Question

i have an interesting question
let's see which could solve it without trignometery
I promise to follow u if you could solve it♥️♥️✌✌

Answer 1

11 square unit will be the answer

Answer 2

IN ∆ ABC,

AB = 6 , AC = 10, BC = 8

IT SEEMS IT IS RIGHT TRIANGLE,
So, if 10² = 6² + 8²
if possible then it should right angles triangle,

100 = 36 + 64

100 = 100

Hence, <B = 90°

Now,

area of triangle ABC = 1/2 * base * height

where, base = 6 , height = 8

area = 1/2*6 * 8

= 6 * 4

area = 24unit²

Now,

area of triangle PQR = 1/4(area of triangle ABC)

= 1/4(24)

= 6unit²

One more solutions,

semi - Perimeter of triangle ABC = (6 + 8 + 10)/2

s = 24/2

s = 12,

now,

By Heron's formula,

area of triangle = $\bold{\large{\sqrt{s(s - a)(s - b)(s - c)}}}$

= $\bold{\large{\sqrt{12(12 - 6)(12 - 8)(12 - 10)}}}$

= $\bold{\large{\sqrt{12*6*4*2}}}$

= $\bold{\large{6*2*2}}$

area of triangle ABC= 24unit²

Now, area of triangle PQR = 1/4(area of triangle ABC)

= 1/4(24)

= 6unit²

hope this help you....,