Question

In the adjoining do figure, find the area of the shaded region (using Heron's formula).​

Answer 1

Step-by-step explanation:

Solution :

In ΔABD, ∠D = 90°

By Pythagoras theorem,

AB² = AD² + BD²

AB² = (12)² + (16)²

AB² = 144 + 256

AB = √400

AB = 20 cm.

Therefore :

Area of ΔABD = ½ × base × height

= ½ × 16 × 12 cm²

= 96 cm².

In ΔABC, the lengths of the sides are :

a = 52 cm

b = 48 cm

c = 20 cm

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

$s = \frac{52 + 48 + 20}{2} \: \: cm$

$s \: = 60 \: cm.$

By Heron's formula,

Area of ΔABC

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

$= \sqrt{60(60-52)(60 - 48)(60 - 20}$

$= \sqrt{60 \times 8 \times 12 \times 40} \: cm ^{2}$

$= 10 \sqrt{48 \times 48} \: \: cm ^{2}$

$= 10 \times 48 \: \: cm ^{2}$

$= 480 \: \: cm ^{2} .$

Therefore :

The area of shaded region = area of ΔABC - area of ΔABD

=> 480 - 96 cm²

=> 384 cm².

Answer 2

Question :

In the adjoining do figure, find the area of the shaded region (using Heron's formula).

Solution :

In ∆ ABD

Hypotentuse = ?

Base = 12 cm

Perpendicular = 16 cm

Using Pythagoras theorem,

(AB)² = (AD)² + (BD)²

(AB)² = (12)² + (16)²

(AB)² = 144 + 256

(AB)² = 400

AB = √400

AB = 20

So AB is 20 cm

Area :

Area of ∆ = 1/2 × Base × Height

In ∆ ABD

Area = 1/2 × BD × AD

Area = 1/2 × 12 × 16

Area = 96 cm²

Lengths of sides are :

S = 20 + 48 + 52/2

S = 60 cm

Using Heron's Formula we get

$\looparrowright \sf \sqrt{S(60 - 20)(60 - 48)(60 - 52)}$

$\looparrowright \sf \sqrt{60(40 \times 12 \times 8)}$

$\looparrowright \sf \sqrt{60 \times 3840}$

$\looparrowright \sf \sqrt{230400}$

$\looparrowright \sf 480 cm²$

Now,

Area of shaded region = 480 cm² - 96 cm²

= 384 cm²

∴ Area of shaded region is 384 cm²