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Question 1:

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.

Given:
Area of a triangle =

Question 2:

The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.

Let the area of the triangular field be $x$ hectare.

Cost of sowing the field is Rs 58 per hectare.

Cost of sowing it is Rs 783.

Therefore,

Using the relation: , we get:

Area of the triangle:

Therefore, base of the triangular field is 900 m and its height is 300 m.

Question 3:

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Find the height corresponding to the longest side.

Let the sides of the triangle be â€‹a = 20 cm, b = 34 cm and c = 42 cm.
Let s be the semi-perimeter of the triangle.

Length of the longest side is 42 cm.

Area of a triangle =$\frac{1}{2}×b×h$

The height corresponding to the longest side is 16 cm.

Question 4:

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm. Also, find the length of the altitude corresponding to the smallest side of the triangle.

Let the sides of triangle be â€‹a = 18 cm, b = 24 cm and c = 30 cm.
Let s be the semi-perimeter of the triangle.

The smallest side is 18 cm long. This is the base.

Now, area of a triangle =$\frac{1}{2}×b×h$

The length of the altitude corresponding to the smallest side is 24 cm.

Question 5:

The sides of a triangle are in the ratio 5 : 12 : 13, and its perimeter is 150 m. Find the area of the triangle.

Let the sides of a triangle be 5x m ,12x m and 13x m.

Since, perimeter is the sum of all the sides,

The  lengths of the sides are:

Question 6:

The perimeter of a right-triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at Rs 18.80 per 100 m2.

Let the sides of the triangular field be 25x m , 17x m and 12x m.
Perimeter is the sum of all the sides. Therefore,

$25x+17x+12x=540\phantom{\rule{0ex}{0ex}}⇒54x=540\phantom{\rule{0ex}{0ex}}⇒x=\frac{540}{54}=10\phantom{\rule{0ex}{0ex}}$

The lengths of the sides are:

The cost of ploughing is Rs. 18.80 per 100 m2.

The cost of ploughing 9000 m2

Thus, it costs Rs. 1692 to plough a field of area 9000 m2.

Question 7:

The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm. Find the area of the triangle.

The perimeter of a right-angled triangle = 40 cm
Therefore , a+b+c= 40 cm
Hypotenuse = 17 cm
Therefore, c = 17 cm
a+b+c= 40 cm
a+b+17 = 40
a+b = 23
b = 23 $-$a...........(i)
Now, using Pythagoras' theorem, we have:

Substituting the value of a=15, in equation(i) we get:
b = 23-a
= 23 - 15
= 8 cm

If we had chosen , then,

In any case,
â€‹

Question 8:

The difference between the sides at right angle in a right-angled triangle is 7 cm. The area of the triangle is 60 cm2. Find its perimeter.

Given:
Area of the triangle =
Let the sides of the triangle be a, b and c, where a is the height, b is the base and c is hypotenuse of the triangle.
$a-b=7\mathrm{cm}$
a = 7 + b.......(1)

Side of a triangle cannot be negative.
Therefore, b = 8 cm.

Substituting the value of b = 8 cm, in equation (1):
a = 7+8 = 15 cm

Now,  a = 15 cm, b = 8 cm

Now, in the given right triangle, we have to find third side.

So, the  third side is 17 cm.

Perimeter of a triangle = $a+b+c$.
Therefore, required perimeter of the triangle .

Question 9:

The lengths of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is 24 cm2, find the perimeter of the triangle.

Given:
Area of triangle =
Let the sides be a and b, where a is the height and b is the base of triangle.

a = 2 + b.......(1)

Side of a triangle cannot be negative.

Therefore, b = 6 cm.

Substituting the value of b = 6 cm in equation(1), we get:
a = 2+6 = 8 cm

Now,  a = 8 cm, b = 6 cm

In the given right triangle we have to find third side. Using the relation

So, the third side is 10 cm.

So, perimeter of the triangle = a b + c
= 8+6+10
â€‹=24 cm

Question 10:

Each side of an equilateral triangle measures 8 cm. Find (i) the area of the triangle, correct to 2 decimal places, and (ii) the height of the triangle, correct to 2 places of decimal. Take $\sqrt{3}=1.732.$

Sides of triangle = a = 8 cm
â€‹(i) Area of an equilateral triangle = $\frac{\sqrt{3}}{4}×{a}^{2}$

(ii )Height of triangle = $\frac{\sqrt{3}}{2}×a$

Question 11:

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take $\sqrt{3}=1.732.$

Given:
Height of the equilateral triangle= 9 cm

Question 12:

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

Area of equilateral triangle =

Area of equilateral triangle = $\left(\frac{\sqrt{3}}{4}×{a}^{2}\right)$, where a is the length of the side.

Perimeter of a triangle = 3a

Question 13:

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

Area of the equilateral triangle =
Area of an equilateral triangle =$\left(\frac{\sqrt{3}}{4}×{a}^{2}\right)$, where a is the length of the side.

Height of triangle = $\frac{\sqrt{3}}{2}×a$

Question 14:

The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.

Base = 48 cm
Hypotenuse =50 cm
First we will find the height of the triangle; let the height be 'p'.

Area of the triangle=$\frac{1}{2}×\mathrm{base}×\mathrm{height}$

Question 15:

The hypotenuse of a right-angled triangle measures 6.5 m and its base measures 6 m. Find the length of perpendicular and hence, calculate the area of the triangle.

Hypotenuse = 6.5 m
â€‹Base = 6 m
In a right-angled triangle,

Area of triangle = $\frac{1}{2}×\mathrm{base}×\mathrm{perpendicular}$

Question 16:

Find the area of a right-angled triangle, the radius of whose circumcircle measure 8 cm and the altitude drawn to the hypotenuse measures 6 cm.

Given: Radius = 8 cm
â€‹Height = 6 cm
Area=?
In a right-angled triangle, the centre of the circumcircle is the mid-point of the hypotenuse.

