Perimeter & Area of Plane Figures — Exercise 17(B)

Find the area of a quadrilateral one of whose diagonals is 25 cm and the lengths of perpendiculars from the other two vertices are 16.4 cm and 11.6 cm respectively.

Answer

A quadrilateral ABCD is shown in the figure below :

AM and CN are the perpendiculars from A and C respectively to the diagonal BD.

Area of triangle = $\dfrac{1}{2}$ × base × height

For △ABD,

Area = $\dfrac{1}{2}$ × BD × AM

= $\dfrac{1}{2}$ × 25 × 16.4

= 25 × 8.2

= 205 cm².

For △CBD,

Area = $\dfrac{1}{2}$ × BD × CN

= $\dfrac{1}{2}$ × 25 × 11.6

= 25 × 5.8

= 145 cm².

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle CBD

= 205 + 145

= 350 cm².

Hence, area of quadrilateral = 350 cm².

The diagonals of a quadrilateral intersect each other at right angles. If the lengths of these diagonals be 14 cm and 19 cm respectively, find the area of the quadrilateral.

Answer

A quadrilateral ABCD is shown in the figure below:

The diagonals AC and BD intersect at O and are perpendiculars (90°)

Given,

AC = 14 cm

BD = 19 cm

When the diagonals of a quadrilateral intersect at right angles, the formula of area is:

Area = $\dfrac{1}{2}$ × d₁ × d₂

where d₁ and d₂ are diagonals.

Area = $\dfrac{1}{2}$ × 14 × 19

= 7 × 19

= 133 cm².

Hence, area = 133 cm².

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

Answer

From figure,

The diagonal AC divides the quadrilateral ABCD into two triangles : △ABC & △ACD

Area of △ABC :

Let sides AB = a = 29 cm, BC = b = 21 cm, AC = c = 20 cm

s = $\dfrac{29 + 21 + 20}{2}$

= $\dfrac{70}{2}$ = 35.

By formula,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{35 \times (35 - 29) \times (35 - 21) \times (35 - 20)} \\[1em] = \sqrt{35 \times (6) \times (14) \times (15)} \\[1em] = \sqrt{44100} \\[1em] = 210 \text{ cm}^2.$

Area of △ACD

Let sides be AD = a = 42 cm, CD = b = 34 cm, AC = c = 20 cm

s = $\dfrac{42 + 34 + 20}{2}$

= $\dfrac{96}{2}$ = 48.

By formula,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{48 \times (48 - 42) \times (48 - 34) \times (48 - 20)} \\[1em] = \sqrt{48 \times (6) \times (14) \times (28)} \\[1em] = \sqrt{112896} \\[1em] = 336 \text{ cm}^2.$

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

= 210 + 336

= 546 cm².

Hence, area of quadrilateral = 546 cm².

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

Answer

Given,

AB = 9 cm

AD = 12 cm

BD = 15 cm

CD = 17 cm

∠CBD = 90°

Diagonal BD divides the quadrilateral into two triangles: △ABD and △BCD.

For △ABD,

Sides are 9 cm, 12 cm, 15 cm.

Since, 9² + 12² = 81 + 144 = 225 = 15².

So they are the pythagorean triplets.

So triangle △ABD is a right angled triangle.

Area of right angle triangle = $\dfrac{1}{2}$ × Base × Height

Area of triangle ABD = $\dfrac{1}{2}$ × AB × AD

= $\dfrac{1}{2}$ × 9 × 12

= 9 × 6

= 54 cm².

For △BCD,

Since, ∠CBD = 90°

∴ Applying Pythagoras theorem for the △BCD,

⇒ BD² + BC² = DC²

⇒ 15² + BC² = 17²

⇒ BC² = 289 - 225

⇒ BC² = 64

⇒ BC = $\sqrt{64}$

⇒ BC = 8 cm.

Area of △BCD = $\dfrac{1}{2}$ × BC × BD

= $\dfrac{1}{2}$ × 8 × 15

= 4 × 15

= 60 cm².

Area of quadrilateral △ABCD = Area of △ABD + Area of △BCD

= 54 + 60

= 114 cm².

