Perimeter & Area of Plane Figures

Given,

Base = 15 cm

Corresponding height = 9.6 cm

We know that

Area of triangle = $\frac{1}{2}$ × Base × Corresponding height

= $\frac{1}{2}$ × 15 × 9.6

= 15 × 4.8

= 72 cm².

Hence, area of triangle = 72 cm².

Let a = 13 cm, b = 14 cm and c = 15 cm.

Then,

s = $\frac{1}{2}(a + b + c) = \frac{1}{2}(13 + 14 + 15) = \frac{42}{2}$ = 21 cm.

⇒ (s - a) = (21 - 13) cm = 8 cm.

⇒ (s - b) = (21 - 14) cm = 7 cm.

⇒ (s - c) = (21 - 15) cm = 6 cm.

We know that,

$$\begin{aligned}\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] \Rightarrow \sqrt{21 × 8 × 7 × 6} \\[1em] \Rightarrow \sqrt{7056} \\[1em] \Rightarrow 84 \text{ cm}^2.\end{aligned}$$

Length of the longest side = 15 cm.

Let the corresponding height be x cm. Then,

$$\begin{aligned}\Rightarrow Area = \frac{1}{2} \times \text{ Base } \times \text{ Corresponding height } \\[1em] \Rightarrow 84 = \frac{1}{2} \times 15 \times x \\[1em] \Rightarrow x = \frac{168}{15} \\[1em] \Rightarrow 11.2 \text{ cm}. \\[1em]\end{aligned}$$

Hence, area of triangle = 84 cm² and the height of the triangle = 11.2 cm.

Let a = 30 cm, b = 24 cm, c = 18 cm.

Then,

s = $\frac{1}{2}(a + b + c) = \frac{1}{2}(30 + 24 + 18) = \frac{72}{2}$ = 36 cm.

⇒ (s - a) = (36 - 30) cm = 6 cm.

⇒ (s - b) = (36 - 24) cm = 12 cm.

⇒ (s - c) = (36 - 18) cm = 18 cm.

We know that,

$$\begin{aligned}\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] \Rightarrow \sqrt{36 × 6 × 12 × 18} \\[1em] \Rightarrow \sqrt{46656} \\[1em] \Rightarrow 216 \text{ cm}^2. \\[1em]\end{aligned}$$

Length of the smallest side = 18 cm.

Let the length of altitude be x cm. Then,

$$\begin{aligned}\Rightarrow Area = \frac{1}{2} \times \text{ Base } \times \text{ Altitude } \\[1em] \Rightarrow 216 = \frac{1}{2} \times 18 \times x \\[1em] \Rightarrow 216 = 9x \\[1em] \Rightarrow x = \frac{216}{9} \\[1em] \Rightarrow 24 \text{ cm}. \\[1em]\end{aligned}$$

Hence, area of triangle = 216 cm² and altitude = 24 cm.

It is given that the lengths of the sides of a triangle are in the ratio 3 : 4 : 5.

Let the lengths of the sides be 3x, 4x and 5x.

The perimeter of the triangle is 144 cm.

Perimeter = sum of all sides of triangle

⇒ 144 = 3x + 4x + 5x

⇒ 144 = 12x

⇒ x = $\frac{144}{12}$

⇒ x = 12.

So the sides of a triangle are

⇒ 3x = 3 × 12 = 36 cm

⇒ 4x = 4 × 12 = 48 cm

⇒ 5x = 5 × 12 = 60 cm

Let a = 36 cm, b = 48 cm, c = 60 cm.

s = $\frac{1}{2}(a + b + c) = \frac{1}{2}(36 + 48 + 60) = \frac{144}{2}$ = 72 cm.

(s - a) = (72 - 36) cm = 36 cm.

(s - b) = (72 - 48) cm = 24 cm.

(s - c) = (72 - 60) cm = 12 cm.

We know that,

$$\begin{aligned}\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] \Rightarrow \sqrt{72 × 36 × 24 × 12} \\[1em] \Rightarrow \sqrt{746496} \\[1em] \Rightarrow 864 \text{ cm}^2. \\[1em]\end{aligned}$$

Hence, area of triangle = 864 cm².

It is given that the sides of a triangular field are in the ratio 25 : 17 : 12.

Let the lengths of the sides be 25x, 17x and 12x.

Given,

The perimeter of a triangular field is 540 m.

Perimeter = sum of all sides of triangle

⇒ 540 = 25x + 17x + 12x

⇒ 540 = 54x

⇒ x = $\frac{540}{54}$

⇒ x = 10.

So the sides of the triangle are

⇒ 25x = 25 × 10 = 250 m

⇒ 17x = 17 × 10 = 170 m

⇒ 12x = 12 × 10 = 120 m

Let a = 250 m, b = 170 m, c = 120 m.

s = $\frac{1}{2}(a + b + c) = \frac{1}{2}(250 + 170 + 120) = \frac{540}{2}$ = 270 m.

(s - a) = (270 - 250) m = 20 m.

(s - b) = (270 - 170) m = 100 m.

(s - c) = (270 - 120) m = 150 m.

We know that,

$$\begin{aligned}\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] \Rightarrow \sqrt{270 × 20 × 100 × 150} \\[1em] \Rightarrow \sqrt{81000000} \\[1em] \Rightarrow 9000 \text{ m}^2. \\[1em]\end{aligned}$$

Rate = ₹ 24.60 per 100 m²

= ₹ $\frac{24.60}{100}$ per m².

Cost = Area of triangle × Rate

= 9000 × $\frac{24.60}{100}$

= 24.60 × 90 = ₹ 2,214.

Hence, area of field = 9000 m² & cost of cultivating the field = ₹ 2,214.