Now, base = 16 cm and height = 6 cm

Area of the triangle =$\frac{1}{2}×base×height\phantom{\rule{0ex}{0ex}}$

Question 17:

Find the area of an isosceles triangle each of whose equal sides measure 13 cm and whose base measures 20 cm. Write your answer correct to 1 place of decimal.

Given: Equal sides of an isosceles triangle = a = 13 cm
Base = b = 20 cm
Area of an isosceles triangle = $\left(\frac{1}{4}b\sqrt{4{a}^{2}-{b}^{2}}\right)$

Question 18:

Find the length of the hypotenuse of an isosceles right-angled triangle whose area is 200 cm2. Also, find its perimeter. Take $\sqrt{2}$ = 1.414.

In a right isosceles triangle, $base=height=a$

Therefore,

Further, given that area of isosceles right triangle = 200 cm2

In an isosceles right triangle, two sides are equal ('a') and the third side is the hypotenuse, i.e. 'c

Therefore, c$\sqrt{{a}^{2}+{a}^{2}}$

Perimeter of the triangle = $a+a+c$
â€‹                                       $=20+20+28.28$

= 68.28 cm

The length of the hypotenuse is 28.28 cm and the perimeter of the triangle is 68.28 cm.

Question 19:

The base of an isosceles triangle measures 80 cm and its area is 360 cm2. Find the perimeter of the triangle.

Given  :
Base = 80 cm
Area = 360 ${\mathrm{cm}}^{2}$
Area of an isosceles triangle = $\left(\frac{1}{4}b\sqrt{4{a}^{2}-{b}^{2}}\right)$

Squaring both the sides, we get:

Perimeter $=\left(2a+b\right)$

So, the perimeter of the triangle is 162 cm.

Question 20:

The perimeter of an isosceles triangle is 42 cm and its base is $1\frac{1}{2}$ times each of the equal sides. Find (i) the length of each side of the triangle, (ii) the area of the triangle, and (iii) the height of the triangle.

Perimeter of the isosceles triangle = 42 cm
Base = $1\frac{1}{2}a=\frac{3}{2}a$
Here, a is the length of one of the equal sides.
Perimeter$=\left(2a+b\right)$
Here, b is the base and $b=\frac{3}{2}a$.

(i) Length of each equal side of the triangle is 12 cm.

Base = $\frac{3}{2}a$

Base = 18 cm

(ii) Area of an isosceles triangle = $\left(\frac{1}{4}b\sqrt{4{a}^{2}-{b}^{2}}\right)$
Here, a is the length of one of the equal sides of the triangle and b is the base.

Thus, we have:

Area of the triangle = 71.42 cm2

(iii)
$\mathrm{Height}=\frac{\sqrt{4{a}^{2}-{b}^{2}}}{2}\phantom{\rule{0ex}{0ex}}$

Question 21:

Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measures 12 cm. Find the area of the triangle.

Let the height of the triangle be h cm.
Each of the equal sides measures and b = 12 cm (base).

Now,
Area of the triangle = Area of the isosceles triangle
$\frac{1}{2}×\mathrm{base}×\mathrm{height}=\frac{1}{4}×b\sqrt{4{a}^{2}-{b}^{2}}$

Area of the triangle = $\frac{1}{2}×b×h$

Question 22:

Two sides of a triangular field are 85 m and 154 m in length, and its perimeter is 324 m. Find (i) the area of the field, and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m.

(i) Perimeter of the triangular field = 324 m

If the unknown side is x m in length, we have:

$85+154+x=324\phantom{\rule{0ex}{0ex}}⇒x=324-154-85=85$

Also,
Semiperimeter of the triangular field,

Now, the area of the field can be expressed as:

(ii) Area of the triangle can be expressed as:

Thus, the length of the perpendicular from the opposite vertex on the side measuring 154 m is 36 m.

Question 23:

In the given figure, âˆ†ABC is an equilateral triangle the length of whose side is equal to 10 cm, and âˆ†DBC is right-angled at D and BD = 8 cm. Find the area of the shaded region. Take $\sqrt{3}$ = 1.732.

Given:
Side of equilateral triangle ABC = 10 cm
BD = 8 cm
Area of $\mathrm{equilateral}∆ABC=\frac{\sqrt{3}}{4}{\mathrm{a}}^{2}$ (where a = 10 cm)

Area of triangle $△$BCD = $\frac{1}{2}×b×h$

Area of the shaded region =
= 43.30 $-$ 24
= 19.3 cm2

Question 24:

Find the area and perimeter of an isosceles right-angled triangle, each of whose equal sides measures 10 cm. Take $\sqrt{2}$ = 1.414.

Let:
Length of each of the equal sides of the isosceles right-angled triangle = a = 10 cm
And,
Base = Height = a

The hypotenuse of an isosceles right-angled triangle can be obtained using Pythagoras' theorem.

If h denotes the hypotenuse, we have:

∴ Perimeter of â€‹the isosceles right-angled triangle = $2a+\sqrt{2}a$

Question 1:

The perimeter of a rectangular plot of land is 75 m and its breadth is 16 m. Find the length and area of the plot.

Perimeter of the rectangular plot of land = 75 m
We know:
Perimeter of the rectangular plot of land =
Here,
b = 16 m
Thus, we have:

Area of the rectangular plot $=$$l×b$

Length of the plot

Question 2:

The length of a rectangular park is twice its breadth, and its perimeter measures 0.84 km. Find the area of the park in square metres.

Let the breadth of the rectangular park be.
∴ Length of the rectangular park$=l=2b$
Perimeter = 0.84 km

Question 3:

One side of a rectangle is 12 cm long and its diagonal measures 37 cm. Find the other side and the area of the rectangle.

One side of the rectangle = 12 cm
Diagonal of the rectangle = 37 cm

The diagonal of a rectangle forms the hypotenuse of a right-angled triangle. The other two sides of the triangle are the length and the breadth of the rectangle.