Perimeter of quadrilateral ABCD = AB + BC + CD + DA

= 9 + 8 + 17 + 12

= 46 cm.

Hence, area = 114 cm² and perimeter = 46 cm.

Calculate the area of quadrilateral ABCD in which : AB = 24 cm, AD = 32 cm, ∠BAD = 90°, and BC = CD = 52 cm.

Answer

Given,

AB = 24 cm

AD = 32 cm

∠BAD = 90°

BC = 52 cm

CD = 52 cm

Diagonal BD divides the quadrilateral into two triangles: △ABD and △BCD

Area of right angle △ABD = $\dfrac{1}{2}$ × Base × Height

= $\dfrac{1}{2}$ × AB × AD

= $\dfrac{1}{2}$ × 24 × 32

= 12 × 32

= 384 cm².

Since ∠BAD = 90°, by applying pythagoras theorem

⇒ BD² = BA² + AD²

⇒ BD² = 24² + 32²

⇒ BD² = 576 + 1024

⇒ BD² = 1600

⇒ BD = $\sqrt{1600}$

⇒ BD = 40 cm.

Area of △BCD,

Let BC = a = 52 cm, CD = b = 52 cm, BD = c = 40 cm

$\Rightarrow s = \dfrac{a + b + c}{2} \\[1em] \Rightarrow s = \dfrac{52 + 52 + 40}{2} \\[1em] \Rightarrow s = \dfrac{144}{2} \\[1em] \Rightarrow s = 72 \text{ cm}.$

By formula,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{72 \times (72 - 52) \times (72 - 52) \times (72 - 40)} \\[1em] = \sqrt{72 \times (20) \times (20) \times (32)} \\[1em] = \sqrt{921600} \\[1em] = 960 \text{ cm}^2.$

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

= 384 + 960

= 1344 cm².

Hence, area of quadrilateral ABCD = 1344 cm².

Calculate the area of quadrilateral ABCD in which △BCD is equilateral with each side equal to 26 cm, ∠BAD = 90° and AD = 24 cm.

Answer

Given

△BCD is equilateral.

Each side = 26 cm

∠BAD = 90°

AD = 24 cm

$\Rightarrow \text{Area of equilateral △BCD} = \dfrac{\sqrt{3}}{4} \times \text{ (side)}^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 26^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 676 \\[1em] = \sqrt{3} \times 169 \\[1em] = 169 \times 1.732 \approx 292.71 \text{ cm}^2.$

Since ∠BAD = 90°

By applying pythagoras theorem in triangle BAD,

⇒ AB² + AD² = BD²

⇒ AB² + 24² = 26²

⇒ AB² + 576 = 676

⇒ AB² = 676 - 576

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm.

⇒ Area of right angle triangle ABD = $\dfrac{1}{2}$ × Base × Height

= $\dfrac{1}{2}$ × AB × AD

= $\dfrac{1}{2}$ × 10 × 24

= 5 × 24 = 120.

⇒ Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

= 120 + 292.71

= 412.71 cm².

Hence, area = 412.71 cm².

In the adjoining figure, △ABC is right angled at A, BC = 7.5 cm and AB = 4.5 cm. If the area of quad. ABCD is 30 cm² and DL is the altitude of △DAC, calculate the length DL.

Answer

Given,

△ABC is right-angled at A.

BC = 7.5 cm (hypotenuse)

AB = 4.5 cm

Area of quadrilateral ABCD = 30 cm²

In △ABC,

By using pythagoras theorem,

⇒ BC² = AB² + AC²

⇒ (7.5)² = (4.5)² + AC²

⇒ 56.25 = 20.25 + AC²

⇒ AC² = 56.25 - 20.25

⇒ AC² = 36

⇒ AC = $\sqrt{36}$ = 6 cm.

$\text{Area of right angle △ABC} = \dfrac{1}{2} \times \text{Base} \times \text{Height} \\[1em] = \dfrac{1}{2} \times AC \times AB \\[1em] = \dfrac{1}{2} \times 6 \times 4.5 \\[1em] = 3 \times 4.5 \\[1em] = 13.5 \text{ cm}^2.$

From figure,

Area of △DAC = Area of quad. ABCD − Area of △ABC

= 30 - 13.5

= 16.5 cm².

