Exercise 10.5 Practical-Geometry -NCERT Solutions Class 7

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Chapter 10 Ex.10.5 Question 1

Construct the right angled \(ΔPQR\), where \(m∠Q = 90^\circ\), \(QR = 8 \rm{cm}\) and \(PR = 10\, \rm{cm}.\)

Solution

Video Solution

What is known?

Length of two side of a right angled \(ΔPQR\).

To construct:

A right angled \(ΔPQR\), where \(∠Q = 90^\circ\), \(QR = 8 \rm{cm}\) and \(PR = 10\, \rm{cm}.\) 

Reasoning:

To construct a right-angled \(ΔPQR\), where \(∠Q = 90^\circ\), \(QR = 8\,\rm{cm}\) and \(PR = 10 \,\rm{cm}\), draw a rough sketch and mark the measures. Remember to mark the right angle and follow the steps given below.

Steps:

Steps of construction :

  1. Draw \(QR\) of length \( 8\, \rm{cm}.\)
  2. At \(Q\), draw \(QX\,\text{perpendicular}\,QR.\)
  3. With \(R\) as centre, draw an arc of radius \(10\,\rm{ cm}\) which should intersect \(QX\) at point \(P\).
  4. Join \(P\) and \(R\).

\(ΔPQR\) is the required triangle.

Chapter 10 Ex.10.5 Question 2

Construct a right-angled triangle whose hypotenuse is \(6\,\rm{ cm}\) long and one of the legs is \(4\,\rm{ cm}\) long.

Solution

Video Solution

What is known?

Hypotenuse of a right-angled triangle and one of the legs.

To construct:

A right-angled triangle whose hypotenuse is \(6\,\rm{ cm}\) long and one of the legs is \(4\,\rm{ cm}\) long.

Reasoning:

To construct a right-angled triangle whose hypotenuse is \(6\,\rm{ cm}\) long and one of the legs is \(4 \,\rm{cm}\) long, draw a rough sketch and mark the measures. Remember to mark the right angle
and follow the steps given below.

Steps:

Steps of construction

  1. Draw a line segment \(EF\) of length \(4\,\rm{ cm}.\)
  2. At \(E\), draw \( EX\,\text{perpendicular}\, EF.\)
  3. With \(F\) as centre, draw an arc of radius 6cm which should intersect \(EX\) at point \(D.\)
  4. Join \(D\) with \(F\).

\(ΔDEF\) is the required triangle.

Chapter 10 Ex.10.5 Question 3

Construct an isosceles right-angled triangle \(ABC\), where \(m∠ACB = 90^\circ \) and \(AC = 6\,\rm{ cm}.\)

Solution

Video Solution

To construct:

An isosceles right-angled triangle \(ABC\), where \(m∠ACB = 90^\circ \) and \(AC = 6\,\rm{ cm}.\)

What is known?

Length of side of an isosceles right-angled triangle \(ABC\),\( AC = 6\,\rm{cm}\) and \(∠ACB = 90^\circ \).

Reasoning:

Since \(ΔABC\) is an isosceles right-angled triangle, where \(∠ACB = 90^\circ\) and \(AC = 6\,\rm{ cm}\). Therefore, length of the other equal side, we can take is \(BC = 6\,\rm{cm}\). To construct this triangle, follow the steps given below.

Steps:

Steps of construction

  1. Draw a line segment \(CA\) of length \(6\,\rm{cm}.\)
  2. At point \(C,\) draw \(CX\) perpendicular \(CA.\)
  3. With \(C\) as centre, draw an arc of radius \(6\,\rm{cm}\) which should intersect \(CX\) at point \(B.\)
  4. Join \(A\) and \(B.\)

Triangle \(ABC\) is the required isosceles right-angles triangle.

  
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