In the given figure, find the value of x.
${\text{A}}{\text{. }}{40^0}$
${\text{B}}{\text{. }}{50^0}$
${\text{C}}{\text{. }}{60^0}$
${\text{D}}{\text{. }}{25^0}$

Answer
329.1k+ views
Hint: Here we go through the properties of the isosceles triangle and the properties of the sum of all internal angles of a triangle which is ${180^0}$ to find the unknown angle.
Given that ABC is a triangle in which D is the midpoint of side BC.
So we can say that BD=DC.
It is also given that the length AD is equal to the length BD and DC.
$\therefore {\text{AD = BD = DC}}$
Now in $\vartriangle {\text{ABD}}$
${\text{AD = BD}}$ (Given in the question)
It means that $\vartriangle {\text{ABD}}$ is an isosceles triangle.
So the base angle of side BD and AD are equal which means $\angle {\text{DBA = }}\angle {\text{DAB = 5}}{{\text{0}}^0}$ given in the figure.
And we know that the sum of all interior angles of a triangle is ${180^0}$.
$
\therefore \angle {\text{DBA + }}\angle {\text{DAB + }}\angle {\text{BDA}} = {180^0} \\
\Rightarrow {50^0} + {50^0} + \angle {\text{BDA}} = {180^0} \\
\Rightarrow \angle {\text{BDA}} = {180^0} - {100^0} \\
\therefore \angle {\text{BDA}} = {80^0} \\
$
And now BDC is a straight line
$\therefore \angle {\text{BDA}} + \angle {\text{ADC = 18}}{{\text{0}}^0}$ Straight angle.
$ \Rightarrow \angle {\text{ADC = 18}}{{\text{0}}^0} - {80^0} = {100^0}$ $\because \angle {\text{BDA}} = {80^0}$
And now in $\vartriangle {\text{ABC}}$
${\text{AD = CD}}$ (Given in question).
It means that $\vartriangle {\text{ADC}}$ is an isosceles triangle.
So the base angle of side BD and AD are equal which means $\angle {\text{DCA = }}\angle {\text{DAC = }}x$ given in the figure.
And we know that the sum of all interior angles of a triangle is ${180^0}$.
$
\therefore \angle {\text{DCA + }}\angle {\text{DAC + }}\angle {\text{ADC}} = {180^0} \\
\Rightarrow x + x + {100^0} = {180^0} \\
\Rightarrow 2x = {180^0} - {100^0} \\
\therefore x = {40^0} \\
$
Hence, option A is the correct answer.
Note: Whenever we face such a type of question the key concept for solving the question is that you have to first read the diagram what it says and the data which are given in the diagrams. Then by applying the properties of triangles start finding the unknown terms to get the answer.
Given that ABC is a triangle in which D is the midpoint of side BC.
So we can say that BD=DC.
It is also given that the length AD is equal to the length BD and DC.
$\therefore {\text{AD = BD = DC}}$
Now in $\vartriangle {\text{ABD}}$
${\text{AD = BD}}$ (Given in the question)
It means that $\vartriangle {\text{ABD}}$ is an isosceles triangle.
So the base angle of side BD and AD are equal which means $\angle {\text{DBA = }}\angle {\text{DAB = 5}}{{\text{0}}^0}$ given in the figure.
And we know that the sum of all interior angles of a triangle is ${180^0}$.
$
\therefore \angle {\text{DBA + }}\angle {\text{DAB + }}\angle {\text{BDA}} = {180^0} \\
\Rightarrow {50^0} + {50^0} + \angle {\text{BDA}} = {180^0} \\
\Rightarrow \angle {\text{BDA}} = {180^0} - {100^0} \\
\therefore \angle {\text{BDA}} = {80^0} \\
$
And now BDC is a straight line
$\therefore \angle {\text{BDA}} + \angle {\text{ADC = 18}}{{\text{0}}^0}$ Straight angle.
$ \Rightarrow \angle {\text{ADC = 18}}{{\text{0}}^0} - {80^0} = {100^0}$ $\because \angle {\text{BDA}} = {80^0}$
And now in $\vartriangle {\text{ABC}}$
${\text{AD = CD}}$ (Given in question).
It means that $\vartriangle {\text{ADC}}$ is an isosceles triangle.
So the base angle of side BD and AD are equal which means $\angle {\text{DCA = }}\angle {\text{DAC = }}x$ given in the figure.
And we know that the sum of all interior angles of a triangle is ${180^0}$.
$
\therefore \angle {\text{DCA + }}\angle {\text{DAC + }}\angle {\text{ADC}} = {180^0} \\
\Rightarrow x + x + {100^0} = {180^0} \\
\Rightarrow 2x = {180^0} - {100^0} \\
\therefore x = {40^0} \\
$
Hence, option A is the correct answer.

Note: Whenever we face such a type of question the key concept for solving the question is that you have to first read the diagram what it says and the data which are given in the diagrams. Then by applying the properties of triangles start finding the unknown terms to get the answer.
Last updated date: 01st Jun 2023
•
Total views: 329.1k
•
Views today: 4.85k
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India
