50+ Pythagorean Theorem worksheets for 2nd Grade on Quizizz | Free ... - Free Printable

Let's solve each of the 12 problems using Pythagoras' Theorem and related geometry concepts. We'll go step-by-step.

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🔷 Pythagoras’ Theorem:

For a right-angled triangle:
$$
a^2 + b^2 = c^2
$$
where $c$ is the hypotenuse (longest side), and $a$, $b$ are the other two sides.

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✔ Problem 1: Calculate $x$

Right triangle with legs: 8 cm and 15 cm

$$
x^2 = 8^2 + 15^2 = 64 + 225 = 289 \\
x = \sqrt{289} = 17\text{ cm}
$$

✔ Answer: $x = 17$ cm

---

✔ Problem 2: Calculate $x$ to 2 d.p.

Legs: 9 cm and 4 cm

$$
x^2 = 9^2 + 4^2 = 81 + 16 = 97 \\
x = \sqrt{97} \approx 9.85\text{ cm}
$$

✔ Answer: $x = 9.85$ cm

---

✔ Problem 3: Calculate $x$ to 1 d.p.

Legs: 7.4 cm and 8.7 cm

$$
x^2 = 7.4^2 + 8.7^2 = 54.76 + 75.69 = 130.45 \\
x = \sqrt{130.45} \approx 11.42 \Rightarrow \text{to 1 d.p. } 11.4\text{ cm}
$$

✔ Answer: $x = 11.4$ cm

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✔ Problem 4: Calculate $x$, leave as a fraction

Legs: $\frac{4}{5}$ m and $\frac{1}{3}$ m

$$
x^2 = \left(\frac{4}{5}\right)^2 + \left(\frac{1}{3}\right)^2 = \frac{16}{25} + \frac{1}{9}
$$

Find common denominator: $225$

$$
= \frac{16 \times 9}{225} + \frac{1 \times 25}{225} = \frac{144 + 25}{225} = \frac{169}{225} \\
x = \sqrt{\frac{169}{225}} = \frac{13}{15}\text{ m}
$$

✔ Answer: $x = \frac{13}{15}$ m

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✔ Problem 5: Find $x$, leave in exact form

Isosceles right triangle with legs = 6 cm

So both legs are 6 cm

$$
x^2 = 6^2 + 6^2 = 36 + 36 = 72 \\
x = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\text{ cm}
$$

✔ Answer: $x = 6\sqrt{2}$ cm

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✔ Problem 6: Calculate the perimeter to 1 d.p.

We have a right triangle with base = 7 + 3 = 10 cm, height = 5 cm

But we need to find the hypotenuse of the large triangle (from bottom-left to top-right).

Wait — actually, it’s a right triangle with:
- Base: 10 cm (7+3)
- Height: 5 cm

So hypotenuse:

$$
x^2 = 10^2 + 5^2 = 100 + 25 = 125 \\
x = \sqrt{125} = 5\sqrt{5} \approx 11.18\text{ cm}
$$

Perimeter = 10 + 5 + 11.18 ≈ 26.18 → to 1 d.p. = 26.2 cm

✔ Answer: Perimeter = 26.2 cm

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✔ Problem 7: Calculate length BD of rectangle to 3 s.f.

Rectangle ABCD: AD = 4 cm, DC = 11 cm

Diagonal BD is hypotenuse of triangle BCD or ABD

$$
BD^2 = 4^2 + 11^2 = 16 + 121 = 137 \\
BD = \sqrt{137} \approx 11.704 \Rightarrow \text{3 s.f. } 11.7\text{ cm}
$$

✔ Answer: $BD = 11.7$ cm

---

✔ Problem 8: Area = $63\text{ cm}^2$, calculate $x$ to 2 d.p.

Equilateral triangle? Wait — has two equal sides marked (x), and base = 7 cm.

So it’s an isosceles triangle with equal sides $x$, base = 7 cm, area = 63 cm².

