3 Marks Question - MATHS STD 10 Questions - Vidyadip

Questions

3 Marks Question

🎯

Test yourself on this topic

17 questions · timed · auto-graded

Question 13 Marks

A lawn is in the from of a rectangle whose sides are in the ratio 5 : 3 and its area is $3375m^2$. Find the cost of fencing the lawn at ₹ 20 per meter.

Answer

Let the sides of the rectangle be 5x and 3x.
So, $5x \times 3x = 3375$
$\Rightarrow x^2 = 225$
$\Rightarrow x = 15$
So, the side are 75m and 45m.
Perimeter of the rectangle = 2(75 + 45) = 2 × 120 = 240m
Since, the cost of fencing the lawn is 20 per metre,
Cost of fencing = 20 × 240 = ₹ 4800.

View full question & answer→

Question 23 Marks

A rectangular park 35m long 18m wide is to be covered with grass, leaving 2.5m uncovered all around it. Find the area to be laid with grass.

Answer

Lenght of the park = 35m
Breath of the park = 18m

Area of the park $=(3218) m ^2$
Lenght of the park with glass $=(355)=13 m$
Breadth of the park with glass $=(18-5) m =13 m$
Area of park with glass $=(30 \times 13) m ^2=390 m^2$
Area of the path without glass = Area of the whole park area of the park with glass $=630 m^2-390 m^2=240 m^2$
Hence, area of the park to be laid with glass $=240 m^2$

View full question & answer→

Question 33 Marks

Find the area and perimeter of a square plot of land whose diagonal is 24m long. $\big[\text{Take}\sqrt{2}=1.41\big]$

Answer

Diagonal of a square plot = 24m
$\therefore$ Area of a square plot $=\frac12\times(\text{diagonal})^2=\frac12\times24\times24=288\text{m}^2$
Now, area of a square $= (Side)^2 = a^2$
$\Rightarrow 288 = a^2$
$\Rightarrow\text{a}=12\sqrt{2}\text{m}$
$\therefore$ Perimeter of a square plot $=4\text{a}=4\times12\sqrt{2}$
$=48\sqrt{2}\text{m}=48\times1.41=67.68\text{m}$

View full question & answer→

Question 43 Marks

The perimeter of a rectangular plot of land is 80m and its breadth is 16m. Find the length and area of the plot.

Answer

As, perimeter = 80m
$\Rightarrow 2(length + breadth) = 80$
$\Rightarrow 2(length + 16) = 80$
$\Rightarrow 2 \times length + 32 = 80$
$\Rightarrow 2 \times length = 80 - 32$
$\Rightarrow length = 482$
$\therefore$ length = 24m
Now, the area of the plot = length × breadth
$= 24 \times 16$
$= 384m^2$
So, the length of the plot is 24m and its area is $384m^2.$

View full question & answer→

Question 53 Marks

The area of a rectangular plot is $462m^2$ and its length is 28m. Find the perimeter of the plot.

Answer

Let the breadth of the plot be x meter
$\therefore$ Area = Length × Breadth = (28 × x) meter
$= 28x m^2$
$\therefore\ 28\text{x}=462\Rightarrow\text{x}=\frac{462}{28}=16.5\text{m}$
Breadth of plot is = 16.5m
Perimeter of the plot is = 2(length + breadth)
= 2(28 + 16.5)m = (2 × 44.5)m
= 89m.

View full question & answer→

Question 63 Marks

A room is 16m long and 13.5m broad. Find the cost of covering its floor with 75-m-wide carpet at ₹ 60 per metre.

Answer

Length of room = 16m
Breadth of a room = 13.5m
$\therefore$ Area of room $= (16 \times 13.5)m^2$
Now,
$\therefore$ length of the carpet $=\frac{\text{Area of the room}}{\text{Width of the carpet}}$
$=\frac{16\times13.5}{0.75}$
= 288m
Cost of carpet = ₹ 60 per meter
$\therefore$ Cost of 288m long carpet = ₹ 60 × 288 = ₹ 17280

View full question & answer→

Question 73 Marks

Find the length of the diagonal of a square of area $128cm^2$. Also find the perimeter of the square, correct to two decimal places.

