A) \[(5,11,13)\]
B) \[(9,15,18)\]
C) \[(17,24,25)\]
D) \[(20,21,29)\]
Correct Answer: D
Solution :
| According of Pythagoras theorem, |
| \[{{(\text{Longest side},\text{ i}.\text{e}.,\text{ hypotenuse)}}^{2}}\] |
| \[={{(\text{one side i}\text{.e}\text{.},\text{ base)}}^{2}}+{{(\text{other side i}\text{.e}\text{.},\text{ height)}}^{2}}\]From option ; \[(5,11,13)\] |
| Longest side = 13, one side = 5 and other side = 11 |
| Now, \[{{(11)}^{2}}+{{(5)}^{2}}=121+25=146\] |
| and \[{{(13)}^{2}}=169\] |
| \[{{(13)}^{2}}\ne {{(11)}^{2}}+{{(5)}^{2}}\] |
| \[(5,11,13)\] does not form a Pythagoras triplet. |
| From option ; \[(9,15,18)\] |
| longest side =18, one side =9 and other side =15 |
| Now, \[{{(15)}^{2}}+{{(9)}^{2}}=225+81=306\] |
| and \[{{(18)}^{2}}=324\] |
| \[{{(18)}^{2}}\ne {{(15)}^{2}}+{{(9)}^{2}}\] |
| \[\therefore \,\,(9,15,18)\]does not form a Pythagoras triplet. |
| From option ; \[(17,24,25)\] |
| longest side =25, one side =17 and other side =24 |
| Now, \[{{(17)}^{2}}+{{(24)}^{2}}=289+576=865\] |
| and \[{{(25)}^{2}}=625\] |
| \[{{(25)}^{2}}={{(17)}^{2}}+{{(24)}^{2}}\] |
| \[\therefore \,(17,24,15)\] does not form a Pythagoras triplet. |
| From option ; \[(20,21,29)\] |
| longest side =29, one side =21 and other side =20 |
| Now, \[{{(20)}^{2}}+{{(21)}^{2}}=400+441=841\] and \[{{(29)}^{2}}=841\] |
| \[{{(29)}^{2}}={{(20)}^{2}}+{{(21)}^{2}}\] |
| \[\therefore \,(20,21,29)\] forms a Pythagoras triplet. |
| So, option [d] is correct. |
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