Practice Set 2.1 | Q 7 | Page 39
In the given figure, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF
We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.
Here, seg GF ⊥ seg ED
\[\therefore {GF}^2 = EG \times GD\]
\[ \Rightarrow {12}^2 = EG \times 8\]
\[ \Rightarrow 144 = EG \times 8\]
\[ \Rightarrow EG = \frac{144}{8}\]
\[ \Rightarrow EG = 18\]
Hence, EG = 18.
Now,
According to Pythagoras theorem, in ∆DGF
\[{DG}^2 + {GF}^2 = {FD}^2 \]
\[ \Rightarrow 8^2 + {12}^2 = {FD}^2 \]
\[ \Rightarrow 64 + 144 = {FD}^2 \]
\[ \Rightarrow {FD}^2 = 208\]
\[ \Rightarrow FD = 4\sqrt{13}\]
In ∆EGF
\[{EG}^2 + {GF}^2 = {EF}^2 \]
\[ \Rightarrow {18}^2 + {12}^2 = {EF}^2 \]
\[ \Rightarrow 324 + 144 = {EF}^2 \]
\[ \Rightarrow {EF}^2 = 468\]
\[ \Rightarrow EF = 6\sqrt{13}\]
Hence, FD =\[4\sqrt{13}\] and EF=\[6\sqrt{13}\]
Explanation:-
The given problem involves finding the lengths of line segments in a right-angled triangle. We are given that:
- Segment GF is perpendicular to segment ED in a right-angled triangle DGF.
- Side DG is of length 8 units.
- Side GF is of length 12 units.
To find the length of side FD, we use the Pythagoras theorem in triangle DGF, which states that in a right-angled triangle, the sum of the squares of the lengths of the two shorter sides is equal to the square of the length of the longest side.
Therefore, we have:
DG^2 + GF^2 = FD^2
Substituting the values of DG and GF, we get:
8^2 + 12^2 = FD^2
Simplifying the above equation, we get:
FD^2 = 208
Taking the square root of both sides, we get:
FD = 4√13
Thus, the length of segment FD is 4√13 units.
Next, we need to find the length of segment EF. To do this, we use the fact that in a right-angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex is the geometric mean of the segments into which the hypotenuse is divided. Here, segment GF is the perpendicular segment to the hypotenuse DE, which divides DE into segments DG and GE.
Thus, we have:
GF^2 = EG × GD
Substituting the values of GF and GD, we get:
12^2 = EG × 8
Simplifying the above equation, we get:
EG = 18
Thus, the length of segment GE is 18 units.
Now, using the Pythagoras theorem in triangle EGF, we get:
EG^2 + GF^2 = EF^2
Substituting the values of EG and GF, we get:
18^2 + 12^2 = EF^2
Simplifying the above equation, we get:
EF^2 = 468
Taking the square root of both sides, we get:
EF = 6√13
Thus, the length of segment EF is 6√13 units.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Practice Set 2.1 | Q 7 | Page 39