Question

please answer me with solution​

Answer 1

Answer:

For normal incidence on a plane mirror identify which of these statements are true --

(I) Both the angles of incidence and reflection are 0 degree .

(II) Both the angles of incidence and reflection are 90 degree .

(III) The angles that both the incident ray and the reflected ray make with the mirror surface are 0 degree .

(IV) The angles that both the incident ray and the reflected ray make with the mirror surface are 90 degree ..

A) Only (I) is true .

B) Only (II) is true .

C) Both (I) and (III) are true .

D) Both (I) and (IV) are true ✔

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Area of Parallelogram has been used. We know that we can calculate the area of parallelogram using base and height of the parallelogram. Here we are given two base and two height of parallelogram but there is one single parallelogram . So area using both base and height will be equal. Using this concept, we can find our answer.

Let's do it !!

_____________________________________________

★ Formula Used :-

$\\\;\boxed{\sf{\pink{Area\;of\;Parallelogram\;=\;\bf{Base\:\times\:Height}}}}$

_____________________________________________

★ Solution :-

Given,

» Length of AD = Base₁ = 8 cm

» Length of BM = Height₁ = 11.4 cm

» Second Base of Parallelogram = Base₂ = AB

» Second Height of Parallelogram = Height₂ = DL

Since its given that BM is perpendicular to AD this means surely its one of the height of parallelogram.

_____________________________________________

a.) For the Area of Parallelogram ::

Here we shall use the first base and first height of parallelogram since both values are given.

We know that

$\\\;\sf{\rightarrow\;\;Area\;of\;Parallelogram\;=\;\bf{Base\:\times\:Height}}$

By applying values, we get

$\\\;\sf{\rightarrow\;\;Area\;of\;Parallelogram,\;ABCD\;=\;\bf{Base_{1}\:\times\:Height_{1}}}$

$\\\;\sf{\rightarrow\;\;Area\;of\;Parallelogram,\;ABCD\;=\;\bf{AD\:\times\:BM}}$

$\\\;\sf{\rightarrow\;\;Area\;of\;Parallelogram,\;ABCD\;=\;\bf{8\:\times\:11.4}}$

$\\\;\bf{\rightarrow\;\;Area\;of\;Parallelogram,\;ABCD\;=\;\bf{\blue{91.2\;\;cm^{2}}}}$

$\\\;\underline{\boxed{\tt{Area\:\;of\:\;Parallelogram\;\:ABCD\;=\;\bf{\purple{91.2\;\;cm^{2}}}}}}$

_____________________________________________

b.) For the height DL if AB = 12 cm ::

We see that we are given that,

• Base₂ = AB = 12 cm

We also know that if we calculate the area of parallelogram ABCD using base as AB and height as DL, then that area will be equal to the area od parallelogram calculated using base as AD and height as DM.

So using this relationship, we get

$\\\;\bf{\mapsto\;\;\green{Base_{2}\:\times\:Height_{2}\;=\;\bf{Base_{1}\:\times\:Height_{1}}}}$

Now by applying values here, we get

$\\\;\sf{:\Longrightarrow\;\;AB\:\times\:DL\;=\;\bf{AD\:\times\:BM}}$

$\\\;\sf{:\Longrightarrow\;\;12\:\times\:DL\;=\;\bf{8\:\times\:11.4}}$

$\\\;\sf{:\Longrightarrow\;\;12\:\times\:DL\;=\;\bf{91.2}}$

$\\\;\sf{:\Longrightarrow\;\;DL\;=\;\bf{\dfrac{91.2}{12}}}$

$\\\;\sf{:\Longrightarrow\;\;DL\;=\;\bf{\red{7.6\;\:cm}}}$

$\\\;\underline{\boxed{\tt{Length\;\:of\;\:DL\;=\;\bf{\purple{7.6\;\:cm}}}}}$

_____________________________________________

★ More to know :-

$\\\;\sf{\leadsto\;\:Perimeter\;of\;Paralellogram\;=\;Sum\;of\;all\;sides}$

• Parallelogram is a four sided figure.

• Opposite sides of parallelogram are equal and parallel.

• Sum of adjacent angles of parallelogram equals to 180°

• Diagonals of parallelogram bisect each other

• Square and Rectangle are one of the kinds of parallelogram.