Exercise 6.3 (Revised) - Chapter 6 - Triangles & Its Properties - Ncert Solutions class 7 - Maths

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Chapter 6 - Triangles & Its Properties | NCERT Solutions Class 7 Maths

Ex 6.3 Question 1.

$\text {Find the value of unknown } x \text { in the following diagrams: }$

Answer:

(i) In $\triangle \mathrm{ABC}$,
$\angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180^{\circ}$ [By angle sum property of a triangle $]$
$
\begin{aligned}
& \Rightarrow x+50^{\circ}+60^{\circ}=180^{\circ} \\
& \Rightarrow x+110^{\circ}=180^{\circ} \Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}
\end{aligned}
$
(ii) In $\triangle \mathrm{PQR}$,
$\angle \mathrm{RPQ}+\angle \mathrm{PQR}+\angle \mathrm{RPQ}=180^{\circ}$ [By angle sum property of a triangle $]$
$
\begin{aligned}
& \Rightarrow 90^{\circ}+30^{\circ}+x=180^{\circ} \\
& \Rightarrow x+120^{\circ}=180^{\circ} \Rightarrow x=180^{\circ}-120^{\circ}=60^{\circ}
\end{aligned}
$
(iii) In $\triangle X Y Z$,

$\angle \mathrm{ZXY}+\angle \mathrm{XYZ}+\angle \mathrm{YZX}=180^{\circ}$ [By angle sum property of a triangle]
$
\begin{aligned}
& \Rightarrow 30^{\circ}+110^{\circ}+x=180^{\circ} \\
& \Rightarrow x+140^{\circ}=180^{\circ} \Rightarrow x=180^{\circ}-140^{\circ}=40^{\circ}
\end{aligned}
$
(iv) In the given isosceles triangle,
$
\begin{aligned}
& x+x+50^{\circ}=180^{\circ} \text { [By angle sum property of a triangle] } \\
& \Rightarrow 2 x+50^{\circ}=180^{\circ} \\
& \Rightarrow 2 x=180^{\circ}-50^{\circ} \Rightarrow 2 x=130^{\circ} \\
& \Rightarrow x=\frac{130^{\circ}}{2}=65^{\circ}
\end{aligned}
$
(v) In the given equilateral triangle,
$
\begin{aligned}
& x+x+x=180^{\circ} \text { [By angle sum property of a triangle] } \\
& \Rightarrow 3 x=180^{\circ} \\
& \Rightarrow x=\frac{180^{\circ}}{3}=60^{\circ}
\end{aligned}
$
(vi) In the given right angled triangle,

$x+2 x+90^{\circ}=180^{\circ}$ [By angle sum property of a triangle]
$
\begin{aligned}
& \Rightarrow 3 x+90^{\circ}=180^{\circ} \\
& \Rightarrow 3 x=180^{\circ}-90^{\circ} \Rightarrow 3 x=90^{\circ} \\
& \Rightarrow x=\frac{90^{\circ}}{3}=30^{\circ}
\end{aligned}
$

Ex 6.3 Question 2.

Find the values of the unknowns $x$ and $y$ in the following diagrams:

Answer:

(i) $50^{\circ}+x=120^{\circ}$ [Exterior angle property of a $\Delta$ ]
$
\Rightarrow x=120^{\circ}-50^{\circ}=70^{\circ}
$

Now, $50^{\circ}+x+y=180^{\circ}$ [Angle sum property of a $\Delta$ ]
$
\begin{aligned}
& \Rightarrow 50^{\circ}+70^{\circ}+y=180^{\circ} \\
& \Rightarrow 120^{\circ}+y=180^{\circ} \Rightarrow y=180^{\circ}-120^{\circ}=60^{\circ}
\end{aligned}
$
(ii) $y=80^{\circ}$ .(i) [Vertically opposite angle]

Now, $50^{\circ}+x+y=180^{\circ}$ [Angle sum property of a $\Delta$ ]
$
\Rightarrow 50^{\circ}+80^{\circ}+x=180^{\circ}
$
[From eq. (i)]
$
\Rightarrow 130^{\circ}+x=180^{\circ} \Rightarrow x=180^{\circ}-130^{\circ}=50^{\circ}
$
(iii) $50^{\circ}+60^{\circ}=x$ [Exterior angle property of a $\Delta$ ]
$
\Rightarrow x=110^{\circ}
$

Now $50^{\circ}+60^{\circ}+y=180^{\circ}$ [Angle sum property of a $\Delta$ ]

$
\begin{aligned}
& \Rightarrow 110^{\circ}+y=180^{\circ} \\
& \Rightarrow y=180^{\circ}-110^{\circ} \Rightarrow y=70^{\circ}
\end{aligned}
$
(iv) $x=60^{\circ}$ $\qquad$ (i) [Vertically opposite angle]

Now, $30^{\circ}+x+y=180^{\circ}$ [Angle sum property of a $\Delta$ ]
$
\begin{aligned}
& \Rightarrow 30^{\circ}+60^{\circ}+y=180^{\circ} \text { [From eq. (i)] } \\
& \Rightarrow 90^{\circ}+y=180^{\circ} \Rightarrow y=180^{\circ}-90^{\circ}=90^{\circ}
\end{aligned}
$
(v) $y=90^{\circ}$ $\qquad$ (i) [Vertically opposite angle]

Now, $y+x+x=180^{\circ}$ [Angle sum property of a $\Delta$ ]
$
\begin{aligned}
& \Rightarrow 90^{\circ}+2 x=180^{\circ} \text { [From eq. (i)] } \\
& \Rightarrow 2 x=180^{\circ}-90^{\circ} \Rightarrow 2 x=90^{\circ} \\
& \Rightarrow x=\frac{90^{\circ}}{2}=45^{\circ}
\end{aligned}
$
(vi) $x=y$ $\qquad$ (i) [Vertically opposite angle]

Now, $x+x+y=180^{\circ}$ [Angle sum property of a $\Delta$ ]
$
\begin{aligned}
& \Rightarrow 2 x+x=180^{\circ} \text { [From eq. (i)] } \\
& \Rightarrow 3 x=180^{\circ} \Rightarrow x=\frac{180^{\circ}}{3}=60^{\circ}
\end{aligned}
$