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This is one of our many quizzes on Pythagorean theorem proof. Click on this to access more such competitive quizzes on Geometry.

Pythagorean theorem proof Quiz 2

mathematics

 If $b$ is the radius of the given circle and ABC is a right angled triangle then find $a$. $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6a9eedae-b9c0-4d95-bb3d-d1423d7e3f692311206878897538580.png"/>

 For a valid triangle, the sum of two sides should be greater than the third side. Longest side of the triangle is the hypotenuse and the side opposite to the largest angle is the greatest side. In a right angled triangle, $90{}^\circ $ is the greatest angle thus its opposite side which is hypotenuse is the longest side.

$b=c=1$ unit $=$ radius of the circle $ \\ $ For a valid triangle, the sum of any two sides of a triangle is greater than the third side. So, $ a>b+c$ $ \\  \Rightarrow a > 1 + 1 \\  \Rightarrow a > 2 \\ $ As, ABC is a right angled triangle,  $ \\ $ $a > 2$ which is greater than $b, c,$ thus $a$ is hypotenuse. So, by Pythagoras Theorem, $ \\  {{a}^{2}}={{b}^{2}}+{{c}^{2}} \\  a=\sqrt{{{1}^{2}}+{{1}^{2}}} \\  a=\sqrt{2} \\ $ 

 In the figure given below, $BA=B{A}'$ and the lengths of $AC=4cm$ and $\angle CAB=45{}^\circ $, Find $B{A}'$  $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1fb44289-2641-4f19-8258-5fe328f222765880797995470430182.png"/>

 Try to prove $\Delta \text{ABC}$ and $\Delta AB{A}'$ congruent as AB is common in both triangles. Using the Pythagoras’s Theorem, we can find BC as $\angle CAB=45{}^\circ $ 

As, $BA=B{A}'$ $ \\ $ So, $\angle BA{A}'=\angle B{A}'A$ $ \\ $ In $\Delta AB{A}'$, $ \\ \angle BA{A}'+\angle B{A}'A+\angle AB{A}'=180{}^\circ $ $($Sum of interior angles of a triangle$)$ $ \\  2\angle BA{A}'=180{}^\circ -90{}^\circ  \\  \angle BA{A}'=\angle B{A}'A=45{}^\circ  \\ $  Thus, $AB=BC$ $ \\\Delta \text{ABC and }\Delta \text{AB{A}'} \\  \text{AB=AB (Common)} \\  \angle \text{AB{A}'=}\angle \text{ABC=90}{}^\circ  \\  \angle {A}'AB=\angle CAB=45{}^\circ  \\ $ So, by ASA criterion,  $ \\ \Delta ABC\cong \Delta AB{A}' \\  \Rightarrow BC=B{A}' \\ $ $\Delta \text{ABC}$ is a right angled triangle. $ \\ $ So, using Pythagoras Theorem, $ \\  A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}} \\  {{4}^{2}}=B{{C}^{2}}+B{{C}^{2}} \\  16=2B{{C}^{2}} \\  BC=B{A}'=\sqrt{8} \\ $

 What will be the length of the diagonal of a square having sides, 2.75 cm each?

 All angles of the square are right-angled.  All the sides of the square are equal thus by using the Pythagoras theorem, we can find the diagonal. 

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6816c61b-54d8-46f2-a4e8-1965ddb313274232557417803125432.png"/>  $ \\ $ ABCD is a square. All the angles of the square are $90{}^\circ $. $ \\ $ So, in $\Delta ABC$ using Pythagoras Theorem, $ \\ A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}} \\  {{2.75}^{2}}+{{2.75}^{2}}=A{{C}^{2}} \\  A{{C}^{2}}=\sqrt{2\times {{2.75}^{2}}}=2.75\sqrt{2} \\ $

 A mountain is standing parallel and at right angles to a tree with the slant height being 7cm, what will be the length of the tree bark if the base of the mountain 5cm $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4f3e8d68-a195-41bd-a8ea-ed7ea2a73dc8294450311136409699.png"/>

 Find the height of the mountain as it is given that height of mountain is equal to height of tree in the fig. Using Pythagoras’s theorem, with the help of base of mountain and slant height of mountain, we can find perpendicular height.

From the given. fig., it can be concluded that the Perpendicular height of the mountain is the length of the bark of the tree. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/61cc78fe-14ad-47cc-b659-412222a5bc733064546547084862608.png"/> $ \\ $ As given, $\angle ACB=90{}^\circ $ $ \\ $ So, by using Pythagoras Theorem, $ \\   A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}} \\  {{7}^{2}}=A{{C}^{2}}+{{5}^{2}} \\  A{{C}^{2}}={{7}^{2}}-{{5}^{2}} \\  AC=\sqrt{(7-5)(7+5)} \\  AC=\sqrt{2\times 12} \\  AC=2\sqrt{6}cm \\ $

 With the help of fig. 1, find the value of AC in fig. 2: $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5025c496-f44e-450e-a4d8-bcb44fc8d3c06291292005664881369.png"/>

 Prove $\Delta ABC$ and $\Delta \text{PQR,}$ are similar triangles because when all angles are found then we find that all their angles are equal. Then after using similar condition of ratio of corresponding sides and Pythagoras Theorem, we can find the required sides.

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0116eba1-37ef-4486-bbcb-9071976508325540154906176961082.png"/>  $ \\ $  As, $\Delta PQR$ is a right angled triangle, so we can use Pythagoras Theorem to find out the hypotenuse PR. $ \\ $ Therefore, PR $=\sqrt{{{4}^{2}}+{{3}^{2}}}=5$ $ \\ $ In $\Delta PQR,$ $ \\ \Rightarrow \angle PQR+\angle PRQ + \angle RPQ=180{}^\circ  \\  \Rightarrow 90{}^\circ +37{}^\circ +\angle PRQ=180{}^\circ  \\  \Rightarrow \angle PRQ=53{}^\circ  \\ $ In $\Delta ABC,$ $ \\ \Rightarrow \angle ABC+\angle ACB+\angle BAC=180{}^\circ \\  \Rightarrow 90{}^\circ +53{}^\circ +\angle BAC=180{}^\circ  \\  \Rightarrow \angle PRQ=37{}^\circ  \\ $ In $\Delta ABC$ and  $\Delta {PQR,}$ $ \\ $ All three angles are equal, thus they are similar triangles. $ \\ $ Therefore, $\dfrac{PQ}{AB}=\dfrac{QR}{BC}=\dfrac{PR}{AC}$ $ \\ $ Using, $\dfrac{PQ}{AB}=\dfrac{PR}{AC}$ $ \\   \Rightarrow \dfrac{4}{8}=\dfrac{5}{AC} \\  \Rightarrow AC=\dfrac{5\times 8}{4}=10 \\ $