Don't Be a Fool!

Geometry Level 1

Well, there is a triangle with sides $8, 17$ and $15$ centimetres. The only thing you have to do is find its AREA ( in cms ). There you go!

The answer is 60.

3 solutions

Shreya R
May 23, 2014

Since 8, 15 and 17 form a Pythagorean Triple, it means that the triangle is a right triangle. Hence, 8 becomes the base , 15 the height and 17 the hypotenuse . Then we use the general formula for area of a triangle- and get $\frac { 15\times 8 }{ 2 }$ = $60$

8,15 and 17 is a pythagorean triplet i.e area = 0.5 8 15

Devank Yadav - 6 years, 10 months ago

Could you please explain what you mean. Do you mean that all Pythagorean triples have the same area of 0.5815?

Trevor Arashiro - 6 years, 8 months ago

He means 0.5 X 8 X 15

Siddharth G - 6 years, 8 months ago

Siddhartha Srivastava
May 23, 2014

We have $8^2 + 15^2 = 17^2$ which means that the triangle is a right angled triangle with $8$ cm and $15$ cm as its legs.

Therefore area = $\dfrac{8 \times 15}{2} = \boxed{60}$

Well, Quite .....Very easy. but the thing is......

Its is clearly a RIGHT-ANGLED TRIANGLE!. With.. $8^2 + 15^2 = 17^2$ ..

So its AREA $=1/2 \times 8 \times 15$

$\Rightarrow$ $=4 \times 15$

$AREA=\boxed{60}$ sq.cms.

No need to go to HERON's FORMULA . :)

Arya Samanta - 7 years ago

Do yeh know o' a fast way teh test whether the given triangle is' righ angled, excep' the Pythagoras theorem.

Satvik Golechha - 7 years ago

Do you? If so, could ya enlighten us?

Krishna Ar - 7 years ago

@Krishna Ar – It dusn't work evry tyme, but in exams, its qwite helpfull. U can si dhe kalkulashen in mod 10.

Satvik Golechha - 7 years ago

@Satvik Golechha – It never works dude! See here, in mod 10 , 8 is 8, 15 is 5 and 17 is 7...so how would you be able to figure out?

Krishna Ar - 7 years ago

@Krishna Ar – It works most of the times. See that in mod 10, $\large 8^{2}+15^{2}=17^{2}$ Thus you won't need to calculate it fully.

Satvik Golechha - 7 years ago

A Former Brilliant Member
Oct 30, 2017

Solution using the cosine rule assuming that we don't that it is a right triangle.

By cosine rule, we have

$15^2=8^2+17^2-2(8)(17)(\cos a)$

$\cos a=\dfrac{8}{17}$

$a=\cos^{-1}\left(\dfrac{8}{17}\right)$

The area of a triangle is half the product of two adjacent sides multiplied by the sine of the included angle. So

$A=\dfrac{1}{2}(8)(17)\left[\sin \left(\cos^{-1} \left(\dfrac{8}{17}\right)\right)\right]=60$

Don't Be a Fool!

The answer is 60.

3 solutions

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