Area - Study Mode
[#126] A triangle with sides 13 cm, 14 cm and 15 cm is inscribed in a circle. The radius of the circle is :
Correct Answer
(D) 8.125 cm
Explanation
Solution: $$eqalign{
& s = frac{{13 + 14 + 15}}{2} = frac{{42}}{2} = 21 cr
& herefore ,,vartriangle = sqrt {21 imes 8 imes 7 imes 6} cr
& ,,,,,,,,,,, = sqrt {sleft( {s - a}
ight)left( {s - b}
ight)left( {s - c}
ight)} cr
& ,,,,,,,,,,, = left( {2 imes 2 imes 3 imes 7}
ight) cr
& ,,,,,,,,,,, = 84,c{m^2} cr} $$ Radius of circle : $$eqalign{
& = left( {frac{{13 imes 14 imes 15}}{{4 imes 84}}}
ight)c{m^2} cr
& = left( {frac{{65}}{8}}
ight)cm cr
& = 8.125,,cm cr} $$
[#127] A skating champion moves along the circumference of a circle of radius 28 m in 44 sec. How many seconds will it take her to move along the perimeter of a hexagon of side 48 m ?
Correct Answer
(C) 72 sec
Explanation
Solution: Distance mode by the skater in 44 sec = Circumference of the circle = $$left( {2 imes frac{{22}}{7} imes 28}
ight)$$ xa0 xa0 m = 176 m Speed of skater : = $$left( {frac{{176}}{{44}}}
ight)$$ xa0 m/sec = 4 m/sec Perimeter of hexagon : = (6 × 48) m = 288 m ∴ Required difference : = $$left( {frac{{288}}{{4}}}
ight)$$ xa0 sec = 72 sec
[#128] The adjoining figure contains three squares with areas of 100, 16 and 49 lying side by side as shown. By how much should the area of the middle square be reduced in order that the total length PQ of the resulting three squares is 19 ?
Correct Answer
(D) 12
Explanation
Solution: PQ = $$sqrt {100} $$ xa0+ $$sqrt {16} $$ + $$sqrt {49} $$ xa0= (10 + 4 + 7) = 21 Side of middle square = $$sqrt {16} $$ xa0= 4 Reduction in PQ = (21 - 19) = 2 New side of middle square = (4 - 2) = 2 ∴ Reduction in area of middle square = (4 2 - 2 2 ) = 12
[#129] The three sides of a triangular field are 20 metres, 21 metres and 29 metres long respectively. The area of the field is :
Correct Answer
(A) 210 m 2
Explanation
Solution: Since (20) 2 + (21) 2 = (29) 2 So, it is a right-angled triangle with base = 20 m and height = 21 m ∴ Area = $$left( {frac{1}{2} imes 20 imes 21}
ight)$$ xa0 m 2 = 210 m 2
[#130] The diagonal of a square is $$4sqrt 2 $$ cm. The diagonal of another square whose area is double that of the first square, is :
Correct Answer
(A) 8 cm
Explanation
Solution: $$eqalign{
& d = 4sqrt 2 ,cm cr
& Rightarrow { ext{Area }} = frac{1}{2}d_1^2 cr
& ,,,,,,,,,,,,,,,,,,,,,,,, = frac{1}{2} imes {left( {4sqrt 2 }
ight)^2} cr
& ,,,,,,,,,,,,,,,,,,,,,,,{ ext{ = 16}},{ ext{c}}{{ ext{m}}^2} cr} $$ Area of new square = (2 × 16) cm 2 = 32 cm 2 $$eqalign{
& herefore frac{1}{2}d_2^2 = 32 cr
& Rightarrow d_2^2 = 64 cr
& Rightarrow {d_2} = 8,cm cr} $$