Use area formula:
$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
$$

Let $h$ be the height from apex to base (7 cm). Then:
$$
63 = \frac{1}{2} \times 7 \times h \Rightarrow 63 = 3.5h \Rightarrow h = \frac{63}{3.5} = 18\text{ cm}
$$

Now use Pythagoras on half-triangle:
Half-base = $3.5$ cm, height = 18 cm

$$
x^2 = 3.5^2 + 18^2 = 12.25 + 324 = 336.25 \\
x = \sqrt{336.25} \approx 18.336 \Rightarrow \text{to 2 d.p. } 18.34\text{ cm}
$$

✔ Answer: $x = 18.34$ cm

---

✔ Problem 9: Calculate perimeter to 1 d.p.

Shape: Symmetrical pentagon-like figure, with vertical height = 7 cm, horizontal base = 10 cm, left/right sides = 5 cm.

It appears to be a symmetrical trapezoid with two slanted sides and two vertical sides?

Wait — let’s interpret carefully.

From the diagram:
- Bottom base: 10 cm
- Two vertical sides: 5 cm each? No — wait, the marks show that the slanted sides are equal, and there are two right angles at the bottom corners.
- Vertical line from top to bottom is 7 cm
- Horizontal base is 10 cm
- Two equal slanted sides connect top point to ends of base

So this is an isosceles triangle with base 10 cm, height 7 cm, and two equal sides (which we need to find)

But wait — no, the figure has two vertical lines of 5 cm? Wait, the image shows:

- Two right angles at bottom corners
- Left and right sides are 5 cm each (vertical?)
- But then top connects to center?

Actually, it looks like a trapezium or house-shaped figure.

Wait — better interpretation:

It's a symmetrical shape with:
- Bottom base: 10 cm
- Two vertical sides: 5 cm each? But they’re shown as equal and perpendicular to base?
- Top connects them with a horizontal line?

Wait — no, the top is not horizontal — it's a point, so it's a triangle?

Wait — look again: There’s a vertical line of 7 cm, and two equal slanted sides (marked with ticks), and two right angles at the bottom corners.

Ah! It’s a symmetrical shape with:
- Base: 10 cm
- Two equal slanted sides (top edges) meeting at a peak
- The vertical height from peak to base is 7 cm
- And two right angles at the bottom corners?

That can't be — unless it's a rectangle with a triangle on top?

Wait — maybe not.

Alternative: The figure is a kite or arrowhead?

Wait — more likely: It's a symmetrical polygon with:
- Bottom base: 10 cm
- Two equal slanted sides from ends of base to top point
- Height from top to base is 7 cm
- And two right angles at the base?

No — if the height is 7 cm and the base is 10 cm, and the two sides are symmetric, then the horizontal projection of each slanted side is $5$ cm (half of 10).

So each slanted side forms a right triangle with:
- Base: 5 cm
- Height: 7 cm

So length of each slanted side:

$$
x = \sqrt{5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74} \approx 8.602\text{ cm}
$$

Now, total perimeter = 2 × slanted side + base = $2 \times \sqrt{74} + 10$

$$
= 2 \times 8.602 + 10 = 17.204 + 10 = 27.204 \Rightarrow \text{to 1 d.p. } 27.2\text{ cm}
$$

✔ Answer: Perimeter = 27.2 cm

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✔ Problem 10: Calculate shaded area to 2 d.p.

Circle with diameter QR, and triangle PQR inscribed with right angle at P.

Given: $PQ = 10$ cm, $PR = 5$ cm

Since angle at P is 90°, and it's on the circle, then QR must be the diameter.

So triangle PQR is right-angled at P.

Area of triangle:
$$
A_{\text{triangle}} = \frac{1}{2} \times PQ \times PR = \frac{1}{2} \times 10 \times 5 = 25\text{ cm}^2
$$

Now, find radius of circle.

First, find hypotenuse QR:
$$
QR^2 = 10^2 + 5^2 = 100 + 25 = 125 \Rightarrow QR = \sqrt{125} = 5\sqrt{5} \text{ cm}
$$

So radius $r = \frac{QR}{2} = \frac{5\sqrt{5}}{2} \approx \frac{5 \times 2.236}{2} \approx 5.59\text{ cm}$

Area of circle:
$$
A_{\text{circle}} = \pi r^2 = \pi \times \left(\frac{125}{4}\right) = \frac{125\pi}{4} \approx 98.17\text{ cm}^2
$$

Shaded area = Circle - Triangle = $98.17 - 25 = 73.17\text{ cm}^2$

✔ Answer: Shaded area = 73.17 cm²

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✔ Problem 11: Calculate $x$ to 2 d.p.

Right triangle with:
- Vertical leg: 9 cm
- Horizontal leg: 15 cm
- But there’s a perpendicular from top vertex to base: 6 cm

Wait — this is a triangle with a height dropped from the top to the base.

But the triangle has:
- Base = 15 cm
- Height = 6 cm
- But also a vertical side of 9 cm?

Wait — the diagram shows:
- A triangle with base 15 cm, vertical height 9 cm
- But a perpendicular of 6 cm is drawn from the top vertex down to the base?

No — let’s read:

The triangle has:
- One side = 9 cm (vertical)
- Base = 15 cm
- A perpendicular from the top vertex to the base is 6 cm?

Wait — but if one side is 9 cm vertical, and base is 15 cm, then the height from top to base should be 9 cm?

Contradiction.

Wait — perhaps it's a right triangle with legs 9 cm and 15 cm? But no — the diagram shows a perpendicular of 6 cm from the top vertex to the base.

Ah — the triangle has:
- Base = 15 cm
- Height from opposite vertex = 6 cm
- But one side is 9 cm

Wait — maybe it’s a triangle with base 15 cm, and a perpendicular from the top vertex of 6 cm, but the side from top to end is 9 cm?

Wait — let’s interpret:

There’s a triangle with:
- Base = 15 cm
- A perpendicular from top vertex to base = 6 cm
- So height = 6 cm
- And one of the other sides is 9 cm

But the hypotenuse $x$ is labeled from top to the far end?

Wait — the diagram shows:
- Vertical side: 9 cm
- Horizontal base: 15 cm
- A perpendicular of 6 cm is drawn from top to base — so it splits the base into two parts

Let me assume:

Triangle with:
- One vertex at top
- Base = 15 cm
- From top, drop perpendicular to base = 6 cm
- So the height is 6 cm
- But one of the legs is 9 cm?

Wait — the vertical side is labeled 9 cm — so perhaps the triangle has:
- Vertical leg = 9 cm
- Horizontal leg = ??
- But base is 15 cm?

No — the 9 cm is the total vertical, and the height is 6 cm?

Wait — confusion.

Looking at the diagram:
- Vertical side: 9 cm
- Base: 15 cm
- A perpendicular of 6 cm is drawn from the top vertex to the base — so it divides the base into two segments

Let’s suppose:
- Triangle has base 15 cm
- Height from top vertex to base is 6 cm
- But one of the sides (say, from top to left end) is 9 cm

Then we can use Pythagoras.

Let’s say:
- Let the foot of the perpendicular be at distance $a$ from left end
- Then from top to left end: $x = 9$ cm
- Horizontal component: $a$
- Vertical component: 6 cm

So:
$$
a^2 + 6^2 = 9^2 \Rightarrow a^2 + 36 = 81 \Rightarrow a^2 = 45 \Rightarrow a = \sqrt{45} = 3\sqrt{5} \approx 6.708\text{ cm}
$$

Then the other segment is $15 - a \approx 15 - 6.708 = 8.292$ cm

But we are to find $x$ — which is the hypotenuse of the entire triangle?

Wait — the label $x$ is on the long side from top to bottom-right corner.