Answer

Area of the square $=\frac12\times(\text{diagonal})^2\text{sq.unit}$
Let diagonal of square be x
$\frac12\times\big(\text{x}^2\big)=128$
$\Rightarrow\text{x}^2=256\Rightarrow\text{x}=16\text{cm}$
Length of diagonal = 16cm
Side of square $=\sqrt{128}\text{cm}=11.31\text{cm}$
Primeter of square $=[4 \text{ side}] \text{sq. units}$
$= [411.31]\text{cm} = 45.24\text{cm}.$

View full question & answer→

Question 83 Marks

Find the area of an isoscale triangle each of whose equal sides is 13cm and whose base is 24cm.

Answer

Let a = 13cm, b = 13cm and c = 24cm
Now,
Semi-perimeter $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$=\frac{13+13+24}{2}$
$=\frac{50}{2}$
$=25\text{cm}$
Now,
Area of isoscale triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{a})(\text{s}-\text{a})}$
$=\sqrt{25(25-13)(25-13)(25-24)}$
$=\sqrt{25\times12\times12\times1}$
$=5\times12$
$=60\text{cm}^2.$

View full question & answer→

Question 93 Marks

The floor of a rectangular hall is 24m and breadth 80cm, will be required to cover the floor of the hall?

Answer

Area of floor = Length × Breadth
$= (24 \times 18)m^2$
Area of carpet = Length \times Breadth
$= (24 \times 0.8)m^2$
Number of carpets $=\frac{\text{Area of floor}}{\text{Area of the carpet}}$
$=\frac{(24\times18)\text{m}^2}{(2.5\times0.8)\text{m}^2}$
Hence, the number of carpet pieces required = 216

View full question & answer→

Question 103 Marks

If the area of an equilateral triangle is $81\sqrt{3}\text{cm}^2,$ find its height.

Answer

Area of the equilateral triangle $=81\sqrt{3}\text{cm}^2$
Area of an equilateral triangle $=\Big(\frac{\sqrt{3}}{4}\times\text{a}^2\Big),$ where a is the length of the side.
$\Rightarrow81\sqrt{3}=\frac{\sqrt{3}}{4}\times\text{a}^2$
$\Rightarrow324=\text{a}^2$
$\Rightarrow\text{a}=18\text{cm}$
Height of triangle $=\frac{\sqrt{3}}{2}\times\text{a}$
$=\frac{\sqrt{3}}{2}\times18$
$=9 \sqrt{3}\text{cm}.$

View full question & answer→

Question 113 Marks

Find the area of an isoscale triangle each of whose equal sides is 13cm and whose base is 24cm.

Answer

Let a = 13cm, b = 13cm and c = 24cm
Now,
Semi-perimeter $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$=\frac{13+13+24}{2}$
$=\frac{50}{2}$
$=25\text{cm}$
Now,
Area of isoscale triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{a})(\text{s}-\text{a})}$
$=\sqrt{25(25-13)(25-13)(25-24)}$
$=\sqrt{25\times12\times12\times1}$
$=5\times12$
$=60\text{cm}^2.$

View full question & answer→

Question 123 Marks

In the given figure, ABCD is a quadrilateral in which diagonal BD = 24cm,$\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD}$ such that AL = 9cm and CM = 12cm. Calculate the area of the quadrilateral.

