TRIGONOMETRY TRIANGLES - 2 (Properties of Triangles)
Examples
1. In a $\triangle \mathrm{ABC}$ the inradius and three exradii are $\mathrm{r}, \mathrm{r} _{1}, \mathrm{r} _{2}$ and $\mathrm{r} _{3}$ respectively. In usual notations the value of $\mathrm{r} \cdot \mathrm{r} _{1} \cdot \mathrm{r} _{2} \cdot \mathrm{r} _{3}$ is equal to
(a). $2 \Delta$
(b). $\Delta^{2}$
(c). $\frac{a b c}{4 R}$
(d). None of these
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Solution:
$ \begin{aligned} \mathrm{r} \cdot \mathrm{r} _{1} \cdot \mathrm{r} _{2} \mathrm{r} _{3} & =\frac{\Delta}{\mathrm{s}} \cdot \frac{\Delta}{\mathrm{s}-\mathrm{a}} \cdot \frac{\Delta}{\mathrm{s}-\mathrm{b}} \frac{\Delta}{\mathrm{s}-\mathrm{c}} \\ & =\frac{\Delta^{4}}{\Delta^{2}} \\ & =\Delta^{2} \end{aligned} $
Correct option is ‘b’
2. The distance between the circumcentre and the orthocenter of a triangle $\mathrm{ABC}$ is
(a). $\mathrm{R} \sqrt{1-8 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}}$
(b). $\mathrm{R} \sqrt{1+8 \cos \mathrm{A} \cos B \cos \mathrm{C}}$
(c). $\mathrm{R} \sqrt{1-4 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}}$
(d). None of these
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Solution:
Let $\mathrm{O} \& \mathrm{P}$ be circumcenter and orthocentre respectively of $\triangle \mathrm{ABC}$
$\mathrm{OF} \perp \mathrm{AB}$.
We have $\angle \mathrm{OAF}=90^{\circ}-\mathrm{C}(\because \angle \mathrm{AOF}=\angle \mathrm{C})$
$ =\angle \mathrm{PAE}(\operatorname{In} \triangle \mathrm{ADC}) $
$ \begin{aligned} & \angle \mathrm{OAP}=\angle \mathrm{A}-\angle \mathrm{OAF}-\angle \mathrm{PAE} \\ & =\angle \mathrm{A}-\left(90^{\circ} \angle \mathrm{C}\right)-\left(90^{\circ} \angle \mathrm{C}\right) \\ & =\angle \mathrm{A}+2 \angle \mathrm{C}-180^{\circ} \\ & =\angle \mathrm{A}+2 \angle \mathrm{C}-(\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}) \\ & =\angle \mathrm{C}-\angle \mathrm{B} \end{aligned} $
$\mathrm{OA}=\mathrm{R}$ (circum radius) and $\mathrm{PA}=2 \mathrm{R} \cos \mathrm{A}$
In $\triangle$ OAP
$ \mathrm{OP}^{2}=\mathrm{OA}^{2}+\mathrm{PA}^{2}-2(\mathrm{OA})(\mathrm{PA}) \operatorname{COS}(\mathrm{C}-\mathrm{B}) $
$ =R^{2}+4 R^{2} \cos ^{2} A-4 R^{2} \cos A \cos (C-B) $
$ \begin{aligned} & =R^{2}+4 R^{2} \cos A(\cos A-\cos (C-B) \\ & =R^{2}+4 R^{2} \cos A(-\cos (B+C)-\cos (C-B)) \\ & =R^{2}-4 R^{2} \cos A(\cos (B+C)+\cos (C-B)) \\ & =R^{2}-4 R^{2} \cos A(2 \cos B \cos C) \\ & =R^{2}-8 R^{2} \cos A \cos B \cos C \end{aligned} $
$\mathrm{OP}^{2}=\mathrm{R}^{2}(1-8 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C})$
$\mathrm{OP}=\mathrm{R} \sqrt{1-8 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}}$
Hence ‘a’ is the correct option
3. The radii $r _{1}, r _{2}, r _{3}$, of escribed circles of the triangle $A B C$ are in HP. If its area is $24 \mathrm{sq} . \mathrm{cm}$ and its perimeter is $24 \mathrm{~cm}$, then the lengths of its sides are
(a). $4, 6, 8$
(b). $3, 9, 11$
(c). $6, 8,1 0$
(d). None of these
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Solution:
Given that $r _{1}, r _{2} r _{3}$ are in HP