$\text{Area of △DAC} = \dfrac{1}{2} \times AC \times DL \\[1em] \Rightarrow 16.5 = \dfrac{1}{2} × 6 × DL \\[1em] \Rightarrow 33 = 6 \times DL \\[1em] \Rightarrow DL = \dfrac{33}{6} \\[1em] \Rightarrow DL = 5.5 \text{ cm}.$

Hence, DL = 5.5 cm.

The perimeter of a rectangle is 81 m and its breadth is 12 m. Find its length and area.

Answer

Given,

Perimeter = 81 m

Breadth (b) = 12 m

Let length of rectangle = l meters

By formula,

Perimeter = 2(l + b)

⇒ 81 = 2(l + 12)

⇒ l + 12 = $\dfrac{81}{2}$

⇒ l + 12 = 40.5

⇒ l = 40.5 - 12 = 28.5 m.

Area = l × b

= 28.5 × 12 = 342 m².

Hence, length = 28.5 m and area = 342 m².

The perimeter of a rectangular field is $\dfrac{3}{5}$ km and its length is twice breadth. Find the area of the field in m².

Answer

Let length = l and breadth = b

Given,

Perimeter = $\dfrac{3}{5} \text{km} = \dfrac{3000}{5}m$

l = 2b

By formula,

Perimeter = 2(l + b)

⇒ $\dfrac{3000}{5}$ = 2(2b + b)

⇒ $\dfrac{1500}{5}$ = 3b

⇒ 3b = 300

⇒ b = 100 m.

⇒ l = 2b = 2 × 100 = 200 m.

Area = l × b

= 200 × 100

= 20000 m².

Hence, area of the field = 20000 m².

A rectangular plot 30 m long and 18 m wide is to be covered with grass leaving 2.5 m all around it. Find the area to be laid with grass.

Answer

ABCD is a rectangular plot and PQRS is an area that has to be laid with grass.

Given,

Length of plot = AD = BC = 30 m

Width of plot = AB = CD = 18 m

From figure,

PQ = AB - 2.5 - 2.5 = 18 - 5 = 13 m

PS = BC - 2.5 - 2.5 = 30 - 5 = 25 m

Area of the rectangle PQRS that has to be laid with grass = length × breadth

= PS × PQ

= 25 × 13

= 325 m².

Hence, area to be laid with grass = 325 m².

A foot path of uniform width runs all around inside of a rectangular field 45 m long and 36 m wide. If the area of the path is 234 m², find the width of the path.

Answer

Consider ABCD as a rectangular field having, length = 45 m and breadth = 36 m.

Let x meters be the width of foot path.

We know that,

Area = length × breadth

From figure,

Area of path = Area of rectangle ABCD - Area of rectangle PQRS

Substituting the values we get,

Area of path = (AB × BC) - (PQ × QR).........(1)

From figure,

PQ = AB - x - x = (45 - 2x) m,

QR = BC - x - x = (36 - 2x) m.

Substituting the values in equation (1) we get,

⇒ 234 = (45 × 36) - (45 - 2x) (36 - 2x)

⇒ 234 = 1620 - [45(36 - 2x) - 2x(36 - 2x)]

⇒ 234 = 1620 - (1620 - 90x - 72x + 4x²)

⇒ 234 = 1620 - 1620 + 90x + 72x - 4x²

⇒ 234 = 162x - 4x²

⇒ 4x² - 162x + 234 = 0

⇒ 4x² - 156x - 6x + 234 = 0

⇒ 4x(x - 39) - 6(x - 39) = 0

⇒ (4x - 6)(x - 39) = 0

⇒ 4x - 6 = 0 or x - 39 = 0

⇒ 4x = 6 or x = 39

⇒ x = $\dfrac{6}{4} = \dfrac{3}{2}$ = 1.5 or x = 39

Since, width of path cannot be greater than breadth of field,

So, x ≠ 39 m.

Hence, width of path = 1.5 m.