Answer

Given,
BD = 24cm
AL = 9cm
CM = 12cm
Area of $\triangle\text{BCD}=\frac{1}{2}\times\text{BD}\times\text{CD}$
$=\frac12\times24\times12=144\text{cm}^2$
Area of $\triangle\text{DAB}=\frac{1}{2}\times\text{BD}\times\text{AL}$
$=\frac12\times24\times9=108\text{cm}^2$
Now, area of quadrilateral ABCD = Area of $\triangle\text{BCD}$ + Area of $\triangle\text{DAB}$
$= (144 + 108)cm^2$
$= 252cm^2$

View full question & answer→

Question 133 Marks

If the area of an equilateral triangle is $36\sqrt{3}\text{cm}^2,$ find its perimeter.

Answer

Area of equilateral triangle $=36\sqrt{3}\text{cm}^2$
Area of equilateral triangle $=\Big(\frac{\sqrt{3}}{4}\times\text{a}^2\Big),$ where a is the length of the side.
$\Rightarrow36\sqrt{3}=\frac{\sqrt{3}}{4}\times\text{a}^2$
$\Rightarrow144=\text{a}^2$
$\Rightarrow\text{a}=12\text{cm}$
Perimeter of a triangle = 3a
$=3\times12$
$=36\text{cm}$

View full question & answer→

Question 143 Marks

A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3. The area of the lawn is $3375m^2$. Find the cost of fencing the lawn at ₹ 65 per metre.

Answer

Let the length and breadth of the rectangular lawn be 5x m and 3x m, respectively.
Area of the rectangular lawn $= 3375m^2$
$\Rightarrow 3375 = 5x \times 3x$
$\Rightarrow 3375 = 15x^2$
$\Rightarrow\frac{3375}{15}=\text{x}^2$
$\Rightarrow 225 = x^2$
$\Rightarrow x = 15$
⇒ Length = 5x = 5 × 15 = 75m and Breadth = 3x = 3 × 15 = 45m
Perimeter of a lawn = 2(75 + 45) = 2 × 120 = 240m.
Cost of fencing the lawn = ₹ 65 per meter
$\therefore$ Cost of fencing 240 m lawn = ₹ (65 × 240)
= ₹ 15,600.

View full question & answer→

Question 153 Marks

The length of the diagonal of a square is 24cm. Find its area.

Answer

We know that, area of the square $= side^2$
Also, the diagonal of a square divides the square into two triangle with equal areas.
Applying Pythagoras theorem to either triangle, we get
$Side^2 + side^2 = diagonal^2$
$\Rightarrow 2side^2 = diagonal^2$
⇒ Area of square $=\frac{1}{2}\times\text{diagonal}^2$
$=\frac{1}{2}\times(24)^2$
$=\frac{1}{2}\times576$
$=288\text{cm}^2$

View full question & answer→

Question 163 Marks

Find the area of a triangle whose sides are 42cm, 34cm and 20cm.

Answer

Let a = 42cm, b = 34cm and c = 20cm
Now,
Semi-perimeter = $\text{s}=\frac{42+34+20}{2}$
$=\frac{96}{2}$
$=48\text{cm}$
Now,
Area of isoscale triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{48(48-42)(48-34)(48-20)}$
$=\sqrt{48\times6\times14\times28}$
$=\sqrt{6\times2\times2\times2\times6\times2\times7\times2\times2\times7}$
$=6\times2\times2\times2\times7$
$=336\text{cm}^2.$

View full question & answer→

Question 173 Marks

The cost of fencing a square lawn at ₹ 14 per metre is ₹ 28000. Find the cost of mowing the lawn at ₹ 54 per $100m^2.$

Answer

₹ 14 is the cost of fencing a length = 1m
₹ 28000 is the cost of fencing the length $=\frac{28000}{14}\text{m}=2000\text{m}$
Perimeter = 4 side = 2000
⇒ side = 500m
Area of a square $= (side)^2 = (500)^2m$
$= 250000m^2$
Cost of moving the lawn $=₹\ \Big(250000\times\frac{54}{100}\Big)=₹\ 135000.$

View full question & answer→

Generate Paper Free All Perimeter and Areas of Plane Figures topics

3 Marks Question - MATHS STD 10 Questions - Vidyadip