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In the above figure, x < y.
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29 Jan 2025, 13:07
29 Jan 2025, 13:07
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In the above figure, x < y.
Quantity A |
Quantity B |
Area of the triangle ABC |
Area of the triangle PQR |
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In the above figure, x < y.
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03 Mar 2025, 10:57
03 Mar 2025, 10:57
Expert Reply
We know that area of triangle, having sides $\(\mathrm{a}, \mathrm{b}\)$ and c opposite angle A , angle B \& angle C respectively (as shown in the figure above), is $\(\frac{1}{2} b c \sin \mathrm{~A}\)$ or $\(\frac{1}{2} \mathrm{ac} \sin \mathrm{B}\)$ or $\(\frac{1}{2} \mathrm{ab} \sin \mathrm{C}\)$
Let us consider the case when $\(x=45^{\circ} \& y=90^{\circ}\)$
So, using the above formula we get the area of triangle ABC as \($\frac{1}{2} \times \mathrm{AB} \times \mathrm{AC} \times \sin \mathrm{x}=\frac{1}{2} \times 1 \times 1 \times \sin 45^{\circ}=\frac{1}{2} \times 1 \times 1 \times \frac{1}{\sqrt{2}}=\frac{1}{2 \sqrt{2}}\)$
And the area of triangle $\(P Q R\)$ is $\(\frac{1}{2} \times P Q \times P R \times \sin y=\frac{1}{2} \times 1 \times 1 \times \sin 90^{\circ}=\frac{1}{2} \times 1 \times 1 \times 1=\frac{1}{2}\)$
Clearly area of triangle $\(\mathrm{ABC}<\)$ area of triangle PQR
But if we consider different values for $\(x\)$ and $\(y\)$, say $\(x=90^{\circ} \& y=135^{\circ}\)$, we get
Area of triangle $\(\mathrm{ABC}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{AC} \times \sin \mathrm{x}=\frac{1}{2} \times 1 \times 1 \times \sin 90^{\circ}=\frac{1}{2} \times 1 \times 1 \times 1=\frac{1}{2} \&\)$ the area
of triangle $\(P Q R\)$ as $\(\frac{1}{2} \times P Q \times P R \times \sin y=\frac{1}{2} \times 1 \times 1 \times \sin 135^{\circ}=\frac{1}{2} \times 1 \times 1 \times \frac{1}{\sqrt{2}}=\frac{1}{2 \sqrt{2}}\)$
Here area of triangle $\(\mathrm{ABC}>\)$ area of triangle PQR .
Hence a unique comparison cannot be formed between column $\(A\)$ and column $\(B\)$ quantities, so the answer is (D).
Let us consider the case when $\(x=45^{\circ} \& y=90^{\circ}\)$
So, using the above formula we get the area of triangle ABC as \($\frac{1}{2} \times \mathrm{AB} \times \mathrm{AC} \times \sin \mathrm{x}=\frac{1}{2} \times 1 \times 1 \times \sin 45^{\circ}=\frac{1}{2} \times 1 \times 1 \times \frac{1}{\sqrt{2}}=\frac{1}{2 \sqrt{2}}\)$
And the area of triangle $\(P Q R\)$ is $\(\frac{1}{2} \times P Q \times P R \times \sin y=\frac{1}{2} \times 1 \times 1 \times \sin 90^{\circ}=\frac{1}{2} \times 1 \times 1 \times 1=\frac{1}{2}\)$
Clearly area of triangle $\(\mathrm{ABC}<\)$ area of triangle PQR
But if we consider different values for $\(x\)$ and $\(y\)$, say $\(x=90^{\circ} \& y=135^{\circ}\)$, we get
Area of triangle $\(\mathrm{ABC}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{AC} \times \sin \mathrm{x}=\frac{1}{2} \times 1 \times 1 \times \sin 90^{\circ}=\frac{1}{2} \times 1 \times 1 \times 1=\frac{1}{2} \&\)$ the area
of triangle $\(P Q R\)$ as $\(\frac{1}{2} \times P Q \times P R \times \sin y=\frac{1}{2} \times 1 \times 1 \times \sin 135^{\circ}=\frac{1}{2} \times 1 \times 1 \times \frac{1}{\sqrt{2}}=\frac{1}{2 \sqrt{2}}\)$
Here area of triangle $\(\mathrm{ABC}>\)$ area of triangle PQR .
Hence a unique comparison cannot be formed between column $\(A\)$ and column $\(B\)$ quantities, so the answer is (D).
Re: In the above figure, x < y.
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03 Mar 2025, 12:43
03 Mar 2025, 12:43
1
We only know that x is less than y, we don't know by what degree. Now since AB = AC = PQ = PR, BC will be smaller than QR but the height of ABC will be greater. With this scenario, we need more information to determine which area is greater. Answer is D.
