so, since the area of the first triangle is 32, with side length 8, we know that
1/2*bh=32
base =8, so... 1/2*8h=32, thus h =8
get the hypotenuse of LNO, which is 8sqrt(2)
that hypotenuse is split in half by the two triangles, each with base length 4sqrt(2)
to get the area of LMN, we see that it's height is 6, and have base 4sqrt(2)
area of that triangle is 1/2*bh === 1/2*4sqrt(2)*6 =24sqrt(2)
the answer is B

First examine the blue triangle.


We're told the area = 32
Area = (base)(height)/2
So, 32 = (x)(8)/2
Solve to get: x = 8

Since the blue triangle is a right triangle, we can apply the Pythagorean Theorem to find the length of the hypotenuse.
We get: 8² + 8² = hypotenuse²
Simplify: 128 = hypotenuse²
So, hypotenuse = √128 = 8√2

We get:


At this point, we can focus on the red triangle


We can see that we now have the length of the base AND the height.
So, area of red triangle = (base)(height)/2
= (8√2)(6)/2
= 24√2

Answer: B


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*First we shouldn’t assume that MNLO is square, because we don’t have any information about 3 angles of it. So don’t judge based on the shape.
Answer: B
Area of triangle LNO = 1/2 * NO * OL = 1/2 * 8 * OL = 4 * OL = 32 -> OL = 8
Area of triangle MNO = 1/2 * MX * LN = 1/2 * 6 * LN = 3 * LN (We call the point which LN and MX cross as X)

In triangle NLO, we have: NO^2 + OL^2 = LN^2
LN = √NO^2 + OL^2 = √8^2 + 8^2 = 8√2

The area of triangle MNO is 1/2 * MX * LN = 1/2 * 6 * 8√2 = 24√2
The answer is B
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