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) = 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     .....