Area Problems
The sides of a triangle area in the ratio of 1/3 : 1/4 : 1/5 and its perimeter is 94 cm. Find the length of the smallest side of the triangle .
Given ratio = 1/3 : 1/4 : 1/5 = 20 : 15 : 12Let length of the sides be 20k, 15k and 12k.Then, according to the question,20k + 15k + 12k = 94? 47k = 94? k = 94/47 = 2Smallest side = 12k = 12 x 2 = 24 cm
Let length of the longer diagonal = d cmThen, length of other diagonal = 0.8 x d cm Area of rhombus = (1/2) x d x 0.8 x d = 2/5 d2= 2/5 d2Area of square of the length of the longer diagonal = d2So the area of the rhombus is 2/5 times the square of the length of the longer diagonal.
Area of the square field = 1 hectare = 10000 m2Side of the square = ? 10000 m = 100 mSide of another square field = 100 + 1 = 101 m? Required difference of area = [(101)2 - (100)2] m2=[(101 + 100 ) (101 - 100) ] m2= 201 m2
Let the diagonal of one square be (2d) cmThen, diagonal of another square = d cm? Area of first square = [ 1/2 x (2d)2] cm2Area of second square = (1/2 x d2) cm2? Ratio of area = (2d)2/ d2= 4/1 = 4: 1
Circular piece is 4 x 11 = 44 cm long, Then Circumference of circle is given by, 44 = pi x D, where D is the diameter D = 44 / pi Take pi = 22 / 7, then D = 44 / (22/7) = (44 x 7) / 22 D = 14 cm.
We know that area of triangle = ( base x height ) / 2So area of triangle = (BC x AD) / 2 = (AC x BE) / 2 ? (10 x 4.4) / 2 = (11 x h) / 2? h= (10 x 4.4)/11 = 4 cm
EXAMIANS