12 multiple-choice questions, progressively harder.
Rectangle ABCDABCDABCD has whole diagonals with these lengths, in centimeters.
Find xxx.
Solution
Correct answer: C
Congruent whole diagonals give an equation.
Both expressions then give 43 cm.
In rhombus PQRSPQRSPQRS, angle QPRQPRQPR measures 27∘27^\circ27∘. What is the measure of the whole angle at QQQ?
Correct answer: A
Diagonal PR bisects angle P, so angle P is 54 degrees.
Consecutive parallelogram angles are supplementary.
The result is 126 degrees.
A parallelogram has perpendicular diagonals of lengths 181818 cm and 303030 cm. Which is the strongest guaranteed classification?
Correct answer: D
Perpendicular diagonals give a rhombus because the shape is a parallelogram.
The diagonals have different lengths, so it cannot be a rectangle. That also rules out a square.
In parallelogram ABCDABCDABCD, diagonal ACACAC bisects angle DABDABDAB. Which triangle fact completes a proof that it is a rhombus?
Correct answer: B
The bisector equates angles DAC and BAC. Parallel sides AD and BC equate angles DAC and ACB.
Thus the angles at A and C of triangle ABC are congruent, so AB and BC are congruent. Equal adjacent sides make this parallelogram a rhombus.
A convex quadrilateral has diagonals meeting at OOO. Its diagonals bisect each other and the whole diagonals are congruent. What is guaranteed?
The bisection condition establishes that the quadrilateral is a parallelogram.
The congruent whole diagonals then establish a rectangle. A square is possible, but perpendicularity or equal adjacent sides has not been established.
In the rectangle proof, triangles DABDABDAB and CBACBACBA have AD=BCAD=BCAD=BC, common side ABABAB, and right included angles. Which congruence test proves them congruent?
The two known side pairs meet at the right angles in the two triangles.
This is SAS. Using diagonal congruence as an SSS premise would assume the conclusion the direct proof is establishing.
Rhombus ABCDABCDABCD has diagonals meeting at OOO. The lengths of AOAOAO and BOBOBO are 777 cm and 242424 cm, respectively. Find ABABAB.
The diagonals are perpendicular, so triangle AOB is right with hypotenuse AB.
Convex quadrilateral ABCDABCDABCD is built from four perpendicular rays at OOO. The lengths AOAOAO, OCOCOC, BOBOBO, and ODODOD are 444, 222, 333, and 333 units, respectively. Which statement is true?
Add the parts of each diagonal.
But AO and OC are unequal, so the diagonals do not bisect each other. It is not a parallelogram and therefore is not a rectangle or a rhombus.
The diagonals of rhombus ABCDABCDABCD meet at OOO. Which comparison proves that angles AOBAOBAOB and AODAODAOD are congruent without assuming perpendicularity?
AB and AD are congruent by the rhombus definition. BO and DO are congruent by diagonal bisection, and AO is common.
SSS gives congruent triangles. The two named angles are adjacent, not vertical; right angles are the conclusion of the proof.
A parallelogram has one right angle and two congruent adjacent sides. What is the strongest guaranteed classification?
The right angle makes the parallelogram a rectangle.
The congruent adjacent sides, combined with congruent opposite sides, make it a rhombus. Being both makes it a square.
Rectangle JKLMJKLMJKLM has diagonals meeting at OOO. The half-diagonal lengths, in centimeters, are given below.
The whole diagonals are equal, and O bisects each. Their halves JO and KO are therefore equal.
Both halves measure 10 cm.
Which evidence guarantees a square for a convex quadrilateral?
Diagonal bisection first proves parallelogram.
Congruence then proves rectangle, and perpendicularity proves rhombus. Both establish square. Dropping the bisection evidence leaves counterexamples.
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