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3 years ago

Find the area of an equilateral triangle with a side of 10cm

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

8 0

Answer:

Since it is an equilateral triangle, all sides are 10cm

Step-by-step explanation:

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goblinko [34]

First find the area of the rectangle by multiplying the length and width of the rectangle, so 12 x25=300. Then find the area of the triangle, which is length x height/ 2. So 8 x 25=200/2=100. Then add the two answers together to get 400.

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3 years ago

Solve. Show all steps. 1/8 x = 1/2

Law Incorporation [45]

Answer:

$\frac{1}{8} x = \frac{1}{2} \\ x = \frac{8}{2} \\ \boxed{x = 4}$

<h3>x=4 is the right answer.</h3>

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3 years ago

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QveST [7]

I want to say A because 168 tenths should equal 168 thousandths

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3 years ago

The slope-intercept form of a linear equation is y=mx + b, where x and y are

HACTEHA [7]

Answer:

m = $\frac{y-b}{x}$

Step-by-step explanation:

Given

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4 years ago

Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!

blagie [28]

Question 4

1) $\overline{BD}$ bisects $\angle ABC$ , $\overline{EF} \perp \overline{AB}$ , and $\overline{EG} \perp \overline{BC}$ (given)

2) $\angle FBE \cong \angle GBE$ (an angle bisector splits an angle into two congruent parts)

3) $\angle BFE$ and $\angle BGE$ are right angles (perpendicular lines form right angles)

4) $\triangle BFE$ and $\triangle BGE$ are right triangles (a triangle with a right angle is a right triangle)

5) $\overline{BE} \cong \overline{BE}$ (reflexive property)

6) $\triangle BFE \cong \triangle BGE$ (HA)

Question 5

1) $\angle AXO$ and $\angle BYO$ are right angles, $\angle A \cong \angle B$ , $O$ is the midpoint of $\overline{AB}$ (given)

2) $\triangle AXO$ and $\triangle BYO$ are right triangles (a triangle with a right angle is a right triangle)

3) $\overline{AO} \cong \overline{OB}$ (a midpoint splits a segment into two congruent parts)

4) $\triangle AXO \cong \triangle BYO$ (HA)

5) $\overline{OX} \cong \overline{OY}$ (CPCTC)

Question 6

1) $\angle B$ and $\angle D$ are right angles, $\overline{AC}$ bisects $\angle BAD$ (given)

2) $\overline{AC} \cong \overline{AC}$ (reflexive property)

3) $\angle BAC \cong \angle CAD$ (an angle bisector splits an angle into two congruent parts)

4) $\triangle BAC$ and $\triangle CAD$ are right triangles (a triangle with a right angle is a right triangle)

5) $\triangle BAC \cong \triangle DCA$ (HA)

6) $\angle BCA \cong \angle DCA$ (CPCTC)

7) $\overline{CA}$ bisects $\angle ACD$ (if a segment splits an angle into two congruent parts, it is an angle bisector)

Question 7

1) $\angle B$ and $\angle C$ are right angles, $\angle 4 \cong \angle 1$ (given)

2) $\triangle BAD$ and $\triangle CAD$ are right triangles (definition of a right triangle)

3) $\angle 1 \cong \angle 3$ (vertical angles are congruent)

4) $\angle 4 \cong \angle 3$ (transitive property of congruence)

5) $\overline{AD} \cong \overline{AD}$ (reflexive property)

6) $\therefore \triangle BAD \cong \triangle CAD$ (HA theorem)

7) $\angle BDA \cong \angle CDA$ (CPCTC)

8) $\therefore \vec{DA}$ bisects $\angle BDC$ (definition of bisector of an angle)

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2 years ago