 km, find the hypotenuse.

Solution:
The cost of leveling = `2,700
The rate of leveling = 
`2/km2
Area of the park = Cost/Rate = 2700/2 = 1350 km2                                      
One side of the park = 45 km
Let the other side = x km
Area of the right angled triangular park =  
× one side × other side

... 1350 =  
× 45 × x x = 60 km
Again (Hypotenuse)2 = (Ist  side)2+(IInd side)2
                            = 452 + 602 = 5625

... Hypotenuse = 75 km

Question-3

The perimeter of a triangular field is 240 m. It's two sides are 78 m and 50 m. Find the length of the altitude on the side of 50 m length from its opposite vertex.

Solution:
Let the three sides of the triangle be a, b, c, then perimeter = a + b + c.
... 240 = 50 + 78 + c or c = 112 m

Area of the triangle =

Or  
 ´ base ´ height =

Þ  ´ 50 ´ h =

 h = 67.2 m
 

Question-4

(i) The area of an equilateral triangle is 100m2. Find the perimeter of the triangle.
(ii)  Find the base of an isosceles triangle whose area is 60 sq.cm and length of equal sides is 13 cm.
(iii) The area of an isosceles right triangle is 200 m2. Find its hypotenuse.
(iv) The ratio of the base to the altitude of a triangular field is 3 : 1. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10. Find its base and height.

Solution:
(i) Area of the equilateral triangle = (side)2.

                                   100 = (a)2

                                                                  Þ a = 20 m
∴ Perimeter of the equilateral triangle = 3a = 3 ´ 20 = 60 m
(ii) We are given an isosceles
ΔABC in which    




AB = AC = 13 cm.
Let AD
BC and BC = 2x
BD = DC = x cm  

AD =

Area of
ΔABC =   ´ base ´ height

60 =

x = 12 or 5   
... base = 24 cm or 10 cm

(iii)



Let AB=BC=x cm
    AC2 = AB2 + BC2     
          = x2 + x2
  
AC = √2x
Area of the triangle = 
 ´ base ´ height
                 Or 200 =  
× x × x x = 20
... Hypotenuse
√2x = √2 ×20 = 28.28 m

(iv) Let the altitude be = x m
Then the base of the triangle = 3x m

Area of the field =

                       = 13.5 hectare
                       = 135000 m2

Þ  ´ base × height = 135000
Þ  ´ 3x × x = 135000
x = 300

Base(3x) = 900 m

Altitude (x) = 300 m 

Question-5

(i) The cross section of a canal is in a trapezium shape. If the canal is 10 m wide at the top, 6 m wide at the bottom and area of cross section is 72 sq. m, find the depth of the canal.
(ii) Find the area of a trapezium whose parallel sides are 12 cm and 16 cm and the distance between the parallel sides in 10 cm.

Solution:
(i) Area of a trapezium = (sum of parallel sides) × height
                            or 72 = (6 + 10) 
´ h
                                                 Þ h = 9 m
(ii) Area of trapezium = (a + b)
×h
                             = (12 + 16)
×10
                             = 140 cm2 

Question-6

The area of a quadrilateral is 360 m2 and the perpendiculars drawn to one of the diagonal from the opposite vertices are 10 m and 8 m. Find the length of the diagonal.

Solution:


In the given Fig. AE = h1 = 8 cm
CF = h2 = 10 cm and area of quadrilateral ABCD = 360 m2
Area of a quadrilateral = (Diagonal)(h1 + h2)
                             360 =  
´ DB ´ (10 + 8)
                              
DB = 40 m

Question-7

The area of a trapezium is 1700 m2. If one of the parallel sides is 64 m and the distance between the parallel sides is 34 m, find the length of the other parallel side.


Solution:
ABCD is a trapezium in which DC = 64 m and altitude
AE = 34 m
Let other parallel side AB = x m
Area (ABCD) = (AB + DC)
× AE       
1700 = (x + 64)
×34
or     x = 36 m.

Question-8

The parallel sides of a trapezium is 77 m and 60 m and its non-parallel sides are 26 m and 25 m. Find the area of the trapezium.

Solution:


Let ABCD be a trapezium in which
AB = 60 m, CD = 77 m, BC = 26 m, AD = 25 m
and
Draw BE
|| to AD
Thus ABED is parallelogram
 
Hence AB = DE = 60 m.
AD = BE = 25 m
and EC = DC - DE = 77 - 60 = 17 m
and
Now in
ΔBEC,
BE = 25 m, BC = 26m, EC = 17m
Let the height of B from EC be = h m

S =  = 34

Area of
ΔBEC =

 
´ EC ´ h =

 ´ 17 ´ h =
or h = 24 m.
Now area of trapezium = (sum of parallel
sides) × height
Area of trapezium (60 + 77) ´ 24 = 1644 m2.
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