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
Given that, area = 40 sq cm, base = 28 cm and height = perpendicular = ?Area = (base x perpendicular) / 2 ? 40 = (28 x perpendicular) / 2 ? perpendicular = 40/14 = 20/7 = 26/7 cm
Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. ∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 ⟹ x2 - 100x + 291 = 0 ⟹ (x - 97)(x - 3) = 0 ⟹ x = 3
Original breadth of rectangle = 720/30 = 24 cmNow , area of rectangle = (5/4) x 720 = 900 cm2? New length of rectangle = 900/24 = 37.5 cm? New perimeter of rectangle = 2(l+ b)= 2(37.5 + 24 ) = 2 x 61.5 = 123 cm
Area of square = (Side)2 = 202= 400 sq cm ? Area of rectangle = 1.8 x 400 = 720 sq cm Let length and breadth of rectangle be 5k and k respectively.Then, according to the question, 5k x k = 720? 5k2 = 720 ? k2 = 720/5 = 144? k = ?144 = 12 cm Perimeter of rectangle = 2(5k + k) = 12k= 12 x12 = 144 cm
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
EXAMIANS