Mensuration - Exercise 23(E)

Find the area of the triangle, having

(i) base = 16 cm, height = 7.5 cm

(ii) base = 5.6 m, height = 3.5 m

(iii) base = 6.4 m, height = 8 dm

(iv) base = 9.5 cm, height = 6 mm

Answer

(i)

Given:

base = 16 cm, height = 7.5 cm

We have the formula,

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

= $\dfrac{1}{2}$ x 16 cm x 7.5 cm

= $\dfrac{1}{1}$ x 8 cm x 7.5 cm

= 60 cm²

Area of the triangle = 60 cm²

(ii)

Given:

base = 5.6 m, height = 3.5 m

We have the formula,

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

= $\dfrac{1}{2}$ x 5.6 m x 3.5 m

= $\dfrac{1}{1}$ x 2.8 m x 3.5 m

= 9.8 m²

Area of the triangle = 9.8 m²

(iii)

Given:

base = 6.4 m, height = 8 dm

Converting base into dm:

1 m = 10 dm

∴ 6.4 m = 64 dm

We have the formula,

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

= $\dfrac{1}{2}$ x 64 dm x 8 dm

= $\dfrac{1}{1}$ x 32 dm x 8 dm

= 256 dm²

Area of the triangle = 256 dm²

(iv)

Given:

base = 9.5 cm, height = 6 mm

Converting height into cm:

1 cm = 10 mm

∴ 6 mm = 0.6 cm

We have the formula,

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

= $\dfrac{1}{2}$ x 0.6 cm x 9.5 cm

= $\dfrac{1}{1}$ x 0.3 cm x 9.5 cm

= 2.85 cm²

Area of the triangle = 2.85 cm²

Find the height of the triangle whose :

(i) area = 28.9 m² base = 8.5 m

(ii) area = 56 dm² base = 2.8 m

Answer

(i)

Given:

area = 28.9 m² base = 8.5 m

We have the formula,

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

$\phantom{=} \text{height} = \dfrac{2 \times \text{Area}}{\text{base}} \\[1em] = \dfrac{2 \times 28.9 \text{ m}^2}{8.5 \text{ m}} \\[1em] = \dfrac{57.8 \text{ m}^2}{8.5 \text{ m}} \\[1em] = 6.8 \text{ m}$

The height of the triangle = 6.8 m

(ii)

Given:

area = 56 dm² base = 2.8 m

Converting base into dm:

1 m = 10 dm

∴ 2.8 m = 28 dm

We have the formula,

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

$\phantom{=} \text{height} = \dfrac{2 \times \text{Area}}{\text{base}} \\[1em] = \dfrac{2 \times 56 \text{ dm}^2}{28 \text{ dm}} \\[1em] = \dfrac{2 \times 2 \text{ dm}^2}{1 \text{ dm}} \\[1em] = 4 \text{ dm}$

The height of the triangle = 4 dm

Find the base of the triangle whose :

(i) area = 4.2 m² height = 2.4 m

(ii) area = 2.4 dm², height = 80 cm

Answer

(i)

Given:

area = 4.2 m² height = 2.4 m

We have the formula,

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

$\phantom{=} \text{base} = \dfrac{2 \times \text{Area}}{\text{height}} \\[1em] = \dfrac{2 \times 4.2 \text{ m}^2}{2.4 \text{ m}} \\[1em] = \dfrac{1 \times 4.2 \text{ m}^2}{1.2 \text{ m}} \\[1em] = 3.5 \text{ m} \\[1em]$

= 3.5 m

The base of the triangle = 3.5 m

(ii)

Given:

area = 2.4 dm², height = 80 cm

Converting area into cm:

1 dm² = 100 cm²

∴ 2.4 dm² = 240 cm²

We have the formula,

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

$\phantom{=} \text{base} = \dfrac{2 \times \text{Area}}{\text{height}} \\[1em] = \dfrac{2 \times 240 \text{ cm}^2}{80 \text{ cm}} \\[1em] = \dfrac{2 \times 3 \text{ cm}^2}{1 \text{ cm}} \\[1em] = 6 \text{ cm}$

The base of the triangle = 6 cm

Find the area of the triangle whose sides are 13 cm, 20 cm and 21 cm. Also find the altitude of the triangle corresponding to the largest side.