i.e. $\frac{\mathrm{s}-\mathrm{a}}{\Delta}, \frac{\mathrm{s}-\mathrm{b}}{\Delta}, \frac{\mathrm{s}-\mathrm{c}}{\Delta}$ are in A.P
or $\mathrm{s}-\mathrm{a}, \mathrm{s}-\mathrm{b}, \mathrm{s}-\mathrm{c}$ are in A.P
or a, b, c are in AP
$\mathrm{a}+\mathrm{c}=2 \mathrm{~b}$
$\mathrm{a}+\mathrm{b}+\mathrm{c}=24$
$3 \mathrm{~b}=24$
$\mathrm{b}=8$
$\mathrm{s}=\frac{24}{2}=12$
$\mathrm{a}+\mathrm{c}=16$
$\mathrm{c}=16-\mathrm{a}$
$\Delta^{2}=\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})$
$24 \times 24=12(12-\mathrm{a})(12-8)(12-(16-\mathrm{a}))$
$24 \times 24=12 \times 4(12-a)(a-4)$
$12=(12-a)(a-4)$
$a^{2}-16 a+60=0$
$a^{2}-10 a-6 a+60=0$
$(a-10)(a-6)=0$
$\mathrm{a}=10$ or $\mathrm{a}=6$
$\mathrm{a}=10, \mathrm{c}=6, \mathrm{~b}=8$ or $\mathrm{a}=6, \mathrm{~b}=8, \mathrm{c}=10$
Correct option is ‘c’
4. In $\triangle \mathrm{ABC}$ the value of $\frac{\mathrm{r} _{1}+\mathrm{r} _{2}}{1+\cos \mathrm{c}}$ is always equal to
(a). $2 \mathrm{r}$
(b). $2 \mathrm{R}$
(c). $\frac{2 r^{2}}{R}$
(d). $\frac{2 R^{2}}{r}$
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Solution:
We know
$ \begin{aligned} & r _{1}+r _{2}=4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}+4 R \sin \frac{B}{2} \cos \frac{C}{2} \cos \frac{A}{2} \\ &=4 R \cos \frac{C}{2}\left(\sin \frac{A}{2} \cos \frac{B}{2}+\cos \frac{A}{2} \sin \frac{B}{2}\right) \\ &=4 R \cos \frac{C}{2}\left(\sin \left(\frac{A+B}{2}\right)\right) \\ &=4 R \cos \frac{C}{2}\left(\cos \frac{C}{2}\right) \\ &=4 R \cos ^{2} \frac{C}{2} \\ &=2 R(1+\cos C) \\ & \quad\left(1+\cos 2 \theta=2 \cos ^{2} \theta\right) \end{aligned} $
$\therefore \frac{\mathrm{r} _{1}+\mathrm{r} _{2}}{1+\cos \mathrm{C}}=2 \mathrm{R}$
5. Let $\mathrm{ABCD}$ be a quadrilateral with area 18 with side $\mathrm{AB}$ parallel to the side $\mathrm{CD}$ and $\mathrm{AB}=2 \mathrm{CD}$. Let $A D$ be perpendicular to $A B$ and $C D$.If a circle is drawn inside the quadrilaterals $A B C D$ touching all the sides, then its radius is
(a). $3$
(b). $2$
(c). $\frac{3}{2}$
(d). $1$
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Solution:
Given that $\mathrm{AB}=2 \mathrm{CD}$
Let $\mathrm{CD}=\mathrm{a} \quad \therefore \mathrm{AB}=2 \mathrm{a}$
$\therefore \mathrm{B}(2 \mathrm{a}, 0)$
and $\mathrm{c}(\mathrm{a}, 2 \mathrm{r})$
Let the centre of in circle be $( \mathrm{r}, \mathrm{r})$
Where $r$ is the radius of the circle
Since $A B \| ^\ell$ to $C D$
$\therefore \mathrm{ABCD}$ is a trapezium
$\operatorname{ar}(\mathrm{ABCD})=\frac{1}{2} \mathrm{~h}$ (sum of parallel sides)
$=\frac{1}{2} \times 2 \mathrm{r}(\mathrm{a}+2 \mathrm{a})$
$18=3$ ar
$ar =6$
Now $\mathrm{BC}$ is a tangent to the circle
$ (x-r)^{2}+(y-r)^{2}=r^{2} $
Equation of tangent is $y=-\frac{2 r}{2}(x-2 a)$
$2\mathrm{rx}+\mathrm{ay}-4 \mathrm{ar}=0$
$\therefore \mathrm{r}=\left|\frac{2 \mathrm{r}^{2}+\mathrm{ar}-4 \mathrm{ar}}{\sqrt{4 \mathrm{r}^{2}+\mathrm{a}^{2}}}\right|$
$r=\left|\frac{2 r^{2}-3 a r}{\sqrt{4 r^{2}+a^{2}}}\right|$
squaring
$4 r^{4}+a^{2} r^{2}=4 r^{4}+9 a^{2} r^{2}-12 a r^{3}$
$12 \mathrm{ar}^{3}=8 \mathrm{a}^{2} \mathrm{r}^{2}$
