Answer
Verified
430.2k+ views
Hint: First draw the diagram based on the question. Then, use the cosine formula in $\Delta ABC$ and $\Delta ABD$ and compare the values to get the value of BD. After that use the properties of the isosceles triangle to get the value of ED. Then use the Pythagoras theorem in $\Delta BED$ to get the value of BE.
Complete step-by-step solution:
The diagram of the above equation is,
In $\Delta ABC$, the cosine formula for A is given by,
$\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
Substitute the values,
$ \Rightarrow \cos A = \dfrac{{{8^2} + {9^2} - {7^2}}}{{2 \times 8 \times 9}}$
Simplify the terms,
$ \Rightarrow \cos A = \dfrac{{64 + 81 - 49}}{{144}}$
Add and subtract the terms in the numerator,
$ \Rightarrow \cos A = \dfrac{{96}}{{144}}$
Cancel out the common terms,
$ \Rightarrow \cos A = \dfrac{2}{3}$.............….. (1)
In $\Delta ABD$, the cosine formula for A is given by,
$\cos A = \dfrac{{A{B^2} + A{D^2} - B{D^2}}}{{2 \times AB \times AD}}$
Substitute the values,
$ \Rightarrow \cos A = \dfrac{{{9^2} + {4^2} - B{D^2}}}{{2 \times 9 \times 4}}$
Simplify the terms,
$ \Rightarrow \cos A = \dfrac{{81 + 16 - B{D^2}}}{{72}}$
Add and subtract the terms in the numerator,
$ \Rightarrow \cos A = \dfrac{{97 - B{D^2}}}{{72}}$................….. (2)
Compare the equation (1) and (2),
$ \Rightarrow \dfrac{{97 - B{D^2}}}{{72}} = \dfrac{2}{3}$
Cross-multiply the terms,
$ \Rightarrow 291 - 3B{D^2} = 144$
Move constant part on one side,
$ \Rightarrow 3B{D^2} = 291 - 144$
Subtract the values,
$ \Rightarrow 3B{D^2} = 147$
Divide both sides by 3,
$ \Rightarrow B{D^2} = 49$
Take the square root on both sides,
$\therefore BD = 7$
In $\Delta BCD$, $BC = 7$ and $BD = 7$.
So, $\Delta BCD$ is an isosceles triangle.
We know that the altitude on the non-equal side of the isosceles triangle is the perpendicular bisector. So,
$ \Rightarrow CE = ED$...................….. (3)
The length of the CD is 4. Then,
$ \Rightarrow CE + ED = 4$
Substitute the value from equation (3),
$ \Rightarrow ED + ED = 4$
Add the terms,
$ \Rightarrow 2ED = 4$
Divide both sides by 2,
$\therefore ED = 2$
In $\Delta BED$, apply Pythagoras theorem,
$B{E^2} = B{D^2} - E{D^2}$
Substitute the values,
$ \Rightarrow B{E^2} = {7^2} - {2^2}$
Simplify the terms,
$ \Rightarrow B{E^2} = 49 - 4$
Subtract the values,
$ \Rightarrow B{E^2} = 45$
Take the square root on both sides,
$\therefore BE = \sqrt {45} $
Hence, option (4) is correct.
Note: A line segment drawn from the vertex of a triangle on the opposite side of a triangle which is perpendicular to it is said to be the altitude of a triangle.
A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side.
Complete step-by-step solution:
The diagram of the above equation is,
In $\Delta ABC$, the cosine formula for A is given by,
$\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
Substitute the values,
$ \Rightarrow \cos A = \dfrac{{{8^2} + {9^2} - {7^2}}}{{2 \times 8 \times 9}}$
Simplify the terms,
$ \Rightarrow \cos A = \dfrac{{64 + 81 - 49}}{{144}}$
Add and subtract the terms in the numerator,
$ \Rightarrow \cos A = \dfrac{{96}}{{144}}$
Cancel out the common terms,
$ \Rightarrow \cos A = \dfrac{2}{3}$.............….. (1)
In $\Delta ABD$, the cosine formula for A is given by,
$\cos A = \dfrac{{A{B^2} + A{D^2} - B{D^2}}}{{2 \times AB \times AD}}$
Substitute the values,
$ \Rightarrow \cos A = \dfrac{{{9^2} + {4^2} - B{D^2}}}{{2 \times 9 \times 4}}$
Simplify the terms,
$ \Rightarrow \cos A = \dfrac{{81 + 16 - B{D^2}}}{{72}}$
Add and subtract the terms in the numerator,
$ \Rightarrow \cos A = \dfrac{{97 - B{D^2}}}{{72}}$................….. (2)
Compare the equation (1) and (2),
$ \Rightarrow \dfrac{{97 - B{D^2}}}{{72}} = \dfrac{2}{3}$
Cross-multiply the terms,
$ \Rightarrow 291 - 3B{D^2} = 144$
Move constant part on one side,
$ \Rightarrow 3B{D^2} = 291 - 144$
Subtract the values,
$ \Rightarrow 3B{D^2} = 147$
Divide both sides by 3,
$ \Rightarrow B{D^2} = 49$
Take the square root on both sides,
$\therefore BD = 7$
In $\Delta BCD$, $BC = 7$ and $BD = 7$.
So, $\Delta BCD$ is an isosceles triangle.
We know that the altitude on the non-equal side of the isosceles triangle is the perpendicular bisector. So,
$ \Rightarrow CE = ED$...................….. (3)
The length of the CD is 4. Then,
$ \Rightarrow CE + ED = 4$
Substitute the value from equation (3),
$ \Rightarrow ED + ED = 4$
Add the terms,
$ \Rightarrow 2ED = 4$
Divide both sides by 2,
$\therefore ED = 2$
In $\Delta BED$, apply Pythagoras theorem,
$B{E^2} = B{D^2} - E{D^2}$
Substitute the values,
$ \Rightarrow B{E^2} = {7^2} - {2^2}$
Simplify the terms,
$ \Rightarrow B{E^2} = 49 - 4$
Subtract the values,
$ \Rightarrow B{E^2} = 45$
Take the square root on both sides,
$\therefore BE = \sqrt {45} $
Hence, option (4) is correct.
Note: A line segment drawn from the vertex of a triangle on the opposite side of a triangle which is perpendicular to it is said to be the altitude of a triangle.
A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE