Someone help with this please

You know the formula to find the area of a triangle is B*h/2. If the area is $15m^{2}$ . Multiply 15 by 2...........which gives you 30 divide 30 by 3 which gives you 10.

Answer: 10

3 0

3 years ago

Which of the following shows that EFGH is a parallelogram for y=7 and z=9?

Anna35 [415]

Option B: $m \angle F=m \angle G=120^{\circ}$ and $m \angle E=60^{\circ}$ , so $\angle E$ is supplementary to both $\angle F$ and $\angle G$ , so EFGH is a parallelogram.

Option C: $m \angle F=m \angle G=120^{\circ}$ so EFGH is a parallelogram.

Option D: $m \angle E+m \angle G=180^{\circ}$ so EFGH is a parallelogram.

Explanation:

Option A: $m \angle E=m \angle F=60^{\circ}$ and $m \angle G=120^{\circ}$ so $\angle G$ is supplementary to both $\angle E$ and $\angle F$ , so EFGH is a parallelogram

Let us substitute $y=7$ and $z=9$ in $m \angle E=(7y+11)^{\circ}$ , $m \angle F=(17y+1)^{\circ}$ and $m \angle G=(14z-6)^{\circ}$ to determine the exact measures the angles of the parallelogram.

Substituting, we get, $m \angle E=60^{\circ}$ , $m \angle F=m \angle G=120^{\circ}$

Thus, $m \angle E\neq m \angle F$ because the measures of these angles are not equal.

Hence, Option A is not the correct answer.

Option B: $m \angle F=m \angle G=120^{\circ}$ and $m \angle E=60^{\circ}$ , so $\angle E$ is supplementary to both $\angle F$ and $\angle G$ , so EFGH is a parallelogram.

Let us substitute $y=7$ and $z=9$ in $m \angle E=(7y+11)^{\circ}$ , $m \angle F=(17y+1)^{\circ}$ and $m \angle G=(14z-6)^{\circ}$ to determine the exact measures the angles of the parallelogram.

Thus, substituting, we have, $m \angle E=60^{\circ}$ , $m \angle F=m \angle G=120^{\circ}$

Hence, Option B is the correct answer.

Option C: $m \angle F=m \angle G=120^{\circ}$ so EFGH is a parallelogram.

To determine the angles, let us substitute $z=9$ in $m \angle F=(17y+1)^{\circ}$ and $m \angle G=(14z-6)^{\circ}$

Thus, $m \angle F=m \angle G=120^{\circ}$

Since, the opposite angles of a parallelogram are equal, EFGH is a parallelogram.

Hence, Option C is the correct answer.

Option D: $m \angle E+m \angle G=180^{\circ}$ so EFGH is a parallelogram.

Let us substitute $y=7$ and $z=9$ in $m \angle E=(7y+11)^{\circ}$ , $m \angle F=(17y+1)^{\circ}$ and $m \angle G=(14z-6)^{\circ}$ to determine the exact measures the angles of the parallelogram.

Substituting, we have, $m \angle E=60^{\circ}$ , $m \angle F=m \angle G=120^{\circ}$

Adding the angles E and G, we have,

$m \angle E+m \angle G=60^{\circ}+120^{\circ}=180^{\circ}$

By the property of parallelogram, any two adjacent angles add upto 180.

Thus, the adjacent angles E and G add upto 180.

Hence, Option D is the correct answer.

3 0

4 years ago