Area Problems
The perimeter of an isosceles triangle is equal to 14 cm . The lateral side is to the base in the ratio 5 : 4 the area of the triangle is?
Let lateral side = (5y) cm and base = (4y) cm ? perimeter = 5y + 5y + 4y = 14 ?y = 1So, the sides are 5 cm , 5 cm and 4 cm Now s= 1/2 (5 + 5 + 4) cm = 7 cm (s-a) = 2 cm (s-b) = 2 cm and (s-c) = 3 cm? Required Area = ? (7 x 2 x 2 x 3) cm2=2?21 cm2
In a triangleSum of two sides is always greater than 3rd side i.e., x < 25 + 15 = 40 .....(i)Difference of two sides is always less than 3rd side i.e., 25 - 15 = 10 < x ...(ii) From Eqs. (i) and (ii) , we get 10 < x < 40
Let the radius of the park be r, then?r + 2r = 288(? + 2)r = 288? [(22/7) + 2]r = 288? r = (288 x 7)/36 = 56 ? Area of the park = (1/2)?r2= (1/2) x (22/7) x 56 x 56= 4928
a1 = 68/4 = 17 cmand a2 = 60/4 = 15 cm [ where a1 and a2 are sides]According to the question,Area of the third square = [(17)2 - (15)2 ] = (17 + 15) (17 - 15) = 32 x 2 = 64 sq cmLet a3 = Side of the third square.According to the question, (a3)2 = 64 sq cm ? a3 = ?64 = 8 cm? Perimeter of the third square = 4 x a3 = 4 x 8 = 32 cm.
Let the parallel sides be 3a and 5a.So Area of trapezium = 1/2 x sum of parallel side x perpendicular distance between them.? 1/2 (3a +5a) x 12 = 384? 8a = 64? a =8? Smaller side = 3x = 3 x 8 = 24 cm.
EXAMIANS