The adjoining diagram shows two cross paths drawn inside a rectangular field 45 m long and 38 m wide, one parallel to length and the other parallel to breadth. The width of each path is 4 m. Find the cost of gravelling the paths at ₹ 5.60 per m².

Answer

Given,

Length of the field = 45 m

Width of the field = 38 m

Width of the each path = 4 m

Cost of gravelling = ₹ 5.60 per m².

Area of path parallel to length:

⇒ Area = 45 × 4 = 180 m².

Area of path parallel to breadth:

⇒ Area = 38 × 4 = 152 m².

Since, both the path overlap in the middle :

⇒ So, the overlapped area = 4 × 4 = 16 m².

⇒ Total area of paths = Area of first path + Area of second path - Overlapped area

= 180 + 152 - 16 = 316 m².

⇒ Total cost = Area of paths × Cost per m²

= 316 × 5.60 = 1769.60

Hence, cost of gravelling the paths = ₹ 1,769.60

A rectangle of area 144 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that its perimeter is 52 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.

Answer

Given,

Area of rectangle = 144 cm²

Length = x cm

Perimeter = 52 cm

Area = length × breadth

⇒ 144 = x × breadth

⇒ Breadth = $\dfrac{144}{x}$ cm.

Perimeter = 2(l + b)

⇒ 52 = 2(x + $\dfrac{144}{x}$)

⇒ x + $\dfrac{144}{x}$ = 26

⇒ $\dfrac{x^2 + 144}{x} = 26$

⇒ x² + 144 = 26x

⇒ x² - 26x + 144 = 0

⇒ x² - 18x - 8x + 144 = 0

⇒ x(x - 18) - 8(x - 18) = 0

⇒ (x - 18)(x - 8) = 0

⇒ x = 18 or x = 8

If x = 18 then,

Length = x = 18 cm.

Breadth = $\dfrac{144}{x} = \dfrac{144}{18}$ = 8 cm.

If x = 8 then,

Length = x = 8 cm.

Breadth = $\dfrac{144}{x} = \dfrac{144}{8}$ = 18 cm.

Generally we consider length > breadth

Hence, length = 18 cm and breadth = 8 cm.

The perimeter of a rectangular plot is 130 m and its area is 1000 m². Take the length of the plot as x metres. Use the perimeter to write the value of breadth in terms of x. Use the values of length, breadth and area to write an equation in x. Solve the equation and calculate the length and breadth of the plot.

Answer

Given,

Perimeter = 130 m

Area = 1000 m²

Length = x meters

Let breadth = b meters

Perimeter = 2(length + breadth)

⇒ 130 = 2(x + b)

⇒ x + b = 65

⇒ b = 65 - x

Area = length × breadth

⇒ 1000 = x × (65 - x)

⇒ 1000 = 65x - x²

⇒ x² - 65x + 1000 = 0

⇒ x² - 40x - 25x + 1000 = 0

⇒ x(x - 40) - 25(x - 40) = 0

⇒ (x - 40) (x - 25) = 0

⇒ x = 40 or x = 25

If x = 40 then,

Length = x = 40 m

Breadth = 65 - x = 65 - 40 = 25 m

If x = 25 then,

Length = x = 25 m

Breadth = 65 - x = 65 - 25 = 40 m

Generally we consider length > breadth

Hence, length = 40 m and breadth = 25 m.

If the length of a rectangle is increased by 10 cm and the breadth decreased by 5 cm, the area remains unchanged. If the length is decreased by 5 cm and the breadth is increased by 4 cm, even then the area remains unchanged. Find the dimensions of the rectangle.

Answer

Let length be = l cm and Breadth be = b cm

Area = length × breadth = lb

In first condition :

Length is increased by 10 cm

Breadth is decreased by 5 cm.

But area remains unchanged:

∴ lb = (l + 10)(b - 5)

⇒ lb = lb - 5l + 10b - 50

⇒ 0 = 10b - 5l - 50

⇒ 50 = 10b - 5l

Dividing by 5,

⇒ 10 = 2b - l

⇒ l = 2b - 10 ..........(1)

In second condition:

Length is decreased by 5 cm

Breadth is increased by 4 cm

Here also area remains unchanged.

