Construct a quadrilateral PQRS in which PQ = 5 cm,

Question:

Construct a quadrilateral PQRS in which PQ = 5 cm, QR = 6.5 cm, ∠P = ∠R = 100° and ∠S = 75°.

Solution:

Steps of construction:

Step 1: Draw PQ5cm">5cm5cm

Step 2: $\angle P+\angle Q+\angle R+\angle S=360^{\circ}$

$100^{\circ}+\angle Q+100^{\circ}+75^{\circ}=360^{\circ}$

$275^{\circ}+\angle Q=360^{\circ}$

$\angle Q=360^{\circ}-275^{\circ}$

$\angle Q=85^{\circ}$

Step 3: Make $\angle P=100^{\circ}$ and $\angle Q=85^{\circ}$

Step 3: With $Q$ as the centre, draw an $\operatorname{arc}$ of $6.5 \mathrm{~cm}$.

Step 4: Make $\angle R=100^{\circ}$

Step 6: Join $R$ and $S$.

Step 7: Measure $\angle S=75^{\circ}$

Then, $P Q R S$ is the required quadrilateral.

 

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