Now, using Pythagoras' theorem, we have:

Thus, we have:
Length  = 35 cm
Breadth = 12 cm

Question 4:

The area of a rectangular plot is 462 m2 and its length is 28 m. Find the perimeter of the plot.

Area of the rectangular plot = 462 m2
Length (l) = 28 m

Perimeter of the plot = $2\left(l+b\right)$

Question 5:

The length of a rectangular hall is 5 m more than its breadth. The area of the hall is 750 m2. Find the perimeter of the hall.

Let breadth of a rectangular hall be b m
Then, the length of the rectangular hall
Area of the hall $=l×b$

Perimeter of the hall $=2\left(l+b\right)$

Question 6:

A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3. The area of the lawn is 3375 m2. Find the cost of fencing the lawn at Rs 8.50 per metre.

Let the length and breadth of the rectangular lawn be 5x m and 3x m, respectively.

Given:
Area of the rectangular lawn =

Thus, we have:

Cost of fencing 1 m lawn = Rs 8.50
∴ Cost of fencing 240 m lawn =

Question 7:

A room is 16 long and 13.5 m broad. Find the cost of covering its floor with 75 cm-wide carpet at Rs 15 per metre.

Given:
Length of the room = 16 m
Breadth of the room = 13.5 m
Thus, we have:

Width of the carpet = 75 cm = 0.75 m

Cost of 75 cm carpet is Rs 15.

∴ Cost of covering the entire floor = $15×288$ = Rs 4320

Question 8:

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

Given:
Length = 24 m
Breadth = 18 m
Thus, we have:

Length of each carpet = 2.5 m
Breadth of each carpet = 80 cm = 0.80 m

Area of one carpet =

Number of carpets required =

Therefore, 216 carpets will be required to cover the floor of the hall.

Question 9:

A 36 m-long, 15 broad verandah is to be paved with stones, each measuring 6 dm by 5 cm. How many stones will be required?

Length of the stone = 6 dm = 0.6 m
Breadth of the stone = 5 dm = 0.5 m

Thus, 1800 stones will be required to pave the verandah.

Question 10:

The area of a rectangle is 192 cm2 and its perimeter is 56 cm. Find the dimensions of the rectangle.

Area of the rectangle = 192 cm2
Perimeter of the rectangle = 56 cm

$\mathrm{Perimeter}=2\left(\mathrm{length}+\mathrm{breadth}\right)\phantom{\rule{0ex}{0ex}}⇒56=2\left(l+b\right)\phantom{\rule{0ex}{0ex}}⇒l+b=28\phantom{\rule{0ex}{0ex}}⇒l=28-b$

Thus, we have;

We will take length as 16 cm and breath as 12 cm because length is greater than breadth by convention.

Question 11:

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

The field is planted with grass, with 2.5 m uncovered on its sides.

The field is shown in the given figure.

Thus, we have;
Length of the area planted with grass =

Width of the area planted with grass =

Area of the rectangular region planted with grass =

Question 12:

A rectangular plot measures 125 m by 78 m. It has gravel path 3 m wide all around on the outside. Find the area of the path and the cost of gravelling it at Rs 75 per m2.

The plot with the gravel path is shown in the figure.

Area of the rectangular plot = $l×b$
Area of the rectangular plot =
Length of the park including the path = 125 + 6 = 131 m
Breadth of the park including the path = 78 + 6 = 84 m
Area of the plot including the path

Area of the path = $11004-9750$
= 1254 m2
Cost of gravelling 1 m2of the path = Rs 75
∴ Cost of gravelling 1254 m2of the path = $1254×75$
= Rs 94050

Question 13:

A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide. If the area of the path is 420 m2., find the width of the path.

Area of the rectangular field =

Let the width of the path be x m. The path is shown in the following diagram:

Length of the park excluding the path = (54 $-$ 2x) m
Breadth of the park excluding the path = (35 $-$ 2x ) m

Thus, we have:
Area of the path = 420 m2

The width of the path cannot be more than the breadth of the rectangular field.
∴ x = 2.5 m

â€‹Thus, the path is 2.5 m wide.

Question 14:

The length and the breadth of a rectangular garden are in the ratio 9 : 5. A path 3.5 m wide, running all around inside it has an area of 1911 m2. Find the dimensions of the garden.

Let the length and breadth of the garden be 9x m and 5x m, respectively,
Now,
Area of the garden = $\left(9x×5x\right)=45{x}^{2}$
Length of the garden excluding the path = (
Breadth of the garden excluding the path = $\left(5x-7\right)$
Area of the path = $45{x}^{2}-\left[\left(9x-7\right)\left(5x-7\right)\right]$

Thus, we have:
Length =

Question 15:

A room 4.9 m long and 3.5 m broad is covered with carpet, leaving an uncovered margin of 25 cm all around the room. If the breadth of the carpet is 80 cm, find its cost at Rs 40 per meter.

Width of the room left uncovered = 0.25 m
Now,
Length of the room to be carpeted =
Breadth of the room be carpeted =

Area to be carpeted =

Breadth of the carpet = 80 cm = 0.8 m
We know:
Area of the room = Area of the carpet

Cost of 1 m carpet = Rs 40
Cost of 16.5 m carpet =

Question 16:

A carpet is laid on the floor of a room 8 m by 5 m. There is a border of constant width all around the carpet. If the area of the border is 12 m2, find its width.

Let the width of the border be x m.
The length and breadth of the carpet are 8 m and 5 m, respectively.
Area of the carpet =
Length of the carpet without border = $\left(8-2x\right)$
Breadth of carpet without border = $\left(5-2x\right)$
Area of the border = 12 m2
Area of the carpet without border = $\left(8-2x\right)\left(5-2x\right)$

Because the border cannot be wider than the entire carpet, the width of the carpet is $\frac{1}{2}$ m, i.e., 50 cm.

Question 17:

A 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running through its   middle, one parallel to its length and the other parallel to its breadth. Find the cost of gravelling the roads at Rs 24 per m2.