Given,

Rate of cultivation = ₹14.50 per 100 m²

= ₹ $\frac{14.50}{100}$ per m².

Total cost = ₹52,200

$$\begin{aligned}\Rightarrow \text{Total cost} = \text{Area } \times \text{ Rate of cultivation} \\[1em] \Rightarrow 52,200 = \text{Area } \times \frac{14.50}{100} \\[1em] \Rightarrow Area = \frac{52200 × 100}{14.50} \\[1em] \Rightarrow Area = \frac{5220000}{14.50} \\[1em] \Rightarrow Area = 360000 \text{ m}^2. \\[1em]\end{aligned}$$

Let altitude be x meters and base be 2x meters.

We know that,

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

⇒ 360000 = $\frac{1}{2}$ × 2x × x

⇒ x² = 360000

⇒ x = 600.

∴ Altitude = x = 600 m and Base = 2x = 2 × 600 = 1200 m.

Hence, altitude = 600 m and base = 1200 m.

Let △ABC be the right triangle.

We know that,

Perimeter of a right-angled triangle = 60 cm

Hypotenuse = 25 cm

So, the sum of other two sides of triangle = 60 – 25 = 35 cm

Let, base (BC) = x cm

So, AB = (35 - x) cm

Using the Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ 25² = (35 - x)² + x²

⇒ 625 = 1225 + x² - 70x + x²

⇒ 2x² - 70x + 600 = 0

Dividing by 2 on both sides,

⇒ x² - 35x + 300 = 0

⇒ x² - 15x - 20x + 300 = 0

⇒ x(x – 15) - 20(x - 15) = 0

⇒ (x - 15)(x - 20) = 0

⇒ x - 15 = 0 or x - 20 = 0

⇒ x = 15 or x = 20.

If x = 15, then 35 - x = 35 - 15 = 20 cm.

If x = 20, then 35 - x = 35 - 20 = 15 cm.

So, length of other two sides apart from hypotenuse are 15 cm and 20 cm.

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

Substituting the values we get,

A = $\frac{1}{2}$ × 15 × 20 = 150 cm².

Hence, area of triangle = 150 cm².

Let ABC be an isosceles right-angled triangle.

Area = 200 cm²

Let, side AB = BC = x cm and hypotenuse AC = h cm

We know that,

$$\begin{aligned}\Rightarrow \text{Area of triangle } = \frac{1}{2} \times base \times height \\[1em] \Rightarrow \text{Area of triangle } = \frac{1}{2} \times BC \times AB \\[1em] \Rightarrow \text{Area of triangle } = \frac{1}{2} \times x \times x \\[1em] \Rightarrow 200 = \frac{1}{2} \times x^2 \\[1em] \Rightarrow x^2 = 400 \\[1em] \Rightarrow x = \sqrt{400} \\[1em] \Rightarrow x = 20 \text{ cm}.\end{aligned}$$

Now, using the Pythagoras Theorem for the △ABC

⇒ AC² = AB² + BC²

⇒ AC² = (20)² + (20)²

⇒ AC² = 400 + 400

⇒ AC² = 800

⇒ AC = $\sqrt{800}$

⇒ AC = $\sqrt{400 × 2}$

⇒ AC = $20\sqrt{2}$

⇒ AC = 20 × 1.414

⇒ AC = 28.28 cm.

Hence, length of hypotenuse = 28.28 cm.

Given:

Perimeter = 60 cm

Let the length of each side of an equilateral triangle be 'a' cm.

Perimeter = Sum of all sides

⇒ 60 = a + a + a

⇒ 60 = 3a

⇒ a = $\frac{60}{3}$ = 20 cm.

$$\begin{aligned}\Rightarrow \text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} \times \text{ side}^2 \\[1em] = \frac{\sqrt{3}}{4} \times 20^2 \\[1em] = \frac{\sqrt{3}}{4} \times 400 \\[1em] = 100 \sqrt{3} \\[1em] = 173.2 \text{ cm}^2.\end{aligned}$$

By formula,

$$\begin{aligned}\text{Height of an equilateral triangle} = \frac{\sqrt{3}}{2} \times \text{ side} \\[1em] = \frac{\sqrt{3}}{2} \times 20 \\[1em] = 10 \sqrt{3} \\[1em] = 17.32 \text{ cm}.\end{aligned}$$

Hence, the area of triangle = 173.2 cm² and the height = 17.32 cm.

Given,

Height (h) = 12 cm

Let the length of the side of an equilateral triangle be a cm.

We know that for an equilateral triangle,

$$\begin{aligned}\Rightarrow \text{Height} = \frac{\sqrt{3}}{2}a \\[1em] \Rightarrow 12 = \frac{\sqrt{3}}{2}a \\[1em] \Rightarrow a = \frac{12 \times 2}{\sqrt{3}} = \frac{24}{\sqrt{3}} \\[1em] \Rightarrow a = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ cm}\end{aligned}$$

Perimeter of an equilateral triangle = 3 × side

= 3 × 8 $\sqrt{3}$

= 3 × 8 × 1.732

= 41.568 ≈ 41.57 cm.

$$\begin{aligned}\Rightarrow \text{Area of an equilateral triangle} = \frac{\sqrt{3}}{4} \times (side)^2 \\[1em] \Rightarrow \frac{\sqrt{3}}{4} \times (8 \sqrt{3})^2 \\[1em] \Rightarrow \frac{\sqrt{3}}{4} \times 64 \times 3 \\[1em] \Rightarrow \sqrt{3} \times 48 \\[1em] \Rightarrow 1.732 \times 48 \\[1em] \Rightarrow 83.136 \approx 83.14 \text{ cm}^2. \\[1em]\end{aligned}$$

Hence, the perimeter = 41.57 cm and area = 83.14 cm².