∴ lb = (l - 5)(b + 4)

⇒ lb = lb + 4l - 5b - 20

⇒ 0 = 4l - 5b - 20

⇒ 4l - 5b = 20 .......(2)

Substituting the value of l from equation (1) in (2), we get :

⇒ 4(2b - 10) - 5b = 20

⇒ 8b - 40 - 5b = 20

⇒ 3b = 20 + 40

⇒ 3b = 60

⇒ b = $\dfrac{60}{3}$ = 20 cm.

⇒ l = 2b - 10

= 2(20) - 10

= 40 - 10 = 30 cm.

Hence, length = 30 cm and breadth = 20 cm.

A room is 13 m long and 9 m wide. Find the cost of carpeting the room with a carpet 75 cm wide at ₹ 12.50 per metre.

Answer

Given,

Room length = 13 m

Room width = 9 m

Carpet width = 75 cm = 0.75 m

Cost of carpet = ₹ 12.50 per metre.

Floor area = Room length × Room width

= 13 × 9 = 117 m².

Floor area = Carpet length × Carpet width

$\Rightarrow 117 = \text{Carpet length} \times 0.75 \\[1em] \Rightarrow \text{Carpet length} = \dfrac{117}{0.75} \\[1em] \Rightarrow \text{Carpet length} = \dfrac{117 \times 100}{75} \\[1em] \Rightarrow \text{Carpet length} = \dfrac{117 \times 4}{3} \\[1em] \Rightarrow \text{Carpet length} = 39 \times 4 \\[1em] \Rightarrow \text{Carpet length} = 156 \text{ m}.$

Total cost = Carpet length × Rate per meter

= 156 × ₹ 12.50

= ₹ 1,950.

Hence, total cost = ₹ 1,950.

A rectangular courtyard 3.78 m long and 5.25 m broad is to be paved exactly with square tiles, all of the same size. What is the largest size of such a tile? Also, find the number of tiles.

Answer

Given,

Length = 3.78 m = 378 cm

Breadth = 5.25 m = 525 cm

The side of the largest square tile that can pave the floor exactly must be the HCF of the length and breadth of the courtyard.

HCF of 378 and 525 = 21 cm.

Number of tiles = $\dfrac{\text{Area of courtyard}}{\text{Area of each tile}} = \dfrac{378 \times 525}{21 \times 21}$ = 450.

Hence, largest size of tile = 21 cm × 21 cm and total number of tiles = 450.

The cost of cultivating a square field at the rate of ₹ 160 per hectare is ₹ 1,440. Find the cost of putting a fence around it at the rate of 75 paise per metre.

Answer

Given,

Rate of cultivation = ₹ 160 per hectare

Total cost = ₹ 1,440

Area of cultivation = $\dfrac{\text{Total cost}}{\text{Rate}}$

= $\dfrac{1440}{160}$ = 9 hectares.

1 hectare = 10,000 m²

9 hectares = 9 × 10,000 = 90,000 m².

Area of square = (side)²

90,000 = (side)²

Side = $\sqrt{90,000}$

Side = 300 m.

Perimeter of square = 4 × side

= 4 × 300 = 1200 m

Total cost = Perimeter × Fencing rate

= 1200 × ₹ 0.75

= ₹ 900.

Hence, cost of fencing around the square field = ₹ 900.

Find the area of parallelogram, if its two adjacent sides are 12 cm and 14 cm, and if the diagonal connecting their ends is 18 cm.

Answer

By formula,

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

In triangle ABC,

Let, AB = a = 12 cm, BC = b = 14 cm, AC = c = 18 cm

s = $\dfrac{a + b + c}{2} = \dfrac{12 + 14 + 18}{2}$ = 22 cm.

Substituting values we get :

$\text{Area of triangle ABC} = \sqrt{22 \times (22 - 12) \times (22 - 14) \times (22 - 18)} \\[1em] = \sqrt{22 \times 10 \times 8 \times 4} \\[1em] = \sqrt{7040} \\[1em] = 83.9 \text{ cm}^2.$

A diagonal of //gm divides it into triangles of equal area.

Thus, Area of //gm ABCD = 2.Area of triangle ABC

= 2 × 83.9

= 167.80 cm².

