# NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 4 Chapter Name Practical Geometry Exercise Ex 4.5 Number of Questions Solved 1 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5

Question 1.
Draw the following:
1. The square READ with RE = 5.1 cm.
2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
3. A rectangle with adjacent sides of lengths 5 cm and 4 cm.
4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique?
Solution.
1. Steps of Construction

1. Draw RE = 5.1 cm.
2. At R, draw a ray RX such that ∠ERX 3. From ray RX, cut RD = 5.1 cm.
4. At E, draw a ray EY such that ∠REY = 90°.
5. From ray EY, cut EA = 5.1 cm.

Then, READ is the required square.

2. Steps of Construction
[We know that the diagonals of a rhombus bisect each other at right angles. So in rhombus ABCD, the diagonals AC and BD will bisect each other at right angles.]

1. Draw AC = 5.2 cm.
2. Construct its perpendicular bisector. Let it intersect AC at O. 3. Cut off $$\frac { 6.4 }{ 2 }$$= 3.2 cm lengths on either side of the bisector drawn in step 2, we get B and D.
4. Join AB, BC, CD, and DA.

Then, ABCD is the required rhombus.

3. Steps of Construction
[We know that each angle of a rectangle is 90°. So, in rectangle PQRS,
∠P=∠Q=∠R=∠S= 90°.
Also, opposite sides of a rectangle are parallel.
So, in rectangle PQRS,
PQ || SR and PS || QR]

1. Draw PQ = 5 cm.
2. At Q, draw a ray QX such that ∠PQX = 90°. 3. From ray QX, cut QR = 4 cm.
4. At P, draw a ray PY parallel to QR.
5. At R, draw a ray RZ parallel to QP to meet the ray drawn in step 4 at S.

Then, PQRS is the required rectangle.

4. Steps of Construction
[We know that in a parallelogram, opposite sides are parallel and equal. So,
OK = YA and OK || YA;
KA = OY and KA || OY]

1. Draw OK = 5.5 cm. 2. At K, draw a ray KX at any suitable angle from OK.
3. From ray KX, cut KA = 4.2 cm.
4. A, draw a ray AT parallel to KO.
5. At O, draw a ray OZ parallel to KA to cut the ray drawn in step 4 at Y.

Then, OKAY is the required parallelogram.
This is not unique.
Note: We can construct countless parallelograms with these dimensions by varying ∠OKA

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