Question

** this is a 2-page document
directions: find the value of x.
1.
right triangle with legs 10 and 7, hypotenuse x
2.
right triangle with legs x and 19, hypotenuse 21
segment with tick marks
3.
right triangle with leg 16, hypotenuse 27, leg x
4.
right triangle with leg 5.3, hypotenuse 12.8, leg x
5.
isosceles triangle with height 18, base segment 20, equal sides x
6.
isosceles trapezoid with bases x and 31, legs 19, height 17
7.
isosceles triangle with base 48, left segment 16, left leg 29, right leg x

1. scott is using a 12-foot ramp to help load furniture into the back of a moving truck. if the back of the truck is 3.5 feet from the ground, what is the horizontal distance from where the ramp reaches the ground to the truck?

Explanation:

Step1: Identify the triangle type (Right Triangle)

All triangles here are right - angled, so we use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse, and \(a,b\) are the legs.

Problem 1:

Step2: Apply Pythagorean theorem

For the first right triangle with legs \(a = 10\) and \(b=7\), we want to find the hypotenuse \(x\).
Using \(x^{2}=10^{2}+7^{2}\)
\(x^{2}=100 + 49=149\)
\(x=\sqrt{149}\approx12.21\)

Problem 2:

Step2: Apply Pythagorean theorem (find leg)

For the right triangle with hypotenuse \(c = 21\) and leg \(a = 19\), we want to find leg \(x\).
Using \(x^{2}+19^{2}=21^{2}\)
\(x^{2}=21^{2}-19^{2}=(21 + 19)(21 - 19)=40\times2 = 80\)
\(x=\sqrt{80}\approx8.94\)

Problem 3:

Step2: Apply Pythagorean theorem (find leg)

For the right triangle with hypotenuse \(c = 27\) and leg \(a = 16\), we want to find leg \(x\).
Using \(x^{2}+16^{2}=27^{2}\)
\(x^{2}=27^{2}-16^{2}=(27 + 16)(27 - 16)=43\times11 = 473\)
\(x=\sqrt{473}\approx21.75\)

Problem 4:

Step2: Apply Pythagorean theorem (find leg)

For the right triangle with hypotenuse \(c = 12.8\) and leg \(a = 5.3\), we want to find leg \(x\).
Using \(x^{2}+5.3^{2}=12.8^{2}\)
\(x^{2}=12.8^{2}-5.3^{2}=(12.8 + 5.3)(12.8 - 5.3)=18.1\times7.5 = 135.75\)
\(x=\sqrt{135.75}\approx11.65\)

Problem 5:

The triangle is isoceles (two equal sides marked). The height is \(18\), and the base of the right - triangle formed (by the height) is \(20\). Let the equal sides be \(x\).

Step2: Apply Pythagorean theorem

Using \(x^{2}=20^{2}+9^{2}\) (since the height is \(18\), half of it is \(9\)? Wait, no. Wait, the height is \(18\), and the segment from the vertex to the base is \(20\)? Wait, maybe I misread. Wait, the triangle has a height of \(18\) (the vertical segment) and a horizontal segment of \(20\) (the base of the right - triangle). So the equal sides \(x\):
\(x^{2}=20^{2}+9^{2}\)? No, wait, if the total height is \(18\), and the triangle is isoceles, the height bisects the base? Wait, the diagram shows a triangle with a height of \(18\) (the vertical line) and a horizontal segment of \(20\) (from the foot of the height to the vertex). So the leg of the right triangle is \(20\), the other leg is \(9\) (half of \(18\))? No, maybe the height is \(18\), and the horizontal segment is \(20\). So \(x^{2}=20^{2}+18^{2}\)? Wait, no, the right triangle has legs \(20\) and \(9\) (if the height is \(18\), half is \(9\))? Wait, maybe the diagram is a triangle with a height of \(18\), and the base of the right - triangle (formed by the height) is \(20\), and the height is \(18\). So \(x^{2}=20^{2}+9^{2}\) is wrong. Wait, perhaps the height is \(18\), and the horizontal segment is \(20\), so the right triangle has legs \(20\) and \(18\). Then \(x^{2}=20^{2}+18^{2}=400 + 324 = 724\), \(x=\sqrt{724}\approx26.91\). Wait, maybe I made a mistake. Let's re - examine. If the triangle is isoceles with two equal sides \(x\), and a height of \(18\) (the altitude), and the length of the base segment (from the foot of the altitude to the vertex) is \(20\). Then the right triangle has legs \(20\) and \(9\) (since the altitude is \(18\), half of it is \(9\))? No, that doesn't make sense. Maybe the altitude is \(18\), and the horizontal segment is \(20\), so the right triangle has legs \(20\) and \(18\), so \(x=\sqrt{20^{2}+18^{2}}=\sqrt{400 + 324}=\sqrt{724}\approx26.91\).

Problem 6:

The figure is a trapezoid. The formula for the area of a trapezoid is \(A=\frac{(a + b)h}{2}\), but we need to find \(x\). The trapezoid has the two non - parallel sides equal (marked with the same tick). So it is an isoceles trapezoid. The difference in the lengths of the two bases is \(31 - x\). Half of that difference is \(\frac{31 - x}{2}\). Then we can form a right triangle with legs \(17\) and \(\frac{31 - x}{2}\), and hypotenuse \(19\).
So \((\frac{31 - x}{2})^{2}+17^{2}=19^{2}\)
\((\frac{31 - x}{2})^{2}=19^{2}-17^{2}=(19 + 17)(19 - 17)=36\times2 = 72\)
\(\frac{31 - x}{2}=\sqrt{72}\approx8.485\)
\(31 - x\approx16.97\)
\(x\approx31 - 16.97 = 14.03\)

Problem 7:

The triangle has a base of \(44\), and a segment of \(16\) from the left vertex to the foot of the altitude. So the other segment of the base is \(44 - 16=28\). The left side of the triangle is \(29\). Let the altitude be \(h\). Then \(h^{2}+16^{2}=29^{2}\), \(h^{2}=29^{2}-16^{2}=(29 + 16)(29 - 16)=45\times13 = 585\), \(h=\sqrt{585}\approx24.19\). Then for the right side \(x\), \(x^{2}=h^{2}+28^{2}=585+784 = 1369\), \(x = 37\)

Problem 8:

This is a right triangle problem. The ramp is the hypotenuse \(c = 12\) feet, the height \(a = 3.5\) feet, and we need to find the horizontal distance \(b\).
Using \(b^{2}=c^{2}-a^{2}\)
\(b^{2}=12^{2}-3.5^{2}=144 - 12.25 = 131.75\)
\(b=\sqrt{131.75}\approx11.48\) feet

Answer:

1. \(\sqrt{149}\approx12.21\)
2. \(\sqrt{80}\approx8.94\)
3. \(\sqrt{473}\approx21.75\)
4. \(\sqrt{135.75}\approx11.65\)
5. \(\sqrt{724}\approx26.91\) (or correct based on diagram re - interpretation)
6. \(\approx14.03\)
7. \(37\)
8. \(\approx11.48\) feet