So from top to right end: horizontal = $15 - a$, vertical = 6 cm

So:
$$
x^2 = (15 - a)^2 + 6^2
$$

We already have $a = \sqrt{45}$, so $15 - a = 15 - \sqrt{45}$

But better to compute numerically:

$a \approx 6.708$, so $15 - a \approx 8.292$

Then:
$$
x^2 = 8.292^2 + 6^2 \approx 68.77 + 36 = 104.77 \\
x \approx \sqrt{104.77} \approx 10.236 \Rightarrow \text{to 2 d.p. } 10.24\text{ cm}
$$

Wait — but is this correct?

Alternatively, maybe the triangle has:
- Right angle at bottom-left
- Vertical leg = 9 cm
- Horizontal leg = 15 cm
- But a perpendicular of 6 cm is drawn from top to base?

No — that doesn’t make sense.

Wait — the diagram shows:
- A triangle with vertical side 9 cm, base 15 cm
- A perpendicular from the top vertex to the base = 6 cm
- But the side from top to right end is labeled $x$

So yes — we can use coordinates.

Let’s place:
- Point A at origin (0,0)
- Point C at (15,0)
- Point B at (0,9) — but then height from B to AC would be 9 cm, not 6 cm

But the diagram shows a perpendicular of 6 cm from the top vertex to the base — so height is 6 cm.

So the triangle has:
- Base = 15 cm
- Height = 6 cm
- But one side is 9 cm

So let’s suppose:
- Let the top vertex be at height 6 cm above base
- Let the foot of perpendicular be at distance $a$ from left
- Then distance from top to left end = 9 cm

So:
$$
a^2 + 6^2 = 9^2 \Rightarrow a^2 = 81 - 36 = 45 \Rightarrow a = \sqrt{45} \approx 6.708
$$

Then distance from top to right end:
- Horizontal = $15 - a \approx 8.292$
- Vertical = 6 cm

So:
$$
x^2 = (15 - a)^2 + 6^2 = (8.292)^2 + 36 \approx 68.77 + 36 = 104.77 \\
x \approx \sqrt{104.77} \approx 10.236 \Rightarrow \boxed{10.24}\text{ cm}
$$

✔ Answer: $x = 10.24$ cm

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✔ Problem 12: Area of hexagon is $24\text{ cm}^2$, calculate $x$ to 2 d.p.

Regular hexagon? All sides equal, marked with ticks.

Yes — regular hexagon with side length $x$

Area of regular hexagon:
$$
A = \frac{3\sqrt{3}}{2} x^2
$$

Set equal to 24:
$$
\frac{3\sqrt{3}}{2} x^2 = 24 \\
x^2 = \frac{24 \times 2}{3\sqrt{3}} = \frac{48}{3\sqrt{3}} = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3}
$$

$$
x = \sqrt{\frac{16\sqrt{3}}{3}} \quad \text{— but better to compute numerically}
$$

First, approximate:
$$
\frac{3\sqrt{3}}{2} \approx \frac{3 \times 1.732}{2} = \frac{5.196}{2} = 2.598
$$

So:
$$
2.598 x^2 = 24 \Rightarrow x^2 = \frac{24}{2.598} \approx 9.237 \\
x \approx \sqrt{9.237} \approx 3.039 \Rightarrow \text{to 2 d.p. } 3.04\text{ cm}
$$

✔ Answer: $x = 3.04$ cm

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📌 Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $17$ cm |
| 2 | $9.85$ cm |
| 3 | $11.4$ cm |
| 4 | $\frac{13}{15}$ m |
| 5 | $6\sqrt{2}$ cm |
| 6 | $26.2$ cm |
| 7 | $11.7$ cm |
| 8 | $18.34$ cm |
| 9 | $27.2$ cm |
| 10 | $73.17$ cm² |
| 11 | $10.24$ cm |
| 12 | $3.04$ cm |

Let me know if you’d like a visual explanation for any!