Question

directions: find each missing measure
1
2.
3.
4.
directions: find the value of each variable.
5.
6.
7.
8.
9.
10.

Explanation:

Step1: Analyze problem 7

Since the triangle has two equal sides (marked with the same tick - marks), it is an isosceles triangle. In an isosceles triangle, the base angles are equal. So, \(5x - 7=8x - 55\).

Step2: Solve the equation for \(x\)

Subtract \(5x\) from both sides: \(-7 = 3x-55\).
Add \(55\) to both sides: \(3x=48\).
Divide both sides by \(3\): \(x = 16\).

Step3: Analyze problem 8

Since the triangle has three equal sides (marked with the same tick - marks), it is an equilateral triangle. In an equilateral triangle, each angle is \(60^{\circ}\). So, \(4x + 8=60\).

Step4: Solve the equation for \(x\)

Subtract \(8\) from both sides: \(4x=52\).
Divide both sides by \(4\): \(x = 13\).

Step5: Analyze problem 9

Since the triangle has two equal angles (\(60^{\circ}\) each), it is an equilateral triangle. So, \(7x + 19=11x-89\).

Step6: Solve the equation for \(x\)

Subtract \(7x\) from both sides: \(19 = 4x-89\).
Add \(89\) to both sides: \(4x=108\).
Divide both sides by \(4\): \(x = 27\).

Step7: Analyze problem 10

Since the triangle has two equal sides (marked with the same tick - marks), it is an isosceles triangle. The sum of the interior angles of a triangle is \(180^{\circ}\). Let the equal angles be \(y\). Then \(y + y+29^{\circ}=180^{\circ}\), so \(2y = 151^{\circ}\), \(y=(11x - 65)^{\circ}\). Also, using the exterior angle property (not necessary here, but for an isosceles triangle with base angles equal).
Since it is isosceles, \(11x-65=(180 - 29)\div2=75.5\).
\(11x=75.5 + 65\).
\(11x=140.5\).
\(x = 12.77\) (approx). But if we use the property that in an isosceles triangle (two equal sides), the base angles are equal. The angle adjacent to \((11x - 65)^{\circ}\) (using the fact that the sum of angles in a triangle is \(180^{\circ}\)):
\(29+(11x - 65)+(11x - 65)=180\) (if we assume the non - equal angle is \(29^{\circ}\)).
\(29+22x-130 = 180\).
\(22x=180 + 101\).
\(22x=281\).
\(x=\frac{281}{22}\approx12.77\). But if we use the exterior angle property (the exterior angle is equal to the sum of the two non - adjacent interior angles. But since it's isosceles, assume the two base angles are equal. Let's re - check:
The sum of angles in a triangle \(A + B + C=180\). Let \(A = 29^{\circ}\), \(B = C=(11x - 65)^{\circ}\).
\(29+2(11x - 65)=180\).
\(29+22x-130 = 180\).
\(22x=180 + 101\).
\(22x=281\).
\(x = 12.77\) (approx). But if we consider the side - angle relationship for isosceles triangle (two equal sides), we can also use the law of sines. \(\frac{\sin29}{a}=\frac{\sin(11x - 65)}{b}\), and \(a = b\) (equal sides), so \(\sin29=\sin(11x - 65)\). Then \(11x-65 = 29\) (since \(11x-65\) and \(29\) are angles in a triangle, \(11x-65
eq180 - 29\) as \(11x-65+29<180\)).
\(11x=29 + 65\).
\(11x=94\).
\(x=\frac{94}{11}\approx8.55\) (wrong, because if \(x = 8.55\), \(11x-65=11\times8.55-65=94 - 65 = 29\), but then the third angle would be \(180-(29 + 29)=122\) which is not possible as the side opposite \(122\) would be the longest, but we have two equal sides. Wait, no, if \(x = 8.55\), \(11x-65 = 29\), then the triangle has angles \(29^{\circ},29^{\circ},122^{\circ}\) which is valid for an isosceles triangle (two equal sides opposite the \(29^{\circ}\) angles).

Answer:

For problem 7: \(x = 16\)
For problem 8: \(x = 13\)
For problem 9: \(x = 27\)
For problem 10: \(x=\frac{94}{11}\approx8.55\)