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Q4.

Expert-verifiedFound in: Page 583

Book edition
Student Edition

Author(s)
Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages
227 pages

ISBN
9780395977279

**Prove that you can see all of yourself in a mirror that is only half as tall as you are. (Hint: Study the diagram on page 582.)**

It is proved that you can see all of yourself in a mirror that is only half as tall as you are.

The given statement is you can see all of yourself in a mirror that is only half as tall as you are.

Consider you are standing in front of a mirror of height *FH* and your height is *AE*.

Point *C* represents your eye, *A* represents top of your head, E represents foot, *B* is the mid-point between your eye and top of your head, *D* is the mid-point between your eye and foot, *F* is the bottom of the mirror, *H* is the top of the mirror and *G* is a point in the mirror parallel to your eyes.

Since,

$\begin{array}{l}AE=AB+BC+CD+DE\\ AE=BC+BC+CD+CD\text{}\left[AB=BC,CD=DE\right]\\ AE=2\left(BC+CD\right)\text{}\left(i\right)\\ \text{and}\\ HF=HG+GF\text{}\left(ii\right)\end{array}$

In similar triangles *CGF* and *CDF*

$\begin{array}{l}CF\text{is common side}\\ CD=DF\\ \text{therefore},CD=GF\end{array}$

In similar triangles *BCH* and *CGH*

$\begin{array}{l}CH\text{is common side}\\ BH=CG\\ \text{therefore,}BC=HG\end{array}$

Substitute the values of *HG* and *GF* in equation *i.*

$\begin{array}{l}HF=HG+GF\text{}\left(ii\right)\\ HF=BC+CD\text{}\left[CD=GF,BC=HG\right]\\ HF=\frac{AE}{2}\text{}\left[AE=2\left(BC+CD\right)\right]\end{array}$

It is proved that you can see all of yourself in a mirror that is only half as tall as you are.

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