Answer

Given:

Sides: a = 13 cm, b = 20 cm, c = 21 cm

s = $\dfrac{1}{2}$ x (a + b + c)

= $\dfrac{1}{2}$ x (13 + 20 + 21) cm

= $\dfrac{1}{2}$ x 54 cm

= $\dfrac{1}{1}$ x 27 cm

= 27 cm

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

$= \sqrt{27(27 - 13)(27 - 20)(27 - 21)} \text{ cm}^2 \\[1em] = \sqrt{27 \times 14 \times 7 \times 6} \text{ cm}^2 \\[1em] = \sqrt{(3 \times 3 \times 3) \times (2 \times 7) \times 7 \times (2 \times 3)} \text{ cm}^2 \\[1em] = \sqrt{3^4 \times 2^2 \times 7^2} \text{ cm}^2 \\[1em] = 3^2 \times 2 \times 7 \text{ cm}^2 \\[1em] = 9 \times 14 \text{ cm}^2 \\[1em] = 126 \text{ cm}^2$

Largest side = 21 cm

Let the altitude corresponding to the largest side be h cm. Then

$\phantom{=} \text{Area} = \dfrac{1}{2} \times \text{base} \times h \\[1em] 126 \text{ cm}^2 = \dfrac{1}{2} \times 21 \times h \\[1em] \Rightarrow h = \dfrac{126 \times 2}{21} \text{ cm} \\[1em] \Rightarrow h = 6 \times 2 \text{ cm} \\[1em] \Rightarrow h = 12 \text{ cm}$

Area = 126 cm², The required height = 12 cm

Find the area of the triangle whose sides are 50 cm, 48 cm and 14 cm. Find the height of the triangle corresponding to the side measuring 48 cm.

Answer

Given:

Sides: a = 50 cm, b = 48 cm, c = 14 cm

s = $\dfrac{1}{2}$ x (a + b + c)

= $\dfrac{1}{2}$ x (50 + 48 + 14) cm

= $\dfrac{1}{2}$ x 112 cm

= $\dfrac{1}{1}$ x 56 cm

= 56 cm

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

$= \sqrt{56(56 - 50)(56 - 48)(56 - 14)} \text{ cm}^2 \\[1em] = \sqrt{56 \times 6 \times 8 \times 42} \text{ cm}^2 \\[1em] = \sqrt{(8 \times 7) \times (6) \times 8 \times (6 \times 7)} \text{ cm}^2 \\[1em] = \sqrt{8^2 \times 7^2 \times 6^2} \text{ cm}^2 \\[1em] = 8 \times 7 \times 6 \text{ cm}^2 \\[1em] = 336 \text{ cm}^2 \\[1em]$

Let the altitude corresponding to the side of 48 cm be h cm. Then

$\phantom{=} \text{Area} = \dfrac{1}{2} \times \text{base} \times h \\[1em] 336 \text{ cm}^2 = \dfrac{1}{2} \times 48 \times h \\[1em] \Rightarrow h = \dfrac{336 \times 2}{48} \text{ cm} \\[1em] \Rightarrow h = 7 \times 2 \text{ cm} \\[1em] \Rightarrow h = 14 \text{ cm}$

Area = 336 cm², The required height = 14 cm

Find the area of an isosceles triangle in which each of the equal sides measures 30 cm and the third side is 48 cm long.

Answer

Given:

Sides: a = 30 cm, b = 30 cm, c = 48 cm

s = $\dfrac{1}{2}$ x (a + b + c)

= $\dfrac{1}{2}$ x (30 + 30 + 48) cm

= $\dfrac{1}{2}$ x 108 cm

= $\dfrac{1}{1}$ x 54 cm

= 54 cm

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

$= \sqrt{54(54 - 30)(54 - 30)(54 - 48)} \text{ cm}^2 \\[1em] = \sqrt{54 \times 24 \times 24 \times 6} \text{ cm}^2 \\[1em] = \sqrt{(6 \times 9) \times 24^2 \times 6} \text{ cm}^2 \\[1em] = \sqrt{6^2 \times 3^2 \times 24^2} \text{ cm}^2 \\[1em] = 6 \times 3 \times 24 \text{ cm}^2 \\[1em] = 432 \text{ cm}^2 \\[1em]$

The area of an isosceles triangle = 432 cm²

The base and the height of a triangle are in the ratio 5 : 3 and its area is 43.2 m². Find the base and the height of the triangle.