ABC is a right angled triangle with a right angle at B.

The area of the triangle is 24 cm².

Let the lengths of BC and AB be x and y, respectively.

Given,

The difference between the two perpendicular sides is 2 cm.

x - y = 2

∴ y = x - 2

Area = $\frac{1}{2}$ × base × height

⇒ 24 = $\frac{1}{2}$ × BC × AB

⇒ 24 = $\frac{1}{2}$ × x × (x - 2)

⇒ x × (x - 2) = 48

⇒ x² - 2x = 48

⇒ x² - 2x - 48 = 0

⇒ x² - 8x + 6x - 48 = 0

⇒ x(x - 8) + 6(x - 8) = 0

⇒ (x - 8)(x + 6) = 0

⇒ (x - 8) = 0 or (x + 6) = 0

⇒ x = 8 or x = -6

Since length cannot be negative, ∴ x = 8 cm.

y = x - 2 = 8 - 2 = 6 cm.

Thus, AB = 6 cm and BC = 8 cm.

By using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AC² = BC² + AB²

⇒ AC² = 8² + 6²

⇒ AC² = 64 + 36

⇒ AC² = 100

⇒ AC = $\sqrt{100}$

⇒ AC = 10 cm.

Perimeter of a triangle = Sum of all the sides of a triangle

= AB + BC + AC

= 10 + 8 + 6

= 24 cm.

Hence, perimeter of a triangle = 24 cm.

Let ABC be a right-angled triangle,

AB = 5x cm and BC = (3x – 1) cm

We know that,

Area of △ ABC = $\frac{1}{2}$ × base × height

Substituting the values we get,

⇒ 60 = $\frac{1}{2}$ × (3x - 1) × 5x

⇒ 120 = 5x(3x – 1)

⇒ 120 = 15x² - 5x

⇒ 15x² - 5x - 120 = 0

⇒ 5(3x² - x - 24) = 0

⇒ 3x² - x - 24 = 0

⇒ 3x² – 9x + 8x – 24 = 0

⇒ 3x(x – 3) + 8(x - 3) = 0

⇒ (3x + 8)(x - 3) = 0

⇒ 3x + 8 = 0 or x - 3 = 0

⇒ 3x = -8 or x = 3

⇒ x = $\frac{-8}{3}$ or x = 3.

Since, x cannot be negative. So, x = 3.

⇒ AB = 5 × 3 = 15 cm

⇒ BC = (3 × 3 – 1) = 9 – 1 = 8 cm

In right angled △ABC,

Using Pythagoras theorem,

AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64

⇒ AC² = 289

⇒ AC = $\sqrt{289}$

⇒ AC = 17 cm.

Perimeter of a triangle = Sum of all the sides of a triangle

= AB + BC + AC

= 15 + 8 + 17

= 40 cm.

Hence, perimeter of the triangle = 40 cm.

Let ABC be an isosceles triangle in which AB = AC, AD ⊥ BC and BC is the base.

Given,

Each of the equal sides of the isosceles triangle is 2 cm more than its height.

Let the height of the triangle be h cm.

Equal sides: AB = AC = h + 2

Base: BC = 12 cm

In Δ ABD and Δ ACD,

AD = AD [Common Side]

∠ADB = ∠ADC [Both equal to 90°]

AB = AC [Δ ABC is an isosceles triangle]

∴ Δ ABD ≅ Δ ACD [By R.H.S. axiom]

∴ BD = CD [C.P.C.T.C.]

∴ BD = CD = $\frac{BC}{2} = \frac{12}{2}$ = 6 cm.

By using the Pythagoras theorem in Δ ABD,

⇒ BD² + AD² = AB²

⇒ 6² + h² = (h + 2)²

⇒ 36 + h² = h² + 4 + 2 × h × 2

⇒ 36 = 4 + 4h

⇒ 4h = 32

⇒ h = $\frac{32}{4}$

⇒ h = 8 cm.

$$\begin{aligned}\Rightarrow \text{Area of triangle ABC} = \frac{1}{2} \times \text{ base } \times \text{ height } \\[1em] = \frac{1}{2} \times BC \times AD \\[1em] = \frac{1}{2} \times 12 \times 8 \\[1em] = 6 \times 8 \\[1em] = 48 \text{ cm}^2.\end{aligned}$$

Hence, area of the triangle = 48 cm².

Each of equal sides (a) = 13 cm and base (b) = 24 cm

By formula,

$$\begin{aligned}\Rightarrow \text{Area of an isosceles triangle} = \frac{1}{4}b\sqrt{4a^2 - b^2} \\[1em] \Rightarrow \frac{1}{4} \times 24 \times \sqrt{4 × 13^2 - 24^2} \\[1em] \Rightarrow 6 \times \sqrt{676 - 576} \\[1em] \Rightarrow 6 \times \sqrt{100} \\[1em] \Rightarrow 6 \times 10 \\[1em] \Rightarrow 60 \text{ cm}^2.\end{aligned}$$

Hence, area of the triangle = 60 cm².

Given,

Base (b) = 18 cm

Area = 108 cm².

Let the length of equal sides of the isosceles triangle be a cm.

By formula,

$$\begin{aligned}\Rightarrow \text{Area of an isosceles triangle} = \frac{1}{4}b\sqrt{4a^2 - b^2} \\[1em] \Rightarrow 108 = \frac{1}{4} \times 18 \times \sqrt{4a^2 - 18^2} \\[1em] \Rightarrow 432 = 18 \times \sqrt{4a^2 - 324} \\[1em] \Rightarrow \sqrt{4a^2 - 324} = \frac{432}{18} \\[1em] \Rightarrow \sqrt{4a^2 - 324} = 24 \\[1em] \Rightarrow 4a^2 - 324 = 24^2 \\[1em] \Rightarrow 4a^2 - 324 = 576 \\[1em] \Rightarrow 4a^2 = 576 + 324 \\[1em] \Rightarrow 4a^2 = 900 \\[1em] \Rightarrow a^2 = \frac{900}{4} \\[1em] \Rightarrow a^2 = 225 \\[1em] \Rightarrow a = \sqrt{225} \\[1em] \Rightarrow a = 15 \text{ cm}.\end{aligned}$$

Perimeter = Sum of all sides of triangle

= 15 + 15 + 18

= 48 cm.

