Question

ᴘʟᴇᴀsᴇ ʜᴇʟᴘ ᴍᴇ ᴡɪᴛʜ ᴛʜɪs !!!

ᴋɪɴᴅʟʏ ᴅᴏɴᴛ sᴘᴀᴍ

sᴛᴇᴘ ʙʏ sᴛᴇᴘ ᴇxᴘʟᴀɴᴀᴛɪᴏɴ ɴᴇᴇᴅᴇᴅ

ᴛʜᴀɴᴋ ᴜʜ :)​

Answer 1

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

• ABCD is a Trapezium.
• In Which, AB || CD & ${\sf{ \angle D = 70^{ \circ } }}$ and ${\sf{ \angle C = 40^{ \circ } }}$

$\\ {\pmb{\underline{\sf{According \ to \ Postulates ... }}}}$

If there is two lines which are Parallel to each other but when any transversal line bisect ( or cross) both line then Sum of angle on Same side will be 180° .

In this Way,

• ${\sf{ \angle A + \angle D = 180^{ \circ } }}$
• ${\sf{ \angle B + \angle C = 180^{ \circ } }}$

These Pairs of angles is also known as Corresponding sides of each other.

$\\ {\pmb{\underline{\sf{Final \ Solution ... }}}}$

$\colon\implies{\sf{ \angle A + \angle D = 180^{ \circ } }} \\ \\ \colon\implies{\sf{ \angle A = 180^{ \circ } - \angle D}} \\ \\ \colon\implies{\sf{ \angle A = 180^{ \circ } - 70^{ \circ } }} \\ \\ \colon\implies{\sf\red{ \angle A = 110^{ \circ } }}$

Similarly,

$\colon\implies{\sf{ \angle B + \angle C = 180^{ \circ } }} \\ \\ \colon\implies{\sf{ \angle B = 180^{ \circ } - \angle C}} \\ \\ \colon\implies{\sf{ \angle B = 180^{ \circ } - 40^{ \circ } }} \\ \\ \colon\implies{\sf\red{ \angle B = 140^{ \circ } }}$

$\\ {\pmb{\underline{\sf{Verification ... }}}}$

As We know that:

The Sum of all the angles of any Quadrilateral ( Four Sides) is 360° .

We can also Apply these Formula to Find the Total angles of Any Quadrilateral of any sides as:

$\circ \ {\underline{\boxed{\sf\green{ Total _{(Angle)} = (Side - 2) \times 180^{ \circ } }}}}$

According to this Method,

We can find the total Angle of Trapezium as:

$\colon\implies{\sf{ Total _{(angle)} = (4 - 2) \times 180^{ \circ } }} \\ \\ \colon\implies{\sf{ = 2 \times 180^{ \circ } }} \\ \\ \colon\implies{\sf{ 360^{ \circ } }}$

Now, Substituting values as:

$\colon\implies{\sf{ \angle A + \angle B + \angle C + \angle D = 360^{ \circ } }} \\ \\ \colon\implies{\sf{ 110^{ \circ } + 140^{ \circ } + 40^{ \circ } + 70^{ \circ } = 360^{ \circ } }} \\ \\ \colon\implies{\sf\green{ 360^{ \circ } = 360^{ \circ } }}$

Hence,

• ${\sf{ \angle A = 110^{ \circ } }}$
• ${\sf{ \angle B = 140^{ \circ } }}$