# Exercise 10.4 Practical-Geometry -NCERT Solutions Class 7

## Question 1

Construct \(ΔABC\), given \(m∠A = 60^\circ\), \(m∠B = 30^\circ \) and \(AB = 5.8\,\rm{ cm}\).

### Solution

**Video Solution**

**What is known?**

Length of side of a \(ΔABC\), \(AB = 5.8\,\rm{ cm}\) and \(∠A = 60^\circ\), \(∠B = 30^\circ \).

**To construct:**

A triangle \(ΔABC\), given \(∠A = 60^\circ\), \(∠B = 30^\circ \) and \(AB = 5.8\,\rm{ cm}\).

**Reasoning:**

To Construct a \(ΔABC\) such that \(∠A = 60^\circ, ∠B = 30^\circ\) and \(AB = 5.8 \,\rm{cm}\)., follow the steps given below.

**Steps:**

__Steps of construction:__

- Draw a line segment \(AB\) of length \(5.8\,\rm{ cm}.\)
- At \(A\), draw ray \(AY\) making \(60^\circ\) with \(AB\).
- At \(B\), draw ray \(BX\) making \(30^\circ\) with \(AB\).
- Rays \(BX\) and \(AY\) will intersect at point \(C\).

Triangle \(ABC\) is now completed.

## Question 2

Construct \(ΔPQR\) if \(PQ = 5 \rm{cm}\), \(m∠PQR = 105^\circ \) and \(m∠QRP = 40^\circ \). (Hint: Recall angle-sum property of a triangle).

### Solution

**Video Solution**

**What is known?**

In triangle \(PQR\) , \(PQ = 5 \rm{cm}\), \(∠PQR = 105^\circ \) and \(∠QRP = 40^\circ \).

**To construct:**

A triangle \(ΔPQR\) if \(PQ = 5 \rm{cm}\), \(∠PQR = 105^\circ \) and \(∠QRP = 40^\circ \).

**Reasoning:**

We will use angle-sum property of a triangle to find the measure of \(∠RPQ\)

By angle sum property of a triangle,

\(∠PQR + ∠QRP + ∠RPQ = 180^\circ.\)

\(105^\circ + 40^\circ + ∠RPQ = 180^\circ.\)

So, \(∠RPQ = 35^\circ\)

Now, let’s Construct \(ΔPQR\) such that \(PQ = 5\, {\rm{cm}} , ∠PQR = 105^\circ\) and \(∠RPQ = 35^\circ\), with the steps given below

**Steps:**

** Steps of construction** –

- Draw a line segment \(PQ\) of length \(5 \, \rm{cm}.\)
- At \(P\), draw \(PX\) making \(35^\circ\) with \(PQ.\)
- At \( Q,\) draw \(QY\) making an angle of \(105^\circ\) with \(PQ.\)
- \(PX\) and \(QY\) will intersect at point \(R.\)

\(PQR\) is the required triangle.

## Question 3

Examine whether you can construct \(ΔDEF\) such that \(EF = 7.2\,\rm{ cm},\) \(m∠E = 110^\circ\) and \(m∠F = 80^\circ. \)Justify your answer.

### Solution

**Video Solution**

**What is known?**

Length of one side of a triangle and measure of two angles.

**What is unknown?**

Whether a triangle be constructed with the given values of length and angles.

**Reasoning:**

By using the angle sum property, we can find out the third angle. If the angle sum property is followed, then it is possible to construct a triangle and if not then we cannot construct a triangle.

**Steps:**

We will use angle-sum property of a triangle to find the measure of \(∠D\)

By angle sum property,

\(∠E +∠F + ∠D =\) \(180^\circ .\)

\(110^\circ + 80^\circ + ∠D =\) \(180^\circ. \)

\(190^\circ + ∠D\) \(=\) \(180^\circ .\)

So, \(∠D = -10^\circ\)

Angle of \(-10^\circ\) is not possible, thus we cannot construct triangle \(ΔDEF.\)

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