The length and breadth of the lawn are 80 m and 64 m, respectively.
The layout of the roads is shown in the figure below:

Area of the road ABCD =
Area of the road PQRS =
Clearly, the area EFGH is common in both the roads.
Area EFGH
Area of the roads = $400+320-25$
=
Given:
Cost of gravelling 1 m2 area = Rs 24
∴ Cost of gravelling 695 m2 area = $695×24$
= Rs 16680

Question 18:

The dimensions of a room are 14 m × 10 m × 6.5 m. There are two doors and 4 windows in the room. Each door measures 2.5 m × 1.2 m and each window measures 1.5 × 1 m. Find the cost of painting the four walls of the room at Rs 38 per m2.

The room has four walls to be painted.

Now,
Area of the two doors =

Area of the four windows =

The walls have to be painted; the doors and windows are not to be painted.

∴ Total area to be painted
Cost for painting 1 m2 = Rs 38
Cost for painting 300 m2 =

Question 19:

The cost of preparing the walls of a room 12 m long at the rate of Rs 30 per m2 is Rs 7560, and the cost of covering the floor with mat at Rs 15 per m2 is 1620. Find the height of the room

Area of the floor =

∴ Width of the room =

Now, let the height of the room be $h$ m.

Thus, the height of the room is 6 m.

Question 20:

Find the area and perimeter of a square plot of land whose diagonal is 24 m long. Find the answers correct to two decimal places.

Area of the square = $\frac{1}{2}×{\mathrm{Diagonal}}^{2}$

=$\frac{1}{2}×24×24$
=
Now, let the side of the square be x m.
Thus, we have:
$\mathrm{Area}={\mathrm{Side}}^{2}\phantom{\rule{0ex}{0ex}}⇒288={x}^{2}\phantom{\rule{0ex}{0ex}}⇒x=16.97$

Perimeter =$4×\mathrm{Side}$

Thus, the perimeter of the square plot is 67.88 m.

Question 21:

Find the length of the diagonal of a square of area 128 cm2. Also find the perimeter of the square, correct to two decimal places.

Area of the square = 128 cm2

Now,
Area = ${\mathrm{Side}}^{2}$

Perimeter = 4(Side)

Question 22:

The area of a square field is 8 hectares. How long would a man take to cross it diagonally by waling at the rate of 4 km per hour?

1 hectare = 0.01 km2
∴ 8 hectares = 0.08 km2
Area of the square field = $\frac{1}{2}{d}^{2}$ (where d is the diagonal of the square)

Or,
d = 400 m

Now,

â€‹

We know:
$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$

A man would take 6 minutes to walk across the field diagonally.

Question 23:

The cost of harvesting a square field at the rate of Rs 180 per hectare is Rs 1620. Find the cost of putting a fence around it at the rate of Rs 6.75 per metre.

Let the side of the field be $l$ m.

Cost of harvesting

Question 24:

The cost of fencing a square lawn at Rs 14 per metre is Rs 28000. Find the cost of mowing the lawn at Rs 54 per 100 m2.

Cost of fencing the lawn = Rs 28000
Let l be the length of each side of the lawn. Then, the perimeter is 4l.
We know:
$\mathrm{Cost}=\mathrm{Rate}×\mathrm{Perimeter}\phantom{\rule{0ex}{0ex}}⇒28000=14×4\mathrm{l}$

Area of the square lawn

Cost of mowing 100 m2 of the lawn = Rs 54
Cost of mowing 1 m2 of the lawn

∴ Cost of mowing 250000 m2 of the lawn

Question 25:

A 5.25 m by 3.78 m rectangular courtyard is to be paved with square tiles of the same size such that only tiles are used. What is the largest possible size of such a tile? Also find the number of tiles required.

Question 26:

Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.

Area of $∆$ABD = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Area of $∆$ABD = $\sqrt{48\left(48-42\right)\left(48-20\right)\left(48-34\right)}$

Area of $∆$BDC = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Area of $∆$BDC = $\sqrt{35\left(35-29\right)\left(35-20\right)\left(35-21\right)}$

∴ Area of quadrilateral ABCD = Area of $∆$ABD + Area of $∆$BDC
â€‹                                                  = 336 + 210
= 546 cm2

Question 27:

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ∠ACB = 90° and AC = 15 cm.

In the right-angled $∆$ACB:

Perimeter = $AB+BC+CD+AD$
= $17+8+12+9$
= 46 cm

Area of $∆ABC=\frac{1}{2}\left(b×h\right)\phantom{\rule{0ex}{0ex}}$

=$\frac{1}{2}\left(8×15\right)$
=

Area of $∆ADC=\frac{1}{2}×b×h$

∴ Area of the quadrilateral = Area of $∆$ABC + Area of $∆$ADC
= 60 + 54
= 114 cm2

Question 28:

Find the area of the quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. Also find the perimeter of the quadrilateral. Take $\sqrt{3}$ = 1.73.

$∆$BDC is an equilateral triangle with side a= 26 cm.
Area of

By using Pythagoras' theorem in the right-angled triangle $∆$DAB, we get:

Area of $∆ABD=\frac{1}{2}×b×h$

Area of the quadrilateral = Area of $∆$BCD + Area of $∆$ABD

= 412.37 cm2

Perimeter of the quadrilateral = AB + BC + CD + AD
= 24 + 10 + 26 + 26
= 86 cm

Question 29:

In the given figure, ABCD is a quadrilateral in which diagonal BD = 64 cm, AL ⊥ BD and CM ⊥ BD such that AL = 13.2 cm and CM = 16.8 cm Calculate the area of the quadrilateral.

Given:
BD = 64 cm
AL = 13.2 cm
CM = 16.8 cm

Question 30:

Find the area of a parallelogram with base equal to 25 cm and the corresponding height measuring 16.8 cm.

Given:
Base = 25 cm
Height = 16.8 cm
∴ Area of the parallelogram

Question 31:

The adjacent sides of a parallelogram are 32 cm and 24 cm. If the distance between the longer sides is 17.4 cm, find hte distance between the shorter sides.

