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 rectangular field = 7k m and breadth of the rectangular field = 2k mAccording to the question,Area of a rectangular field = Length x Breadth? 3584 = 7k x 2k ? 14 x k2 = 3584 ? k2 = 3584/14 = 256? k2 = 256 = 16 m? Length of rectangular field = 7k = 7 x 16 = 112 mAnd breadth of rectangular field = 2 x 16 = 32 m? Perimeter of rectangle = 2(Length x Breadth)= 2(112 + 32) = 2 x 144 = 288 m
Let circumference = 100 cm . Then, ? 2?r = 100? r = 100/2? =50/?? New circumference = 105 cm Then, 2?R = 105? R = 105 / (2?)&rArr Original area = [ ? x (50/?) x (50/?) ] = 2500/? cm2? New Area = [? x (105/2?) x (105/2?)]= 11025 / (4?) cm2? Increase in area = [11025/(4?)] - 2500/? cm2= 1025 / 4? cm2Required increase percent [1025/(4?)] x 2500/? x 100 = 41/4%= 10.25%
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 one diagonal be k.Then, other diagonal = (60k/100) = 3k/5 cmArea of rhombus =(1/2) x k x (3k/5) = (3/10) = 3/10 (square of longer diagonal)Hence, area of rhombus is 3/10 times.
EXAMIANS