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2 years ago

Can someone please help as soon as possible!!

Mathematics

1 answer:

sergiy2304 [10]2 years ago

4 0

Answer:

1) triangles are similar

Step-by-step explanation:

The height from vertex X of isosceles ∆WXY is 4 units. The width WY is also 4 units. In isosceles ∆UVW, the height from vertex V is 6 units, and the width UW is also 6 units.

The height ratios are ...

∆WXY/∆UVW = 4/6

The width ratios are ...

∆WXY/∆UVW = 4/6

The measures of ∆WXY are proportional to those of ∆UVW, so the triangles are similar.

_____

Strictly speaking, you cannot go by triangle height and width alone. That is why we made not of the fact that the triangles are <em>isosceles</em>. When base and height of an isosceles triangle are proportional, the Pythagorean theorem guarantees that side lengths are proportional. Trigonometry can also be invoked to support the claim that angles are congruent.

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3 years ago

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If you can solve all parts I will give brainliest (also give strategy)

Alexxx [7]

The Halloween conical hat, with given height, circular base and brim

extension has the following calculated parameters;

Part a. The slant height is <u>18.2 inches</u>

Part b. The volume of the cone is $37\frac{1}{2} \cdot \pi$ in.³

Part c. The area of the brim, <em>A</em> = <u>36·π in.²</u>

Part d. The area of the brim is found by <u>subtracting the area of the base of the cone from the area covered by the perimeter of the brim</u>

Reasons:

Known parameters;

Height of the conical portion, h = 18 inches

Base circumference, C = 5·π inches

Part a. Slant height of the conical portion; Required

Solution:

The circumference of a circle, C = 2·π·r

Therefore;

$r = \dfrac{C}{2 \cdot \pi}$

Which gives;

$r = \dfrac{5 \cdot \pi}{2 \cdot \pi} = \dfrac{5}{2} = 2.5$

Radius, r = 2.5 inches

According to Pythagoras's theorem, we have; s² = r² + h²

Where;

s = The slant height of the cone

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s = √(330.25) ≈ 18.2

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Part b. The measure in cubic inches of candy that exactly fills the conical portion of the hat is the volume of the cone.

$Volume \ of \ a \ cone = \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot h$

Therefore;

$V = \dfrac{1}{3} \times \pi \times 2.5^2 \times 18 = 37\frac{1}{2} \cdot \pi$

The volume of the cone, V = $37\frac{1}{2}$ ·π in.³

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The radius of the brim, R = Radius of the base of the cone + 4 inches

∴ <em>R</em> = 2.5 inches + 4 inches = 6.5 inches

Area of the brim, <em>A</em> = Area of the 6.5 inch circle - Area of the circular base of the cone

∴ A = π × 6.5² - π × 2.5² = 36·π

The area of the brim, <em>A</em> = <u>36·π in.²</u>

Part d. The procedure for solving the question in part c, is described as follows;

The area of the brim can be found by finding the entire area of the circle formed by the perimeter of the brim, then subtracting the area of the base of the cone from that area.

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