Question

Please help me please​

Answer 1

• Answer :-

The altitude of the triangle is 12cm and the base is 20cm

• Given :-

The altitude of a triangle is three-fifth of the length of the corresponding base .

When the altitude is decreased by 4cm and the corresponding base is increased by 10cm , the area of the triangle remains the same.

• To Find :-

The base and the altitude of the triangle.

• Formula to be used :-

$\sf{Area \: \: of \: \: a \: \: triangle = \dfrac{1}{2}\times Base \times Altitude \: \: (or Height) }$

• Solution :-

Let us consider the altitude of the triangle be x cm and corresponding base be y cm

Therefore,

$\bf{The \: \: area \: \: of \: \: triangle = \dfrac{1}{2} }$

• According to Question :-

The altitude of a triangle is three-fifth of the length of the corresponding base.

$\sf{\implies x = \dfrac{3}{5}\times y ...........(1)}$

When the altitude is decreased by 4cm and the corresponding base is increased by 10cm , the area of the triangle remains the same.

$\begin{gathered}\sf \dfrac{1}{2} \times (x - 4)(y + 10) = \dfrac{1}{2} xy \\ \\ \implies \sf(x - 4)(y + 10) = xy \\ \\ \implies \sf xy + 10x - 4y - 40 = xy \\ \\ \sf \implies10x - 4y - 40 = 0.......(2)\end{gathered}$

Using the value of (1) in (2) we have :-

$\begin{gathered}\sf10\{ \dfrac{3}{5} \times y \} - 4y - 40 = 0 \\ \\ \sf \implies6y - 4y - 40 = 0 \\ \\ \sf \implies2y = 40 \\ \\ \bf \implies y = 20\end{gathered}$

And Now using the value of y in (1) :-

$\begin{gathered}\sf{\implies x = \dfrac{3}{5}\times 20}\\\\ \implies \sf{x = 3\times4}\\\\ \implies\bf{x= 12}\end{gathered}$

∴$\small \underline \bold \red {The\:altitude\:is\:12cm\:and\:the\:base\:is\:20cm}$

Answer 2

ANSWER

∣z1+z2∣2=∣z1∣2+∣z2∣2

z1z2ˉ+z2z1ˉ=0

⇒z2z1=−z2ˉz1ˉ

⇒z2z1+(z2z1)ˉ=0⇒z2z1 is purely imaginary

so amp (z2z1) is may be 2π or −2π