Question

solve me some sums of triangle... tysm and plz *send a pic of ur answer*

Answer 1

hope you understand and please keep your promise

Answer 2

Solution-

In ΔABC and ΔADE,

• ∠A is common to both the triangles,
• ∠ABC = ∠ADE (As they are the corresponding angles),
• ∠ACB = ∠AED (As they are the corresponding angles),

So, ΔABC ≈ ΔADE (AAA similarity)

$\Rightarrow \frac{AD}{AB}=\frac{AE}{AC}$

$\Rightarrow \frac{AD}{AD+DB}=\frac{AE}{AC}$

$\Rightarrow \frac{AD+DB}{AD}=\frac{AC}{AE}$

$\Rightarrow 1+\frac{DB}{AD}=\frac{AC}{AE}$

1.

$\Rightarrow 1+\frac{3}{1}=\frac{AC}{6}$

$\Rightarrow AC=4\times 6$

$\Rightarrow AC=24$

2.

$\Rightarrow 1+\frac{4}{3}=\frac{28}{AE}$

$\Rightarrow \frac{7}{3}=\frac{28}{AE}$

$\Rightarrow AE=\frac{3\times 28}{7}$

$\Rightarrow AE=12$

3.

$\Rightarrow \frac{4x-3}{3x-1}=\frac{8x-7}{5x-3}$

$\Rightarrow 24x^2-21x-8x+7=20x^2-12x-15x+9$

$\Rightarrow 24x^2-29x+7=20x^2-27x+9$

$\Rightarrow 4x^2-2x-2=0$

$\Rightarrow 4x^2-4x+2x-2=0$

$\Rightarrow 4x(x-1)+2(x-1)=0$

$\Rightarrow (4x+2)(x-1)=0$

$\Rightarrow x=-\frac{1}{2} ,\ 1$