Hence, perimeter of the isosceles triangle = 48 cm.

Given,

△ABC is an equilateral triangle.

Each side = 10 cm

By formula,

$$\begin{aligned}\Rightarrow \text{Area of equilateral △ABC} = \frac{\sqrt{3}}{4} \times \text{ (side)}^2 \\[1em] \Rightarrow \frac{\sqrt{3}}{4} \times 10^2 \\[1em] \Rightarrow \frac{\sqrt{3}}{4} \times 100 \\[1em] \Rightarrow 25\sqrt{3} \\[1em] \Rightarrow 43.3 \text{ cm}^2.\end{aligned}$$

Given,

PB = 8 cm and BC = 10 cm

By using the Pythagoras theorem in △PBC,

⇒ BC² = PB² + PC²

⇒ 10² = 8² + PC²

⇒ 100 = 64 + PC²

⇒ PC² = 36

⇒ PC = $\sqrt{36}$

⇒ PC = 6 cm.

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

Area of △PBC = $\frac{1}{2}$ × PB × PC

= $\frac{1}{2}$ × 8 × 6

= 4 × 6 = 24 cm².

Shaded region = Area of △ABC − Area of △PBC

= 43.3 - 24

= 19.3 cm².

Hence, area of shaded region = 19.3 cm².

Let the length of the side of an equilateral triangle be a cm.

$$\begin{aligned}\Rightarrow \text{Area of triangle} = \frac{\sqrt{3}}{4} \times a^2 \\[1em] \Rightarrow 81\sqrt{3} = \frac{\sqrt{3}}{4} \times a^2 \\[1em] \Rightarrow a^2 = 81 \times 4 \\[1em] \Rightarrow a^2 = 324 \\[1em] \Rightarrow a = \sqrt{324} \\[1em] \Rightarrow a = 18. \\[1em]\end{aligned}$$

Perimeter = Sum of all sides of a triangle

= 3a

= 3 × 18

= 54 cm.

Hence, perimeter of the triangle = 54 cm.

Given,

Hypotenuse = 25 cm

Base = 24 cm

By using the Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ 25² = 24² + Height²

⇒ 625 = 576 + Height²

⇒ Height² = 625 - 576

⇒ Height² = 49

⇒ Height = $\sqrt{49}$

⇒ Height = 7 cm.

Area = $\frac{1}{2}$ × Base × Height

= $\frac{1}{2}$ × 24 × 7

= 12 × 7 = 84 cm².

Hence, area of the triangle = 84 cm².

Let ABC be an isosceles triangle with AB = AC = a cm and BC = b cm.

Altitude (AD) = 8 cm

Perimeter = 32 cm

Perimeter = sum of all sides of a triangle

⇒ 32 = a + a + b

⇒ 32 = 2a + b

⇒ b = 32 - 2a .........(1)

In an isosceles triangle, the altitude drawn from the common vertex bisects the base.

Thus, AD bisects BC.

So, BD = DC = $\frac{b}{2}$

∴ ∠ADC = ∠ADB = 90°.

In triangle ADB,

By pythagorean theorem,

$$\begin{aligned}\Rightarrow AB^2 = AD^2 + DB^2 \\[1em] \Rightarrow a^2 = 8^2 + \Big(\frac{b}{2}\Big)^2 \\[1em] \Rightarrow a^2 = 64 + \frac{b^2}{4} \\[1em] \Rightarrow a^2 = \frac{256 + b^2}{4} \\[1em] \Rightarrow 4a^2 = 256 + b^2\end{aligned}$$

Substituting the value of b from equation (1) in above equation, we get :

$$\begin{aligned}\Rightarrow 4a^2 = 256 + (32 - 2a)^2 \\[1em] \Rightarrow 4a^2 = 256 + 1024 -128a + 4a^2 \\[1em] \Rightarrow 0 = 1280 - 128a \\[1em] \Rightarrow 128a = 1280 \\[1em] \Rightarrow a = \frac{1280}{128} \\[1em] \Rightarrow a = 10 \text{ cm}.\end{aligned}$$

∴ b = 32 - 2(10)

⇒ b = 32 - 20

⇒ b = 12 cm.

Area of triangle ABC = $\frac{1}{2}$ × Base × Height

= $\frac{1}{2}$ × BC × AD

= $\frac{1}{2}$ × 12 × 8

= 6 × 8 = 48 cm².

Hence, area of triangle = 48 cm².

Given,

Area = 216 cm²

Sides = 3 : 4 : 5

Let the sides of a triangle be 3x, 4x and 5x.

Since 3 : 4 : 5 is a pythagorean triplet (3² + 4² = 5²)

Thus, the triangle is a right angled triangle, and sides containing right angle are 3x cm and 4x cm.

$$\begin{aligned}\Rightarrow \text{Area of right angled triangle} = \frac{1}{2} \times \text{ (product of sides containg right angle)} \\[1em] \Rightarrow 216 = \frac{1}{2} \times 3x \times 4x \\[1em] \Rightarrow 432 = 12x^2 \\[1em] \Rightarrow x^2 = 36 \\[1em] \Rightarrow x = \sqrt{36} \\[1em] \Rightarrow x = 6.\end{aligned}$$

∴ Sides of a triangle are

⇒ 3x = 3 × 6 = 18 cm

⇒ 4x = 4 × 6 = 24 cm

⇒ 5x = 5 × 6 = 30 cm

Perimeter = Sum of all sides of a triangle

= 18 + 24 + 30

= 72 cm.

