# Ex.10.5 Q3 Practical Geometry Solution- NCERT Maths Class 7

## Question

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

Video Solution

Practical Geometry

Ex 10.5 | Question 3

## Text 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.

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