Practical Geometry - Solutions 4

CBSE Class –VII Mathematics
NCERT Solutions
Chapter 10 Practical Geometry (Ex. 10.4)

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Question 1. Construct Δ$\mathrm{\Delta }$ABC, given m∠$m\mathrm{\angle }$A = 60∘,${60}^{\circ },$m∠$m\mathrm{\angle }$B = 30∘${30}^{\circ }$ and AB = 5.8 cm.

Answer: To construct: Δ$\mathrm{\Delta }$ABC where m∠$m\mathrm{\angle }$A = 60∘,${60}^{\circ },$m∠$m\mathrm{\angle }$B = 30∘${30}^{\circ }$ and AB = 5.8 cm.

Steps of construction:

(a) Draw a line segment AB = 5.8 cm.

(b) At point A, draw an angle ∠$\mathrm{\angle }$YAB = 60∘${60}^{\circ }$ with the help of compass.

(c) At point B, draw ∠$\mathrm{\angle }$XBA = 30∘${30}^{\circ }$ with the help of compass.

(d) AY and BX intersect at the point C.

It is the required triangle ABC.

Question 2. Construct Δ$\mathrm{\Delta }$PQR if PQ = 5 cm, m∠$m\mathrm{\angle }$PQR = 105∘${105}^{\circ }$ and m∠$m\mathrm{\angle }$QRP = 40∘.${40}^{\circ }.$

Answer: Given: m∠$m\mathrm{\angle }$PQR = 105∘${105}^{\circ }$ and m∠$m\mathrm{\angle }$QRP = 40∘${40}^{\circ }$

We know that sum of angles of a triangle is 180∘.${180}^{\circ }.$

∴$\therefore$ m∠$m\mathrm{\angle }$PQR + m∠$m\mathrm{\angle }$QRP + m∠$m\mathrm{\angle }$QPR = 180∘${180}^{\circ }$

⇒$\Rightarrow$ 105∘+40∘+m∠${105}^{\circ }+{40}^{\circ }+m\mathrm{\angle }$QPR = 180∘${180}^{\circ }$

⇒$\Rightarrow$ 145∘${145}^{\circ }$ + m∠$m\mathrm{\angle }$QPR = 180∘${180}^{\circ }$

⇒$\Rightarrow$ m∠$m\mathrm{\angle }$QPR = 180∘${180}^{\circ }$ – 145∘${145}^{\circ }$

⇒$\Rightarrow$ m∠$m\mathrm{\angle }$QPR = 35∘${35}^{\circ }$

To construct: Δ$\mathrm{\Delta }$PQR where m∠$m\mathrm{\angle }$P = 35∘${35}^{\circ }$, m∠$m\mathrm{\angle }$Q = 105∘${105}^{\circ }$ and PQ = 5 cm.

Steps of construction:

(a) Draw a line segment PQ = 5 cm.

(b) At point P, draw ∠$\mathrm{\angle }$XPQ = 35∘${35}^{\circ }$ with the help of protractor.

(c) At point Q, draw ∠$\mathrm{\angle }$YQP = 105∘${105}^{\circ }$ with the help of protractor.

(d) XP and YQ intersect at point R.

It is the required triangle PQR.

Question 3. Examine whether you can construct Δ$\mathrm{\Delta }$DEF such that EF = 7.2 cm, m∠$m\mathrm{\angle }$E = 110∘${110}^{\circ }$ and m∠$m\mathrm{\angle }$F = 80∘.${80}^{\circ }.$ Justify your answer.

Answer: Triangel DEF cannot be constructed since Angle E + Angle F = 110o + 80o = 190o 

and the sum of all the angles of triangle is 180o.