Pythagoras - Finding the hypotenuse: A worksheet with 12 problems using the Pythagorean theorem to calculate unknown sides in triangles and related shapes.
Worksheet with 12 Pythagoras problems involving finding the hypotenuse and other geometric calculations.
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Step-by-step solution for: Pythagoras and Right-Angled Trigonometry Practice
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Show Answer Key & Explanations
Step-by-step solution for: Pythagoras and Right-Angled Trigonometry Practice
Here are the step-by-step solutions for each problem on the worksheet.
Task: Calculate $x$.
* This is a right-angled triangle. We use Pythagoras' theorem: $a^2 + b^2 = c^2$.
* The sides next to the right angle are $8$ and $15$. The hypotenuse (longest side) is $x$.
* $8^2 + 15^2 = x^2$
* $64 + 225 = x^2$
* $289 = x^2$
* $\sqrt{289} = 17$
Answer: $x = 17 \text{ cm}$
---
Task: Calculate $x$ to 2 decimal places (d.p.).
* Sides are $9$ and $4$. Hypotenuse is $x$.
* $9^2 + 4^2 = x^2$
* $81 + 16 = x^2$
* $97 = x^2$
* $x = \sqrt{97} \approx 9.8488...$
* Rounding to 2 d.p.: The third decimal is 8, so we round up.
Answer: $x = 9.85 \text{ cm}$
---
Task: Calculate $x$ to 1 decimal place (d.p.).
* Sides are $7.4$ and $8.7$. Hypotenuse is $x$.
* $7.4^2 + 8.7^2 = x^2$
* $54.76 + 75.69 = x^2$
* $130.45 = x^2$
* $x = \sqrt{130.45} \approx 11.421...$
* Rounding to 1 d.p.: The second decimal is 2, so we keep it as is.
Answer: $x = 11.4 \text{ cm}$
---
Task: Calculate $x$, leave as a fraction.
* Sides are $\frac{1}{3}$ and $\frac{4}{5}$. Hypotenuse is $x$.
* $(\frac{1}{3})^2 + (\frac{4}{5})^2 = x^2$
* $\frac{1}{9} + \frac{16}{25} = x^2$
* Find a common denominator ($9 \times 25 = 225$):
* $\frac{1 \times 25}{225} + \frac{16 \times 9}{225} = x^2$
* $\frac{25}{225} + \frac{144}{225} = x^2$
* $\frac{169}{225} = x^2$
* Take the square root of the top and bottom separately:
* $x = \sqrt{\frac{169}{225}} = \frac{13}{15}$
Answer: $x = \frac{13}{15} \text{ m}$
---
Task: Find $x$ in exact form.
* This is an isosceles right-angled triangle. The two shorter sides are both $6$.
* $6^2 + 6^2 = x^2$
* $36 + 36 = x^2$
* $72 = x^2$
* $x = \sqrt{72}$
* Simplify the surd: $\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$
Answer: $x = 6\sqrt{2} \text{ cm}$
---
Task: Calculate the perimeter to 1 d.p.
* First, find the missing diagonal side using Pythagoras. Let's call it $h$.
* Base $= 7 + 3 = 10$. Height $= 5$.
* $10^2 + 5^2 = h^2$
* $100 + 25 = 125$
* $h = \sqrt{125} \approx 11.18$
* Next, find the other slanted side. Let's call it $s$. It sits over the base of $3$ and height of $5$.
* $3^2 + 5^2 = s^2$
* $9 + 25 = 34$
* $s = \sqrt{34} \approx 5.83$
* Perimeter $= \text{Bottom} + \text{Right Vertical} + \text{Long Slant} + \text{Short Slant}$?
* Wait, looking at the diagram, the shape is composed of two triangles joined together or one large triangle split. The perimeter is the outside boundary.
* Bottom side $= 7 + 3 = 10$.
* Right vertical side $= 5$.
* Long hypotenuse (left side) connects the far left point to the top. Base is $10$, height is $5$. Length $= \sqrt{10^2+5^2} = \sqrt{125} \approx 11.18$.
* The internal line is not part of the perimeter.
* Wait, let me look closer at the shape. It looks like a large triangle with a line drawn inside. The perimeter is the sum of the three outer sides.
* Side 1 (Bottom): $7 + 3 = 10$.
* Side 2 (Right): $5$.
* Side 3 (Left/Hypotenuse): Connects the end of the 7cm segment to the top. The total base is $10$, height is $5$. So hypotenuse $= \sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.18$.
* There is another side labeled? No, the diagram shows a triangle split into two. The perimeter is the outer boundary.
* Let's re-read the diagram carefully. It shows a large triangle. The base is split into $7$ and $3$. The height is $5$ on the right side? No, the right angle is on the bottom right of the smaller triangle.
* Okay, let's trace the perimeter:
1. Bottom horizontal line: $7 + 3 = 10$.
2. Right vertical line: $5$.
3. The long slanted line on the left: This is the hypotenuse of the large right triangle with base $10$ and height $5$. Length $= \sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.18$.
4. Is there a fourth side? The diagram shows a line splitting the triangle. That internal line is NOT part of the perimeter.
5. Wait, is the shape a quadrilateral? No, it looks like a single large triangle. But usually "perimeter" questions involve finding all outer edges.
6. Let's check if the left part is a separate triangle. If the vertex is shared, the left side is just the hypotenuse calculated above.
7. Let's assume the question asks for the perimeter of the entire large triangle.
8. Perimeter $= 10 + 5 + 11.18 = 26.18$.
9. Rounding to 1 d.p.: $26.2$.
*Alternative Interpretation:* Is it a trapezoid or irregular quad? The arrows indicate parallel lines? No. It looks like a standard right-angled triangle problem where you calculate the hypotenuse.
Let's double check the side labeled "7cm". It points to the segment on the left. The segment on the right is "3cm". The height is "5cm".
The perimeter consists of:
- Base: $7 + 3 = 10$
- Height: $5$
- Hypotenuse: $\sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.18$
Total $= 10 + 5 + 11.18 = 26.18 \rightarrow 26.2$.
Answer: $26.2 \text{ cm}$
---
Task: Calculate the length BD to 3 significant figures (s.f.).
* BD is the diagonal of the rectangle.
* Sides are $4$ and $11$.
* $4^2 + 11^2 = BD^2$
* $16 + 121 = 137$
* $BD = \sqrt{137} \approx 11.7046...$
* To 3 s.f.: The first three digits are 1, 1, 7. The next digit is 0, so we do not round up.
Answer: $11.7 \text{ cm}$
---
Task: Area is $63 \text{ cm}^2$, calculate $x$ to 2 d.p.
* This is an isosceles triangle (marks on the sides show they are equal length $x$).
* Base $= 7$.
* To find the area, we need the height. Let's split the triangle down the middle to create two right-angled triangles.
* The base of the small right triangle is half of $7$, which is $3.5$.
* The hypotenuse is $x$.
* Let height be $h$. By Pythagoras: $h^2 + 3.5^2 = x^2 \Rightarrow h = \sqrt{x^2 - 3.5^2}$.
* Area formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
* $63 = \frac{1}{2} \times 7 \times h$
* $63 = 3.5 \times h$
* $h = \frac{63}{3.5} = 18$.
* Now substitute $h$ back into the Pythagoras equation:
* $18^2 + 3.5^2 = x^2$
* $324 + 12.25 = x^2$
* $336.25 = x^2$
* $x = \sqrt{336.25} \approx 18.337...$
* Rounding to 2 d.p.: $18.34$.
Answer: $x = 18.34 \text{ cm}$
---
Task: Calculate the perimeter to 1 d.p.
* This is a pentagon shaped like a house.
* Bottom width $= 10$. Side walls $= 5$.
* The roof consists of two sloped sides. We need to find their length.
* Draw a vertical line from the peak to the base. This splits the roof into two right-angled triangles.
* The total height of the shape is $7$. The wall height is $5$. So the height of the triangular roof part is $7 - 5 = 2$.
* The base of the roof triangle is half the total width: $10 / 2 = 5$.
* Let the slope length be $s$.
* $s^2 = 5^2 + 2^2$
* $s^2 = 25 + 4 = 29$
* $s = \sqrt{29} \approx 5.385$
* There are two slopes, so $2 \times 5.385 = 10.77$.
* Perimeter $= \text{Bottom} + \text{Left Wall} + \text{Right Wall} + \text{Two Slopes}$
* Note: The diagram shows tick marks indicating the two roof slopes are equal, and the two vertical walls are equal.
