Question 12 Marks
The shape of a park is a rectangle bounded by semicircles at the ends, each of radius 17.5 m, as shown in the adjoining figure. Find the area and the perimeter of the park.
Answer
View full question & answer→$2362.5 m^2, 190 m$
Questions
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34 questions · timed · auto-graded
Question 22 Marks
A wire when bent in the form of a square encloses an area of $756.25 cm^2$. If the same wire is bent to form a semi-circle, what will be the radius of the semi-circle so formed?
Answer
View full question & answer→35 cm
Question 32 Marks
A wire is in the form of a square of side 49.5 cm. It is straightened and bent into a semicircle. What is the radius of the semi-circle so formed?
Answer
View full question & answer→63 cm
Question 42 Marks
A wire is in the form of a semi-circle of radius 28 cm. It is straightened and bent into the form of a square. What is the length of the side of the square?
Answer
View full question & answer→22 cm
Question 52 Marks
Find the area of the shaded region in the figure :
Answer
View full question & answer→$1039.5 cm^2$
Question 62 Marks
Find the area of the shaded region in the figure :
Answer
View full question & answer→$1134 cm^2$
[Hint : (i) $a+\frac{a}{2}+\frac{a}{2}=42 \Rightarrow a=21 . \therefore$ Side of the square $\left.=21 cm.\right]$
[Hint : (i) $a+\frac{a}{2}+\frac{a}{2}=42 \Rightarrow a=21 . \therefore$ Side of the square $\left.=21 cm.\right]$
Question 72 Marks
Find the area of the shaded region in the figure :
Answer
View full question & answer→$134.75 m^2$
Question 82 Marks
Find the area of the shaded region in the figure :
Answer
View full question & answer→$259.875 m^2$
Question 92 Marks
Find the area of the shaded region in the figure :
Answer
View full question & answer→$619.5 m^2$
Question 102 Marks
Find the area of the shaded region in the figure :
Answer
View full question & answer→$141.75 m^2$
Question 112 Marks
Find the area of the figure ABCDE , it being given that:
$AE \| BD , AF \perp BD , CG \perp BD , AE =12 cm, BD =16 cm$, $AF =6.5 cm$ and $CG =8.5 cm$.
$AE \| BD , AF \perp BD , CG \perp BD , AE =12 cm, BD =16 cm$, $AF =6.5 cm$ and $CG =8.5 cm$.
Answer
View full question & answer→$159 cm^2$
Question 122 Marks
Find the area of the given hexagon LMNPQR in which each side measures 5 cm, height $MQ =11 cm$ and width $RP =8 cm$.
Answer
View full question & answer→$64 cm^2$
[Hint : Required area $=2 \times$ area of trapezium MNPQ.]
[Hint : Required area $=2 \times$ area of trapezium MNPQ.]
Question 132 Marks
ABCD is a parallelogram having adjacent sides $AB =16 cm$ and $BC =14 cm$. If its area is $168 cm^2$, find the distance between its longer sides and that between its shorter sides.
Answer
View full question & answer→$10.5 cm, 12 cm$
[Hint: $A B \times$ Distance between longer sides $=168 ; B C \times$ Distance between shorter sides $=168$ ]
[Hint: $A B \times$ Distance between longer sides $=168 ; B C \times$ Distance between shorter sides $=168$ ]
Question 142 Marks
$P Q R S$ is a parallelogram with $P Q=26 cm$ and $Q R=20 cm$. If the distance between its longer sides is 12.5 cm, find
(i) the area of the parallelogram;
(ii) the distance between its shorter sides.
(i) the area of the parallelogram;
(ii) the distance between its shorter sides.
Answer
View full question & answer→(i) $325 cm^2$$\quad$(ii) 16.25 cm
Question 152 Marks
In the adjoining figure, ABCD is a rectangle in which $AB =18 cm, BC =8 cm$ and $DE =10 cm$. Find the area of the shaded region EBCD.
Answer
View full question & answer→$120 cm^2$
Question 162 Marks
The area of a trapezium is $198 cm^2$ and its height is 9 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.
Answer
View full question & answer→$18 cm, 26 cm$
Question 172 Marks
The parallel sides of a trapezium are 20 cm and 10 cm. Its non-parallel sides are both equal, each being 13 cm. Find the area of the trapezium.
Answer
View full question & answer→$180 cm^2$
[Hint : Draw $C E \perp A B$, and $D F \perp A B$
So, $A F=B E=5 cm$
In $\triangle A D F$, use Pythagoras theorem to find DF, which is the distance between the parallel sides $A B$ and $D C$ ]
[Hint : Draw $C E \perp A B$, and $D F \perp A B$
So, $A F=B E=5 cm$
In $\triangle A D F$, use Pythagoras theorem to find DF, which is the distance between the parallel sides $A B$ and $D C$ ]
Question 182 Marks
In the adjoining figure, ABCD is a trapezium in which parallel sides are $AB =78 cm, DC =52 cm$ and the nonparallel sides are $BC =30 cm$ and $AD =28 cm$. Find the area of the trapezium.
Answer
View full question & answer→$1680 cm^2$
Question 192 Marks
The lengths of parallel sides of a trapezium are in the ratio $7: 5$ and the distance between them is 14 cm. If the area of the trapezium is $252 cm^2$, find the lengths of its parallel sides.
Answer
View full question & answer→$21 cm, 15 cm$
Question 202 Marks
The length of a rectangle is 30 cm and one of its diagonals measures 34 cm. Find the breadth, perimeter and area of the rectangle.
Answer
View full question & answer→$16 cm, 92 cm, 480 cm^2$
Question 212 Marks
The perimeter of a rectangle is 68 m and its length is 24 m. Find its breadth, area and diagonal.