Hence, perimeter of the triangle = 72 cm.

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 = $\frac{1}{2}$ × base × height

For △ABD,

Area = $\frac{1}{2}$ × BD × AM

= $\frac{1}{2}$ × 25 × 16.4

= 25 × 8.2

= 205 cm².

For △CBD,

Area = $\frac{1}{2}$ × BD × CN

= $\frac{1}{2}$ × 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.

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 = $\frac{1}{2}$ × d₁ × d₂

where d₁ and d₂ are diagonals.

Area = $\frac{1}{2}$ × 14 × 19

= 7 × 19

= 133 cm².

Hence, area = 133 cm².

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 = $\frac{29 + 21 + 20}{2}$

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

By formula,

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

Area of △ACD

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

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

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

By formula,

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

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

= 210 + 336

= 546 cm².

Hence, area of quadrilateral = 546 cm².

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 = $\frac{1}{2}$ × Base × Height

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

= $\frac{1}{2}$ × 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 = $\frac{1}{2}$ × BC × BD

= $\frac{1}{2}$ × 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.

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 = $\frac{1}{2}$ × Base × Height

= $\frac{1}{2}$ × AB × AD

= $\frac{1}{2}$ × 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

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

By formula,

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

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

= 384 + 960

= 1344 cm².

Hence, area of quadrilateral ABCD = 1344 cm².

Given

△BCD is equilateral.

Each side = 26 cm

∠BAD = 90°

AD = 24 cm

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

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 = $\frac{1}{2}$ × Base × Height

= $\frac{1}{2}$ × AB × AD

= $\frac{1}{2}$ × 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.

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.

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

From figure,

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

= 30 - 13.5

= 16.5 cm².

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

Hence, DL = 5.5 cm.

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 = $\frac{81}{2}$

⇒ 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.

Let length = l and breadth = b

Given,

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

l = 2b

By formula,

Perimeter = 2(l + b)

⇒ $\frac{3000}{5}$ = 2(2b + b)

⇒ $\frac{1500}{5}$ = 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.

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.

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 = $\frac{6}{4} = \frac{3}{2}$ = 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.

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

Given,

Area of rectangle = 144 cm²

Length = x cm

Perimeter = 52 cm

Area = length × breadth

⇒ 144 = x × breadth

⇒ Breadth = $\frac{144}{x}$ cm.

Perimeter = 2(l + b)

⇒ 52 = 2(x + $\frac{144}{x}$)

⇒ x + $\frac{144}{x}$ = 26

⇒ $\frac{x^2 + 144}{x} = 26$

⇒ 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 = $\frac{144}{x} = \frac{144}{18}$ = 8 cm.

If x = 8 then,

Length = x = 8 cm.

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

Generally we consider length > breadth

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

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.

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 = $\frac{60}{3}$ = 20 cm.

⇒ l = 2b - 10

= 2(20) - 10

= 40 - 10 = 30 cm.

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

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

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

Total cost = Carpet length × Rate per meter

= 156 × ₹ 12.50

= ₹ 1,950.

Hence, total cost = ₹ 1,950.

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

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

Given,

Rate of cultivation = ₹ 160 per hectare

Total cost = ₹ 1,440

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

= $\frac{1440}{160}$ = 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.

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

Substituting values we get :

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

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.

ABCD is a square with diagonal BD = d cm.

We know that,

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

Hence, length of diagonal = 20 cm.

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

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

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

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

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

We know that,

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

1 hr = 60 min

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

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

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 = $\frac{225}{2}$

⇒ 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.

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 = $\frac{338}{2}$

⇒ 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.

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 = $\frac{BD}{2} = \frac{24}{2}$ = 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 = $\frac{1}{2}$ × (product of diagonals)

= $\frac{1}{2}$ × 24 × 32

= 12 × 32

= 384 cm².

Hence, area of rhombus = 384 cm².

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

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

Hence, area = 750 m².

(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 = $\frac{52}{4}$

⇒ a = 13 cm.

BD = 24 cm

Since the diagonals of a rhombus bisect at 90°.

Then, OB = OD = $\frac{24}{2}$ = 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 = $\frac{1}{2}$ x product of diagonals

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

= 12 x 10 = 120 cm².

Hence, area of the rhombus = 120 cm².

(i) Given:

Area of rhombus = 216 cm²

One diagonal = 24 cm

Let 'd' be the other diagonal of rhombus.

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

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 = $\frac{24}{2}$ = 12 cm

Diagonal, BD = 18 cm

Then, OB = OD = $\frac{18}{2}$ = 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.

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 = $\frac{540}{25}$ = 21.6 cm.

Hence, the distance between shorter sides = 21.6 cm.

(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 = $\frac{1}{2}$ × (sum of parallel sides) × distance between them

= $\frac{1}{2}$ × (AD + BC) × EC

= $\frac{1}{2}$ × (13 + 5) × 15

= $\frac{1}{2}$ × 18 × 15

= 9 × 15 = 135 cm².

Hence, area of trapezium = 135 cm².

Calculating the area of triangle DXC,

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

Calculating the area of trapezium CXZB,

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

Calculating the area of triangle AZB,

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

From figure,

AD = 12 + 15 + 10 = 37 m.

Calculating the area of triangle AED,

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

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

Hence, area of the figure = 1087.5 m².

Let each side of equilateral triangle be a cm.