Answer

Given:

The ratio of height and base of a triangle = 5 : 3

Area of triangle = 43.2 m²

Let the Base (b) = 5x

Let the Height (h) = 3x

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

43.2 m² = $\dfrac{1}{2}$ x (5x) x (3x)

43.2 m² = $\dfrac{1}{2}$ x (15x²)

43.2 m² = $\dfrac{15 x^2}{2}$

43.2 m² x 2 = 15x²

86.4 m² = 15x²

⇒ x² = $\dfrac{86.4}{15}$ m²

⇒ x² = 5.76 m²

⇒ x = $\sqrt{5.76}$ m²

⇒ x = 2.4 m

Base = 5x = 5 x 2.4 m = 12 m

Height = 3x = 3 x 2.4 m = 7.2 m

Base = 12 m, Height = 7.2 m

Find the area and the height of an equilateral triangle whose each side measures. (Take √3 = 1.73 in each case) :

(i) 12 cm

(ii) 10 m

(iii) 6.4 m (Take √3 = 1.73 in each case)

Answer

(i)

Given:

Side(a) = 12 cm

Area of the triangle = $\dfrac{\sqrt{3}}{4}a^2$

$= \dfrac{\sqrt{3}}{4}(12^2) \text{ cm}^2 \\[1em] = \dfrac{\sqrt{3}}{4}(144) \text{ cm}^2 \\[1em] = \dfrac{\sqrt{3}}{1}(36) \text{ cm}^2 \\[1em] = 36\sqrt{3} \text{ cm}^2 \\[1em] = 36 \times 1.73 \text{ cm}^2 \\[1em] = 62.28 \text{ cm}^2$

Height of the triangle = $\dfrac{\sqrt{3}}{2}a$

$= \dfrac{\sqrt{3}}{2} \times 12 \text{ cm} \\[1em] = \dfrac{\sqrt{3}}{1} \times 6 \text{ cm} \\[1em] = 6\sqrt{3} \text{ cm} \\[1em] = 6 \times 1.73 \text{ cm} \\[1em] = 10.38 \text{ cm}$

Area = 62.28 cm², Height = 10.38 cm

(ii)

Given:

Side(a) = 10 m

Area of the triangle = $\dfrac{\sqrt{3}}{4}a^2$

$= \dfrac{\sqrt{3}}{4}(10^2) \text{ m}^2 \\[1em] = \dfrac{\sqrt{3}}{4}(100) \text{ m}^2 \\[1em] = \dfrac{\sqrt{3}}{1}(25) \text{ m}^2 \\[1em] = 25\sqrt{3} \text{ m}^2 \\[1em] = 25 \times 1.73 \text{ m}^2 \\[1em] = 43.25 \text{ m}^2$

Height of the triangle = $\dfrac{\sqrt{3}}{2}a$

$= \dfrac{\sqrt{3}}{2} \times 10 \text{ m} \\[1em] = \dfrac{\sqrt{3}}{1} \times 5 \text{ m} \\[1em] = 5\sqrt{3} \text{ m} \\[1em] = 5 \times 1.73 \text{ m} \\[1em] = 8.65 \text{ m}$

Area = 43.25 m², Height = 8.65 m

(iii)

Given:

Side(a) = 6.4 m

Area of the triangle = $\dfrac{\sqrt{3}}{4}a^2$

$= \dfrac{\sqrt{3}}{4}(6.4^2) \text{ m}^2 \\[1em] = \dfrac{\sqrt{3}}{4}(40.96) \text{ m}^2 \\[1em] = \dfrac{\sqrt{3}}{1}(10.24) \text{ m}^2 \\[1em] = 10.24\sqrt{3} \text{ m}^2 \\[1em] = 10.24 \times 1.73 \text{ m}^2 \\[1em] = 17.7152 \text{ m}^2$

Height of the triangle = $\dfrac{\sqrt{3}}{2}a$

$= \dfrac{\sqrt{3}}{2} \times 6.4 \text{ m} \\[1em] = \dfrac{\sqrt{3}}{1} \times 3.2 \text{ m} \\[1em] = 3.2\sqrt{3} \text{ m} \\[1em] = 3.2 \times 1.73 \text{ m} \\[1em] =5.536 \text{ m}$

Area = 17.7152 m², Height = 5.536 m

Find the area of a right triangle whose hypotenuse is 26 cm long and one of the sides containing the right angle measures 10 cm.

