Mensuration 2D - Study Mode

[#71] The area of a field in the shape of a triangle with each side $$x$$ metres is equal to the area of another triangular field having sides 50 m, 70 m and 80 m. The value of $$x$$ is closest to:
Correct Answer

(B) 63.2

Explanation

Solution: $$eqalign{
& 2S = 50 + 70 + 80 cr
& S = 100 cr
& A = sqrt {Sleft( {S - a}
ight)left( {S - b}
ight)left( {S - c}
ight)} cr
& A = sqrt {100 imes 50 imes 30 imes 20} cr
& A = 1000sqrt 3 cr
& frac{{sqrt 3 }}{4}{x^2} = 1000sqrt 3 cr
& {x^2} = 4000 cr
& x = 63.2 cr} $$

[#72] From a point within an equilateral triangle, perpendiculars drawn to the three sides are 6 cm, 7 cm and 8 cm respectively, the length of the side of the triangle is:
Correct Answer

(C) $$14sqrt 3 { ext{ cm}}$$

Explanation

Solution: $$eqalign{
& { ext{Length of side}} = frac{2}{{sqrt 3 }}left( {{P_1} + {P_2} + {P_3}}
ight) cr
& = frac{2}{{sqrt 3 }}left( {6 + 7 + 8}
ight) cr
& = frac{2}{{sqrt 3 }} imes 21 cr
& = frac{{42}}{{sqrt 3 }} imes frac{{sqrt 3 }}{{sqrt 3 }} cr
& = frac{{42sqrt 3 }}{3} cr
& = 14sqrt 3 { ext{ cm}} cr} $$

[#73] In the given figure, AB, AE, EF, FG and GB are semicircles. AB = 56 cm and AE = EF = FG = GB. What is the area (in cm 2 ) of the shaded region?
Correct Answer

(D) 394.24

Explanation

Solution: AB = 56 cm AE = EF = FG = GB = 14 cm ∴ EM = MF = 7 cm PF = $$frac{{56}}{2}$$ = 28 cm OF = 28 - r OM = 7 + r MF = 7 In ΔOMF, OM 2 = OF 2 + MF 2 (7 + r) 2 = (28 - r) 2 + 7 2 49 + r 2 + 14r = 784 + r 2 - 56r + 49 70r = 784 r = $$frac{{784}}{{70}}$$ = 11.2 Area of circle = πr 2 = $$frac{{22}}{7} imes frac{{112}}{{10}} imes frac{{112}}{{10}}$$ = 394.24 cm 2

[#74] Two equal maximum sized circular plates are cut off from a circular paper sheet of circumference 352 cm. Then the circumference of each circular plate is
Correct Answer

(A) 176 cm

Explanation

Solution: Circumference of paper sheet = 352 $$eqalign{
& 2pi R = 352 cr
& R = frac{{352}}{{2pi }} = frac{{352 imes 7}}{{2 imes 22}} = 56{ ext{ cm}} cr
& r = frac{R}{2} = frac{{56}}{2} = 28{ ext{ cm}} cr} $$ ∴ Circumference of circular plate $$eqalign{
& = 2pi r cr
& = 2 imes frac{{22}}{7} imes 28 cr
& = 176{ ext{ cm}} cr} $$

[#75] The area of a square park is 16x 2 + 8x + 1. What is the length of the park?
Correct Answer

(B) (4x + 1) units

Explanation

Solution: Let the length of square park = L Area = L 2 L 2 = 16x 2 + 8x + 1 Length of the park $$eqalign{
& = sqrt {16{x^2} + 8x + 1} cr
& = sqrt {{{left( {4x}
ight)}^2} + 2 imes 4x + {1^2}} cr
& = sqrt {{{left( {4x + 1}
ight)}^2}} cr
& = left( {4x + 1}
ight){ ext{ units}} cr} $$