# Ex.10.3 Q1 Practical Geometry Solution- NCERT Maths Class 7

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## Question

Construct $$ΔDEF$$ such that $$DE = 5\,\rm{ cm}$$, $$DF = 3\,\rm{ cm}$$ and $$m∠EDF = 90^\circ$$.

Video Solution
Practical Geometry
Ex 10.3 | Question 1

## Text Solution

What is known?

Lengths of sides of a $$ΔDEF$$ and measure of one of the angles.

To construct:

A triangle $$ΔDEF$$ such that $$DE = 5\,\rm{ cm}$$, $$DF = 3\,\rm{ cm}$$ and $$∠EDF = 90^\circ$$.

Reasoning:

To construct a $$ΔDEF$$ first, we draw a rough sketch with the given measure such that $$DE = 5\,\rm{ cm}, DF = 3\,\rm{ cm}$$ and $$∠EDF = 90^\circ$$, then follow the steps given below.

Steps:

Steps of construction:

1. Draw a line segment $$DE$$ of length $$5\,\rm{cm}.$$
2. At $$D,$$ draw $$DX$$ making $$90^\circ$$ with $$DE$$.
3. With $$D$$ as centre, draw an arc of radius $$3\,\rm{cm}$$. It cuts $$DX$$ at the point $$F.$$
4. Join $$EF$$ to get the required triangle.

Thus,​​​​​​$$ΔDEF$$  is the required triangle.

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