Area of equilateral triangle = $\frac{\sqrt{3}}{4}$ a²

⇒ $16\sqrt{3} = \frac{\sqrt{3}}{4}$ a²

⇒ 64 = a²

⇒ a = $\sqrt{64}$

⇒ a = 8 cm.

Hence, option 3 is the correct option.

ABC is an isosceles triangle with AB = AC and base (BC) = 8 cm.

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

Area of triangle ABC = $\frac{1}{2}$ × BC × AD

⇒ 12 = $\frac{1}{2}$ × 8 × AD

⇒ AD = $\frac{24}{8}$ = 3 cm.

In an isosceles triangle, the perpendicular from the common vertex to the base, bisects the base.

∴ BD = DC = $\frac{BC}{2} = \frac{8}{2}$ = 4 cm.

In triangle ABD,

⇒ Hypotenuse² = Base² + Height²

⇒ AB² = BD² + AD²

⇒ AB² = 4² + 3²

⇒ AB² = 16 + 9

⇒ AB² = 25

⇒ AB = $\sqrt{25}$

⇒ AB = 5 cm.

∴ AB = AC = 5 cm.

Hence, option 4 is the correct option.

Given,

Length of each equal side (a) = 13 cm and base (b) = 24 cm.

By formula,

$$\begin{aligned}\Rightarrow \text{Area of isosceles triangle} = \frac{1}{4}b \sqrt{4a^2 - b^2} \\[1em] = \frac{24}{4} \sqrt{4(13)^2 - 24^2} \\[1em] = 6 \sqrt{4 × 169 - 576} \\[1em] = 6 \sqrt{676 - 576} \\[1em] = 6 \sqrt{100} \\[1em] = 6 × 10 \\[1em] = 60 \text{ cm}^2.\end{aligned}$$

Hence, option 1 is the correct option.

Area of //gm = Base × Height

⇒ 30 = 6 × Height

⇒ Height = $\frac{30}{6}$ = 5 m.

Hence, option 2 is the correct option.

Area of square = $\frac{1}{2}$ × (diagonals)²

= $\frac{1}{2}$ × (8)²

= $\frac{1}{2}$ × 64

= 32 cm².

Hence, option 2 is the correct option.

By formula,

Area of rhombus = $\frac{1}{2}$ × d₁ × d₂

⇒ 24 = $\frac{1}{2}$ × 6 × d₂

⇒ d₂ = $\frac{48}{6}$ = 8 cm.

Hence, option 2 is the correct option.

Let the length of other parallel side be a cm.

By formula,

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

⇒ 27.5 = $\frac{1}{2}$ × (7.5 + a) × 5

⇒ 55 = (7.5 + a) × 5

⇒ 7.5 + a = $\frac{55}{5}$

⇒ 7.5 + a = 11

⇒ a = 11 - 7.5 = 3.5 cm.

Hence, option 1 is the correct option.

Let a = 13 cm, b = 14 cm and c = 15 cm.

s = $\frac{a + b + c}{2}$

= $\frac{13 + 14 + 15}{2}$

= $\frac{42}{2}$ = 21.

By formula,

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

$$\begin{aligned}= \sqrt{21(21 - 13) \times (21 - 14) \times (21 - 15)} \\[1em] = \sqrt{21 \times (8) \times (7) \times (6)} \\[1em] = \sqrt{7056} \\[1em] = 84 \text{ cm}^2.\end{aligned}$$

Hence, option 2 is the correct option.

Side (a) = 4 cm.

Area of an equilateral triangle = $\frac{\sqrt{3}}{4}$ × a²

= $\frac{\sqrt{3}}{4}$ × (4)²

= $\frac{\sqrt{3}}{4}$ × 16

= $4\sqrt{3}$

= 4 × 1.732 = 6.928 cm².

Hence, option 1 is the correct option.

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

We know that,

Height of an equilateral triangle = $\frac{\sqrt{3}}{2}$ side

⇒ $12\sqrt{3} = \frac{\sqrt{3}}{2}a$

⇒ a = 24 cm.

$$\begin{aligned}\Rightarrow \text{Area of an equilateral triangle} = \frac{\sqrt{3}}{4} \times a^2 \\[1em] = \frac{\sqrt{3}}{4} \times (24)^2 \\[1em] = \frac{\sqrt{3}}{4} \times 24 \times 24 \\[1em] = \sqrt{3} \times 24 \times 6 \\[1em] = 144 \sqrt{3} \text{ cm}^2.\end{aligned}$$

Hence, option 2 is the correct option.

Longer side = 12 cm

Shorter side = 10 cm

Distance between longer sides (height corresponding to 12 cm) = 8 cm

Area of //gm = Base × Height

= 12 × 8 = 96 cm².

Distance between the shorter side be h cm.

Area of //gm = Base × Height

⇒ 96 = 10 × h

⇒ h = $\frac{96}{10}$ = 9.6 cm.

Hence, option 4 is the correct option.

Let b₁ and b₂ be the bases of the triangles.

Heights (h₁ : h₂) = 2 : 3

Areas (A₁ : A₂) = 3 : 2

Area of a triangle = $\frac{1}{2}$ × Base × Height

$$\begin{aligned}\Rightarrow \frac{A_1}{A_2} = \frac{\frac{1}{2} \times b_1 × h_1}{\frac{1}{2} \times b_2 × h_2} \\[1em] \Rightarrow \frac{3}{2} = \frac{b_1}{b_2} \times \frac{h_1}{h_2} \\[1em] \Rightarrow \frac{3}{2} = \frac{b_1}{b_2} \times \frac{2}{3} \\[1em] \Rightarrow \frac{b_1}{b_2} = \frac{3 × 3}{2 × 2} \\[1em] \Rightarrow \frac{b_1}{b_2} = \frac{9}{4} \\[1em] \Rightarrow b_1 : b_2 = 9 : 4.\end{aligned}$$

Hence, option 3 is the correct option.