Answer
View full question & answer→$10 m, 240 m^2, 26 m$
Question 222 Marks
The cost of ploughing a square field at ₹ 13.50 per square metre is ₹ 5400. Find the cost of fencing the field at ₹ 28.50 per metre.
Answer
View full question & answer→₹ 2280
Question 232 Marks
A square field has an area of 6.25 ares. Find the cost of putting a fence round it at ₹ 32.50 per metre.
Answer
View full question & answer→₹ 3250
$\left[\right.$Hint : Area $\left.=(6.25 \times 100) m ^2=625 m^2\right]$
$\left[\right.$Hint : Area $\left.=(6.25 \times 100) m ^2=625 m^2\right]$
Question 242 Marks
Find the perimeter, area and length of diagonal of a rectangle, having :
length = 3.2 m, breadth = 2.4 m
length = 3.2 m, breadth = 2.4 m
Answer
View full question & answer→$11.2 m, 7.68 m^2, 4 m$
Question 252 Marks
Find the perimeter, area and length of diagonal of a rectangle, having :
length = 20 m, breadth = 15 m
length = 20 m, breadth = 15 m
Answer
View full question & answer→$70 m, 300 m^2, 25 m$
Question 262 Marks
Find the perimeter, area and length of diagonal of a rectangle, having :
length = 15 cm, breadth = 8 cm
length = 15 cm, breadth = 8 cm
Answer
View full question & answer→$46 cm, 120 cm^2, 17 cm$
Question 272 Marks
The length and breadth of a rectangular park are in the ratio $5: 2$. A 2.5 m wide path running all around the outside of the park has an area of $305 m^2$. Find the dimensions of the park.
Answer
View full question & answer→40 m by 16 m
[Hint : Let the dimensions of the park be $5 x$ and $2 x$.
Then, dimensions of the outer rectangle $=(5 x+2 \times 2 \cdot 5) m$ and $(2 x+2 \times 2 \cdot 5) m$
i.e $(5 x+5) m$ and $(2 x+5) m$
But, $(5 x+5) \times(2 x+5)-5 x \times 2 x=305$ ]
[Hint : Let the dimensions of the park be $5 x$ and $2 x$.
Then, dimensions of the outer rectangle $=(5 x+2 \times 2 \cdot 5) m$ and $(2 x+2 \times 2 \cdot 5) m$
i.e $(5 x+5) m$ and $(2 x+5) m$
But, $(5 x+5) \times(2 x+5)-5 x \times 2 x=305$ ]
Question 282 Marks
A rectangular lawn 75 m by 60 m has two roads each 4 m wide running in the middle of it, one parallel to length and the other parallel to breadth. Find the cost of gravelling the roads at ₹ 14.50 per sq. metre.
Answer
View full question & answer→₹ 7598
Question 292 Marks
A rectangular hall is 22 m long and 15.5 m broad. A carpet is laid inside the hall leaving all around a margin of 75 cm from the walls. Find the area of the carpet and the area of the strip left uncovered. If the width of the carpet is 82 cm, find its cost at the rate of ₹ 124 per metre.
Answer
View full question & answer→$287 m^2, 54 m^2$, ₹ 43400
$ \text {[Hint: } \text { Area of the carpet }=\left(22-2 \times 0.75 m \times(15.5-2 \times 0.75) m =20.5 m \times 14 m=287 m^2\right. $
$ \text { Area of strip }=(22 \times 15.5-287) m ^2=54 m^2 $
$ \left.\text { Length of carpet needed }=\frac{287}{0.82} m\right]$
$ \text {[Hint: } \text { Area of the carpet }=\left(22-2 \times 0.75 m \times(15.5-2 \times 0.75) m =20.5 m \times 14 m=287 m^2\right. $
$ \text { Area of strip }=(22 \times 15.5-287) m ^2=54 m^2 $
$ \left.\text { Length of carpet needed }=\frac{287}{0.82} m\right]$
Question 302 Marks
A rectangular plot of land measures 95 m by 72 m. Inside the plot, a path of uniform width 3.5 m is to be constructed all around. The rest of the plot is to be laid with grass. Find the total expense involved in constructing the path at ₹ 46.50 per $m ^2$ and laying the grass at ₹ 3.75 per $m ^2$.
Answer
View full question & answer→₹ 73530
Question 312 Marks
A rectangular grassy plot is 125 m long and 74 m broad. It has a path 2.5 m wide all round it on the inside. Find the cost of levelling the path at ₹ 6.80 per $m ^2$.
Answer
View full question & answer→₹ 6596
Question 322 Marks
A room 9.5 m long and 6 m wide is surrounded by a verandah 1.25 m wide. Calculate the cost of cementing the floor of this verandah at ₹28 per sq. metre.
Answer
View full question & answer→₹ 1260
Question 332 Marks
The length and breadth of a rectangular field are in the ratio $7: 5$ and its perimeter is 384. Find the cost of reaping the field at ₹ 1.25 per sq. metre.
Answer
View full question & answer→₹ 11200
Question 342 Marks
A verandah 50 m long and 12 m broad is to be paved with tiles, each measuring 6 dm by 5 dm. Find the number of tiles needed.
Answer
View full question & answer→2000
[Hint : Area of 1 tile $=\frac{6}{10} m \times \frac{5}{10} m$. No. of tiles required $=\frac{\text { Area of Verandah }}{\text { Area of } 1 \text { tile }}$ ]
[Hint : Area of 1 tile $=\frac{6}{10} m \times \frac{5}{10} m$. No. of tiles required $=\frac{\text { Area of Verandah }}{\text { Area of } 1 \text { tile }}$ ]