* Perimeter $= 10 + 5 + 5 + \sqrt{29} + \sqrt{29}$
* Perimeter $= 20 + 2(5.385)$
* Perimeter $= 20 + 10.77 = 30.77$
* Rounding to 1 d.p.: $30.8$.
Answer: $30.8 \text{ cm}$
---
Task: Calculate the shaded area to 2 d.p.
* The shape is a circle with a triangle removed (or rather, the shaded area is the circle minus the white triangle? No, the white triangle is inside, and the rest is shaded).
* Actually, looking closely, $QPR$ is a triangle inscribed in a semi-circle? Or is $QR$ the diameter?
* Angle $P$ is marked as a right angle ($90^\circ$). A triangle inscribed in a circle with a right angle means the side opposite the right angle ($QR$) is the diameter.
* We are given $PQ = 10$ and $PR = 5$.
* First, find the diameter $QR$ using Pythagoras on triangle $PQR$:
* $QR^2 = 10^2 + 5^2 = 100 + 25 = 125$.
* $QR = \sqrt{125}$.
* The radius $r$ is half the diameter: $r = \frac{\sqrt{125}}{2}$.
* Area of the full circle $= \pi r^2 = \pi \left(\frac{\sqrt{125}}{2}\right)^2 = \pi \left(\frac{125}{4}\right) = 31.25\pi$.
* Area of the white triangle $PQR = \frac{1}{2} \times \text{base} \times \text{height}$. Since it's a right triangle at $P$, the legs are base and height.
* $\text{Area}_{\text{triangle}} = \frac{1}{2} \times 10 \times 5 = 25$.
* Shaded Area $= \text{Area of Circle} - \text{Area of Triangle}$.
* $\text{Shaded Area} = 31.25\pi - 25$.
* $31.25 \times 3.14159... \approx 98.1747...$
* $98.1747 - 25 = 73.1747...$
* Rounding to 2 d.p.: $73.17$.
Answer: $73.17 \text{ cm}^2$
---
Task: Calculate $x$ to 2 d.p.
* This shows two similar right-angled triangles. One is inside the other.
* Small triangle: Height $= 6$, Base $= ?$
* Large triangle: Height $= 9$, Base $= 15$.
* Wait, the diagram labels the *entire* base as $15$ and the *entire* height as $9$. The inner vertical line is $6$. The hypotenuse of the large triangle is labeled $x$.
* Let's check if the triangles are similar. Yes, they share the bottom-left angle and both have right angles.
* We can just use Pythagoras on the large triangle directly because we are given its full height and full base.
* Height $= 9$.
* Base $= 15$.
* Hypotenuse $= x$.
* $x^2 = 9^2 + 15^2$
* $x^2 = 81 + 225$
* $x^2 = 306$
* $x = \sqrt{306} \approx 17.4928...$
* Rounding to 2 d.p.: $17.49$.
*(Note: The inner dimension '6' is extra information or used to find the position of the vertical line, but isn't needed to find the outer hypotenuse $x$ if 9 and 15 are the total dimensions.)*
Answer: $x = 17.49 \text{ cm}$
---
Task: Area of hexagon is $24 \text{ cm}^2$, calculate $x$ to 2 d.p.
* A regular hexagon can be split into 6 equilateral triangles.
* Let the side length be $x$.
* The area of one equilateral triangle with side $x$ is $\frac{\sqrt{3}}{4}x^2$.
* Total Area of Hexagon $= 6 \times \left( \frac{\sqrt{3}}{4}x^2 \right) = \frac{3\sqrt{3}}{2}x^2$.
* We know Area $= 24$.
* $24 = \frac{3\sqrt{3}}{2}x^2$
* Multiply by 2: $48 = 3\sqrt{3}x^2$
* Divide by 3: $16 = \sqrt{3}x^2$
* Divide by $\sqrt{3}$: $x^2 = \frac{16}{\sqrt{3}}$
* $x^2 \approx \frac{16}{1.73205} \approx 9.2376$
* $x = \sqrt{9.2376} \approx 3.0393...$
* Rounding to 2 d.p.: $3.04$.
Answer: $x = 3.04 \text{ cm}$
──────────────────────────────────────
Final Answer:
1. 17 cm
2. 9.85 cm
3. 11.4 cm
4. 13/15 m
5. 6√2 cm
6. 26.2 cm
7. 11.7 cm
8. 18.34 cm
9. 30.8 cm
10. 73.17 cm²
11. 17.49 cm
12. 3.04 cm
Problem 1
Task: Calculate $x$.
