Worksheet with four multi-step Pythagorean problems requiring students to find lengths using the Pythagorean theorem.

I will solve each of the four multi-step Pythagorean problems shown in the image. I'll work through them one by one, explaining the solution for each.

Problem 1: Find the length of x

First, look closely:
- The diagram shows a large triangle split into two right-angled triangles by a perpendicular line.
- The left right-angled triangle has a hypotenuse of 13 and a base of 5.
- The right right-angled triangle has a base of 35 and we need to find the hypotenuse, labeled x.
- Both right-angled triangles share the same height (the perpendicular line).

Next, find information:
I need to use the Pythagorean theorem twice.
1. First, find the height of the left triangle using its known sides.
2. Then, use that height with the base of 35 to find x.

Then, review the findings:
The Pythagorean theorem states that for a right-angled triangle, $a^2 + b^2 = c^2$, where c is the hypotenuse.

For the left triangle:
- Let h be the height.
- $h^2 + 5^2 = 13^2$
- $h^2 + 25 = 169$
- $h^2 = 144$
- $h = 12$

Now for the right triangle:
- $h^2 + 35^2 = x^2$
- $12^2 + 35^2 = x^2$
- $144 + 1225 = x^2$
- $x^2 = 1369$
- $x = \sqrt{1369} = 37$

Solution for Problem 1: x = 37

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Problem 2: Find the length of y (to 1 d.p.)

First, look closely:
- The diagram shows a quadrilateral made of two right-angled triangles sharing a common side.
- The left triangle has sides 11 and 16.
- The right triangle has sides y and 24, with the hypotenuse being the shared side from the left triangle.

Next, find information:
I need to use the Pythagorean theorem twice.
1. First, find the length of the shared hypotenuse from the left triangle.
2. Then, use that length as the hypotenuse for the right triangle to find y.

Then, review the findings:
For the left triangle:
- Let d be the shared diagonal (hypotenuse).
- $d^2 = 11^2 + 16^2$
- $d^2 = 121 + 256 = 377$
- $d = \sqrt{377}$

For the right triangle:
- $d^2 = y^2 + 24^2$
- $377 = y^2 + 576$
- $y^2 = 377 - 576 = -199$

Wait, this is impossible. I must have misinterpreted the diagram.

Let me re-examine the diagram. The right triangle has sides y and 24, and the shared side is one of its legs, not the hypotenuse. The right angle is at the top, so the shared side is the hypotenuse of the left triangle and a leg of the right triangle.

Actually, looking again, the right triangle has sides y and 24, with the shared side being the hypotenuse of the left triangle. But if 24 is a leg and y is the other leg, then the hypotenuse would be longer than 24. However, the shared side is $\sqrt{377} \approx 19.4$, which is less than 24. This suggests my interpretation is wrong.

Let me try another interpretation: perhaps the 24 is the hypotenuse of the right triangle, and y and the shared side are the legs.

So for the right triangle:
- $y^2 + (\sqrt{377})^2 = 24^2$
- $y^2 + 377 = 576$
- $y^2 = 576 - 377 = 199$
- $y = \sqrt{199} \approx 14.1$ (to 1 d.p.)

This makes sense because 14.1 and ~19.4 are both less than 24.

Solution for Problem 2: y ≈ 14.1

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Problem 3: Find the length of z in the form $a\sqrt{b}$

First, look closely:
- The diagram shows two right-angled triangles sharing a common vertex.
- The larger triangle has sides 18 and 20, with 20 being the hypotenuse.
- The smaller triangle has sides z and 2, with the shared side being the base of the larger triangle.

Next, find information:
I need to use the Pythagorean theorem twice.
1. First, find the base of the larger triangle.
2. Then, use that base and the side of length 2 to find z.

Then, review the findings:
For the larger triangle:
- Let b be the base.
- $18^2 + b^2 = 20^2$
- $324 + b^2 = 400$
- $b^2 = 76$
- $b = \sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19}$

For the smaller triangle:
- $z^2 + 2^2 = b^2$
- $z^2 + 4 = 76$
- $z^2 = 72$
- $z = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

Solution for Problem 3: z = $6\sqrt{2}$

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Problem 4: Find the length of a

First, look closely:
- The diagram shows a shape made of four similar isosceles triangles.
- One of the smallest triangles has a leg of length 1.
- We need to find the length of a, which is the hypotenuse of the largest triangle.

Next, find information:
Since the triangles are similar isosceles right-angled triangles, each subsequent triangle's legs are the hypotenuse of the previous triangle.

Let me trace the pattern:
- Triangle 1: legs = 1, hypotenuse = $\sqrt{1^2 + 1^2} = \sqrt{2}$
- Triangle 2: legs = $\sqrt{2}$, hypotenuse = $\sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2$
- Triangle 3: legs = 2, hypotenuse = $\sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
- Triangle 4: legs = $2\sqrt{2}$, hypotenuse = $\sqrt{(2\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$

So a = 4.

Solution for Problem 4: a = 4

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## Final Answers:

1) x = 37
2) y ≈ 14.1
3) z = $6\sqrt{2}$
4) a = 4