A solid right pyramid has a regular hexagonal base with an area of 7.4 units2. The pyramid has a height of 6 units.

Mathematics

2 answers:

Delvig [45]4 years ago

6 1

b . im writing this for minium characters limit

sergejj [24]4 years ago

3 0

Answer:

Option B. 14.8 units³ is the answer.

Step-by-step explanation:

It is given in the question A solid right pyramid which has a regular hexagonal base with area = 7.4 unit²

Height of the pyramid = 6 units.

We have to calculate the volume of the pyramid.

Since volume of the pyramid = (1/3)×Area of the base × height

Volume = (1/3)×7.4×6 = 7.4 × 2 = 14.8 units³

Therefore option B. 14.8 units³ is the answer.

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5 0

4 years ago

Read 2 more answers

Given: ΔPSQ, PS = SQ

horrorfan [7]

<h2>Answer:</h2>

To solve this problem we will use Heron's formula:

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Where $a, \ b \ and \ c$ are the side lengths of the triangle and $s$ is the semiperimeter (half the perimeter of the triangle). We know that:

$Perimeter \ P=\triangle PSQ=PS+PQ+SQ: \\ \\ \triangle PSQ=P=50 \\ \\ Semiperimeter \ s: \\ \\ s=\frac{P}{2}=25$

Also:

$(I) \ PS=SQ \\ \\ (II) \ SQ-PQ = 1 \\ \\ (III) \ PS+PQ+SQ=50 \\ \\ \\ (I) \ into \ (III): \\ \\ SQ+PQ+SQ=50 \\ \\ \therefore (IV) \ 2SQ+PQ=50 \\ \\ From \ (II): \\ \\ PQ=SQ-1 \\ \\ (II) \ into \ (IV): \\ \\ 2SQ+(SQ-1)=50 \\ 3SQ-1=50 \\ 3SQ=51 \\ \\ \boxed{SQ=17} \\ \\ \boxed{PS=17} \\ \\ PQ=SQ-1=17-1 \therefore \boxed{PQ=16}$

Finally:

$A=\sqrt{s(s-a)(s-b)(s-c)} \\ \\ A=\sqrt{s(s-PS)(s-SQ)(s-PQ)} \\ \\ A=\sqrt{s(s-17)(s-17)(s-16)} \\ \\ A=\sqrt{25(25-17)(25-17)(25-16)} \\ \\ \boxed{A=120}$