NIMCET 2027 Properties Of Triangle Previous Year Questions (PYQs) – Page 1 of 3
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If $(4,-3)$ and $(-9,7)$ are two vertices of a triangle and $(1,4)$ is its centroid, find the area of the triangle.
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Solution
Let the third vertex be $(x,y)$.
Centroid formula:
$\displaystyle \left(\frac{4 + (-9) + x}{3}, \frac{-3 + 7 + y}{3}\right) = (1,4)$
From x–coordinate:
$\dfrac{4 - 9 + x}{3} = 1$
$x - 5 = 3$
$x = 8$
From y–coordinate:
$\dfrac{-3 + 7 + y}{3} = 4$
$y + 4 = 12$
$y = 8$
So third vertex is:
$(8,8)$
Vertices:
$A(4,-3), B(-9,7), C(8,8)$
Area:
$\displaystyle \text{Area}
= \frac{1}{2}\left| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right|$
Substitute:
$\displaystyle
= \frac{1}{2}\left|4(7-8) + (-9)(8+3) + 8(-3-7) \right|$
$\displaystyle
= \frac{1}{2}\left| 4(-1) - 9(11) + 8(-10) \right|$
$\displaystyle
= \frac{1}{2}\left| -4 - 99 - 80 \right|$
$\displaystyle
= \frac{1}{2} \times 183$
$\displaystyle
= \frac{183}{2}$
In $\triangle ABC$, $B = 45^\circ, C = 105^\circ, c=\sqrt{2}$.
Find side $a$ and $b$.
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Solution
$A = 180 - (45 + 105) = 30^\circ$
Using Law of Sines:
$ \displaystyle \frac{a}{\sin A} = \frac{c}{\sin C} $
$ \displaystyle a = \frac{\sin 30^\circ}{\sin 105^\circ}\cdot \sqrt{2} $
Compute:
$\sin 30^\circ = \frac12$
$\sin 105^\circ = \sin(60+45)=\frac{\sqrt{6}+\sqrt{2}}{4}$
$ \displaystyle a = \sqrt{2}\cdot \frac{1/2}{(\sqrt{6}+\sqrt{2})/4}
= \frac{\sqrt{2}}{(\sqrt{6}+\sqrt{2})/2}
= \frac{2\sqrt{2}}{\sqrt{6}+\sqrt{2}} $
After rationalizing:
$ \displaystyle a=\sqrt{3}-1 $
Similarly,
$ \displaystyle b=\sqrt{2}(\sqrt{3}-1) $
Let $\Delta ABC$ be a triangle whose area is $10\sqrt{3}$ units with side lengths $|AB|= 8$ units and $|AC|=5$
units. Find possible values of the angle A
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Solution
Triangle sides:
$x^{2}+x+1,\ 2x+1,\ x^{2}-1$.
Largest angle?
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Solution
Largest side ≈ $x^{2}+x+1$.
Use cosine rule, solve quadratic relationships → angle opposite largest side = $150^\circ$
In a $\Delta ABC$ , $(c+a+b)(a+b-c)=ab$ The measure of the angle C is.
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Solution
If $\Delta=a^2-(b-c)^2$, where $\Delta$ is the are of the triangle ABC, then $tanA=$
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Solution
The equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, –1). The length of the side of the triangle is.
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Solution
In a triangle ABC, if the tangent of half the difference of two angles is equal to one third of the tangent of half the sum of the angles, then the ratio of the sides opposite to the angles is
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Solution
Let $M$ be a point inside the triangle $ABC$. Then which one of the following is true?
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Solution
By triangle inequality in $\triangle ABM$ and $\triangle ACM$,
$AB>MB-AM,\ AC>MC-AM$
Adding,
$AB+AC>MB+MC$
Answer: $\boxed{AB+AC>MB+MC}$
In an acute-angled ΔABC the least value of secA+secB+secC is
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Solution
NIMCET 2027
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and More.
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