Answer

Let ABC be the right angle triangle.

Hypotenuse AC = 26 cm

One side AB = 10 cm

BC² = AC² - AB²

BC² = 26² - 10²

BC² = 676 - 100

BC² = 576

BC = $\sqrt{576}$

BC = 24 cm

∴ Area of the triangle = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x BC x AB

= $\dfrac{1}{2}$ x (24 x 10) cm²

= $\dfrac{1}{2}$ x 240 cm²

= $\dfrac{1}{1}$ x 120 cm²

= 120 cm²

Area of the triangle = 120 cm²

The area of a right triangle is 240 cm² and one of its legs is 16 cm long. Find the length of the other leg.

Answer

Given:

Area = 240 cm²

One leg (AB) = 16 cm

Let the other leg (BC) be x.

We know the formula:

∴ Area of the triangle = $\dfrac{1}{2}$ x base x height

240 cm² = $\dfrac{1}{2}$ x BC x AB

240 cm² = $\dfrac{1}{2}$ x (x x 16 cm)

240 cm² = $\dfrac{1}{1}$ x (x x 8 cm)

⇒ x = $\dfrac{240 \text{ cm}^2}{8 \text{ cm}}$

⇒ x = 30 cm

∴ BC = 30 cm

Length of the other leg = 30 cm

The legs of a right triangle are in the ratio 3 : 4 and its area is 1014 cm². Find its hypotenuse.

Answer

Given:

Area = 1014 cm²

Let the legs be 3x and 4x

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

1014 cm² = $\dfrac{1}{2}$ x (3x) x (4x)

1014 cm² = $\dfrac{1}{2}$ x (12x²)

1014 cm² = $\dfrac{1}{1}$ x 6x²

⇒ x² = $\dfrac{1014}{6}$ cm²

⇒ x² = 169 cm²

⇒ x = $\sqrt{169}$ cm²

⇒ x = 13 cm

AB = 3x = 3 x 13 = 39 cm

BC = 4x = 4 x 13 = 52 cm

Hypotenuse² = AB² + BC²

AC² = (39² + 52²) cm²

AC² = (1521 + 2704) cm²

AC² = 4225 cm²

AC = $\sqrt{4225}$ cm²

AC = 65 cm

Hypotenuse = 65 cm

The sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find the area of the triangle.

Answer

Let the sides be 13x, 14x and 15x

Perimeter of the triangle = 84 cm

Perimeter of the triangle = AB + BC + AC

84 cm = 13x + 14x + 15x

84 cm = 42x

⇒ x = $\dfrac{84}{42}$ cm

⇒ x = 2 cm

∴ AB = 13x = 13 x 2 = 26 cm

BC = 14x = 14 x 2 = 28 cm

AC = 15x = 15 x 2 = 30 cm

s = $\dfrac{AB + BC + AC}{2}$

s = $\dfrac{(26 + 28 + 30)}{2}$ cm

s = $\dfrac{84}{2}$ cm

s = 42 cm

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

$= \sqrt{42(42 - 26)(42 - 28)(42 - 30)} \text{ cm}^2 \\[1em] = \sqrt{42 \times 16 \times 14 \times 12} \text{ cm}^2 \\[1em] = \sqrt{(14 \times 3) \times 16 \times 14 \times (4 \times 3)} \text{ cm}^2 \\[1em] = \sqrt{14^2 \times 3^2 \times 4^2 \times 4^2} \text{ cm}^2 \\[1em] = 14 \times 3 \times 4 \times 4 \text{ cm}^2 \\[1em] = 336 \text{ cm}^2$

Area of the triangle = 336 cm²

The base of an isosceles triangle is 12 cm and its perimeter is 32 cm. Find its area.

Answer

In an isosceles triangle, two sides are equal.

Let the equal sides be a and the base be b = 12 cm.

Perimeter = 32 cm

Perimeter of the triangle = AB + AC + BC

32 cm = a + a + 12

(32 - 12) cm = 2a

20 cm = 2a

⇒ a = $\dfrac{20}{2}$ cm

⇒ a = 10 cm

∴ AB = AC = 10 cm

The altitude to the base of an isosceles triangle bisects the base. Using the Pythagorean theorem on one of the right-angled halves:

$\phantom{=} h^2 + \left(\dfrac{b}{2}\right)^2 = a^2 \\[1em] h^2 + 6^2 = 10^2 \\[1em] h^2 + 36 = 100 \\[1em] h^2 = 100 - 36 \\[1em] h^2 = 64 \\[1em] h = \sqrt{64} \\[1em] h = 8$

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

= $\dfrac{1}{2}$ x 12 x 8 cm²

= $\dfrac{1}{2}$ x 96 cm²

= $\dfrac{1}{1}$ x 48 cm²

= 48 cm²

Area of the triangle = 48 cm²

The cost of painting the top surface of a triangular board at 80 paise per square metre is ₹ 176.40. If the height of the board measures 24.5 m, find its base.

