Mensuration 2D - Study Mode

[#51] In the given figure, ABCD and BEFG are squares of sides 8 cm and 6 cm respectively, what is the area (in cm 2 ) of the shaded region?
Correct Answer

(B) 12

Explanation

Solution: Diagonal of square ABCD = 8√2 Diagonal of square EBGF = 6√2 Height of triangle EDG = 8√2 - 3√2 = 5√2 Area of triangle EDG = $$frac{1}{2}$$ × 6√2 × 5√2 = 30 Area of triangle EFG = (3√2) 2 = 18 Area of shaded region = 30 - 18 = 12

[#52] In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45°, then area of the triangle is
Correct Answer

(C) $$25sqrt 2 { ext{ c}}{{ ext{m}}^2}$$

Explanation

Solution: Remember: area of isosceles triangle $$ = frac{1}{2}{a^2}sin heta $$ (θ is angle between equal sides) $$eqalign{
& = frac{1}{2}{left( {10}
ight)^2}sin {45^ circ } cr
& = frac{{100}}{2} imes frac{1}{{sqrt 2 }} cr
& = frac{{50}}{{sqrt 2 }} imes frac{{sqrt 2 }}{{sqrt 2 }} cr
& = 25sqrt 2 { ext{ c}}{{ ext{m}}^2} cr} $$

[#53] In the given figure, ABCDEF is a regular hexagon of side 12 cm P, Q and R are the mid points of the sides AB, CD and EF respectively. What is the area (in cm 2 ) of triangle PQR?
Correct Answer

(B) 81√3

Explanation

Solution: So, If OF = OE = ED So, FC = 24 cm So, RQ $$ = frac{{{ ext{FC}} + { ext{ED}}}}{2} = frac{{36}}{2}$$ RQ = 18 cm Area of $$Delta { ext{PQR}} = frac{{sqrt 3 }}{4} imes 18 imes 18 = 81sqrt 3 { ext{ c}}{{ ext{m}}^2}$$

[#54] A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope streched and describes 88 metres when it has traced out 72° at the centre, the length of the rope is $$left( {{ ext{Take }}pi = frac{{22}}{7}}
ight)$$
Correct Answer

(A) 70 m

Explanation

Solution: $$eqalign{
& { ext{Length of arc}} = frac{ heta }{{360}} imes 2pi r cr
& frac{{72}}{{360}} imes 2 imes frac{{22}}{7} imes r = 88 cr
& r = frac{{88 imes 7 imes 360}}{{72 imes 2 imes 22}} cr
& r = 70{ ext{ m}} cr} $$

[#55] Three circles of radius 21 cm are placed in such a way that each circle touches the other two. What is the area of the portion enclosed by the three circles?
Correct Answer

(A) 441√3 - 693

Explanation

Solution: Area of portion enclosed by the three circles = Area of Equilateral triangle - 3 × Area of sector $$eqalign{
& = frac{{sqrt 3 }}{4} imes {left( {{ ext{side}}}
ight)^2} - left( {pi {r^2}left( {frac{ heta }{{{{360}^ circ }}}}
ight) imes 3}
ight) cr
& = frac{1}{2}left( {frac{{sqrt 3 }}{4} imes 42 imes 42 - frac{{22}}{7} imes 21 imes 21 imes frac{1}{3} imes 3}
ight) cr
& = 441sqrt 3 - 693 cr} $$