By formula,

Perimeter of a rectangle = 2(l + b)

⇒ 26 = 2(3 + b + b)

⇒ 26 = 2(3 + 2b)

⇒ 3 + 2b = $\frac{26}{2}$

⇒ 3 + 2b = 13

⇒ 2b = 13 - 3

⇒ 2b = 10

⇒ b = $\frac{10}{2}$ = 5 m.

⇒ l = 3 + b = 3 + 5 = 8 m.

Area of rectangle = l × b

= 8 × 5 = 40 m².

Hence, option 3 is the correct option.

Area of rhombus = $\frac{1}{2}$ × d₁ × d₂

= $\frac{1}{2}$ × 6 × 4

= 3 × 4 = 12 cm².

Hence, option 2 is the correct option.

1. Let a = 122, b = 120, c = 22.

s = $\frac{a + b + c}{2} = \frac{122 + 120 + 22}{2}$

= $\frac{264}{2}$ = 132 m.

Hence, option (b) is the correct option.

2. By formula,

$$\begin{aligned}\text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{132(132 - 122)(132 - 120)(132 - 22)} \\[1em] = \sqrt{132(10)(12)(110)} \\[1em] = \sqrt{1742400} \\[1em] = 1320 \text{ m}^2.\end{aligned}$$

Hence, option (d) is the correct option.

3. Rate per year = ₹ 1,000 per m²

Cost = Area × Rate per m²

= 1320 × 1000 = ₹ 13,20,000 per year.

For 4 months, cost will be :

⇒ 13,20,000 × $\frac{4}{12}$

⇒ 13,20,000 × $\frac{1}{3}$

⇒ ₹4,40,000.

Hence, option (c) is the correct option.

4. Yearly rent = ₹13,20,000

Monthly rent = $\frac{13,20,000}{12}$ = ₹1,10,000

∴ No. of months = $\frac{\text{Amount paid}}{\text{Charge per month}} = \frac{5,50,000}{1,10,000}$ = 5.

Hence, option (a) is the correct option.

5. By formula,

For a triangle with sides x, y, z and semi perimeter p.

Area = $\sqrt{p(p - x)(p - y)(p - z)}$

Hence, option (d) is the correct option.

Diagonal length = $15\sqrt{2}$ m

For square :

Side = $\frac{\text{diagonal}}{\sqrt{2}}$

⇒ Side = $\frac{15\sqrt{2}}{\sqrt{2}}$

⇒ Side = 15 m.

Area = (side)²

= 15² = 225 m².

Cost of ploughing = ₹20 per m²

Total cost = Area × Cost per m²

= 225 × 20 = ₹4,500.

∴ Assertion (A) is true.

By formula,

$\text{Area of a Square} = \frac{(\text{Diagonal})^2}{2}$

∴ Reason (R) is false.

Assertion (A) is true, reason (R) is false.

Hence, option 1 is the correct option.

Let base and height of isosceles right triangle be a cm.

Area of triangle = $\frac{1}{2} \times \text{ base} \times \text{ height}$

⇒ 8 = $\frac{1}{2} \times a \times a$

⇒ 8 = $\frac{1}{2} \times a^2$

⇒ a² = 16

⇒ a = 4 cm.

By pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ Hypotenuse² = 4² + 4²

⇒ Hypotenuse² = 16 + 16

⇒ Hypotenuse² = 32

⇒ Hypotenuse = $\sqrt{32}$

⇒ Hypotenuse = $4\sqrt{2}$ cm.

∴ Assertion (A) is false.

Area of an isosceles triangle with sides a, a and b is given by $\frac{1}{4}b\sqrt{4a^2 - b^2}$.

∴ Reason (R) is false.

Hence, option 4 is the correct option.

By formula,

$$\begin{aligned}\Rightarrow \text{Area of rhombus} = \frac{1}{2} \times d_1 \times d_2 \\[1em] \Rightarrow 84 = \frac{1}{2} × 7 × d_2 \\[1em] \Rightarrow d_2 = \frac{168}{7} \\[1em] \Rightarrow d_2 = 24 \text{ cm}.\end{aligned}$$

Now in rhombus diagonals bisect each other at right angles.

Thus,

By pythagras theorem,

(Side)² = $\Big(\frac{d_1}{2}\Big)^2 + \Big(\frac{d_2}{2}\Big)^2$

(Side)² = $\Big(\frac{7}{2}\Big)^2 + \Big(\frac{24}{2}\Big)^2$

(Side)² = (3.5)² + (12)²

(Side)² = 12.25 + 144

(Side)² = 156.25

Side = $\sqrt{156.25}$

Side = 12.5 cm

∴ Assertion (A) is false.

By formula,

Area of rhombus = $\frac{1}{2}$ × product of its diagonals

∴ Reason (R) is true.

Assertion (A) is false, reason (R) is true.

Hence, option 2 is the correct option.

Given,

Area of rectangle = 3 × Area of square ........(1)

Length of rectangle = 40 cm

Let side of square be x cm.

Breadth of rectangle = $\frac{3}{2}$ × side of a square

= 1.5 × x = 1.5x

Area of square = side² = x² cm.

Area of rectangle = length × breadth

= 40 × 1.5x

= 60x.

Substituting values in equation (1), we get :

⇒ 60x = 3 × x²

⇒ 60 = 3x

⇒ x = $\frac{60}{3}$ = 20 cm.

Hence, option 3 is the correct option.

Let ABCD be a field in the shape of a || gm .