Longer side = 32 cm
Shorter side = 24 cm
Let the distance between the shorter sides be x cm.
Area of a parallelogram =
=
or, $32×17.4=24×x$

or,

∴ Distance between the shorter sides = 23.2 cm

Question 32:

The area of a parallelogram is 392 m2. If its altitude is twice the corresponding base, determine the base and the altitude.

Area of the parallelogram = 392 m2
Let the base of the parallelogram be b m.
Given:
Height of the parallelogram is twice the base.
∴ Height = 2b m
â€‹Area of a parallelogram =

∴ Base = 14 m
Altitude =

Question 33:

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.

Parallelogram ABCD is made up of congruent $∆$ABC and $∆$ADC.

Area of triangle ABC$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$   (Here, s is the semiperimeter.)

Thus, we have:

Now,
Area of the parallelogram =

Question 34:

Find the area of the rhombus, the lengths of whose diagonals are 30 cm and 16 cm. Also find the perimeter of the rhombus.

Area of the rhombus = $\frac{1}{2}×{d}_{1×}{d}_{2}$, where d1 and d2 are the lengths of the diagonals.

Side of the rhombus = $\frac{1}{2}\sqrt{{{d}_{1}}^{2}+{{d}_{2}}^{2}}$

Perimeter of the rhombus = $4a$
= $4×17$
= 68 cm

Question 35:

The perimeter of a rhombus is 60 cm. If one of its diagonals is 18 cm long, find (i) the length of the other diagonal, and (ii) the area of the rhombus.

Perimeter of a rhombus = 4a    (Here, a is the side of the rhombus.)

(i) Given:
One of the diagonals is 18 cm long.

Thus, we have:

∴ Length of the other diagonal = 24 cm

(ii) Area of the rhombus$=\frac{1}{2}{d}_{1}×{d}_{2}$

â€‹

Question 36:

The area of a rhombus is 480 cm2, and one of its diagonals measures 48 cm. Find (i) the length of the other diagonal, (ii) the length of each of its sides, and (iii) its perimeter.

i) Area of a rhombus$=\frac{1}{2}×{d}_{1}×{d}_{2}$, where d1 and d2 are the lengths of the diagonals.

∴ Length of the other diagonal = 20 cm

(ii) Side = $\frac{1}{2}\sqrt{{{d}_{1}}^{2}+{{d}_{2}}^{2}}$

∴ Length of the side of the rhombus = 26 cm

(iii) Perimeter of the rhombus = $4×\mathrm{Side}$

Question 37:

Find the area of a rhombus one side of which measures 20 cm and one of whose diagonals is 24 cm.

Side of the rhombus = 20 cm

It is given that one of the diagonals is 24 cm long.

Thus, we have:

∴ Area of the rhombus = $\frac{1}{2}{d}_{1}×{d}_{2}$

Question 38:

A parallelogram and a rhombus are equal in area. The diagonals of the rhombus measure 120 m and 44 m. If one of the sides of the parallelogram measures 66 m, find its corresponding altitude.

Area of a rhombus = $\frac{1}{2}{d}_{1}×{d}_{2}$, where d1 and d2are the diagonals of the rhombus.

Given:
Area of the parallelogram = Area of the rhombus
Thus, we have:
Area of the parallelogram =
We know:
Area of the parallelogram =

Therefore, the corresponding altitude is 40 m.

Question 39:

A parallelogram and a square have the same area. If the sides of the square measure 40 m, and altitude of the parallelogram measures 25 m, find the length of the corresponding base of the parallelogram.

Area of a square = ${\mathrm{Side}}^{2}$
Area of the given square = ${40}^{2}$
= 1600 m2
Area of the square = Area of the parallelogram
∴ Area of the parallelogram =
Area of the parallelogram = $\mathrm{Base}×\mathrm{Height}$

Therefore, the length of the corresponding base of the parallelogram is 64 m.

Question 40:

The parallel sides of a trapezium measure 9.7 cm and 6.3 cm, and the distance between them is 6.5 cm. Find the area of the trapezium.

Area of trapezium =

Question 41:

The shape of the cross section of a canal is a trapezium. If the canal is 10 m wide at the top, 6 m wide at the bottom and the area of its cross section is 640 m2, find the depth of the canal.

Area of the canal =
Area of trapezium =

Therefore, the depth of the canal is 80 m.

Question 42:

Find the area of a trapezium whose parallel sides are 11 ma and 25 m long, and the nonparallel sides are 15 m and 13 m long.

Draw $DE\parallel BC$ and DL perpendicular to AB.
The opposite sides of quadrilateral DEBC are parallel. Hence, DEBC is a parallelogram.
DE = BC = 13 m
Also,

For $∆DAE$:
Let:
AE = a =14 m
DE = b = 13 m
DA = c =15 m

Thus, we have:

Area of

Area of $∆DAE=\frac{1}{2}×AE×DL$

Area of trapezium =

Question 43:

The difference between the lengths of the parallel sides of a trapezium is 8 cm, the perpendicular distance between these sides is 24 cm and the area of the trapezium is 312 cm2. Find the length of each of the parallel sides.

Given:
Area = 312 cm2

$a-b=8\phantom{\rule{0ex}{0ex}}a=8+b$

Height = 24 cm
Area of trapezium =

Thus, the lengths of parallel sides are 17 cm and 9 cm.

Question 44:

The adjoining figure shows a field, with the measurements given in metres. Calculate the area of the field.

