value of x for which 7 by 12 to the power -4 x 7 by 12 to the power 3x equal to 7/12 the power ^5
Answers
Step-by-step explanation:
Correct question:-
In an equilateral triangle ABC, D is a point on side BC such that BD=⅓BC. prove that 9AD²=7AB².
Given:-
Equilateral triangle ABC D is a point an BC.
To prove:-
9AD²=7AB².
Construct:-
Lets draw AE _|_ BC.
Proof:-
All sides of equilateral triangle is equal.
\sf \: AB=BC=ACAB=BC=AC
\sf \: Let \: AB=BC=AC = xLetAB=BC=AC=x
Given
\sf \: BD = \frac{1}{3}BCBD=
3
1
BC
\sf \: BD = \frac{x}{3}BD=
3
x
In ∆AEB and ∆AEC
AE=AE (common)
AB=AC. ( Both x as it is equilateral triangle)
∠AEB= ∠AEC. ( both 90° as AE_|_ BC)
Hence by RHS congruency
∆AEB \cong ∆AEC∆AEB≅∆AEC
\sf \: \therefore \: BE= EC \: \: \: \: \: \: \: \: \: \: \: ( By \: CPCT )∴BE=EC(ByCPCT)
So, BE=EC= 1/2 BC
BE=EC=x/2
So,
\sf \: BE = \frac{ x }{2}BE=
2
x
\sf \: BD + DE = \frac{x}{2}BD+DE=
2
x
\sf \: \frac{x}{3} + DE = \frac{x}{2}
3
x
+DE=
2
x
\sf \: DE = \frac{x}{2} - \frac{x}{3}DE=
2
x
−
3
x
\sf \: DE = \frac{3x - 2x}{6}DE=
6
3x−2x
\sf \: DE = \frac{x}{6}DE=
6
x
Using pythagoras theorem
⇒(Hypotenuse)²= (Height)²+ (Base)²
Now in right ∆ AEB
⇒AB²=AE²+(BE)²
\sf \: ⇒x²= (AE)² +(\frac{x}{2})^2⇒x²=(AE)²+(
2
x
)
2
\sf \: ⇒x² =(AE)² + (\frac{x}{2})^2⇒x²=(AE)²+(
2
x
)
2
\sf \: ⇒x²-\frac{x}{4}^2=AE²⇒x²−
4
x
2
=AE²
\sf \: AE ^{2} = \frac{3x^{2} }{4} \: \: \: \: \: \: \: \: \: (1)AE
2
=
4
3x
2
(1)
Similarly,
In right ∆AED
\sf \: AD²=AE²+DE²AD²=AE²+DE²
\sf \: AD² = {3x}^{2} + ( \frac{x }{6} )^{2}AD²=3x
2
+(
6
x
)
2
\sf \: AD²= \frac{3 {x}^{2} }{4} + \frac{ {x}^{2} }{36}AD²=
4
3x
2
+
36
x
2
\sf \: AD²= \frac{( {3x}^{2}) \times 9 \times {x}^{2} }{36}AD²=
36
(3x
2
)×9×x
2
\sf \: AD²= \frac{27 {x}^{2} + {x}^{2} }{36}AD²=
36
27x
2
+x
2
\sf \: AD²= \frac{28 {x}^{2} }{36}AD²=
36
28x
2
\sf \: AD²= \frac{7 {x}^{2} }{9}AD²=
9
7x
2
\sf \: 9AD²=7 {x}^{2}9AD²=7x
2
\sf \: 9AD²=7AB ^{2}9AD²=7AB
2