Question

When the two triangle are similar the ratio of those triangle is equal to the ratio of the aquares of thier coresponding side

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Answer 1

Answer:

Given:△ABC∼△PQR

To prove:

$ar \frac{(△abc)}{(△pqr)} = (\frac{ab}{pq})^{2} = (\frac{bc}{qr} ) ^{2} = (\frac{ac}{pr}) ^{2}$

Construction:Draw AM⊥BC and PN⊥QR

Proof:area(△ABC)= 1/2×base×height=

1/2×BC×AM ...(1)

area(△PQR)=1/2×base×height=1/2×QR×PN ..(2)

Dividing (1) by (2) we get

$ar \frac{(△abc)}{(△pqr)} = \frac{ \frac{1}{2} \times bc \times am}{ \frac{1}{2} \times qr \times pn}$

$= \frac{bc \times am}{qr \times pn}...(a)$

In △ABM and △PQN

∠B=∠Q as △ABC∼△PQR and angles of similar triangles are equal.

∠M=∠N (both 90°)

△ABM∼△PQN by AA similarity

$\frac{ab}{pq} = \frac{am}{pn}(corresponding \: sides \: of \: similar \: triangles \: are \: proportional)...(b)$

From (A)

$ar \frac{(△abc)}{(△pqr)} = \frac{bc \times am}{qr \times pn} = \frac{bc}{qr} \times \frac{ab}{pq} from(b)...(c)$

Now, Given:△ABC∼△PQR

$= \frac{ab}{pq} = \frac{bc}{qr} = \frac{ac}{pr}$

Putting in (C) we get

$ar \frac{(△abc)}{(△pqr)} = \frac{ab}{pq} \times \frac{ab}{pq} = ( \frac{ab}{pq})^{2}$

$now \: again \: using \: \frac{ab}{pq} = \frac{bc}{qr} = \frac{ac}{pr}$

ar (△ABC)/(△PQR) = (AB)/(PQ)^2 = (BC)/(QR)^2 = (AC)/(PR)^2

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Answer 2

Answer:

Review. Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing. ... Inductive reactance is the reduction in current flow in a circuit due to induction.