Hence, area of parallelogram = 167.80 cm².

Find the length of a diagonal of a square of area 200 cm².

Answer

ABCD is a square with diagonal BD = d cm.

We know that,

$\Rightarrow \text{Area of a square} = \dfrac{1}{2} \times (d)^2 \\[1em] \Rightarrow 200 = \dfrac{1}{2} \times (d)^2 \\[1em] \Rightarrow d^2 = 400 \\[1em] \Rightarrow d = \sqrt{400} \\[1em] \Rightarrow d = 20 \text{ cm}.$

Hence, length of diagonal = 20 cm.

The area of a square field is 8 hectare. How long would a man take to cross it diagonally by walking at the rate of 4 kmph?

Answer

Given,

Area of square field = 8 hectares

Speed of man = 4 kmph

We know,

1 hectare = 10,000 m²

8 hectares = 8 × 10,000 = 80,000 m²

$\Rightarrow \text{Area of a square} = \text{ (side)}^2 \\[1em] \Rightarrow \text{(side)}^2 = 80,000 \\[1em] \Rightarrow \text{ side} = \sqrt{80000} \\[1em] \Rightarrow \text{ side} = \sqrt{8 × 10000} \\[1em] \Rightarrow \text{ side} = \sqrt{8} \times \sqrt{10000} \\[1em] \Rightarrow \text{ side} = 2 \sqrt{2} \times 100 \\[1em] \Rightarrow \text{ side} = 200 \sqrt{2} \text{ m}.$

Diagonal of a square = side × $\sqrt{2}$

= 200 $\sqrt{2} \times \sqrt{2}$

= 200 × 2 = 400 m = $\dfrac{400}{1000}$ = 0.4 km

We know that,

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{0.4}{4} = \dfrac{1}{10}$ hr.

1 hr = 60 min

∴ $\dfrac{1}{10} \text{ hr} = \dfrac{1}{10}$ × 60 min = 6 minutes.

Hence, a man takes 6 min to cross it diagonally.

Find the area and perimeter of a square plot of land whose diagonal is 15 m. Give your answer correct to two decimal places.

Answer

Given:

Diagonal of the square = 15 m

Let 'a' be the length of side of the square.

Using the Pythagoras theorem in square,

⇒ Diagonal² = side² + side²

⇒ 15² = 2 × side²

⇒ 225 = 2 × a²

⇒ a² = $\dfrac{225}{2}$

⇒ a² = 112.5

⇒ a = $\sqrt{112.5}$ = 10.606 m.

Area of square = side² = a²

= 112.50 m².

Perimeter = 4 × side

= 4 × 10.606 = 42.42 ≈ 42.42 m.

Hence, area = 112.50 m² and perimeter = 42.42 m.

The area of a parallelogram is 338 m². If its altitude is twice the corresponding base, determine the base and the altitude.

Answer

ABCD is a parallelogram given below:

Given,

Area of parallelogram = 338 m²

Let base BC = x

∴ Altitude (height) = AE = 2x

Area of parallelogram = Base × Height

⇒ 338 = x × 2x

⇒ 2x² = 338

⇒ x² = $\dfrac{338}{2}$

⇒ x² = 169

⇒ x = $\sqrt{169}$ = 13 m

∴ Base (BC) = x = 13 m.

Altitude (AE) = 2x = 2 × 13 = 26 m.

Hence, base = 13 m and altitude = 26 m.

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

Answer

ABCD is a rhombus with diagonals BD and AC.

DC = 20 cm

BD = 24 cm

Diagonals of a rhombus bisect each other at 90°.

DE = $\dfrac{BD}{2} = \dfrac{24}{2}$ = 12 cm.

In a right triangle DEC,

Hypotenuse (DC) = 20 cm

DE = 12 cm.

By using pythagoras theorem for the right triangle DEC,

⇒ DC² = DE ² + EC²

⇒ 20² = 12² + EC²

⇒ EC² = 20² - 12²

⇒ EC² = 400 - 144

⇒ EC² = 256

⇒ EC = $\sqrt{256}$

⇒ EC = 16 cm.

