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serg [7]
3 years ago
10

An isosceles triangle has

Mathematics
1 answer:
Nadya [2.5K]3 years ago
5 0

Answer:

Step-by-step explanation:

The area of an isoceles triangle is:

A = base x height / 2

The base is the unequal side in the isosceles triangle.  You have to calculate the height using the Pythagorean Theorem.

Since two side are equal in length, if you draw a vertical lined from where these two sides meet and bisect the base, you will create two right triangles.

So a² + b² = c² where c is the hypotenuse of the right triangle.  This hypotenuse is also the two equal sides of the isosceles triangle.  They told you the length of the base but you need to divide that in half since your line bisected the base.

So now you know one side of the right triangle is 8cm which is half of 16cm.  You also know the hypotenuse is 17cm.  Solve the Pythagorean Theorem for the other side and that will give you the height of your bisecting line.

8² + b² = 17²  OR  64 + b² = 289

Solve for b

b² = 289 - 64  OR  b² = 225  and finally    b = \sqrt{ 225  which is 15

So now you can calculate the area of the triangle:

A = 16 x 15 / 2   or   240 / 2  which is   120

So the area is 120 cm²

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Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

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Whatever you do to the denominator, you have to do to the numerator too. So

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2

Expand the numerator:

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}

So we have

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2

But √12 = √(3•4) = 2√3, so

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}

7 0
3 years ago
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