The Triangle and its Properties - Solutions 3

CBSE Class –VII Mathematics
NCERT Solutions
Chapter 6 The Triangle and its Properties (Ex. 6.3)

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Question 1. Find the value of unknown x$x$ in the following diagrams:

Answer: (i) In ΔABC,$\mathrm{\Delta }\text{ABC},$

∠$\mathrm{\angle }$BAC + ∠$\mathrm{\angle }$ABC + ∠$\mathrm{\angle }$ACB = 180∘${180}^{\circ }$ [By angle sum property of a triangle]

⇒$\Rightarrow$ x+50∘+60∘=180∘$x+{50}^{\circ }+{60}^{\circ }={180}^{\circ }$

⇒$\Rightarrow$ x+110∘=180∘$x+{110}^{\circ }={180}^{\circ }$ ⇒$\Rightarrow$ x=180∘−110∘=70∘$x={180}^{\circ }-{110}^{\circ }={70}^{\circ }$

(ii) In Δ$\mathrm{\Delta }$PQR,

∠$\mathrm{\angle }$RPQ + ∠$\mathrm{\angle }$PQR + ∠$\mathrm{\angle }$RPQ = 180∘${180}^{\circ }$ [By angle sum property of a triangle]

⇒$\Rightarrow$ 90∘+30∘+x=180∘${90}^{\circ }+{30}^{\circ }+x={180}^{\circ }$

⇒$\Rightarrow$ x+120∘=180∘$x+{120}^{\circ }={180}^{\circ }$ ⇒$\Rightarrow$ x=180∘−120∘=60∘$x={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

(iii) In Δ$\mathrm{\Delta }$XYZ,

∠$\mathrm{\angle }$ZXY + ∠$\mathrm{\angle }$XYZ + ∠$\mathrm{\angle }$YZX = 180∘${180}^{\circ }$ [By angle sum property of a triangle]

⇒$\Rightarrow$ 30∘+110∘+x=180∘${30}^{\circ }+{110}^{\circ }+x={180}^{\circ }$

⇒$\Rightarrow$ x+140∘=180∘$x+{140}^{\circ }={180}^{\circ }$ ⇒$\Rightarrow$ x=180∘−140∘=40∘$x={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

(iv) In the given isosceles triangle,

x+x+50∘=180∘$x+x+{50}^{\circ }={180}^{\circ }$ [By angle sum property of a triangle]

⇒$\Rightarrow$ 2x+50∘=180∘$2x+{50}^{\circ }={180}^{\circ }$

⇒$\Rightarrow$ 2x=180∘−50∘$2x={180}^{\circ }-{50}^{\circ }$ ⇒$\Rightarrow$ 2x=130∘$2x={130}^{\circ }$

⇒$\Rightarrow$ x=130∘2=65∘$x=\frac{{130}^{\circ }}{2}={65}^{\circ }$

(v) In the given equilateral triangle,

x+x+x=180∘$x+x+x={180}^{\circ }$ [By angle sum property of a triangle]

⇒$\Rightarrow$ 3x=180∘$3x={180}^{\circ }$

⇒$\Rightarrow$ x=180∘3=60∘$x=\frac{{180}^{\circ }}{3}={60}^{\circ }$

(vi) In the given right angled triangle,

x+2x+90∘=180∘$x+2x+{90}^{\circ }={180}^{\circ }$ [By angle sum property of a triangle]

⇒$\Rightarrow$ 3x+90∘=180∘$3x+{90}^{\circ }={180}^{\circ }$

⇒$\Rightarrow$ 3x=180∘−90∘$3x={180}^{\circ }-{90}^{\circ }$ ⇒$\Rightarrow$ 3x=90∘$3x={90}^{\circ }$

⇒$\Rightarrow$ x=90∘3=30∘$x=\frac{{90}^{\circ }}{3}={30}^{\circ }$

Question 2. Find the values of the unknowns x$x$ and y$y$ in the following diagrams:


Answer: (i) 50∘+x=120∘${50}^{\circ }+x={120}^{\circ }$ [Exterior angle property of a Δ$\mathrm{\Delta }$ ]

⇒$\Rightarrow$ x=120∘−50∘=70∘$x={120}^{\circ }-{50}^{\circ }={70}^{\circ }$

Now, 50∘+x+y=180∘${50}^{\circ }+x+y={180}^{\circ }$ [Angle sum property of a Δ$\mathrm{\Delta }$ ]

⇒$\Rightarrow$ 50∘+70∘+y=180∘${50}^{\circ }+{70}^{\circ }+y={180}^{\circ }$

⇒$\Rightarrow$ 120∘+y=180∘${120}^{\circ }+y={180}^{\circ }$ ⇒$\Rightarrow$ y=180∘−120∘=60∘$y={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

(ii) y=80∘$y={80}^{\circ }$ ……….(i) [Vertically opposite angle]

Now, 50∘+x+y=180∘${50}^{\circ }+x+y={180}^{\circ }$ [Angle sum property of a Δ$\mathrm{\Delta }$ ]

⇒$\Rightarrow$ 50∘+80∘+x=180∘${50}^{\circ }+{80}^{\circ }+x={180}^{\circ }$

[From eq. (i)]

⇒$\Rightarrow$ 130∘+x=180∘${130}^{\circ }+x={180}^{\circ }$ ⇒$\Rightarrow$ x=180∘−130∘=50∘$x={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

(iii) 50∘+60∘=x${50}^{\circ }+{60}^{\circ }=x$ [Exterior angle property of a Δ$\mathrm{\Delta }$ ]

⇒$\Rightarrow$ x=110∘$x={110}^{\circ }$

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

⇒$\Rightarrow$ 110∘+y=180∘${110}^{\circ }+y={180}^{\circ }$

⇒$\Rightarrow$ y=180∘−110∘$y={180}^{\circ }-{110}^{\circ }$ ⇒$\Rightarrow$ y=70∘$y={70}^{\circ }$

(iv) x=60∘$x={60}^{\circ }$ ……….(i) [Vertically opposite angle]

Now, 30∘+x+y=180∘${30}^{\circ }+x+y={180}^{\circ }$ [Angle sum property of a Δ$\mathrm{\Delta }$ ]

⇒$\Rightarrow$ 30∘+60∘+y=180∘${30}^{\circ }+{60}^{\circ }+y={180}^{\circ }$ [From eq. (i)]

⇒$\Rightarrow$ 90∘+y=180∘${90}^{\circ }+y={180}^{\circ }$ ⇒$\Rightarrow$ y=180∘−90∘=90∘$y={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

(v) y=90∘$y={90}^{\circ }$ ……….(i) [Vertically opposite angle]

Now, y+x+x=180∘$y+x+x={180}^{\circ }$ [Angle sum property of a Δ$\mathrm{\Delta }$ ]

⇒$\Rightarrow$ 90∘+2x=180∘${90}^{\circ }+2x={180}^{\circ }$ [From eq. (i)]

⇒$\Rightarrow$ 2x=180∘−90∘$2x={180}^{\circ }-{90}^{\circ }$ ⇒$\Rightarrow$ 2x=90∘$2x={90}^{\circ }$

⇒$\Rightarrow$ x=90∘2=45∘$x=\frac{{90}^{\circ }}{2}={45}^{\circ }$

(vi) x=y$x=y$ ……….(i) [Vertically opposite angle]

Now, x+x+y=180∘$x+x+y={180}^{\circ }$ [Angle sum property of a Δ$\mathrm{\Delta }$ ]

⇒$\Rightarrow$ 2x+x=180∘$2x+x={180}^{\circ }$ [From eq. (i)]

⇒$\Rightarrow$ 3x=180∘$3x={180}^{\circ }$ ⇒$\Rightarrow$ x=180∘3=60∘