$3 \mathrm{r}=2 \mathrm{a} \quad[\therefore \mathrm{ar}=6]$
$3 \mathrm{r}^{2}=2 \mathrm{ar}$
$3 \mathrm{r}^{2}=12$
$\mathrm{r}^{2}=4$
$\mathrm{r}=2$
6. In a triangle $\mathrm{ABC}, \mathrm{a}: \mathrm{b}: \mathrm{c}=4: 5: 6$. The ratio of the radius of the circumcircle to that of the incircle is
(a). $\frac{15}{4}$
(b). $\frac{11}{5}$
(c). $\frac{16}{7}$
(d). $\frac{16}{3}$
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Solution:
$ \begin{aligned} & \mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}, \quad \mathrm{r}=\frac{\Delta}{\mathrm{s}} \\ & \frac{\mathrm{R}}{\mathrm{r}}=\frac{\mathrm{abc}}{4 \Delta} \times \frac{\mathrm{s}}{\Delta} \\ & =\frac{\mathrm{abc} \times \mathrm{s}}{4 \Delta^{2}} \\ & =\frac{\mathrm{abc} \times \mathrm{s}}{4(\mathrm{~s}(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c}))} \\ & =\frac{2 \mathrm{abc}}{(2 \mathrm{~s}-2 \mathrm{a})(2 \mathrm{~s}-2 \mathrm{~b})(2 \mathrm{~s}-2 \mathrm{c})} \\ & =\frac{2 \mathrm{abc}}{(\mathrm{a}+\mathrm{b}-\mathrm{c})(\mathrm{c}+\mathrm{a}-\mathrm{b})(\mathrm{b}+\mathrm{c}-\mathrm{a})} \\ & =\frac{2.4 .5 .6}{3.5 .7} \\ & =\frac{16}{7} \end{aligned} $
7. The value of $\frac{1}{\mathrm{r} _{1}{ }^{2}}+\frac{1}{\mathrm{r} _{2}{ }^{2}}+\frac{1}{\mathrm{r} _{3}{ }^{2}}+\frac{1}{\mathrm{r}^{2}}$ is
(a). $0$
(b). $\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}{\Delta^{2}}$
(c). $\frac{\Delta^{2}}{a^{2}+b^{2}+c^{2}}$
(d). $\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}{\Delta}$
Show Answer
Solution:
$\therefore \frac{1}{\mathrm{r} _{1}{ }^{2}}=\frac{(\mathrm{s}-\mathrm{a})^{2}}{\Delta^{2}} ; \frac{1}{\mathrm{r} _{2}{ }^{2}}=\frac{(\mathrm{s}-\mathrm{b})^{2}}{\Delta^{2}} ; \frac{1}{\mathrm{r} _{3}{ }^{2}}=\frac{(\mathrm{s}-\mathrm{c})^{2}}{\Delta^{2}} ; \frac{1}{\mathrm{r}^{2}}=\frac{\mathrm{s}^{2}}{\Delta^{2}}$
$\therefore \frac{1}{\mathrm{r} _{1}{ }^{2}}+\frac{1}{\mathrm{r} _{2}{ }^{2}}+\frac{1}{\mathrm{r} _{3}{ }^{2}}+\frac{1}{\mathrm{r}^{2}}=\frac{\mathrm{s}^{2}+(\mathrm{s}-\mathrm{a})^{2}+(\mathrm{s}-\mathrm{b})^{2}+(\mathrm{s}-\mathrm{c})^{2}}{\Delta^{2}}$
$=\frac{4 \mathrm{~s}^{2}+\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}-2 \mathrm{~s}(\mathrm{a}+\mathrm{b}+\mathrm{c})}{\Delta^{2}}$
$=\frac{4 \mathrm{~s}^{2}+\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}-4 \mathrm{~s}^{2}}{\Delta^{2}} \quad(\mathrm{a}+\mathrm{b}+\mathrm{c}=2 \mathrm{~s})$
$=\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}{\Delta^{2}}$
Practice questions
1. In $\triangle \mathrm{ABC}, \mathrm{a} \geq \mathrm{b} \geq \mathrm{c}$, if $\frac{\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}}{\sin ^{3} \mathrm{~A}+\sin ^{2} \mathrm{~B}+\sin ^{3} \mathrm{c}}=8$, then the maximum value of $\mathrm{a}$ is
(a). $\frac{1}{2}$
(b). $2$
(c). $8$
(d). $64$
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Answer: (b)2. Sides of triangle $\mathrm{ABC}$ are in $\mathrm{AP}$. If $\mathrm{a}<\min \{\mathrm{b}, \mathrm{c}\}$, then $\cos \mathrm{A}$ may be equal to
(a). $\frac{3 c-4 b}{2 b}$
(b). $\frac{3 \mathrm{c}-4 \mathrm{~b}}{2 \mathrm{c}}$
(c). $\frac{4 \mathrm{c}-3 \mathrm{~b}}{2 \mathrm{~b}}$