AC = 2 × EC

= 2 × 16 = 32 cm.

Area of rhombus = $\dfrac{1}{2}$ × (product of diagonals)

= $\dfrac{1}{2}$ × 24 × 32

= 12 × 32

= 384 cm².

Hence, area of rhombus = 384 cm².

The two parallel sides of a trapezium are 58 m and 42 m long. The other two sides are equal, each being 17 m. Find its area.

Answer

ABCD is a trapezium.

AB = 42 m

CD = 58 m

AD = BC = 17 m

From A and B drop perpendiculars AE and BF respectively to DC.

From figure,

EF = AB = 42 m

Since, AD = BC (given) and AE = BF (perpendicular between same parallels)

Thus,

DE = FC = x (let)

From figure,

⇒ DE + FC + EF = DC

⇒ x + x + 42 = 58

⇒ 2x = 58 - 42

⇒ 2x = 16

⇒ x = 8 meters.

In △ AED,

Using pythagoras theorem,

⇒ AD² = AE² + ED²

⇒ 17² = AE² + 8²

⇒ AE² = 17² - 8²

⇒ AE² = 289 - 64

⇒ AE² = 225

⇒ AE = $\sqrt{225}$ = 15 m.

Height = AE = 15 m.

By formula,

Area of trapezium = $\dfrac{1}{2} \times \text{ (sum of // sides)} \times \text{ (distance between them)}$

$\text{Area of trapezium ABCD } = \dfrac{1}{2} \times (AB + CD) \times AE \\[1em] = \dfrac{1}{2} \times (42 + 58) \times 15 \\[1em] = \dfrac{1}{2} \times 100 \times 15 \\[1em] = 750 \text{ m}^2.$

Hence, area = 750 m².

The perimeter of a rhombus is 52 cm. If one of its diagonals is 24 cm long, find:

(i) the length of the other diagonal,

(ii) the area of the rhombus.

Answer

(i) Let ABCD be a rhombus.

Perimeter of the rhombus = 52 cm

One diagonal BD = 24 cm

Let a be the length of a side of the rhombus.

Perimeter of a rhombus = 4 x Side

⇒ 4 x a = 52

⇒ a = $\dfrac{52}{4}$

⇒ a = 13 cm.

BD = 24 cm

Since the diagonals of a rhombus bisect at 90°.

Then, OB = OD = $\dfrac{24}{2}$ = 12 cm.

Applying pythagoras theorem for △AOB, we get:

⇒ AB² = OA² + OB²

⇒ (13)² = OA² + (12)²

⇒ 169 = OA² + 144

⇒ OA² = 169 - 144

⇒ OA² = 25

⇒ OA = $\sqrt{25}$

⇒ OA = 5 cm.

⇒ AC = 2 x OA = 2 x 5 cm = 10 cm.

Hence, the length of the other diagonal = 10 cm.

(ii) By formula,

Area of rhombus = $\dfrac{1}{2}$ x product of diagonals

= $\dfrac{1}{2}$ x 24 x 10

= 12 x 10 = 120 cm².

Hence, area of the rhombus = 120 cm².

The area of a rhombus is 216 cm² and one of its diagonals measures 24 cm. Find:

(i) the length of the other diagonal,

(ii) the length of each of its sides,

(iii) its perimeter.

Answer

(i) Given:

Area of rhombus = 216 cm²

One diagonal = 24 cm

Let 'd' be the other diagonal of rhombus.

$\Rightarrow \text{Area of rhombus } = \dfrac{1}{2} \times \text{product of diagonals} \\[1em] \Rightarrow 216 = \dfrac{1}{2} \times 24 \times d \\[1em] \Rightarrow 216 = 12 \times d \\[1em] \Rightarrow d = \dfrac{216}{12} \\[1em] \Rightarrow d = 18 \text{ cm}.$

Hence the length of the other diagonal = 18 cm.

(ii) The rhombus is shown in the figure below:

Diagonal AC = 24 cm.

The diagonals of a rhombus bisect each other at right angle.

