MHT CET 2025 23rd April Evening Shift | Properties of Triangles Question 5 | Mathematics

In a triangle with one of the angles $120^{\circ}$, the lengths of the sides form an A.P. If length of the greatest side is 7 m , then the area of the triangle is

$\frac{15 \sqrt{3}}{4} \mathrm{~m}^2$

$\frac{15 \sqrt{3}}{2} \mathrm{~m}^2$

$\frac{15}{2} \mathrm{~m}^2$

$\frac{15}{4} \mathrm{~m}^2$

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

In $\triangle \mathrm{ABC}$, with usual notations, if $\cos \frac{B}{2}=\sqrt{\frac{c+a}{2 a}}$, then $a^2=$

$\mathrm{b}^2-\mathrm{c}^2$

$\mathrm{b}+\mathrm{c}$

$\mathrm{b}^2+\mathrm{c}^2$

$\mathrm{b}-\mathrm{c}$

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

In a triangle ABC with usual notations, $\cot \frac{\mathrm{A}}{2}+\cot \frac{\mathrm{B}}{2}+\cot \frac{\mathrm{C}}{2}=$

$\frac{\mathrm{s}^2}{\Delta}$, where $\Delta$ is the area of the triangle ABC .

$\frac{s}{\Delta}$, where $\Delta$ is the area of the triangle ABC .

$\frac{\Delta}{\mathrm{s}}$, where $\Delta$ is the area of the triangle ABC .

$\Delta$, where $\Delta$ is the area of the triangle ABC .

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

With usual notations, in $\triangle \mathrm{ABC}$, the lengths of two sides are 10 cm and 9 cm respectively. If angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are in A.P. then perimeter of $\triangle \mathrm{ABC}$ is

$24+2 \sqrt{6} \mathrm{~cm}$

$24+\sqrt{6} \mathrm{~cm}$

$24-2 \sqrt{6} \mathrm{~cm}$

$22-\sqrt{6} \mathrm{~cm}$

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