Area of is: - Vidyadip

MCQ

Area of $\triangle\text{ABC}$ is:

A
45 squares units
B
55 squares units
C
65 squares units
D
none of these

✓

Answer

none of these

Solution:

Line PA: $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}-6}{1}$

Line PB: $\frac{\text{x}-1}{-1}=\frac{\text{y}-2}{2}=\frac{\text{z}-6}{1}$

Line PC: $\frac{\text{x}-1}{2}=\frac{\text{y}-2}{-1}=\frac{\text{z}-6}{-2}$

Then $\text{A}\Big(\frac{7}{2},-\frac{1}{2},\frac{17}{2}\Big)$

$\text{B}\Big(\frac{17}{2},-13,-\frac{3}{2}\Big)$

$\text{C}\Big(-14,\frac{19}{2},21\Big)$

Hence area of $\triangle\text{ABC}=\frac{225\sqrt{14}}{8},$ volume of tetrahedron

$\text{PABC}=\frac{125}{8}\text{cubic units}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Three Dimensional Geometry→Three Dimensional Geometry — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $\text{f(x)}=|\sin\text{x}|.$ then,

f(x) is everywhere differentiable.
f(x) is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}$
f(x) is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$
None of these.

The points $(3, 2, 0), (5, 3, 2)$ and $(-9, 6, -3)$, are the vertices of a triangle $ABC$. $AD$ is the internal bisector of $\angle BAC$ which meets $BC$ at $D$, then the co-ordinates of $D$, are

If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=$

$2\overrightarrow{\text{OG}}$
$4\overrightarrow{\text{OG}}$
$5\overrightarrow{\text{OG}}$
$3\overrightarrow{\text{OG}}$

If $\int\frac{\sin^8\text{x}-\cos^8\text{x}}{1-2\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\text{a}\sin2\text{x}+\text{C},$ then a =

$-\frac{1}{2}$
$\frac{1}{2}$
$-1$
$1$

If $a = 3i - 2j + k,\,\,b = 2i - 4j - 3k$ and $c = - i + 2j + 2k,$ then $a + b + c$ is

The solution of the differential equation $x\frac{{{d^2}y}}{{d{x^2}}} = 1$, given that $y = 1,\;\frac{{dy}}{{dx}} = 0$when $x = 1$, is

If $P\left(\frac{A}{B}\right)=0.3, P(A)=0.4$ and $P(B)=0.8$, then $P\left(\frac{B}{A}\right)$ is equal to:

If A is a square matrix, then A – A’ is a:

Diagonal matrix.
Skew-symmetric matrix.
Symmetric matrix.
None of these.

The projection of a line on a co-ordinate axes are $2, 3, 6$. Then the length of the line is

If ${a_{1,}}{a_2},{a_3}.....,{a_n}$ is an $A.P.$ with common difference d then $\tan \left[ {{{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + ...} \right.$ $\left. { + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_{n - 1}}{a_n}}}} \right)} \right] = $