Practice MCQ Questions and Answer on Area

41.

A room 5m × 8 m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is Rs. 18 per sq. meter, the cost of carpeting the room will be :

• (A) Rs. 673.92
• (B) Rs. 682.46
• (C) Rs. 691.80
• (D) Rs. 702.60

Show Answer

 Area of the carpet : = [(5 - 0.20) × (8 - 0.20)] m2 = (4.8 × 7.8) m2 = 37.44 m2 ∴ Cost of carpeting : = Rs. (37.44 × 18) = Rs. 673.92

42.

Two sides of a rectangle were measured. One of the sides (length) was measured 10% more than its actual length and the other side (width) was measured 5% less than its actual length. The percentage error in measure obtained for the area of the rectangle is :

• (A) 4.5%
• (B) 5%
• (C) 7.56%
• (D) 15%

Show Answer

 Let the actual length and width of the rectangle be $$l$$ and b respectively. Then, measured length : $$ = 100\% {\text{ of }}l = \frac{{11l}}{{10}}$$ Measured width : $$ = 95\% {\text{ of }}b = \frac{{19b}}{{20}}$$ Actual area = $$lb$$ Measured area : $$\eqalign{ & = \left( {\frac{{11l}}{{10}} \times \frac{{19b}}{{20}}} \right) \cr & = \frac{{209lb}}{{200}} \cr} $$ Error in measurement : $$\eqalign{ & = \left( {\frac{{209lb}}{{200}} - lb} \right) \cr & = \frac{{9lb}}{{200}} \cr} $$ ∴ Error % :$$\eqalign{ & = \left( {\frac{{9lb}}{{200}} \times \frac{1}{{lb}} \times 100} \right)\% \cr & = 4.5\% \cr} $$

43.

Which of the following figures has the longest perimeter ?

• (A) A square of side 10 cm
• (B) A rectangular of sides 12 cm and 9 cm
• (C) A circle of radius 7 cm
• (D) A rhombus of side 9 cm

Show Answer

 (a) Perimeter = (4 × 10) cm = 40 cm (b) Perimeter = 2(12 + 9) cm = 42 cm (c) Perimeter = $$\left( {2 \times \frac{{22}}{7} \times 7} \right)$$   cm = 44 cm (d) Perimeter = (4 × 9) cm = 36 cm

44.

If a square of area $$\frac{\text{A}}{2}$$ is cut off from a given square of area A, then the ratio of diagonal of the cut off square to that of the given square is :

• (A) 1 : 5
• (B) 1 : $$2\sqrt{5}$$
• (C) 1 : $$\sqrt{5}$$
• (D) 1 : $$\sqrt{2}$$

Show Answer

 Let the length of diagonal of the bigger square be x and that of the smaller square be y. Then, $$A = \frac{1}{2}{x^2}\,\,or\,\,x = \sqrt {2A} $$ And, $$\frac{A}{2} = \frac{1}{2}{y^2}\,\,or\,\,y = \sqrt A $$ $$\eqalign{ & \therefore {\text{ Required ratio :}} \cr & = \frac{y}{x} = \frac{{\sqrt A }}{{\sqrt {2A} }} = 1:\sqrt 2 {\text{ }} \cr} $$

45.

The circumference of the front wheel of a cart is 40 ft long and that of the back wheel is 48 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel ?

• (A) 850 ft
• (B) 950 ft
• (C) 1200 ft
• (D) 1450 ft

Show Answer

 Let the rear wheel make x revolutions Then, the front wheel makes (x + 5) revolutions (x + 5) × 40 = 48x ⇒ 8x = 200 ⇒ x = 25 Distance travelled by the cart : = (48 × 25) ft = 1200 ft

46.