(d). $\frac{4 \mathrm{c}-3 \mathrm{~b}}{2 \mathrm{c}}$
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Answer: (d)3. In a $\triangle A B C$, angles $A, B, C$ are in $A P$. Then $x \rightarrow c \frac{\sqrt{3-4 \sin A \sin C}}{|A-C|}$ is
(a). 1
(b). 2
(c). 3
(d). 4
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Answer: (a)4. In a triangle $\mathrm{ABC}, 2 \mathrm{a}^{2}+4 \mathrm{~b}^{2}+\mathrm{c}^{2}=4 \mathrm{ab}+2 \mathrm{ac}$, then the numerical value of $\cos \mathrm{B}$ is equal to
(a). $0$
(b). $\frac{3}{8}$
(c). $\frac{5}{8}$
(d). $\frac{7}{8}$
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Answer: (d)5. If $a, b, c$ be the sides of a $\triangle A B C$ and if roots of the equation $a(b-c) x^{2}+b(c-a) x+c(a-b)=0$ are equal, then $\sin ^{2} \frac{A}{2}, \sin ^{2} \frac{B}{2}, \sin ^{2} \frac{C}{2}$ are in
(a). $\mathrm{AP}$
(b). $\mathrm{GP}$
(c). $\mathrm{HP}$
(d). $\mathrm{AGP}$
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Answer: (c)6. In a $\triangle \mathrm{ABC}$ sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{AP}$ and $\frac{2}{1 ! 9 !}+\frac{2}{3 ! 7 !}+\frac{1}{5 ! 5 !}=\frac{8^{\mathrm{a}}}{(2 \mathrm{~b}) !}$ then the maximum value of $\tan \mathrm{A} \tan \mathrm{B}$ is equal to
(a). $\frac{1}{2}$
(b). $\frac{1}{3}$
(c). $\frac{1}{4}$
(d). $\frac{1}{5}$
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Answer: (b)7. If $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are the angles of quadrilateral, then $\frac{\sum \tan \mathrm{A}}{\sum \cot \mathrm{A}}$ is equal to
(a). $\pi \tan \mathrm{A}$
(b). $\pi \cot \mathrm{A}$
(c). $ \sum \tan ^{2} \mathrm{~A}$
(d). $ \sum \cot ^{2} \mathrm{~A}$
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Answer: (a)8. With usual notations, if in a $\triangle \mathrm{ABC}, \frac{\mathrm{b}+\mathrm{c}}{11}=\frac{\mathrm{c}+\mathrm{a}}{12}=\frac{\mathrm{a}+\mathrm{b}}{13}$, then $\cos \mathrm{A}: \cos \mathrm{B}$ : $\cos \mathrm{C}$ is equal to
(a). $7: 19: 25$
(b). $19: 7: 25$
(c). $12: 14: 20$
(d). $19: 25: 20$
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Answer: (a)9. In and $\triangle \mathrm{ABC}, \Pi\left(\frac{\sin ^{2} \mathrm{~A}+\sin \mathrm{A}+1}{\sin \mathrm{A}}\right)$ is always greater than
(a). 9
(b). 3
(c). 27
(d). None of these
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Answer: (c)10. The radius of the circle passing through the centre of incircle of $\triangle \mathrm{ABC}$, and through the end points of $\mathrm{BC}$ is given
(a). $\left(\frac{\mathrm{a}}{2}\right) \cos \mathrm{A}$
(b). $\left(\frac{a}{2}\right) \sec \left(\frac{A}{2}\right)$
(c). $\left(\frac{a}{2}\right) \sin \mathrm{A}$
(d). $\operatorname{asec}\left(\frac{\mathrm{A}}{2}\right)$
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Answer: (b)11. If in a $\triangle \mathrm{ABC}, \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{AP}$ and $\mathrm{p} _{1}, \mathrm{p} _{2}, \mathrm{p} _{3}$ are the altitude from the vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively then
(a). $\mathrm{p} _{1}, \mathrm{p} _{2}, \mathrm{p} _{3}$ are in $\mathrm{AP}$
