Please help me with this one. The base of an isosceles triangle and the altitude drawin from one of the congruent sides are equal to 18cm and 15cm, respectively. Find the lengths of the sides of the triangle.
Let's start with the basics, What's our shape?
It's an isosceles triangle.
ok good now...
Sorry if it takes me too long sometimes to reply, connection's kinda crappy at this moment. :/
With an isosceles triangle What's the rules? What do we know about the two sides? And its ok. :)
Ok, we know that An isosceles triangle is a triangle with two equal sides.
Uhm, two sides of that triangle are equal?
That's okay Thanks for telling me what's going on your end My connection sometimes fails as well.
Yup, that's right!
What do we know about the base? What is the length of the base?
The base is 18cm.
Awesome!
And the 2nd number is height Right?
which is 15cm
Right, and it is 15cm. :)
Now, We are asked to solve for the sides Have you learned about the Pythagorean theorem?
\[C^2 \sqrt{a^2+b^2}\]
C (the hypotenuse) does look like that It is normally a^2 + b^2 = c^2 The ^ sign means exponent So with your set up You removed the middle man And put down the square root immediately Now with that in mind We need to set up a right triangle, to solve for c.
Yup, I'm familiar. And, only if it's a right triangle that we can use that equation. :)
What's our height of the right triangle? Keep in mind
The right triangle is 1/2 the isosceles. Now, that base I mentioned that the right triangle is half for a reason. This part is the reason What'll be the base of the right triangle? 18 is the base for the isosceles triangle The right triangle is half of the isosceles triangle It's cut in half, so that would be 9. With that done We can plug in the 9 and 15 into a and b And solve for c
\[\sqrt{15^2+9^2}= 17.50\]
The height is 15. Base is 9. it would be 15^2+9^2=c^2
Then sqrt the answer, right?
Yes. Do you have any questions now?
17.50 is our final answer. :)
The original answer is 17.4928.. round it off= 17.50. Are you clear with that?
Yes, yes. Thank you very much! I'm kinda new here, but people here are awesome. Haha. Thank you for your help. :)
Not a problem, and welcome to open study. I wish you all the best!
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