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This is the second of our many quizzes on Angle bisector theorem. Click on this to access more such competitive quizzes on 2D Coordinate Geometry.

Angle bisector theorem Quiz 2

mathematics

 In the given triangle $ABC$, $AD$ is the angle bisector and $AB = 1,AD = x,BC = 7,CD = 2$, then find $x$. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5bc2c69f-3daa-49f9-9147-531582ea44802243997546964619191.png"/>

 Simply apply angle bisector theorem i.e. $\dfrac{{AB}}{{AC}} = \dfrac{{BD}}{{CD}}$ from the given diagram. 

  Consider the angle bisector theorem $ \\  \dfrac{{AB}}{{AC}} = \dfrac{{BD}}{{CD}}  \\   \Rightarrow \dfrac{1}{x} = \dfrac{7}{2}  \\   \Rightarrow x = \dfrac{2}{7}  \\  $

 In the given triangle  $CAB$, $AD$ is an angle bisector and $CD = 9,AC = 2,AB = 8$ ,find $x$.$ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2ba90e5b-4675-4fa8-b004-8cf7dd017d9a48438861226107664.png"/>

 As we can see $AD$ is angle bisector, so we can use angle bisector theorem $\dfrac{{AC}}{{AB}} = \dfrac{{CD}}{{DB}}$.

  Consider the angle bisector theorem $  \\  \dfrac{{AC}}{{AB}} = \dfrac{{CD}}{{DB}}  \\   \Rightarrow \dfrac{2}{8} = \dfrac{9}{x}  \\   \Rightarrow x = 36  \\  $

 For the given figure, $JL$ is an angle bisector, and $JK =  - x + 1,KL = 2x + 4,JM = x + 3,LM = x + 1$, find $x$. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/bd81457e-a933-414a-884c-aa9db7aa05a25471383629167869443.png"/>

 Find the roots of equation using $x = \dfrac{{ - b \pm \sqrt {{b^2} - \left( { + 4ac} \right)} }}{{2a}}$.

By using angle bisector theorem  $ \\   \Rightarrow \dfrac{{ - x + 1}}{{x + 3}} = \dfrac{{2x + 4}}{{x + 1}}  \\   \Rightarrow 3{x^2} + 10x + 11 = 0  \\   \Rightarrow {x_{1,2}} = \dfrac{{ - 10 \pm \sqrt { - 36} }}{6} \left( \text{not real} \right)  \\  $

 In the given figure, $AO$ is an angle bisector $AM = 5,AO = 3,\angle MAO = {45^\circ },\angle AMO = {45^\circ }$. Calculate the area of the triangle. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0d5b067a-c670-42e4-8cba-0fad74ac15c54428844025413911019.png"/>

 Using angle bisector theorem $\Delta AMO\& \Delta AOJ$  becomes congruent.

 If $\angle M = 45^\circ  = \angle MAO$ then $\angle MOA = 90^\circ $ $ \\ $ By using Pythagoras theorem  $ \\   \Rightarrow MO = 4  \\  $ Now area of triangle $= \dfrac{1}{2}MO \times AO  \\   = \dfrac{1}{2} \times 4 \times 3  \\   \Rightarrow 6  \\ $  As $\Delta AMO$ and $\Delta AOJ $ are congruent so area of triangle $AMT is = 12 $.

 If $\angle YXW = \angle WXZ$ and $XY = \dfrac{x}{2},XZ = x,YW = x + 2,WZ = 8$, solve for $x$.$ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/64e53316-0e17-4b8d-993c-d3ff4c6688662448446896554989086.png"/>

 Use angle bisector theorem, $\dfrac{{XY}}{{XZ}} = \dfrac{{YW}}{{WZ}}$ from the given diagram.

 $  \Rightarrow \dfrac{{\dfrac{x}{2}}}{x} = \dfrac{{x + 2}}{8}  \\  \Rightarrow \dfrac{1}{2} = \dfrac{{x + 2}}{8}  \\  \Rightarrow x + 2 = 4  \\  \Rightarrow x = 2  \\  $