Answer

Given:

Height = 24.5 m

Total Cost = ₹ 176.40

Rate = 80 paise per m²

Converting rate into rupees:

1 rupee = 100 paise

∴ 80 paise = ₹ 0.80 per m²

Area = $\dfrac{\text{Total Cost}}{\text{Rate}}$

= $\dfrac{176.40}{0.80}$ m²

= 220.5 m²

We know the formula:

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

220.5 m² = $\dfrac{1}{2}$ x base x 24.5

220.5 m² = $\dfrac{1}{1}$ x base x 12.25

220.5 m² = 12.25 x base

⇒ base = $\dfrac{220.5 \text{ m}^2}{12.25 \text{ m}}$

base = 18 m

Calculate the area of the quadrilateral ABCD in which AB = BD = AD = 10 cm, ∠BCD = 90° and CD = 8 cm. (Take √3 = 1.732).

Answer

To find the area of the quadrilateral ABCD, we can split it into two triangles: the equilateral triangle △ABD and the right-angled triangle △BCD.

Given:

AB = BD = AD = 10 cm

∠BCD = 90°

CD = 8 cm

Since AB = BD = AD = 10 cm, △ABD is an equilateral triangle.

The formula for the area of an equilateral triangle is:

Area (△ABD) = $\dfrac{\sqrt{3}}{4} \times (\text{side})^2$

$= \dfrac{1.732}{4} \times (10)^2 \text{ cm}^2 \\[1em] = \dfrac{1.732}{4} \times 100 \text{ cm}^2 \\[1em] = \dfrac{1.732}{1} \times 25 \text{ cm}^2 \\[1em] = 43.25 \text{ cm}^2$

Area of △BCD:

△BCD is a right-angled triangle at ∠C = 90°.

We know the hypotenuse BD = 10 cm and one side CD = 8 cm.

Let's find the base (BC) using Pythagoras' theorem:

BC² + CD² = BD²

BC² + (8 cm)² = (10 cm)²

BC² + 64 cm² = 100 cm²

BC² = (100 - 64) cm²

BC² = 36 cm²

BC = $\sqrt{36}$ cm²

BC = 6 cm

Area(△BCD) = $\dfrac{1}{2}$ x base x height

$= \dfrac{1}{2} \times 6 \times 8 \text{ cm}^2 \\[1em] = \dfrac{1}{2} \times 48 \text{ cm}^2 \\[1em] = \dfrac{1}{1} \times 24 \text{ cm}^2 \\[1em] = 24 \text{ cm}^2$

Let's find total area of Quadrilateral ABCD:

Area of Quadrilateral ABCD = Area(△ABD) + Area(△BCD)

= 43.25 cm² + 24 cm²

= 67.25 cm²

The total area of the quadrilateral = 67.25 cm²

Calculate the area of the quadrilateral PQRS shown in the adjoining figure, it being given that PQ = 8 cm, RQ = 17 cm, ∠RPQ = 90°, RS = 6 cm and ∠PRS = 90°.

Answer

To calculate the area of the quadrilateral PQRS, we can split it into two right-angled triangles: △ PRS and △ PRQ.

Given:

PQ = 8 cm

RQ = 17 cm

∠RPQ = 90°

RS = 6 cm

∠PRS = 90°

Let's find PR which acts as the height for one triangle and the base for the other.

In the right-angled triangle △ PRQ, the right angle is at P (∠RPQ = 90°).

