Exercise 10.5 Practical-Geometry -NCERT Solutions Class 7
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 :
- Draw \(QR\) of length \( 8\, \rm{cm}.\)
- At \(Q\), draw \(QX\,\text{perpendicular}\,QR.\)
- With \(R\) as centre, draw an arc of radius \(10\,\rm{ cm}\) which should intersect \(QX\) at point \(P\).
- 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 –
- Draw a line segment \(EF\) of length \(4\,\rm{ cm}.\)
- At \(E\), draw \( EX\,\text{perpendicular}\, EF.\)
- With \(F\) as centre, draw an arc of radius 6cm which should intersect \(EX\) at point \(D.\)
- 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 –
- Draw a line segment \(CA\) of length \(6\,\rm{cm}.\)
- At point \(C,\) draw \(CX\) perpendicular \(CA.\)
- With \(C\) as centre, draw an arc of radius \(6\,\rm{cm}\) which should intersect \(CX\) at point \(B.\)
- Join \(A\) and \(B.\)
Triangle \(ABC\) is the required isosceles right-angles triangle.