* This is a right-angled triangle. We use Pythagoras' theorem: $a^2 + b^2 = c^2$.
* The sides next to the right angle are $8$ and $15$. The hypotenuse (longest side) is $x$.
* $8^2 + 15^2 = x^2$
* $64 + 225 = x^2$
* $289 = x^2$
* $\sqrt{289} = 17$
Answer: $x = 17 \text{ cm}$
---
Problem 2
Task: Calculate $x$ to 2 decimal places (d.p.).
* Sides are $9$ and $4$. Hypotenuse is $x$.
* $9^2 + 4^2 = x^2$
* $81 + 16 = x^2$
* $97 = x^2$
* $x = \sqrt{97} \approx 9.8488...$
* Rounding to 2 d.p.: The third decimal is 8, so we round up.
Answer: $x = 9.85 \text{ cm}$
---
Problem 3
Task: Calculate $x$ to 1 decimal place (d.p.).
* Sides are $7.4$ and $8.7$. Hypotenuse is $x$.
* $7.4^2 + 8.7^2 = x^2$
* $54.76 + 75.69 = x^2$
* $130.45 = x^2$
* $x = \sqrt{130.45} \approx 11.421...$
* Rounding to 1 d.p.: The second decimal is 2, so we keep it as is.
Answer: $x = 11.4 \text{ cm}$
---
Problem 4
Task: Calculate $x$, leave as a fraction.
* Sides are $\frac{1}{3}$ and $\frac{4}{5}$. Hypotenuse is $x$.
* $(\frac{1}{3})^2 + (\frac{4}{5})^2 = x^2$
* $\frac{1}{9} + \frac{16}{25} = x^2$
* Find a common denominator ($9 \times 25 = 225$):
* $\frac{1 \times 25}{225} + \frac{16 \times 9}{225} = x^2$
* $\frac{25}{225} + \frac{144}{225} = x^2$
* $\frac{169}{225} = x^2$
* Take the square root of the top and bottom separately:
* $x = \sqrt{\frac{169}{225}} = \frac{13}{15}$
Answer: $x = \frac{13}{15} \text{ m}$
---
Problem 5
Task: Find $x$ in exact form.
* This is an isosceles right-angled triangle. The two shorter sides are both $6$.
* $6^2 + 6^2 = x^2$
* $36 + 36 = x^2$
* $72 = x^2$
* $x = \sqrt{72}$
* Simplify the surd: $\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$
Answer: $x = 6\sqrt{2} \text{ cm}$
---
Problem 6
Task: Calculate the perimeter to 1 d.p.
* First, find the missing diagonal side using Pythagoras. Let's call it $h$.
* Base $= 7 + 3 = 10$. Height $= 5$.
* $10^2 + 5^2 = h^2$
* $100 + 25 = 125$
* $h = \sqrt{125} \approx 11.18$
* Next, find the other slanted side. Let's call it $s$. It sits over the base of $3$ and height of $5$.
* $3^2 + 5^2 = s^2$
* $9 + 25 = 34$
* $s = \sqrt{34} \approx 5.83$
* Perimeter $= \text{Bottom} + \text{Right Vertical} + \text{Long Slant} + \text{Short Slant}$?
* Wait, looking at the diagram, the shape is composed of two triangles joined together or one large triangle split. The perimeter is the outside boundary.
* Bottom side $= 7 + 3 = 10$.
* Right vertical side $= 5$.
* Long hypotenuse (left side) connects the far left point to the top. Base is $10$, height is $5$. Length $= \sqrt{10^2+5^2} = \sqrt{125} \approx 11.18$.
* The internal line is not part of the perimeter.
* Wait, let me look closer at the shape. It looks like a large triangle with a line drawn inside. The perimeter is the sum of the three outer sides.
* Side 1 (Bottom): $7 + 3 = 10$.
* Side 2 (Right): $5$.
* Side 3 (Left/Hypotenuse): Connects the end of the 7cm segment to the top. The total base is $10$, height is $5$. So hypotenuse $= \sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.18$.
* There is another side labeled? No, the diagram shows a triangle split into two. The perimeter is the outer boundary.