Area of $∆$DGC = $\frac{1}{2}×DG×GC$

Area of trapezium GHBC =

Now,
AD = AH + GH + DG = 8 + 18 + 14 = 40 m
Area of $∆$DEA = $\frac{1}{2}×\left(DA×EH\right)$

∴ Area of quadrilateral EDCBA = Area of $∆$DGC + Area of $∆$AFB + Area of $∆$DEA + Area of trapezium GHBC
â€‹                                                     = 280 + 140 + 520+ 675
= 1615 m2

Question 1:

The length of a rectangular hall is 5 cm more than its breadth. If the area of the hall is 750 m2, then its length is
(a) 15 m
(b) 22.5 m
(c) 25 m
(d) 30 m

Disclaimer : -The length is given in centimetres. It should be in metres.

(d) 30 m

Let the length of the rectangle be x m.
∴ Breadth of the rectangle

Length cannot be negative.
∴ Length = x = 30 m

Question 2:

The length of a rectangular field is 12 m and the length of its diagonal is 15 m. The area of the field is
(a) 180 m2
(b) $30\sqrt{3}{\mathrm{m}}^{2}$
(c) $12\sqrt{15}{\mathrm{m}}^{2}$
(d) none of these

(d) none of these

Length of the rectangular field = 12 m
Diagonal = 15 m

${\mathrm{Diagonal}}^{2}={\mathrm{Length}}^{2}+{\mathrm{Breadth}}^{2}$

∴ Area of the field

Question 3:

The cost of carpeting a room 15 m long with a carpet 75 cm wide at Rs 50 per m is Rs 6000. The width of the room is
(a) 6 m
(b) 8 m
(c) 9 m
(d) 12 m

(a) 6 m

Total cost = Rs 6000
Rate = Rs 50 per metre

∴ Length of the carpet

Width of the carpet = 75 cm = 0.75 m
â€‹Area of the carpet
We know that area of the carpet is equal to the area of the room.
Thus, we have:
Length of the room = 15 m
Width

Question 4:

The length of a rectangular field is 23 m more than its breadth. If the perimeter of the field is 206 m, then its area is
(a) 1520 m2
(b) 2420 m2
(c) 2480 m2
(d) 2520 m2

(d) 2520 m2

Let the breadth of the field be x m.
∴ Length = (x + 23) m

Now,
Perimeter
Thus, we have:
$4x+46=206\phantom{\rule{0ex}{0ex}}⇒4x=206-46=160\phantom{\rule{0ex}{0ex}}⇒x=\frac{160}{4}=40$
∴ Breadth = x = 40 m
Length = x + 23
Area

Question 5:

A rectangular ground 80 m × 50 m has a path 1 m wide outside around it. The area of the path is
(a) 264 m2
(b) 284 m2
(c) 400 m2
(d) 464 m2

(a) 264 m2

Length of the ground including the path
Breadth of the ground including the path

Total area (including the path)

Area of the field

Area of the path

Question 6:

On increasing the length of a rectangle by 20% and decreasing its breadth by 20%, what is the change in its area?
(a) 20% increase
(b) 20% decrease
(c) No change
(d) 4% decrease

(d) 4% decrease

Let:
Length$=x$
breadth$=y$
Area$=xy$

Now,
New length$=x+20%x=x+\frac{1}{5}x=\frac{6}{5}x$

New breadth$=y-20%y=y-\frac{1}{5}y=\frac{4}{5}y$

New area$=\frac{6}{5}x×\frac{4}{5}y=\frac{24}{25}xy$

Difference in the areas$=xy-\frac{24}{25}xy=\frac{1}{25}xy$

Difference in percentage$=\left[\left(\frac{\frac{1}{25}xy}{xy}\right)×100\right]%=4%$

Question 7:

The length of a rectangle is thrice its breadth and the length of its diagonal is $8\sqrt{10}\mathrm{cm}.$. The perimeter of the rectangle is
(a)
(b) $16\sqrt{10}\mathrm{cm}$
(c) $24\sqrt{10}\mathrm{cm}$
(d) 64 cm

(d) 64 cm

Let the breadth of the rectangle be x cm.
∴ Length of the rectangle = 3x cm
We know:
$\mathrm{Diagonal}=\sqrt{\left(\mathrm{Length}{\right)}^{2}+\left(\mathrm{Breadth}{\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒8\sqrt{10}=\sqrt{{x}^{2}+\left(3x{\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒8\sqrt{10}=\sqrt{{x}^{2}+9{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒8\sqrt{10}=x\sqrt{10}\phantom{\rule{0ex}{0ex}}⇒x=8$

Now,
Breadth of the rectangle = x = 8 cm
Length of the rectangle = 3x = 24 cm
Perimeter of the rectangle

Question 8:

The length of the diagonal of a square is $10\sqrt{2}\mathrm{cm}.$ Its area is
(a) 100 cm2
(b) 150 cm2
(c) 200 cm2
(d) 50 cm2

(a) 100 cm2

A diagonal of a square forms the hypotenuse of a right-angled triangle with base and height equal to side a.

∴ Area of the square

Question 9:

The length of a square field is 0.5 hectare. The length of its diagonal is
(a) 50 m
(b) $50\sqrt{2}\mathrm{m}$
(c) 100 m
(d) 150 m

(c) 100 m

Disclaimer :- The length cannot be in hectare So we used is as area of the square.
Area of the square field

The diagonal divides the square into two isosceles right-angled triangles.

Using Pythagoras' theorem, we have:

${\mathrm{Diagonal}}^{2}={a}^{2}+{a}^{2}=2{a}^{2}$
Area of a square = ${a}^{2}$

∴ Diagonalâ€‹

Question 10:

The area of a square field is 6050 m2. The length of its diagonal is
(a) 110 m
(b) 112 m
(c) 120 m
(d) 135 m

(a) 110 m

Let the diagonal of the square field be d m.