The base of triangle is 15 cm and height is 12 cm the height of another triangle of double the area having base 20 cm is :

• (A) 22 cm
• (B) 20 cm
• (C) 18 cm
• (D) 10 cm

Show Answer

 Given base of triangle and its height is 15 cm and 12 cm respectively Area of first triangle : $$\eqalign{ & = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr & = \frac{1}{2} \times 15 \times 12 \cr & = 90{\text{ sq}}{\text{. cm}} \cr} $$ According to the question, Let height of triangle be h cm Area of new triangle = 180 sq. cm Base = 20 sq. cm $$\eqalign{ & \Rightarrow 180 = \frac{1}{2} \times 20 \times {\text{Height}} \cr & \Rightarrow {\text{h}} = \frac{{2 \times 180}}{{20}} \cr & \Rightarrow {\text{h}} = 18{\text{ cm}} \cr} $$

47.

A man is walking in a rectangular field whose perimeter is 6 km. If the area of the rectangular field be 2 sq. km, then what is the difference between the length and breadth of the rectangle ?

• (A) $$\frac{1}{2}km$$
• (B) $$1km$$
• (C) $$1\frac{1}{2}km$$
• (D) $$2km$$

Show Answer

 Let the length and breadth of the field be l and b km respectively Then, 2 (l + b) = 6 ⇒ l + b = 3 And lb = 2 (l - b)2 = (l + b)2 - 4lb ⇒ (l - b)2 = 32 - 4 × 2 ⇒ (l - b)2 = 1 ⇒ (l - b) = 1 km

48.

The breadth of a rectangular field is $$\frac{3}{4}$$ of its length and its area is 300 sq. metres. What will be the area (in sq. metres) of the garden of breath 1.5 metres developed around the field ?

• (A) 96 m2
• (B) 105 m2
• (C) 114 m2
• (D) Cannot be determined

Show Answer

 Let the length of the field be x metres Then, breadth of the field = $$\frac{3x}{4}$$ metres $$\eqalign{ & x \times \frac{{3x}}{4} = 300 \cr & \Rightarrow {x^2} = 300 \times \frac{4}{3} \cr & \Rightarrow {x^2} = 400 \cr & \Rightarrow x = 20 \cr} $$ So, length = 20 m, breadth = 15 m ∴ Area of the garden : $$\eqalign{ & = \left[ {\left\{ {\left( {20 + 3} \right) \times \left( {15 + 3} \right)} \right\} - \left( {20 \times 15} \right)} \right]{m^2} \cr & = \left[ {\left( {23 \times 18} \right) - \left( {20 \times 15} \right)} \right]{m^2} \cr & = \left( {414 - 300} \right){m^2} \cr & = 114\,{m^2} \cr} $$

49.

A diagonal of a rhombus is 6 cm. If its area is 24 cm² then the length of each side of the rhombus is :

• (A) 5 cm
• (B) 6 cm
• (C) 7 cm
• (D) 8 cm

Show Answer

 $$\eqalign{ & \frac{1}{2} \times 6 \times d = 24 \cr & \Rightarrow d = \frac{{24}}{3} \cr & \Rightarrow d = 8\,cm \cr & OA = 4\,cm\,{\text{and}}\,OB = 3\,cm \cr & \therefore AB = \sqrt {{{\left( {OA} \right)}^2} + {{\left( {OB} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{4^2} + {3^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5\,cm \cr} $$

50.

A rectangular carpet has an area of 60 sq.m. If its diagonal and longer side together equal 5 times the shorter side, the length of the carpet is :

• (A) 5 m
• (B) 12 m
• (C) 13 m
• (D) 14.5 m

Show Answer

 We have : $$lb$$ = 60 and $$\sqrt {{l^2} + {b^2}} + l = 5b$$ Now, $$\eqalign{ & {l^2} + {b^2} = {\left( {5b - l} \right)^2} \cr & \Rightarrow 24{b^2} - 10lb = 0 \cr & \Rightarrow 24{b^2} - 600 = 0 \cr & \Rightarrow {b^2} = 25 \cr & \Rightarrow b = 5 \cr & \therefore l = \left( {\frac{{60}}{b}} \right) \cr & \,\,\,\,\,\,\,\, = \left( {\frac{{60}}{5}} \right)m \cr & \,\,\,\,\,\,\,\, = 12\,m \cr} $$ So, length of the carpet = 12 m