Students and Teachers Forum
Given, Diagonal of square (d) = 6 cm Area = ? We know, Area of the square = 12 d2 .....
Here, triangle ABC is an isosceles triangle. Let, the equal sides, a = 8 cm and b = 6 cm So, Area of triangle ABC = b4 √(4a2 - b2 ) .....
Here, height (h) = b cm and base (b) = a cm We have, Area of right-angled triangle (A) = 12 bh .....
Let, a = 20 cm b = 13 cm c = 21 cm Semi-perimeter (s) = 1 2 (a + b + c) .....
Given, Side of the equilateral triangle (a) = 4cm Area = ? We have, Area of equilateral triangle = √34 a2 .....
Given, First diagonal (d1) = 12 cm Second diagonal (d2) = 8 cm We know, Area of kite = 12 × d1 × d2 .....
Given, a = 7.5 cm b = 10 cm and c = 12.5 cm Now, Semi-perimeter of ∆ABC (s) = 12 (a + b + c) .....
Given, a = 6 cm b = 8 cm and c = 10 cm Now, Semi-perimeter of ∆ABC (s) = 12 (a + b + c) .....
Here Sides of the triangular base, a = 5 cm b = 12 cm c = 13 cm Height of the .....
Let a be the side of equilateral triangle. Then, Area of an equilateral triangle = 16√3 cm2 Or, √34 a2 = 16√3 Or, a2 = 16⨉4 Or, a =√(16⨉4) .....
Given, AB (c) = 36 cm BC (a) = 48 cm and AC (b) = 60 cm. Now, Semi-perimeter of ∆ABC (s) = 12 (a + b + c) .....
Given, AB (c) = 11 cm BC (a) = 12 cm and AC (b) = 13 cm Now, Semi-perimeter of ∆ABC (s) = 12 (a + b + c) .....
Here the sides of the triangular base, a = 5 cm b = 12 cm c = 13 cm Height of the prism (h) .....
Let, the equal sides of the isosceles triangle be a and the unequal side be b. Here, perimeter = 25 cm Or, 2a + b = 25 cm Or, 2a + 9 = 25 cm .....
Given, a = 9 cm b = 12 cm and c = 15 cm. Now, Semi-perimeter of ∆ABC (s) = 12 (a + b + c) .....
Given, a = 3 cm b = 4 cm and c = 5 cm Now, Semi-perimeter of ∆ABC (s) = 12 (a + b + c) .....
Let, a, b and c be the three sides of triangle. Given, P = a + b + c = 36cm a = 8cm and b = 12cm Now, c = P - (a + c) = 36 - (8 + .....
Let, The hypotenuse of the right triangle (h) = 74 cm The base of the right-angled triangle (b) = 70 cm Now, The perpendicular of the triangle is given by, p = √(h2- p2 ) = √(742-702 .....
Let a be the length of equal sides and b be the base of the isosceles triangle. Then, The area of a triangle = 60 cm2 Or, b4 √(4a2-b2 ) = 60 Or, 104 √(4a2-102 ) = 60 Or, .....
Here, Let a be the length of a side of the triangular land. The area of an equilateral triangular land = 900√3 m2 Or, √34a2 = 900√3 Or, a2 = 900⨉4 Or, a = .....
Let, x be any constant. Then, the lengths of the edges of a given land is given by 12x, 17x and 25x. According to the question, Perimeter = 540 feet. Or, 12x + .....
Here, Hypotenuse of the right triangle (h) = 74 cm Let, base of the right triangle (b) = 70cm Now, the perpendicular of the triangle is given by, .....
Let, perendicular = base = a = 10 m. Now, the area of the right-angled triangle is given by, Area = 12 ⨉ p ⨉ b = 12 ⨉ a ⨉ a [ since, perendicular = base = a.] .....
Let, the equal sides of the isosceles triangle be a and unequal side be b. Here, Perimeter = 25 cm Or, 2a + b = 25 cm Or, 2a + 9 = 25 cm Or, 2a = 16 cm Or, a = 8 cm Now, the area of the isosceles triangle is given .....
Let, b be the base and a be the equal sides of an isoscales triangle. Here, the area of a triangle = 60 cm2 Or, b44a2-b2 = 60 Or, 1044a2-102 = 60 Or, 4a2-100 = 24 Squaring both .....
