In each figure given below, altitude is drawn on hypotenuse by a right angled triangle. The length of different segments is marked in each figure. Determine x, y, z in each case.

Question

Vedantu Content Team · Accepted Answer

Hint: In both cases there is a right angled triangle (Third side is an altitude) in a right angled triangle.We can use Pythagora's theorem to establish a relation between inner triangle and outer triangle. Then by using substitution we can find x, y, z.Complete step-by-step answer:i.In the given figure,

In triangle DBE by Pythagoras theorem,(In a right triangle square of hypotenuse is equal to the sum of square of height and square of base.) $ \begin{align}  & {{z}^{2}}={{y}^{2}}+{{5}^{2}} \\  & {{z}^{2}}={{y}^{2}}+25 \\ \end{align} $                                        ………………….(1)Now use Pythagoras theorem in triangle ABC $ \begin{align}  & {{(4+5)}^{2}}={{x}^{2}}+{{z}^{2}} \\  & 81={{x}^{2}}+{{z}^{2}} \\  & {{z}^{2}}=81-{{x}^{2}} \\ \end{align} $                                     ………………………….(2)From equation 1 and 2 we can conclude that, $ \begin{align}  & {{y}^{2}}+25=81-{{x}^{2}} \\  & {{x}^{2}}+{{y}^{2}}=81-25 \\  & {{x}^{2}}+{{y}^{2}}=56 \\ \end{align} $                                       …………………..(3)Now, in triangle ADB use Pythagoras theorem, $ \begin{align}  & {{x}^{2}}={{y}^{2}}+16 \\  & {{x}^{2}}-{{y}^{2}}=16 \\ \end{align} $                              …………………….. (4)Now add equation 3 and 4, $ \begin{align}  & {{x}^{2}}+{{y}^{2}}+{{x}^{2}}-{{y}^{2}}=56+16 \\  & 2{{x}^{2}}=72 \\  & {{x}^{2}}=36 \\  & x=6 \\ \end{align} $ We will ignore -6 because the side of a triangle cannot be negative.Now, put the value of x in equation 3, $ \begin{align}  & {{6}^{2}}+{{y}^{2}}=56 \\  & {{y}^{2}}=56-36 \\  & {{y}^{2}}=20 \\  & y=2\sqrt{5} \\ \end{align} $ Ignore negative values for the same reason.Again put the value of x in equation 2, $ \begin{align}  & {{z}^{2}}=81-{{6}^{2}} \\  & {{z}^{2}}=81-36 \\  & {{z}^{2}}=45 \\  & z=3\sqrt{5} \\ \end{align} $

In triangle PSQ, use Pythagoras theorem, $ \begin{align}  & {{6}^{2}}={{4}^{2}}+{{y}^{2}} \\  & 36-16={{y}^{2}} \\  & {{y}^{2}}=20 \\  & y=4\sqrt{5} \\ \end{align} $                          ………………….(1)Use Pythagoras theorem in triangle QSR, $ \begin{align}  & {{z}^{2}}={{x}^{2}}+{{y}^{2}} \\  & {{z}^{2}}-{{x}^{2}}=20 \\ \end{align} $                           (from equation 1)    …………………….. (2)                                          Now, take triangle PQR and apply Pythagoras theorem in it. $ {{6}^{2}}+{{z}^{2}}={{(x+4)}^{2}} $ From equation 2 put the value of z in it. $ \begin{align}  & {{6}^{2}}+{{x}^{2}}+20={{x}^{2}}+16+8x \\  & 56=16+8x \\  & 8x=40 \\  & x=5 \\  &  \\ \end{align} $ Put the value of x in equation 2. $ \begin{align}  & {{z}^{2}}-{{5}^{2}}=20 \\  & {{z}^{2}}=45 \\  & z=3\sqrt{5} \\ \end{align} $ Note:In figure (i) angle D is a right angle that’s why AB and BC is hypotenuse for triangle ADB and BDC respectively. Same for figure (ii) angle S is a right angle that’s why PQ and QR are hypotenuse for triangle PSQ and triangle QSR respectively. Always ignore negative values for a side of the triangle.