Question

In each of the figures given below, an altitude is drawn to the hypotenuse by a right angle triangle. The length of different line segments are marked in each figure. Determine x, y and z in each case.

Answer 1

Hint: In figure there are three right angle triangle using Pythagoras theorem form three equations with respect to three right angle triangles and then solving these three formed equations by using method substitution to find values of x, y and z.
Pythagoras Theorem $ {(Hyp.)^2} = {\left( {Perp.} \right)^2} + {\left( {Base} \right)^2} $

Complete step-by-step answer:
I.Triangle ABC is a right angle triangle at B.
And, also a perpendicular is drawn from B to hypotenuse AC of triangle ABC. Therefore BD is perpendicular to AC.

From figure we have
 $ AB = x,\,\,BC = z,\,\,AD = 4,\,\,DC = 5,\,\,and\,\,BD = y $
Triangle ADB is a right angle triangle as $ \angle ADB = {90^0} $
Therefore, by Pythagoras theorem we have:
 $ {\left( {AB} \right)^2} = {\left( {AD} \right)^2} + {\left( {BD} \right)^2} $
Substituting values in above equation we have
 $
  {(x)^2} = {(y)^2} + {(4)^2} \\
   \Rightarrow {x^2} = {y^2} + 16.....................(i) \;
  $
Also, triangle BDC is a right angle triangle. As $ \angle BDC = {90^0} $
Therefore, by Pythagoras theorem we have:
 $ {\left( {BC} \right)^2} = {\left( {CD} \right)^2} + {\left( {BD} \right)^2} $
Substituting values in above equation we have
 $
  {(z)^2} = {(5)^2} + {(y)^2} \\
   \Rightarrow {z^2} = 25 + {y^2}.....................(ii) \;
  $
Also, triangle ABC is a right angle triangle. As, \[\angle B = {90^0}\].
Therefore, by Pythagoras theorem we have:
 $ {\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2} $
Substituting values in above equation we have
 $
  {(9)^2} = {(x)^2} + {(z)^2} \\
   \Rightarrow 81 = {x^2} + {z^2} .................(iii) \;
  $
Using equation (i) and (ii) in (iii) we have
 $
  81 = {y^2} + 16 + 15 + {y^2} \\
   \Rightarrow 81 = 31 + 2{y^2} \\
   \Rightarrow 81 - 31 = 2{y^2} \\
   \Rightarrow 50 = 2{y^2} \\
   \Rightarrow {y^2} = 25 \\
   \Rightarrow y = 5 \;
  $
Using above calculated value of y in equation (i) we have
 $
  {x^2} = {(5)^2} + 16 \\
   \Rightarrow {x^2} = 25 + 16 \\
   \Rightarrow {x^2} = 41 \\
   \Rightarrow x = \sqrt {41} \;
  $
Using value of y in equation (ii) we have:
 $
  {z^2} = {(5)^2} + 25 \\
   \Rightarrow {z^2} = 25 + 25 \\
   \Rightarrow {z^2} = 50 \\
   \Rightarrow z = \sqrt {50} \\
   \Rightarrow z = 5\sqrt 2 \;
  $
Hence, from above we see that the values of x, y and z are $ \sqrt {41} ,\,\,5\,\,and\,\,5\sqrt 2 $ respectively.

(ii)

From figure we have
 $ PQ = 6,\,\,QR = z,\,\,PS = 4,\,\,SR = x,\,\,and\,\,SQ = y $
Triangle PSQ is a right angle triangle as $ \angle PSQ = {90^0} $
Therefore, by Pythagoras theorem we have:
 $ {\left( {PQ} \right)^2} = {\left( {PS} \right)^2} + {\left( {QS} \right)^2} $
Substituting values in above equation we have
 $
  {(6)^2} = {(4)^2} + {(y)^2} \\
   \Rightarrow 36 = {y^2} + 16 \\
   \Rightarrow {y^2} = 36 - 16 \\
   \Rightarrow {y^2} = 20 \\
   \Rightarrow y = \sqrt {20} \\
   \Rightarrow y = 2\sqrt 5 \;
  $
Also, triangle QSR is a right angle triangle. As $ \angle QSR = {90^0} $
Therefore, by Pythagoras theorem we have:
 $ {(QR)^2} = {\left( {QS} \right)^2} + {\left( {SR} \right)^2} $
Substituting values in above equation we have
\[{(z)^2} = {(y)^2} + {(x)^2}\]
Or
 $ {z^2} = {y^2} + {x^2} $
Substituting value of y above. We have,
 $
  {z^2} = {(2\sqrt 5 )^2} + {x^2} \\
   \Rightarrow {z^2} = 20 + {x^2}....................(i) \;
  $
Also, triangle PQR is a right angle triangle. As, \[\angle Q = {90^0}\].
Therefore, by Pythagoras theorem we have:
 $ {\left( {PR} \right)^2} = {\left( {PQ} \right)^2} + {\left( {QR} \right)^2} $
Substituting values in above equation we have
 $
  {(4 + x)^2} = {(6)^2} + {(z)^2} \\
   \Rightarrow {\left( {4 + x} \right)^2} = 36 + {z^2}...............(ii) \;
  $
Using equation (i) in (ii) we have
 $
  {\left( {4 + x} \right)^2} = 36 + 20 + {x^2} \\
   \Rightarrow 16 + {x^2} + 8x = 56 + {x^2} \\
   \Rightarrow 8x = 56 - 16 \\
   \Rightarrow 8x = 40 \\
   \Rightarrow x = 5 \;
  $
Using above calculated value of x in equation (i) we have
 $
  {z^2} = 20 + {\left( 5 \right)^2} \\
   \Rightarrow {z^2} = 20 + 25 \\
   \Rightarrow {z^2} = 45 \\
   \Rightarrow z = \sqrt {45} \\
   \Rightarrow z = 3\sqrt 5 \;
  $
Hence, from above we see that the values of x, y and z are $ 5,\,\,2\sqrt 5 \,\,and\,\,3\sqrt 5 $ respectively.

Note: On drawing an altitude to hypotenuse divide the given right angle triangle into two small right angle triangles. Applying Pythagora's theorem to each right triangle to form three different equations and hence solving them together to find values of x, y and z by using method substitution