Question

i have an amazing question
and I a person who solves it could be a great mathematician ✌✌ ...without using trigonometry

Answer 1

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....,

Answer 2

Yeah sure for your help

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Nice question

Here is your answer which you are searching for

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way to solve

ar ( triangle PQR ) = ar ( triangle BQR ) + ar ( triangle BPR ) - ar ( traingle BPQ )

ar ( traingle BPQ ) = $\frac{1}{2} . 4 .3 = 6 ( angle B = 90° )$

ar ( traingle BQR ) = $\frac{4}{15} .A ( traingle ABC ) = 6.4$

ar ( triange BPR ) = $\frac{9}{40} . A ( traingle ABC ) = 5.4$

ar ( traingle ABC ) = $\frac{1}{2} . 6 .8 = 24$ ( angle ABC = 90°)

ar ( traingle PQR ) = 6.4 + 5.4

=> 5.8

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Hope it helpful to you !!

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@ Brainlestuser