Test your knowledge with AI-generated multiple choice questions
When lines are parallel, alternate interior angles are congruent. Therefore, the alternate interior angle has the same measure as the given angle, $68^\circ$, not its supplement.
By the exterior angle theorem, the exterior angle equals the sum of the two remote interior angles: $128^\circ=\angle A+\angle B=51^\circ+\angle B$. Therefore, $\angle B=128^\circ-51^\circ=77^\circ$.
Angles in a triangle sum to $180^\circ$. So $\angle C=180^\circ-47^\circ-56^\circ=77^\circ$.
By Pythagoras, $c=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\,\text{cm}$.
For parallel lines, alternate interior angles are equal. So the required angle is $68^\circ$. $112^\circ$ would be a linear-pair supplement, $34^\circ$ incorrectly halves the angle, and $180^\circ$ is a straight angle, not the interior angle asked for.
By the Pythagorean Theorem, $c=\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}=25\,\text{m}$. The other options do not satisfy $a^2+b^2=c^2$.
Area of the large rectangle is $8\times 6=48\ \text{m}^2$. The removed notch is $4\times 2=8\ \text{m}^2$. Net area $=48-8=40\ \text{m}^2$.
