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3 years ago

Find the area of this please help

Mathematics

1 answer:

Harrizon [31]3 years ago

6 0

We can split this figure up into two shapes - a rectangle and a triangle.

The rectangle has a width of 9 cm and a length of 12 cm. The triangle has a height of 15 - 9 = 6 cm and a base length of 12 - 3 - 3 = 6 cm.

Rectangle:

A = 12 x 9

A = 108 cm^2

Triangle:

A = 1/2 x 6 x 6

A = 1/2 x 36

A = 18 cm^2

Total figure:

108+ 18 = 126

The area of the figure is 126 cm^2.

Hope this helps!! :)

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FInd the GCF of the numbers using prime factorizations.

lesya [120]

Answer:

GCF(330, 75, 450, 225) = 15

Steps:

Prime factorization of the numbers:

330 = 2 × 3 × 5 × 11

75 = 3 × 5 × 5

450 = 2 × 3 × 3 × 5 × 5

225 = 3 × 3 × 5 × 5

GCF(330, 75, 450, 225)

= 3 × 5

= 15

3 0

3 years ago

50 POINTS!!! In rectangle ABCD, AB = 6 cm, BC = 8 cm, and DE = DF. The area of triangle DEF is one-fourth the area of rectangle

aalyn [17]

Answer:

$EF=4\sqrt{3}$

Step-by-step explanation:

In rectangle ABCD, AB = 6, BC = 8, and DE = DF.

ΔDEF is one-fourth the area of rectangle ABCD.

We want to determine the length of EF.

First, we can find the area of the rectangle. Since the length AB and width BC measures 6 by 8, the area of the rectangle is:

$A_{\text{rect}}=8(6)=48\text{ cm}^2$

The area of the triangle is 1/4 of this. Therefore:

$\displaystyle A_{\text{tri}}=\frac{1}{4}(48)=12\text{ cm}^2$

The area of a triangle is half of its base times its height. The base and height of the triangle is DE and DF. Therefore:

$\displaystyle 12=\frac{1}{2}(DE)(DF)$

Since DE = DF:

$24=DF^2$

Thus:

$DF=\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}=DE$

Since ABCD is a rectangle, ∠D is a right angle. Then by the Pythagorean Theorem:

$(DE)^2+(DF)^2=(EF)^2$

Therefore:

$(2\sqrt6)^2+(2\sqrt6)^2=EF^2$

Square:

$24+24=EF^2$

Add:

$EF^2=48$

And finally, we can take the square root of both sides:

$EF=\sqrt{48}=\sqrt{16\cdot 3}=4\sqrt{3}$

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