Calculating area of triangle BCD,

CD = a = 120 m, BC = b = 170 m, BD = c = 250 m

$$\begin{aligned}\text{Semi-perimeter (s)} = \frac{a + b + c}{2} \\[1em] = \frac{120 + 170 + 250}{2} \\[1em] = \frac{540}{2} \\[1em] = 270 \text{ m.}\end{aligned}$$

By formula,

$$\begin{aligned}\Rightarrow \text{Area of triangle} = \sqrt{s(s - a) (s - b) (s - c)} \\[1em] \Rightarrow \text{Area of triangle BCD} = \sqrt{270(270 - 120) (270 - 170) (270 - 250)} \\[1em] = \sqrt{270(150) (100) (20)} \\[1em] = \sqrt{81000000} \\[1em] = 9000 \text{ m}^2.\end{aligned}$$

We know that,

In a parallelogram, the diagonal divides the parallelogram into two congruent triangles.

Thus, area of triangle ABD = area of triangle BCD = 9000 m².

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

= 9000 + 9000 = 18000 m².

Given,

Cost of ploughing the field = ₹20 per sq m.

Total cost = Area × Cost per sq m = 18000 × ₹20 = ₹3,60,000.

Hence, option 4 is the correct option.

From figure,

ABC is an isosceles triangle with AM as the height and BC as the base.

Base (BC) = 16 cm

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

⇒ 48 = $\frac{1}{2}$ × 16 × AM

⇒ 48 = 8 × AM

⇒ AM = $\frac{48}{8}$ = 6 cm.

In an isosceles triangle,

The altitude from the common vertex bisects the base.

In triangle ABC,

The altitude AM bisects the base BC.

So, BM = MC = $\frac{BC}{2} = \frac{16}{2}$ = 8 cm.

Using Pythagoras theorem for the △AMB,

Hypotenuse² = Base² + Height²

⇒ AB² = BM² + AM²

⇒ AB² = 8² + 6²

⇒ AB² = 64 + 36

⇒ AB² = 100

⇒ AB = $\sqrt{100}$ = 10 cm.

∴ AC = AB = 10 cm.

Perimeter of triangle = Sum of all sides of the triangle

= AB + AC + BC

= 10 + 10 + 16 = 36 cm.

Hence, option 3 is the correct option.

Since, AB || DC.

∴ Triangles BEC and EDC lie between the same parallel lines.

Both triangles have the same height (distance between AB and CD).

There bases are also equal:

Base of △BEC = EB = 6 cm

Base of △EDC = DC = 6 cm

We know that,

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

Since both triangles (△BEC and △EDC) have same height and base.

So, there areas are also equal.

∴ Area of △BEC = Area of △EDC = 12 cm².

Hence, option 1 is the correct option.

We know that,

For sides with a, b, c and semi-perimeter 's', Heron's Formula for area of triangle is :

Area of triangle = $\sqrt{s(s - a) (s - b) (s - c)}$ $$\begin{aligned}\Rightarrow \sqrt{\Big (\frac{a + b + c}{2}\Big)\Big(\frac{a + b + c}{2} - a\Big)\Big(\frac{a + b + c}{2} - b\Big)\Big(\frac{a + b + c}{2} - c\Big)} \\[1em] \Rightarrow \sqrt{\Big (\frac{a + b + c}{2}\Big)\Big(\frac{a + b + c - 2a}{2}\Big)\Big(\frac{a + b + c - 2b}{2}\Big)\Big(\frac{a + b + c - 2c}{2}\Big)} \\[1em] \Rightarrow \sqrt{\Big (\frac{a + b + c}{2}\Big)\Big(\frac{b + c - a}{2}\Big)\Big(\frac{a + c - b}{2}\Big)\Big(\frac{a + b - c}{2}\Big)} \\[1em]\end{aligned}$$

But the given formula is :

$\sqrt{\Big (\frac{a + b + c}{2}\Big)\Big(\frac{a + b}{2}\Big)\Big(\frac{b + c}{2}\Big)\Big(\frac{a + c}{2}\Big)}$.

Hence, the student wrote the incorrect formula.

In triangle I,

Only 2 sides are given, but with the help of semi-perimeter from triangle I we can find 3rd side of the triangle, which is the common side for both the triangles.

In triangle II,

2 sides are already given and we got the 3rd side, which is common side for both triangles.

Since, the length of all the sides of the triangle are known. Thus, by using Heron's Formula we can find the area of triangle II.

Hence, yes we can find area of triangle II using Heron's Formula.

From figure,

Perimeter of △ABC = AB + BC + CA = 5 + 5 + 6 = 16 cm

Perimeter of △XYZ = XY + YZ + ZX = 5 + 5 + 8 = 18 cm.

By formula,

Area of an isosceles triangle = $\frac{1}{4}b\sqrt{4a^2 - b^2}$, where a is the length of equal sides and b is the length of base.

Area of △ABC = $\frac{1}{4} \times 6 \times \sqrt{4(5)^2 - 6^2}$

$$\begin{aligned}= \frac{1}{4} \times 6 \times \sqrt{100 - 36} \\[1em] = \frac{1}{4} \times 6 \times \sqrt{64} \\[1em] = \frac{1}{4} \times 6 \times 8 \\[1em] = 12 \text{ cm}^2.\end{aligned}$$

Area of △XYZ = $\frac{1}{4} \times 8 \times \sqrt{4(5)^2 - 8^2}$

$$\begin{aligned}= \frac{1}{4} \times 8 \times \sqrt{100 - 64} \\[1em] = \frac{1}{4} \times 8 \times \sqrt{36} \\[1em] = \frac{1}{4} \times 8 \times 6 \\[1em] = 12 \text{ cm}^2.\end{aligned}$$

Although, perimeter of △XYZ is more than that of △ABC, but area of both the triangles are equal.

Hence, the student is wrong.