(b). $\mathrm{p} _{1}, \mathrm{p} _{2}, \mathrm{p} _{3}$ are in $\mathrm{HP}$
(c). $\mathrm{p} _{1}+\mathrm{p} _{2}+\mathrm{p} _{3} \leq \frac{3 \mathrm{R}}{\Delta}$
(d). $\frac{1}{\mathrm{p} _{1}}+\frac{1}{\mathrm{p} _{2}}+\frac{1}{\mathrm{p} _{3}} \leq \frac{3 \mathrm{R}}{\Delta}$
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Answer: (b, d)12. If $\tan A, \tan B$ are the roots of the quadratic $a b x^{2}-c^{2} x+a b=0$, where $a, b, c$ are the sides of $a$ triangle, then
(a). $\tan \mathrm{A}=\frac{\mathrm{a}}{\mathrm{b}}$
(b). $\tan \mathrm{B}=\frac{\mathrm{b}}{\mathrm{a}}$
(c). $ \cos \mathrm{C}=0$
(d). $\tan \mathrm{A}+\tan \mathrm{B}=\frac{\mathrm{c}^{2}}{\mathrm{ab}}$
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Answer: (a, b, c, d)13. If $\sin \beta$ is the GM between $\sin \alpha$ and $\cos \alpha$, then $\cos 2 \beta$ is equal to
(a). $2 \sin ^{2}\left(\frac{\pi}{4}-\alpha\right)$
(b). $2 \cos ^{2}\left(\frac{\pi}{4}-\alpha\right)$
(c). $2 \cos ^{2}\left(\frac{\pi}{4}+\alpha\right)$
(d). $2 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right)$
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Answer: (a, c)14. Passage - I
If $\mathrm{p} _{1}, \mathrm{p} _{2}, \mathrm{p} _{3}$ are altitudes of a triangle $\mathrm{ABC}$ from the vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively and $\Delta$ is the area of the triangle and $\mathrm{s}$ is the semipermanent of the triangle. On the basis of above information, answer the following questions :
(i). If $\frac{1}{p _{1}}+\frac{1}{p _{2}}+\frac{1}{p _{3}}=\frac{1}{2}$ then the least value of $p _{1}, p _{2}, p _{3}$ is
(a). 8
(b). 27
(c). 125
(c). 216
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Answer: (b)(ii). The value of $\frac{\cos A}{p _{1}}+\frac{\cos B}{p _{2}}+\frac{\cos C}{p _{3}}$ is
(a). $\frac{1}{r}$
(b). $\frac{1}{R}$
(c). $\frac{a^{2}+b^{2}+C^{2}}{2 R}$
(d). $\frac{1}{\Delta}$
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Answer: (b)(iii). The minimum value of $\frac{b^{2} p _{1}}{c}+\frac{c^{2} p _{2}}{a}+\frac{a^{2} p _{3}}{b}$ is
(a). $\Delta$
(b). $2 \Delta$
(c). $3 \Delta$
(d). $6 \Delta$
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Answer: (d)(iv). The value of $\mathrm{p} _{1}{ }^{-2}+\mathrm{p} _{2}{ }^{-2}+\mathrm{p} _{3}{ }^{-2}$ is
(a). $\frac{\left(\sum \mathrm{a}\right)^{2}}{4 \Delta^{2}}$
(b). $\frac{(\pi \mathrm{a})^{3}}{8 \Delta^{3}}$
(c). $\frac{\sum \mathrm{a}^{2}}{4 \Delta^{2}}$
(d). $\frac{\pi \mathrm{a}^{2}}{8 \Delta^{2}}$
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Answer: (c)(v). In the triangle $A B C$, the altitudes are in $A P$, then
(a). a, b, c are in AP
(b). a, b, c are in HP
(c). a, b, c are in GP
(d). angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are in $\mathrm{AP}$
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Answer: (b)15. Passage II
In a triangle if the sum of two sides is $x$ and this product is $y(x \geq 2 \sqrt{y})$ such that $(\mathrm{x}+\mathrm{z})(\mathrm{x}-\mathrm{z})=\mathrm{y}$ where $\mathrm{z}$ is the third side of the triangle.
