Area Problems The area of a right angled triangle is 10 sq cm. If its perpendicular is equal to 20 cm, find its base. 1 cm 2 cm 3 cm 4 cm 1 cm 2 cm 3 cm 4 cm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Given that, area = 10 sq cm, perpendicular = 20 cm and base = ? area = (base x Perpendicular) / 2 ? 10 = (base x 20) / 2 ? base = 1 cm
Area Problems The area of the sector of a circle, whose radius is 12 meters and whose angle at the centre is 42, is? 79.2 sq. meters 52.8 sq. meters 39.6 sq. meters 26.4 sq. meters 79.2 sq. meters 52.8 sq. meters 39.6 sq. meters 26.4 sq. meters ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Area of the sector = [(22/7) x 12 x 12] x [42°/360°] m2 = 52.8 m2
Area Problems The perimeter of two square is 12 cm and 24 cm. The area of the bigger square is how many times that of the smaller? 3 times 5 times 2 times 4 times 3 times 5 times 2 times 4 times ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Ratio of area of 2 squares = (ratio of perimeter of 2 squares)2= (24/12)2= 4
Area Problems The area of rectangle whose length is 5 more than twice its width is 75 sq units. What is the perimeter of the rectangle? 20 unit 30 unit 24 unit 40 unit 20 unit 30 unit 24 unit 40 unit ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let the width of the rectangle = k units? Length = (2k + 5) units.According to the question,Area = k(2k + 5)? 75 = 2k2 + 5k? 2k2 + 5k - 75 = 0 ? 2k2 + 15k - 10k - 75 = 0 ? k(2k + 15) - 5(2k + 15) = 0 ? (2k + 15) (k - 5) = 0 ? k = 5 and -15/2Width cannot be negative. ? Width = 5 units? Length = 2x + 5 = 2 x 5 + 5 = 15 unit? perimeter of the rectangle = 2(15 + 5) = 40 units
Area Problems A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? 40 88 34 68 40 88 34 68 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft
Area Problems If the base of a rectangular is increased by 10% and the area is unchanged , then the corresponding altitude must to be decreased by? 10% 11% 111/9 % 91/11 % 10% 11% 111/9 % 91/11 % ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let base = b and altitude = h Then, Area = b x h But New base = 110b / 100 = 11b / 10Let New altitude = HThen, Decrese = (h - 10h /11 )= h / 11? Required decrease per cent = (h/11) x (1 / h ) x 100 %= 91/11 %
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