In case of a square field, ${d}^{2}=2{a}^{2}$, where a is the side of the square field.
Now,

∴ $d=\sqrt{2×6050}=\sqrt{12100}=110$

Question 11:

The area of an equilateral triangle is $4\sqrt{3}{\mathrm{cm}}^{2}$. Its perimeter is
(a) 9 cm
(b) 12 cm
(c) $4\sqrt{3}$ cm
(d) $\frac{3\sqrt{3}}{4}$ cm

(b) 12 cm

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^{2}$ (where a is the length of the side)
Thus, we have:

Perimeter of the equilateral triangle = 3a

Question 12:

Each side of an equilateral triangle measures 8 cm. Its area is
(a) $2\sqrt{3}{\mathrm{cm}}^{2}$
(b) $16\sqrt{3}{\mathrm{cm}}^{2}$
(c) 32 cm2
(d) 64 cm2

(b) $16\sqrt{3}{\mathrm{cm}}^{2}$

Let the side of the equilateral triangle be a.
Given:
a = 8 cm
Now,
Area of the equilateral triangle

Question 13:

The length of each side of an equilateral triangle is . Its altitude is
(a) $\frac{3}{2}$ cm
(b) 3 cm
(c) $\frac{\sqrt{3}}{4}$ cm
(d) $\frac{\sqrt{3}}{2}$ cm

(b) 3 cm

Let the length of each side of the equilateral triangle be .

∴ Height of the equilateral triangle

Question 14:

The height of an equilateral triangle is $\sqrt{6}$ cm. Its area is
(a)
(b)
(c) $2\sqrt{3}{\mathrm{cm}}^{2}$
(d) $3\sqrt{3}{\mathrm{cm}}^{2}$

(c)

Height of an equilateral triangle$=\frac{\sqrt{3}}{2}a$ (where a is the side of the equilateral triangle)

∴ Area of the triangle

Question 15:

The length of the side of a square is equal to the length of the side of an equilateral triangle. The ratio of their areas is
(a) 2 : 1
(b) $2:\sqrt{3}$
(c) 4 : 3
(d) $4:\sqrt{3}$

(d) $4:\sqrt{3}$

Let:
Length of the side of the square = Length of the side of the equilateral triangle = a unit

Now,

Area of the square

Area of the equilateral triangle

Question 16:

The side of an equilateral triangle is the same as the radius of a circle whose area is 154 cm2. The area of the triangle is
(a) $\frac{7\sqrt{3}}{4}{\mathrm{cm}}^{2}$
(b) $\frac{49\sqrt{3}}{4}{\mathrm{cm}}^{2}$
(c) 35 cm2
(d) 49 cm2

(b)

Area of a circle$=\mathrm{\pi }{r}^{2}$

The radius of the circle is equal to the side of the equilateral triangle.

r = a (Here, a is the side of the equilateral triangle.)

∴ Area of the equilateral triangle

Question 17:

The area of a triangle is 1176 cm2. If the base and height of the triangle are in the ratio 3 : 4, then the height of the triangle is
(a) 42 cm
(b) 52 cm
(c) 54 cm
(d) 56 cm

(d) 56 cm

Let the base and height of the triangle be $3x$ cm and $4x$ cm, respectively.

Now,

$⇒x=\sqrt{\frac{1176}{6}}=\sqrt{196}=14$

∴ Height of the triangle

Question 18:

The lengths of the sides of a triangular field measure 20 m, 21 m and 29 m. The cost of cultivating the field at Rs 4.50 per m2 is
(a) Rs 900
(b) Rs 945
(c) Rs 1305
(d) Rs 1890

(b) Rs 945

We will find the semiperimeter, s, of the field.

Now,

∴ Total cost of cultivating 210 m2 of the field at the rate of Rs 4.50 per metre =

Question 19:

The length of one side of a parallelogram is 18 cm and the length of perpendicular on it from its opposite side is 8 cm. The area of the parallelogram is
(a) 144 cm2
(b) 72 cm2
(c) 100 cm2
(d) 48 cm2

(a) 144 cm2

Area of a parallelogram = $\left(b×h\right)$ (Where b is the length of one side of the parallelogram and h is the perpendicular distance of the opposite side)

Thus, we have:

Question 20:

Two adjacent sides of a parallelogram are 30 m and 14 m and the diagonal joining the end points of these sides is 40 m. The area of the parallelogram is
(a) 168 m2
(b) 336 m2
(c) 372 m2
(d) 480 m2

(b) 336 m2

Parallelogram ABCD is shown in the figure. Diagonal AC divides the parallelogram into two congruent triangles ABC and ADC.

Now,

Because $∆$ABC and $∆$ADC are congruent, .

Using Hero's formula for the area of triangle ABC, we get:

Semiperimeter, s

Area of triangle ABC of the parallelogram:

∴ Area of parallelogram ABCD

Question 21:

The length of a diagonal of a parallelogram is 70 cm and lengths of perpendiculars on this diagonal from its opposite vertices are 27 cm each. The area of the parallelogram is
(a) 1800 cm2
(b) 1836 cm2
(c) 1890 cm2
(d) 1980 cm2

(c) 1890 cm2

A diagonal of a parallelogram divides it into two congruent triangles, as shown in the figure.

Here, the area of the parallelogram is twice the area of each triangle.

Thus, we have:

Question 1:

In the given figure ABCD is a quadrilateral in which ∠ABC = 90°, ∠BDC = 90°, AC = 17 cm, BC = 15 cm, BD = 12 cm and CD = 9 cm. The area of quad. ABCD is

(a) 102 cm2
(b) 114 cm2
(c) 95 cm2
(d) 57 cm2

(b) 114 sq cm

Using Pythagoras' theorem in $∆$ABC, we get:

∴ Area of quadrilateral ABCD

Question 2:

In the given figure ABCD is a trapezium in which AB = 40 m, BC = 15 m, CD = 28 m, AD = 9 m and CEAB. Area of trap. ABCD is

(a) 306 m2
(b) 316 m2
(c) 296 m2
(d) 284 m2

(a) 306 m2

In the given figure, AECD is a rectangle.

Length AE = Length CD = 28 m

Now,

Also,
AD = CE = 9 m

In the given figure, if DA is perpendicular to AE, then it can be solved, otherwise it cannot be solved.