Then, OA = OC = $\dfrac{24}{2}$ = 12 cm

Diagonal, BD = 18 cm

Then, OB = OD = $\dfrac{18}{2}$ = 9 cm

Applying pythagoras theorem for △AOB, we get:

⇒ AB² = OA² + OB²

⇒ AB² = (12)² + (9)²

⇒ AB² = 144 + 81

⇒ AB² = 225

⇒ AB = $\sqrt{225}$

⇒ AB = 15 cm.

Hence, the length of the each of its side = 15 cm.

(iii) Perimeter of rhombus = 4 × side

= 4 × 15

= 60 cm.

Hence, perimeter of the rhombus = 60 cm.

Two adjacent sides of a parallelogram are 36 cm and 25 cm. If the distance between longer sides is 15 cm, find the distance between the shorter sides.

Answer

Let ABCD be a parallelogram with side AB = 36 cm and side BC = 25 cm.

Distance between longer side DM = 15 cm

Consider DN as the distance between the shorter sides.

Area of parallelogram ABCD = base × height

= AB × DM = 36 × 15 = 540 cm².

Considering base BC and height DN.

Area of parallelogram ABCD = BC × DN

⇒ 540 = 25 × DN

⇒ DN = $\dfrac{540}{25}$ = 21.6 cm.

Hence, the distance between shorter sides = 21.6 cm.

In the given figure, ABCD is a trapezium in which AD = 13 cm, BC = 5 cm, CD = 17 cm and ∠A = ∠B = 90°. Calculate :

(i) AB

(ii) Area of trap. ABCD

Answer

(i) Draw CE ⊥ AD.

∴ DE = 13 - 5 = 8 cm.

In triangle DEC,

By pythagoras theorem,

⇒ CD² = EC² + DE²

⇒ 17² = EC² + 8²

⇒ EC² = 17² - 8²

⇒ EC² = 289 - 64

⇒ EC² = 225

⇒ EC = $\sqrt{225}$

⇒ EC = 15 cm

From figure,

AB = EC = 15 cm.

Hence, AB = 15 cm.

(ii) Area of trapezium = $\dfrac{1}{2}$ × (sum of parallel sides) × distance between them

= $\dfrac{1}{2}$ × (AD + BC) × EC

= $\dfrac{1}{2}$ × (13 + 5) × 15

= $\dfrac{1}{2}$ × 18 × 15

= 9 × 15 = 135 cm².

Hence, area of trapezium = 135 cm².

The adjoining figure shows a field with the measurement given in metres. Find the area of the field.

Answer

Calculating the area of triangle DXC,

$\text {Area of △DXC} = \dfrac{1}{2} \times CX \times DX \\[1em] = \dfrac{1}{2} \times 30 \times 12 \\[1em] = 15 \times 12 \\[1em] = 180 \text{ m}^2.$

Calculating the area of trapezium CXZB,

$\text {Area of trapezium CXZB} = \dfrac{1}{2} \times \text{(sum of parallel lines)} \times \text{distance between them} \\[1em] = \dfrac{1}{2} \times (CX + BZ) \times XZ \\[1em] = \dfrac{1}{2} \times (30 + 25) \times 15 \\[1em] = \dfrac{1}{2} \times 55 \times 15 \\[1em] = 27.5 \times 15 \\[1em] = 412.5 \text{ m}^2.$

Calculating the area of triangle AZB,

$\text {Area of △ AZB} = \dfrac{1}{2} \times \text{ base } \times \text{ height } \\[1em] = \dfrac{1}{2} \times BZ \times AZ \\[1em] = \dfrac{1}{2} \times 25 \times 10 \\[1em] = 25 \times 5 \\[1em] = 125 \text{ m}^2.$

From figure,

AD = 12 + 15 + 10 = 37 m.

Calculating the area of triangle AED,

$\text {Area of △AED} = \dfrac{1}{2} \times \text{ base } \times \text{ height } \\[1em] = \dfrac{1}{2} \times AD \times EY \\[1em] = \dfrac{1}{2} \times 37 \times 20 \\[1em] = 37 \times 10 \\[1em] = 370 \text{ m}^2.$

Total area = 180 + 412.5 + 125 + 370 = 1087.5 m².

Hence, area of the figure = 1087.5 m².