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belka [17]
2 years ago
10

A triangle prism has a base with an area of 45 square inches.The prism also has a length of 61/2 inches.Find the volume of this

triangular prism.
Mathematics
1 answer:
Anastaziya [24]2 years ago
8 0
9 times 5 = 45 thats the bases area so now use 9 and 5 to multiply 6.5 in which gets the volume answer wich is 292.5 you are welcome please mark me as brainliest
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The fraction computed shows that the cups of sugar that can be used based on the flour will be B. 4 1/2 cups.

<h3>How to calculate the fraction?</h3>

From the information, Jonah used 1 1/2 cups of brown sugar and 2 1/3 cups for flour.

Therefore, the cups of sugar needed will be:

= 1.5/2.33 × 7

= 4.5 = 4 1/2.

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How to solve problems on this website
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What is the area of this triangle?​
Burka [1]

Answer:it’s 21 square units

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8 0
3 years ago
How do you prove each of the following theorems using either a two-column, paragraph, or flow chart proof?
lilavasa [31]

All the theorems are proved as follows.

<h3>What is a Triangle ?</h3>

A triangle is a polygon with three sides , three vertices and three angles.

1. The Triangle sum Theorem

According to the Triangle Sum Theorem, the sum of a triangle's angles equals 180 degrees.

To create a triangle ABC, starting at point A, move 180 degrees away from A to arrive at point B.

We turn 180 degrees from B to C and 180 degrees from C to return to A, giving a total turn of 360 degrees to arrive to A.

180° - ∠A + 180° - ∠B + 180° - ∠C = 360°

- ∠A - ∠B  - ∠C = 360° - (180°+ 180°+ 180°) = -180°

∠A + ∠B  + ∠C = 180°

(Hence Proved)

2. Isosceles Triangle Theorem

Considering an isosceles triangle ΔABC

with AB = AC, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as AB = AC

sin B = sin C

angle B = angle C

3.Converse of the Isosceles theorem

Consider an isosceles triangle ΔABC with ∠B= ∠C, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as  ∠B= ∠C ,

AB = AC

4. Midsegment of a triangle theorem

It states that the midsegment of two sides of a triangle is equal to (1/2)of the third side parallel to it.

Given triangle ABC with midsegment at D and F of AB and AC respectively, DF is parallel to BC

In ΔABC and ΔADF

∠A ≅ ∠A

BA = 2 × DA, BC = 2 × FA

Hence;

ΔABC ~ ΔADF (SAS similarity)

BA/DA = BC/FA = DF/AC = 2

Hence AC = 2×DF

5.Concurrency of Medians Theorem

A median of a triangle is a segment whose end points are on vertex of the triangle and the middle point of the side ,the medians of a triangle are concurrent and  the point of intersection is inside the triangle known as Centroid .

Consider a triangle ABC , X,Y and Z are the midpoints of the sides

Since the medians bisect the segment AB into AZ + ZB

BC into BX + XB

AC into AY + YC

Where:

AZ = ZB

BX = XB

AY = YC

AZ/ZB = BX/XB = AY/YC = 1

AZ/ZB × BX/XB × AY/YC = 1 and

the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).

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