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Two sides of triangle DEF are equal to 3
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11 May 2026, 15:04
11 May 2026, 15:04
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Expert Reply
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Question Stats:
22% (02:01) correct
77% (02:19) wrong based on 18 sessions
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Two sides of triangle DEF are equal to 3 . Which of the following, taken alone, would be sufficient in finding the area of triangle DEF?
Indicate all such statements
□ The ratio of \(D E\) to \(E F=1: \sqrt{ } 2\)
□ The sum of angles DEF and EFD is 135 degrees
□ The sum of angles DEF and FDE is 90 degrees
Indicate all such statements
□ The ratio of \(D E\) to \(E F=1: \sqrt{ } 2\)
□ The sum of angles DEF and EFD is 135 degrees
□ The sum of angles DEF and FDE is 90 degrees
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Re: Two sides of triangle DEF are equal to 3
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07 Jul 2026, 11:19
07 Jul 2026, 11:19
2
A is not correct because the third side could be 1.5root 2 or 3 root 2 based on the ratio given.
B could give us angles
45, 45, 90 or 45, 67.5, 67.5
For option C, we definitely know that one of the angles is 90 and the only 2 equal sides has to be the legs so we have a definite area in this case
Hope this clarifies the issue KeshabThapa?
B could give us angles
45, 45, 90 or 45, 67.5, 67.5
For option C, we definitely know that one of the angles is 90 and the only 2 equal sides has to be the legs so we have a definite area in this case
Hope this clarifies the issue KeshabThapa?
Re: Two sides of triangle DEF are equal to 3
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07 Jul 2026, 12:12
07 Jul 2026, 12:12
1
Expert Reply
Let \(D E=x \) and \(E F=x \sqrt{2}\) . Because the ratio between these two sides is not 1: 1 , they cannot be the two equal sides. Therefore, the third side, D F , must be equal to one of them. This creates two distinct possibilities:
- Case A: The equal sides are D E and D F (both equal 3). Then \(E F=3 \sqrt{2}\) . Since \(3^2+ 3^2=(3 \sqrt{2})^2\) , this forms a right isosceles triangle with an area of:
\(\text { Area }=\frac{1}{2} \times 3 \times 3=4.5\)
- Case B: The equal sides are E F and D F (both equal 3). Then \(D E=\frac{3}{\sqrt{2}}\) . This forms an acute isosceles triangle with a completely different height and area (Area \(\approx 2.98\) ).
Because Statement 1 yields two different possible areas, A is wrong.
Since the sum of all interior angles in a triangle is \(180^{\circ} \), the third angle must be:
\(\angle F D E=180^{\circ}-135^{\circ}=45^{\circ}\)
Knowing one angle is \(45^{\circ}\) and two sides are 3 still leaves room for multiple configurations:
- Case A: The \(45^{\circ}\) angle is enclosed by the two sides of length 3.
\(\text { Area }=\frac{1}{2} \times 3 \times 3 \times \sin \left(45^{\circ}\right) \approx 3.18\)
- Case B: The triangle is a right isosceles triangle ( $45^{\circ}-45^{\circ}-90^{\circ}$ ), where the sides of length 3 are the legs enclosing the $90^{\circ}$ angle.
\(\text { Area }=\frac{1}{2} \times 3 \times 3=4.5\)
Because Statement 2 yields different possible areas, B is wrong.
If the sum of these two angles is \(90^{\circ}\) , the remaining angle must be a right angle:
\(\angle E F D=180^{\circ}-90^{\circ}=90^{\circ}\)
In any right triangle, the hypotenuse is strictly the longest side. Therefore, the two equal sides of length 3 must be the two perpendicular legs ( D F and E F ).
This locks in a single, unique triangle configuration where the area can be definitively calculated:
\(\text { Area }=\frac{1}{2} \times \text { base × height }=\frac{1}{2} \times 3 \times 3=4.5\)
Since this statement guarantees exactly one clear area value, C is correct.
- Case A: The equal sides are D E and D F (both equal 3). Then \(E F=3 \sqrt{2}\) . Since \(3^2+ 3^2=(3 \sqrt{2})^2\) , this forms a right isosceles triangle with an area of:
\(\text { Area }=\frac{1}{2} \times 3 \times 3=4.5\)
- Case B: The equal sides are E F and D F (both equal 3). Then \(D E=\frac{3}{\sqrt{2}}\) . This forms an acute isosceles triangle with a completely different height and area (Area \(\approx 2.98\) ).
Because Statement 1 yields two different possible areas, A is wrong.
Since the sum of all interior angles in a triangle is \(180^{\circ} \), the third angle must be:
\(\angle F D E=180^{\circ}-135^{\circ}=45^{\circ}\)
Knowing one angle is \(45^{\circ}\) and two sides are 3 still leaves room for multiple configurations:
- Case A: The \(45^{\circ}\) angle is enclosed by the two sides of length 3.
\(\text { Area }=\frac{1}{2} \times 3 \times 3 \times \sin \left(45^{\circ}\right) \approx 3.18\)
- Case B: The triangle is a right isosceles triangle ( $45^{\circ}-45^{\circ}-90^{\circ}$ ), where the sides of length 3 are the legs enclosing the $90^{\circ}$ angle.
\(\text { Area }=\frac{1}{2} \times 3 \times 3=4.5\)
Because Statement 2 yields different possible areas, B is wrong.
If the sum of these two angles is \(90^{\circ}\) , the remaining angle must be a right angle:
\(\angle E F D=180^{\circ}-90^{\circ}=90^{\circ}\)
In any right triangle, the hypotenuse is strictly the longest side. Therefore, the two equal sides of length 3 must be the two perpendicular legs ( D F and E F ).
This locks in a single, unique triangle configuration where the area can be definitively calculated:
\(\text { Area }=\frac{1}{2} \times \text { base × height }=\frac{1}{2} \times 3 \times 3=4.5\)
Since this statement guarantees exactly one clear area value, C is correct.