On the passes of above information, answer the following questions.
(i). Greatest angle of the triangle is
(a). 105
(b). 120
(c). 135
(d). 150
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Answer: (b)(ii). Cerium radius of the triangle is
(a). $\mathrm{x}$
(b). $y$
(c). $\mathrm{z}$
(d). None of these
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Answer: (d)(iii). In radius of the triangle is
(a). $\frac{y}{2(z+x)}$
(b). $\frac{z}{2(x+y)}$
(c). $\frac{y \sqrt{3}}{z+x}$
(d). $\frac{z \sqrt{3}}{x+y}$
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Answer: (c)(iv). Area of the triangle is
(a). $\frac{\mathrm{y} \sqrt{3}}{4}$
(b). $\frac{x \sqrt{3}}{4}$
(c). $\frac{\mathrm{z} \sqrt{3}}{4}$
(d). None of these
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Answer: (a)(v). The sides of the triangle are
(a). $\frac{x \pm \sqrt{x^{2}-4 y}}{2}, z$
(b). $\frac{\mathrm{y} \pm \sqrt{\mathrm{y}^{2}-4 \mathrm{z}}}{2}, \mathrm{z}$
(c). $\frac{z \pm \sqrt{z^{2}-4 x}}{2}, z$
(d). None of these
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Answer: (a)Assertion and Reason
16. Assertion (A) : In any $\triangle A B C$, the minimum value of $\frac{r _{1}+r _{2}+r _{3}}{r}$ is
Reason (R) : $\mathrm{AM} \geq \mathrm{GM}$
(a). $\mathrm{A}$
(b). $\mathrm{B}$
(c). $\mathrm{C}$
(d). $\mathrm{D}$
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Answer: (b)17. Assertion (A) : If A, B, C D are angle of a cyclic quadrilateral then $\sum \sin A=0$
Reason (R) : If A, B, C, D are angles of cyclic quadrilateral then $\sum \cos A=0$
(a). $\mathrm{A}$
(b). $\mathrm{B}$
(c). $\mathrm{C}$
(d). $\mathrm{D}$
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Answer: (d)18. Assertion (A) : In any $\triangle \mathrm{ABC}$, the square of the length of the bisector $\mathrm{AD}$ is $\mathrm{bc}\left(\frac{1-\mathrm{a}^{2}}{(\mathrm{~b}+\mathrm{c})^{2}}\right)$
Reason (R) : In any $\triangle \mathrm{ABC}$, length of bisector $\mathrm{AD}$ is $\frac{2 \mathrm{bc}}{\mathrm{b}+\mathrm{c}} \cos \frac{\mathrm{A}}{2}$
(a). $\mathrm{A}$
(b). $\mathrm{B}$
(c). $\mathrm{C}$
(d). $\mathrm{D}$
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Answer: (a)Integer Type Questions
19. If the radius of the circumcircle of a triangle is 12 and that of the incircle is 4 , then the square of the sum of radii of the escribed cirle must be
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Answer: 270420. In a $\triangle \mathrm{ABC}$, the maximum value of $1000\left(\frac{\sum \mathrm{a} \cos ^{2} \frac{\mathrm{A}}{2}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\right)$ must be
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Answer: 075021. Matrix Match Type
| Column I | Column II |
|---|---|
| (a). In a $\triangle A B C$, if $2 a^{2}+b^{2}+c^{2}=2 a c+2 a b$, then | (p). $\triangle \mathrm{ABC}$ is equilateral $\Delta$ |
| (b). In a $\triangle A B C$, if $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=\sqrt{2} \mathrm{~b}(\mathrm{c}+\mathrm{a})$, then | (q). $\triangle \mathrm{ABC}$ is right angled $\Delta$ |
| (c). In a $\triangle \mathrm{ABC}$,if $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=\mathrm{bc}+\mathrm{ca} \sqrt{3}$, then | (r). $\triangle \mathrm{ABC}$ is scalene $\Delta$ |
| (s). $\triangle \mathrm{ABC}$ is scalene right angled $\Delta$ | |
| (t). Angles $B, C, A$ are in $AP$ |