* Let's re-read the diagram carefully. It shows a large triangle. The base is split into $7$ and $3$. The height is $5$ on the right side? No, the right angle is on the bottom right of the smaller triangle.
* Okay, let's trace the perimeter:
1. Bottom horizontal line: $7 + 3 = 10$.
2. Right vertical line: $5$.
3. The long slanted line on the left: This is the hypotenuse of the large right triangle with base $10$ and height $5$. Length $= \sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.18$.
4. Is there a fourth side? The diagram shows a line splitting the triangle. That internal line is NOT part of the perimeter.
5. Wait, is the shape a quadrilateral? No, it looks like a single large triangle. But usually "perimeter" questions involve finding all outer edges.
6. Let's check if the left part is a separate triangle. If the vertex is shared, the left side is just the hypotenuse calculated above.
7. Let's assume the question asks for the perimeter of the entire large triangle.
8. Perimeter $= 10 + 5 + 11.18 = 26.18$.
9. Rounding to 1 d.p.: $26.2$.
*Alternative Interpretation:* Is it a trapezoid or irregular quad? The arrows indicate parallel lines? No. It looks like a standard right-angled triangle problem where you calculate the hypotenuse.
Let's double check the side labeled "7cm". It points to the segment on the left. The segment on the right is "3cm". The height is "5cm".
The perimeter consists of:
- Base: $7 + 3 = 10$
- Height: $5$
- Hypotenuse: $\sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.18$
Total $= 10 + 5 + 11.18 = 26.18 \rightarrow 26.2$.
Answer: $26.2 \text{ cm}$
---
Problem 7
Task: Calculate the length BD to 3 significant figures (s.f.).
* BD is the diagonal of the rectangle.
* Sides are $4$ and $11$.
* $4^2 + 11^2 = BD^2$
* $16 + 121 = 137$
* $BD = \sqrt{137} \approx 11.7046...$
* To 3 s.f.: The first three digits are 1, 1, 7. The next digit is 0, so we do not round up.
Answer: $11.7 \text{ cm}$
---
Problem 8
Task: Area is $63 \text{ cm}^2$, calculate $x$ to 2 d.p.
* This is an isosceles triangle (marks on the sides show they are equal length $x$).
* Base $= 7$.
* To find the area, we need the height. Let's split the triangle down the middle to create two right-angled triangles.
* The base of the small right triangle is half of $7$, which is $3.5$.
* The hypotenuse is $x$.
* Let height be $h$. By Pythagoras: $h^2 + 3.5^2 = x^2 \Rightarrow h = \sqrt{x^2 - 3.5^2}$.
* Area formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
* $63 = \frac{1}{2} \times 7 \times h$
* $63 = 3.5 \times h$
* $h = \frac{63}{3.5} = 18$.
* Now substitute $h$ back into the Pythagoras equation:
* $18^2 + 3.5^2 = x^2$
* $324 + 12.25 = x^2$
* $336.25 = x^2$
* $x = \sqrt{336.25} \approx 18.337...$
* Rounding to 2 d.p.: $18.34$.
Answer: $x = 18.34 \text{ cm}$
---
Problem 9
Task: Calculate the perimeter to 1 d.p.
* This is a pentagon shaped like a house.
* Bottom width $= 10$. Side walls $= 5$.
* The roof consists of two sloped sides. We need to find their length.
* Draw a vertical line from the peak to the base. This splits the roof into two right-angled triangles.
* The total height of the shape is $7$. The wall height is $5$. So the height of the triangular roof part is $7 - 5 = 2$.
* The base of the roof triangle is half the total width: $10 / 2 = 5$.
* Let the slope length be $s$.
* $s^2 = 5^2 + 2^2$
* $s^2 = 25 + 4 = 29$
* $s = \sqrt{29} \approx 5.385$
* There are two slopes, so $2 \times 5.385 = 10.77$.
* Perimeter $= \text{Bottom} + \text{Left Wall} + \text{Right Wall} + \text{Two Slopes}$
* Note: The diagram shows tick marks indicating the two roof slopes are equal, and the two vertical walls are equal.
* Perimeter $= 10 + 5 + 5 + \sqrt{29} + \sqrt{29}$
* Perimeter $= 20 + 2(5.385)$
* Perimeter $= 20 + 10.77 = 30.77$
* Rounding to 1 d.p.: $30.8$.
