The Triangle and its Properties - Solutions 2

CBSE Class –VII Mathematics
NCERT Solutions
Chapter 6 The Triangle and its Properties (Ex. 6.2)

---

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

Answer: Since, Exterior angle = Sum of opposite interior angles, therefore

1. x=50∘+70∘=120∘$x={50}^{\circ }+{70}^{\circ }={120}^{\circ }$
2. x=65∘+45∘=110∘$x={65}^{\circ }+{45}^{\circ }={110}^{\circ }$
3. x=30∘+40∘=70∘$x={30}^{\circ }+{40}^{\circ }={70}^{\circ }$
4. x=60∘+60∘=120∘$x={60}^{\circ }+{60}^{\circ }={120}^{\circ }$
5. x=50∘+50∘=100∘$x={50}^{\circ }+{50}^{\circ }={100}^{\circ }$
6. x=60∘+30∘=90∘$x={60}^{\circ }+{30}^{\circ }={90}^{\circ }$

Question 2. Find the value of the unknown interior angle x$x$ in the following figures:

Answer: Since, Exterior angle = Sum of  opposite interior angles, therefore

1. x+50∘=115∘$x+{50}^{\circ }={115}^{\circ }$ ⇒$\Rightarrow$ x=115∘−50∘=65∘$x={115}^{\circ }-{50}^{\circ }={65}^{\circ }$
2. 70∘+x=100∘${70}^{\circ }+x={100}^{\circ }$ ⇒$\Rightarrow$ x=100∘−70∘=30∘$x={100}^{\circ }-{70}^{\circ }={30}^{\circ }$
3. x+90∘=125∘$x+{90}^{\circ }={125}^{\circ }$ ⇒$\Rightarrow$ x=125∘−90∘=35∘$x={125}^{\circ }-{90}^{\circ }={35}^{\circ }$
4. 60∘+x=120∘${60}^{\circ }+x={120}^{\circ }$ ⇒$\Rightarrow$ x=120∘−60∘=60∘$x={120}^{\circ }-{60}^{\circ }={60}^{\circ }$
5. 30∘+x=80∘${30}^{\circ }+x={80}^{\circ }$ ⇒$\Rightarrow$ x=80∘−30∘=50∘$x={80}^{\circ }-{30}^{\circ }={50}^{\circ }$
6. x+35∘=75∘$x+{35}^{\circ }={75}^{\circ }$ ⇒$\Rightarrow$ x=75∘−35∘=40∘