Let x be a constant. Then, the measure of the edges of the given land is given by 12x, 17x and 25x. According to the question, Perimeter = 540 feet Or, 12x + 17x + 25x = 540 .....
Given, Population of city 2 years ago (P) = 40000 Rate of population growth (R) = 2.5% Time (T) = 2 years Increase in population in 2 years (PT) .....
Given, Population of a city (P) = 250000 Annual birth rate (Rb) = 3.5% Annual death rate (Rd) = 1.5% So, Annual growth rate (R) = Rb - Rd .....
Given, Rate of population growth (R) = 5% Population after two years (PT) = 10000 Out migration (O) = 1025 We know, Population after two years (PT) = P (1 + R100)T .....
Here, Original value = Rs. 2400 Depreciated rate = 10% Price after 1 year = ? We know, Value of machine after one year .....
Here, Selling price of motor cycle after a year (PT ) = Rs. 57000 Time period (T) = 1 year Rate of depreciation (R) = 5% Initial price of the motor cycle (P) =? We have, PT = P (1- R100)T Or, Rs. 57000 = P(1- 5100)1 Or, .....
Given, Present population (P) = 30000 Increase Rate (R) = 10% Time period (T) = 2 years Population after two years (P2) = ? Now, .....
Here, Initial cost of bicycle = Rs. 5500 Cost of bicycle after one year = Rs. 5060 So, Amount of depreciation = Rs. 5500 – Rs. 5060 .....
Here, Cost of the motorcycle after 3 years(PT) = Rs.92583 Rate of depreciation (R) = 10% Time (T) = 3 years Original price of the motorcycle (P) = ? We know .....
Given, Original population (P) = 40000 Population after 2 years (PT) = 44100 Time period (T) = 2 years Growth rate (R) = ? We have, .....
Given, Present population (P) = 40000 Increase rate by birth(R1) = 2% Increase rate by immigration(R2) = 3% Total increase rate (R) = 2% + 3% .....
Here, Present population (PT) = 122412 Previous population (P) = 120000 Crude death rate = 5% So, if the birth rate is x%. .....
Here, Present price (P) = Rs. 175000 Rate of depreciation(R) = 20% Time (T) = 3 years We know that, PT = P (1- R100)T = 175000 (1- 20100)3 .....
Here, Population at the end of two years = 24895 No. of people .....
Given, Cost of the sewing machine (P) = Rs. 3200 Rate of depreciation (R) = 5% Time (T) = 2 years We know, Value at the end of 2 years (DT) = P (1 - R100)T .....
Here, The population of the district 3 years ago (P) = 375000 The annual growth rate (R) = 2% No. of people migrated at the end of second year = 1480 No. of people who .....
Given, Cost price of the DVD player (P) = Rs. 4000 Rate of depreciation (R) = ¼ of value at the beginning .....
Given, Cost price of the three fans (C) = Rs. 5250 Cost price of one fan (P) = Rs. 5250 / 3 .....
Given, Cost price of the car two years ago (C) = Rs. 325000 Cost on repairing (R) = Rs. 6630 Time (T) = 2 years Rate of depreciation (R) = 12% .....
Here, The present number of the workers in the industry (P) = 40000 The number of the workers in the industry before 3 years (P3) = ? The rate of cut off (R) = .....
Given, Cost price of the bus (P) = Rs. 2400000 Selling price of the bus (DT) = Rs.2208000 Time (T) = 1 year Rate of depreciation (R) = ? We .....
Given, Population of city 2 years ago (P) = 62500 Present population of district (PT) = 67600 Time (T) = 2 years Rate of growth (R) = ? We .....
Here, Rate = 2%, Initial population = 100000 Now, Population in 2066, Population = P (1+ R100)T = 100000 (1+ 2100)1 .....
Given, Population of the district 2 years ago (P) = 64000 Growth rate of population (R) = 2.5% Number of inhabitants leaving the district (O) = 1025 Number of .....
Given, Rate of population growth (R) = 10% Time (T) = 2 years Population after 2 years (PT) = 30000 In migration (I) = 5800 .....