Question 3:

The sides of a triangle are in the ratio 12 : 14 : 25 and its perimeter is 25.5 cm. The largest side of the triangle is
(a) 7 cm
(b) 14 cm
(c) 12.5 cm
(d) 18 cm

(c) 12.5 cm

Let the sides of the triangle be 12x cm, 14x cm and 25x cm.
Thus, we have:

∴ Greatest side of the triangle

Question 4:

The parallel sides of a trapezium are 9.7 cm and 6.3 cm, and the distance between them is 6.5 cm. The area of the trapezium is
(a) 104 cm2
(b) 78 cm2
(c) 52 cm2
(d) 65 cm2

(c) 52 cm2

Question 5:

Find the area of an equilateral triangle having each side of length 10 cm.

Given:
Side of the equilateral triangle = 10 cm
Thus, we have:

Question 6:

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

Area of an isosceles triangle:

Question 7:

The longer side of a rectangular hall is 24 m and the length of its diagonal is 26 m. Find the area of the hall.

Let the rectangle ABCD represent the hall.

Using the Pythagorean theorem in the right-angled triangle ABC, we have:
${\mathrm{Diagonal}}^{2}={\mathrm{Length}}^{2}+{\mathrm{Breadth}}^{2}$

∴ Area of the hall

Question 8:

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

The diagonal of a square forms the hypotenuse of an isosceles right triangle. The other two sides are the sides of the square of length a cm.

Using Pythagoras' theorem, we have:

$⇒$a$=\frac{24}{\sqrt{2}}$
Area of the square

Question 9:

Find the area of a rhombus whose diagonals are 48 cm and 20 cm long.

Question 10:

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

To find the area of the triangle, we will first find the semiperimeter of the triangle.

Thus, we have:

Now,

Question 11:

A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3 and its area is 3375 m2. Find the cost of fencing the lawn at Rs 20 per metre.

Let the length and breadth of the lawn be , respectively.

Now,

Area of the lawn$=5x×3x=5{x}^{\mathit{2}}$

∴ Perimeter of the lawn
Total cost of fencing the lawn at Rs 20 per metre

Question 12:

Find the area of a rhombus each side of which measures 20 cm and one of whose diagonals is 24 cm.

Given:
Sides are 20 cm each and one diagonal is of 24 cm.
The diagonal divides the rhombus into two congruent triangles, as shown in the figure below.

We will now use Hero's formula to find the area of triangle ABC.
First, we will find the semiperimeter.

Now,
Area of the rhombus

Question 13:

Find the area of a trapezium whose parallel sides are 11 cm and 25 cm long and non-parallel sides are 15 cm and 13 cm.

We will divide the trapezium into a triangle and a parallelogram.

Difference in the lengths of parallel sides
We can represent this in the following figure:

Trapezium ABCD is divided into parallelogram AECD and triangle CEB.

1. Consider triangle CEB.
In triangle CEB, we have:

Using Hero's theorem, we will first evaluate the semiperimeter of triangle CEB and then evaluate its area.

Semiperimeter,

Also,

Area of triangle CEB$=\frac{1}{2}\left(\mathrm{Base}×\mathrm{Height}\right)$

Height of triangle CEB
1. Consider parallelogram AECD.
â€‹â€‹Area of parallelogram AECD

Area of trapezium ABCD

Question 14:

The adjacent sides of a âˆ¥gm ABCD measure 34 cm and 20 cm and the diagonal AC is 42 cm long. Find the area of the âˆ¥gm.

The diagonal of a parallelogram divides it into two congruent triangles. Also, the area of the parallelogram is the sum of the areas of the triangles.

We will now use Hero's formula to calculate the area of triangle ABC.

Area of the parallelogram

Question 15:

The cost of fencing a square lawn at Rs 14 per metre is Rs 2800. Find the cost of mowing the lawn at Rs 54 per 100 cm2.

Given:
Cost of fencing = Rs 2800
Rate of fencing = Rs 14

Now,
Perimeter

Because the lawn is square, its perimeter .

Area of the lawn =

Cost for mowing the lawn per 100 m2= Rs 54

Cost for mowing the lawn per 1 m2

Total cost for mowing the lawn  per 2500 m2

Question 16:

Find the area of quad. ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diag. BD = 20 cm.

Quadrilateral ABCD is divided into triangles $∆$ABD and $∆$BCD.
We will now use Hero's formula.
For $∆$ABD:

For $∆$BCD:

Thus, we have:
Area of quadrilateral ABCD

Question 17:

A parallelogram and a rhombus are equal in area. The diagonals of the rhombus measure 120 m add 44 m. If one of the sides of the âˆ¥gm is 66 m long, find its corresponding altitude.

Area of the rhombus

Area of the parallelogram$=\mathrm{Base}×\mathrm{Height}=66×\mathrm{Height}$

Given:
The area of the rhombus is equal to the area of the parallelogram.

∴ Corresponding height of the parallelogram = 40 m

Question 18:

The diagonals of a rhombus are 48 cm and 20 cm long. Find the perimeter of the rhombus.

Diagonals of a rhombus perpendicularly bisect each other. The statement can help us find a side of the rhombus. Consider the following figure.

ABCD is the rhombus and AC and BD are the diagonals. The diagonals intersect at point O.

We know:

Similarly,

Using Pythagoras' theorem in the right-angled triangle $∆$DOC, we get:

DC is a side of the rhombus.
We know that in a rhombus, all sides are equal.
∴ Perimeter of ABCD

Question 19:

The adjacent sides of a parallelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12 cm, find the distance between the longer sides.

∴ Distance between the longer sides = 9 cm

Question 20:

In a four-sides field, the length of the longer diagonal is 128 m. The lengths of perpendiculars from the opposite vertices upon this diagonal are 22.7 m and 17.3 m. Find the area of the field.

The field, which is represented as ABCD, is given below.

The area of the field is the sum of the areas of triangles ABC and ADC.

Area of the triangle ABC

Area of the triangle ADC

Area of the field = Sum of the areas of both the triangles

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