Answer: $30.8 \text{ cm}$
---
Problem 10
Task: Calculate the shaded area to 2 d.p.
* The shape is a circle with a triangle removed (or rather, the shaded area is the circle minus the white triangle? No, the white triangle is inside, and the rest is shaded).
* Actually, looking closely, $QPR$ is a triangle inscribed in a semi-circle? Or is $QR$ the diameter?
* Angle $P$ is marked as a right angle ($90^\circ$). A triangle inscribed in a circle with a right angle means the side opposite the right angle ($QR$) is the diameter.
* We are given $PQ = 10$ and $PR = 5$.
* First, find the diameter $QR$ using Pythagoras on triangle $PQR$:
* $QR^2 = 10^2 + 5^2 = 100 + 25 = 125$.
* $QR = \sqrt{125}$.
* The radius $r$ is half the diameter: $r = \frac{\sqrt{125}}{2}$.
* Area of the full circle $= \pi r^2 = \pi \left(\frac{\sqrt{125}}{2}\right)^2 = \pi \left(\frac{125}{4}\right) = 31.25\pi$.
* Area of the white triangle $PQR = \frac{1}{2} \times \text{base} \times \text{height}$. Since it's a right triangle at $P$, the legs are base and height.
* $\text{Area}_{\text{triangle}} = \frac{1}{2} \times 10 \times 5 = 25$.
* Shaded Area $= \text{Area of Circle} - \text{Area of Triangle}$.
* $\text{Shaded Area} = 31.25\pi - 25$.
* $31.25 \times 3.14159... \approx 98.1747...$
* $98.1747 - 25 = 73.1747...$
* Rounding to 2 d.p.: $73.17$.
Answer: $73.17 \text{ cm}^2$
---
Problem 11
Task: Calculate $x$ to 2 d.p.
* This shows two similar right-angled triangles. One is inside the other.
* Small triangle: Height $= 6$, Base $= ?$
* Large triangle: Height $= 9$, Base $= 15$.
* Wait, the diagram labels the *entire* base as $15$ and the *entire* height as $9$. The inner vertical line is $6$. The hypotenuse of the large triangle is labeled $x$.
* Let's check if the triangles are similar. Yes, they share the bottom-left angle and both have right angles.
* We can just use Pythagoras on the large triangle directly because we are given its full height and full base.
* Height $= 9$.
* Base $= 15$.
* Hypotenuse $= x$.
* $x^2 = 9^2 + 15^2$
* $x^2 = 81 + 225$
* $x^2 = 306$
* $x = \sqrt{306} \approx 17.4928...$
* Rounding to 2 d.p.: $17.49$.
*(Note: The inner dimension '6' is extra information or used to find the position of the vertical line, but isn't needed to find the outer hypotenuse $x$ if 9 and 15 are the total dimensions.)*
Answer: $x = 17.49 \text{ cm}$
---
Problem 12
Task: Area of hexagon is $24 \text{ cm}^2$, calculate $x$ to 2 d.p.
* A regular hexagon can be split into 6 equilateral triangles.
* Let the side length be $x$.
* The area of one equilateral triangle with side $x$ is $\frac{\sqrt{3}}{4}x^2$.
* Total Area of Hexagon $= 6 \times \left( \frac{\sqrt{3}}{4}x^2 \right) = \frac{3\sqrt{3}}{2}x^2$.
* We know Area $= 24$.
* $24 = \frac{3\sqrt{3}}{2}x^2$
* Multiply by 2: $48 = 3\sqrt{3}x^2$
* Divide by 3: $16 = \sqrt{3}x^2$
* Divide by $\sqrt{3}$: $x^2 = \frac{16}{\sqrt{3}}$
* $x^2 \approx \frac{16}{1.73205} \approx 9.2376$
* $x = \sqrt{9.2376} \approx 3.0393...$
* Rounding to 2 d.p.: $3.04$.
Answer: $x = 3.04 \text{ cm}$
──────────────────────────────────────
Final Answer:
1. 17 cm
2. 9.85 cm
3. 11.4 cm
4. 13/15 m
5. 6√2 cm
6. 26.2 cm
7. 11.7 cm
8. 18.34 cm
9. 30.8 cm
10. 73.17 cm²
11. 17.49 cm
12. 3.04 cm
Parent Tip: Review the logic above to help your child master the concept of pythagoras right triangle worksheet.
