Exercise 10.5 Practical-Geometry -NCERT Solutions Class 7

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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.

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.

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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