Question

how much area does it occupy?

Answer 1

Hlo friend.. Cutiepie Here..

Here is ur answer :

First, the formula used in this question :

Area of right triangle = 1/2 × base × height

Area of triangle ( heron's formula) :

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

Now come to the question,

Given,

BK = 5 m

CK = 6 m

AC = 3 m

AB = 4 m

To find,

Area of equilateral BACK

Solution :

In Triangle ABC,

Angle A is right angle.

ABC is a right angled triangle.

Base b of triangle = 4 m

Height h of triangle = 3 m

Area =
$\frac{1}{2} \times b \times h$

$= \frac{1}{2} \times 4 \times 3$

$= 2 \times 3$

$= 6 \: \: {m}^{2}$

We have to find BC by pythagoras theorem.

( BC)² = ( AB)² + ( AC)²

( BC)² = (3)² + (4)²

( BC)² = 9 + 16

( BC)² = 25

BC =
$\sqrt{25}$

BC = 5 m

In Triangle BKC,

a = 6 m

b = 5 m

c = 5 m

Semi Perimeter s = a + b + c / 2

= 6 + 5 + 5 / 2

= 16 / 2

= 8 m

Area of triangle ( heron's formula)

$= \sqrt{8(8 - 6)(8 - 5)(8 - 5)}$

$= \sqrt{8 \times 2 \times 3 \times 3}$

$= \sqrt { 2 \times 2 \times 2 \times 2 \times 3 \times 3}$


$= 2 \times 2 \times 3$

= 12 m²

Area ( BACK) = ar( ABC) + ar( BKC)

= 6 + 12

= 18 m²

HOPE IT HELPS YOU..

Answer 2

Hay
Dear user -

In Triangle ABC,

We know that,

Angle A is right angle.

ABC is a right angled triangle.

Base of triangle = 4 m

Height of triangle = 3 m

Area =( BC)² = ( AB)² + ( AC)²

( BC)² = (3)² + (4)²

( BC)² = 9 + 16

( BC)² = 25

BC = 5 m

In Triangle BKC,

a = 6 m

b = 5 m

c = 5 m

Semi Perimeter s = A+ B + C / 2

=> 6 + 5 + 5 / 2

=> 11+5/2

=> 16 / 2

=> 8 m

Area of triangle = 12 m²

Area = ( ABC) + ( BKC)

= 6m + 12m

= 18 m²

HOPE IT HELPS YOU..