B] Solver
1. Expand: x + 1/2
Answers
Answer:
❍ The angles of the Quadrilateral are x°, (x – 10)°, (x + 30)° and 2x° respectively.
⠀⠀⠀\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}
†Asweknowthat:
⠀⠀⠀⠀
The sum of all angles of the Quadrilateral is 360°. Therefore,
\begin{gathered}:\implies\sf x + (x - 10)^\circ + (x + 30)^\circ + 2x = 360^\circ \\\\\\:\implies\sf x + x + x + 2x - 10 + 30 = 360^\circ \\\\\\:\implies\sf 3x + 2x - 10 + 30= 360^\circ \\\\\\:\implies\sf 5x + 20 = 360^\circ \\\\\\:\implies\sf 5x = 360^\circ - 20 \\\\\\:\implies\sf 5x = 340^\circ \\\\\\:\implies\sf x = \cancel\dfrac{340^\circ}{5} \\\\\\:\implies{\underline{\boxed{\frak{\purple{x = 68^\circ}}}}}\;\bigstar\end{gathered}
:⟹x+(x−10)
∘
+(x+30)
∘
+2x=360
∘
:⟹x+x+x+2x−10+30=360
∘
:⟹3x+2x−10+30=360
∘
:⟹5x+20=360
∘
:⟹5x=360
∘
−20
:⟹5x=340
∘
:⟹x=
5
340
∘
:⟹
x=68
∘
★
Hence,
First angle, x = 68°
Second angle, (x - 10)° = (68 - 10)° = 58°
Third angle, (x + 30)° = (68 + 30)° = 98°
Fourth angle, 2x = 2(68)° = 136°
\therefore{\underline{\sf{Hence, the\;greatest\;angle\;is\;\bf{136^\circ }.}}}∴
Hence,thegreatestangleis136
∘
.
━━━━━━━━━━━━━━━━━━━⠀⠀⠀
⠀
V E R I F I C A T I O N :
As we know that sum of the all angles of Quadrilateral is 360°. And, we've measure of each angle. So, Let's verify :
\begin{gathered}\dashrightarrow\sf a + b + c + d = 360^\circ \\\\\\\dashrightarrow\sf 68^\circ + 58^\circ + 98^\circ + 136^\circ = 360^\circ \\\\\\\dashrightarrow{\boxed{\underline{\sf{360^\circ = 360^\circ}}}}\end{gathered}
⇢a+b+c+d=360
∘
⇢68
∘
+58
∘
+98
∘
+136
∘
=360
∘
⇢
360
∘
=360
∘
\therefore{\underline{\textsf{\textbf{Hence Verified!}}}}∴
Hence Verified!
❍ The angles of the Quadrilateral are x°, (x – 10)°, (x + 30)° and 2x° respectively.
⠀⠀⠀\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}
†Asweknowthat:
⠀⠀⠀⠀
The sum of all angles of the Quadrilateral is 360°. Therefore,
\begin{gathered}:\implies\sf x + (x - 10)^\circ + (x + 30)^\circ + 2x = 360^\circ \\\\\\:\implies\sf x + x + x + 2x - 10 + 30 = 360^\circ \\\\\\:\implies\sf 3x + 2x - 10 + 30= 360^\circ \\\\\\:\implies\sf 5x + 20 = 360^\circ \\\\\\:\implies\sf 5x = 360^\circ - 20 \\\\\\:\implies\sf 5x = 340^\circ \\\\\\:\implies\sf x = \cancel\dfrac{340^\circ}{5} \\\\\\:\implies{\underline{\boxed{\frak{\purple{x = 68^\circ}}}}}\;\bigstar\end{gathered}
:⟹x+(x−10)
∘
+(x+30)
∘
+2x=360
∘
:⟹x+x+x+2x−10+30=360
∘
:⟹3x+2x−10+30=360
∘
:⟹5x+20=360
∘
:⟹5x=360
∘
−20
:⟹5x=340
∘
:⟹x=
5
340
∘
:⟹
x=68
∘
★
Hence,
First angle, x = 68°
Second angle, (x - 10)° = (68 - 10)° = 58°
Third angle, (x + 30)° = (68 + 30)° = 98°
Fourth angle, 2x = 2(68)° = 136°
\therefore{\underline{\sf{Hence, the\;greatest\;angle\;is\;\bf{136^\circ }.}}}∴
Hence,thegreatestangleis136
∘
.
━━━━━━━━━━━━━━━━━━━⠀⠀⠀
⠀
V E R I F I C A T I O N :
As we know that sum of the all angles of Quadrilateral is 360°. And, we've measure of each angle. So, Let's verify :
\begin{gathered}\dashrightarrow\sf a + b + c + d = 360^\circ \\\\\\\dashrightarrow\sf 68^\circ + 58^\circ + 98^\circ + 136^\circ = 360^\circ \\\\\\\dashrightarrow{\boxed{\underline{\sf{360^\circ = 360^\circ}}}}\end{gathered}
⇢a+b+c+d=360
∘
⇢68
∘
+58
∘
+98
∘
+136
∘
=360
∘
⇢
360
∘
=360
∘
\therefore{\underline{\textsf{\textbf{Hence Verified!}}}}∴
Hence Verified!