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find the value of p that makes quadrilateral bcde a parallelogram. p = …

Question

find the value of p that makes quadrilateral bcde a parallelogram.

p = \square^\circ

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Explanation:

Step1: Sum of parallelogram angles

The sum of interior angles of a quadrilateral is $360^\circ$. For a parallelogram, opposite angles are equal, so we set up the equation:
$$p + (4p-70^\circ) + (3p-100^\circ) + (3p-20^\circ) = 360^\circ$$

Step2: Combine like terms

Add all variable terms and constant terms:
$$p + 4p + 3p + 3p -70^\circ -100^\circ -20^\circ = 360^\circ$$
$$11p - 190^\circ = 360^\circ$$

Step3: Isolate the variable term

Add $190^\circ$ to both sides:
$$11p = 360^\circ + 190^\circ$$
$$11p = 550^\circ$$

Step4: Solve for $p$

Divide both sides by 11:
$$p = \frac{550^\circ}{11}$$

Answer:

$p = 50^\circ$