Using Pythagoras' theorem:

PR² + PQ² = RQ²

PR² + (8 cm)² = (17 cm)²

PR² + 64 cm² = 289 cm²

PR² = 289 cm² - 64 cm²

PR² = 225 cm²

PR = $\sqrt{225}$ cm²

PR = 15 cm

Let's find area of △PRQ:

Area of △PRQ = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x PQ x PR

= $\dfrac{1}{2}$ x 8 cm x 15 cm

= $\dfrac{1}{2}$ x 120 cm²

= $\dfrac{1}{1}$ x 60 cm²

= 60 cm²

Now, let's find area of △PRS:

Area of △PRS = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x RS x PR

= $\dfrac{1}{2}$ x 6 cm x 15 cm

= $\dfrac{1}{2}$ x 90 cm²

= $\dfrac{1}{1}$ x 45 cm²

= 45 cm²

Let's find total Area of Quadrilateral PQRS:

Area of Quadrilateral PQRS = Area of △ PRQ + Area of △ PRS

= 60 cm² + 45 cm²

= 105 cm²

The total area of the quadrilateral = 105 cm²

Find the area of a quadrilateral ABCD whose diagonal AC is 25 cm long and the lengths of perpendiculars from opposite vertices B and D on AC are BE = 3.6 cm and DF = 2.4 cm.

Answer

Given:

AC = 25 cm

BE = 3.6 cm

DF = 2.4 cm

To find the area of this quadrilateral, we treat it as two triangles sharing a common base (the diagonal AC).

Area of △ABC = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x AC x BE

= $\dfrac{1}{2}$ x 25 cm x 3.6 cm

= $\dfrac{1}{2}$ x 90 cm²

= $\dfrac{1}{1}$ x 45 cm²

= 45 cm²

Area of △ACD = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x AC x DF

= $\dfrac{1}{2}$ x 25 cm x 2.4 cm

= $\dfrac{1}{2}$ x 60 cm²

= $\dfrac{1}{1}$ x 30 cm²

= 30 cm²

Let's find the total area of quadrilateral ABCD:

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

= 45 cm² + 30 cm²

= 75 cm²

The area of the quadrilateral ABCD = 75 cm²

Find the area of the quadrilateral ABCD, given in the adjoining figure in which AB = 14 cm, BC = 78 cm, CD = 112 cm, BD = 50 cm and DA = 48 cm.

Answer

Given:

To find the area of the quadrilateral ABCD, we can split it into two triangles △ABD and △DBC.

Area of △ABD:

The sides of the triangle are a = 48 cm, b = 14 cm, c = 50 cm

Semi perimeter (s1) = $\dfrac{a + b + c}{2} = \dfrac{48 + 14 + 50}{2} \text{ cm} = \dfrac{112}{2} \text{ cm} = 56 \text{ cm}$

Area of △ABD = $\sqrt{s1(s1 - a)(s1 - b)(s1 - c)}$

$= \sqrt{56(56 - 48)(56 - 14)(56 - 50)} \text{ cm}^2 \\[1em] = \sqrt{56 \times 8 \times 42 \times 6} \text{ cm}^2 \\[1em] = \sqrt{(7 \times 8) \times 8 \times (7 \times 6) \times 6} \text{ cm}^2 \\[1em] = \sqrt{7^2 \times 8^2 \times 6^2} \text{ cm}^2 \\[1em] = 7 \times 8 \times 6 \text{ cm}^2 \\[1em] = 336 \text{ cm}^2$

Area of △DBC:

The sides of the triangle are a = 50 cm, b = 78 cm, c = 112 cm

Semi perimeter (s2) = $\dfrac{a + b + c}{2} = \dfrac{50 + 78 + 112}{2} \text{ cm} = \dfrac{240}{2} \text{ cm} = 120 \text{ cm}$

Area of △DBC = $\sqrt{s2(s2 - a)(s2 - b)(s2 - c)}$

$= \sqrt{120(120-50)(120-78)(120-112)} \text{ cm}^2 \\[1em] = \sqrt{120 \times 70 \times 42 \times 8} \text{ cm}^2 \\[1em] = \sqrt{(10 \times 12) \times (10 \times 7) \times (6 \times 7) \times 8} \text{ cm}^2 \\[1em] = \sqrt{10^2 \times 7^2 \times (12 \times 6 \times 8)} \text{ cm}^2 \\[1em] = \sqrt{10^2 \times 7^2 \times 576} \text{ cm}^2 \\[1em] = 10 \times 7 \times 24 \text{ cm}^2 \\[1em] = 1680\text{ cm}^2$

Let's find the total area of quadrilateral ABCD:

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

= 336 cm² + 1680 cm²

= 2016 cm²

The area of the quadrilateral ABCD = 2016 cm²