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Luba_88 [7]
3 years ago
10

Find the area of the irregular figure.

Mathematics
2 answers:
Lera25 [3.4K]3 years ago
7 0

Answer:

a

Step-by-step explanation:

PilotLPTM [1.2K]3 years ago
6 0

Answer:

Option A. 432\ units^{2}

Step-by-step explanation:

we know that

The area of the irregular figure is equal to the area of two trapezoids

The area of trapezoid is equal to

A=\frac{1}{2}(b1+b2)h

where

b1,b2 are the parallel bases

h is the perpendicular distance between the parallel bases

<u>Find the left trapezoid area</u>

A1=\frac{1}{2}(4+25)(16)=232\ units^{2}

<u>Find the trapezoid area on the right</u>

A2=\frac{1}{2}(25+15)(10)=200\ units^{2}

The area of the irregular figure is

A=A1+A2=232\ units^{2}+200\ units^{2}=432\ units^{2}

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The triangle below is equilateral. Find the length of side x to the nearest tenth.
Gemiola [76]

Answer:

\sqrt{\frac{15}{2} } or 2.738

Step-by-step explanation:

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There is a small square in the bottom right corner, which indicates that this triangle is a right triangle. This means that we can use the Pythagorean Theorem: a^{2} +b^{2} =c^{2}

We know that \sqrt{10} is our hypotenuse, and therefore our c in our equation. Let’s say that x=a in our equation. Therefore we are left to find b. However, b is half the length of the side of the original equilateral triangle. An equilateral triangle means that all three sides are the same length. Therefore our side would also be \sqrt{10} units long. However we know that b is half of that value, so b=\frac{1}{2}(\sqrt{10}) or \frac{\sqrt{10} }{2}

Plugging these values into the equation:

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