Answer:
For right angle triangle,
we use Pythagoras theorem that is:
![c^(2) =a^(2) +b^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/st598esm7t4fso7obas4von9866sztgafv.png)
c =
![\sqrt{a^(2) +b^(2) }](https://img.qammunity.org/2022/formulas/mathematics/high-school/2s525zoyd63ttz0itbplcbnjcgsnbp5g6u.png)
For question 1:
c = ?
a = 40
b = 9
putting them in formula,
c =
![\sqrt{40^(2) + 9^(2) }](https://img.qammunity.org/2022/formulas/mathematics/high-school/luomqz29a85cn6wa3ho2iwql3ahbl15vuk.png)
c = 41
For question 2:
c = ?
a = 12
b = 13
putting them in formula,
c =
![\sqrt{12^(2) + 13^(2) }](https://img.qammunity.org/2022/formulas/mathematics/high-school/whbkz7xiy4qqwvuqqywoj7gjlu5kifgy6m.png)
c = approximately 17.69181
For question 3:
c = 35
a = 20
b = ?
putting them in formula,
![c^(2) =a^(2) +b^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/st598esm7t4fso7obas4von9866sztgafv.png)
![35^(2) = 20^(2) + b^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/17f8zid9jomz20l89gy8e179w9b0sg05uk.png)
1225 = 400 +
![b^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/nego0i9xhj9ijncn3zrwihgj7ltslb9t0z.png)
= 1225 - 400
= 825
![\sqrt{b^(2) } = √(825)](https://img.qammunity.org/2022/formulas/mathematics/high-school/jt902lrf5i4z81g5gyat7ega7bgtvosm8m.png)
b = 5
![√(33)](https://img.qammunity.org/2022/formulas/mathematics/high-school/rlscigqlzucf51onw3j4bkk6jbdmfj4gik.png)
For question 4:
c = 37
a = 20
b = ?
putting them in formula,
![c^(2) =a^(2) +b^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/st598esm7t4fso7obas4von9866sztgafv.png)
![37^(2) = 20^(2) + b^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/3jyvkc64kqami0s5a2zsafmp0pfk4xmhxf.png)
1369 = 400 +
![b^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/nego0i9xhj9ijncn3zrwihgj7ltslb9t0z.png)
= 1369 - 400
= 969
Taking square root on both sides
b = 31.12
Hope it helps.