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# NMTC Geometry Problems and Solutions

### NMTC 2019 Stage 1 Inter Question 5

The area of the curve enclosed by $|x-2 \sqrt{2}|+|y-\sqrt{5}|=2$ is :

(A) 16
(B) 12
(C) 8
(D) 4

Area of $|x-2 \sqrt{2}|+|y-\sqrt{5}|=2$ is equivalent to area of $|x|+|y|=2$

So it is a rhombus

Hence area will be, $4\times (\frac{1}{2} \times 2 \times 2) = 8$

### NMTC 2019 Inter Stage 1 Question 11

In a rectangle $A B C D$, point $E$ lies on $B C$ such that $\frac{B E}{E C}=2$ and point $F$ lies on $C D$ such that $\frac{C F}{F D}=$ 2. Lines $A E$ and $A C$ intersect $B F$ at $X$ and $Y$ respectively. If $F Y: Y X: X B=a: b: c$, are relatively prime positive integers, then the minimum value of $a+b+c$ is :

$\frac{c}{a+b}=\frac{6}{7} \quad \Rightarrow 76=6 a+6 b \ldots .$ (i)
$\frac{b+c}{a}=\frac{6}{4}=\frac{3}{2} \Rightarrow 2 b+2 c=3 a \ldots$.(ii)

$\Rightarrow \frac{a}{26}=\frac{b}{9}=\frac{c}{30} \quad \Rightarrow a=26, b=9, c=30 \Rightarrow a+b+c=65$

### NMTC 2019 Inter Stage 1 Question 13

A regular polygon has 100 sides each of length. A another regular polygon has 200 sided each of length 2. When the area of the larger polygon is divided by the area of the smaller polygon, the quotient is closest to the integer
(A) 2
(B) 4
(C) 8
(D) 16

\begin{aligned} \frac{\mathrm{AC}}{\mathrm{OC}} &=\tan \frac{\pi}{100} \ \Rightarrow \quad \mathrm{OC} &=\frac{1}{2 \tan \frac{\pi}{100}} \end{aligned}
Area of polygon 1
$$=100 \times \frac{1}{2} \times 1 \times \frac{1}{2 \tan \frac{\pi}{100}}$$

\begin{aligned} &\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{O}^{\prime} \mathrm{C}^{\prime}}=\tan \frac{\pi}{200} \ &\mathrm{O}^{\prime} \mathrm{C}^{\prime}=\frac{1}{\tan \frac{\pi}{200}} \end{aligned}
Area of polygon 2
$$=200 \times \frac{1}{2} \times 2 \times \frac{1}{\tan \frac{\pi}{200}}$$

$\frac{\text { Area of polygon2 }}{\text { Area of polygon1 }}=\frac{200 \cos \frac{\pi}{200}}{\sin \frac{\pi}{200}} \times \frac{\sin \frac{\pi}{100}}{25 \cos \frac{\pi}{100}}=\frac{16 \cos \frac{\pi}{200} \times \sin \frac{\pi}{200} \times \cos \frac{\pi}{200}}{\sin \frac{\pi}{200} \times \cos \frac{\pi}{100}}$
$=\frac{8\left(1+\cos \frac{\pi}{100}\right)}{\cos \frac{\pi}{100}}=8+8 \sec \frac{\pi}{100} \simeq 8+8=16$

### NMTC 2019 Inter Stage 1 Question 18

In a $38 \times 32$ rectangle $A B C D$, points $P, Q, R, S$ are taken on the sides $A B, B C, C D, D A$ respectively such that the lengths $A P, B Q, C R$ and $D S$ are integers and $P Q R S$ is rectangle. The largest possible area of PQRS is

Area of $PQRS= 40\times 10 = 400$

### NMTC 2019 Inter Stage 1 Question 21

$\mathrm{O}$ is a point inside an equilateral triangle $\mathrm{ABC}$. The perpendicular distance $\mathrm{OP}, \mathrm{OQ}, \mathrm{OR}$ to the sides of the triangle are in the ratio $O P: O Q: O R=1: 2: 3$. If $\frac{\text { Area of quadrilateal } O P B R}{\text { Area of triangleABC }}=\frac{a}{b}$ where $a, b$ are co-prime positive integers, then $a+b$ equals

$\frac{1}{2} \times s \times 1+\frac{1}{2} \times s \times 2+\frac{1}{2} \times s \times 3=\frac{\sqrt{3}}{4} s^{2}$
$\frac{1}{2} \times s(6)=\frac{\sqrt{3}}{4} s^{2}$
$s=\frac{12}{\sqrt{3}}=4 \sqrt{3}$

Let $\angle \mathrm{OBP}=\theta$
$\sin \theta=\frac{O P}{O B} \quad \Rightarrow O B=\operatorname{cosec} \theta$
Similarly $O B=3 \operatorname{cosec}(60-\theta)$
$\Rightarrow \frac{1}{\sin \theta}=\frac{3}{\sin (60-\theta)}$
$\Rightarrow 3 \sin \theta=\frac{\sqrt{3}}{2} \cos \theta-\frac{\sin \theta}{2} \Rightarrow \tan \theta=\frac{\sqrt{3}}{7}$
$\sin \theta=\frac{\sqrt{3}}{\sqrt{52}} \quad \Rightarrow B P=\frac{7}{\sqrt{3}}, O B=\frac{\sqrt{52}}{\sqrt{3}}$
$R B=\frac{5}{\sqrt{3}}$

Area of quadrilateral $\mathrm{BPOR}=\frac{1}{2} \times 1 \times \frac{7}{\sqrt{3}}+\frac{1}{2} \times 3 \times \frac{5}{\sqrt{3}}=\frac{11}{\sqrt{3}}$
Area of $\triangle A B C=\frac{\sqrt{3}}{4} \times(4 \sqrt{3})^{2}=12 \sqrt{3}$
$$\frac{a}{b}=\frac{11 / \sqrt{3}}{12 \sqrt{3}}=\frac{11}{36} \Rightarrow a+b=47$$

### NMTC 2019 Inter Stage 1 Question 22

In $\triangle \mathrm{ABC}, \mathrm{AB}=6, \mathrm{BC}=7$ and $\mathrm{CA}=8$. Point $\mathrm{D}$ lies on $\mathrm{BC}$ and $\mathrm{AD}$ bisects $\angle \mathrm{BAC}$. Point $\mathrm{E}$ lies on $\mathrm{AC}$ and $\mathrm{BE}$ bisects $\angle \mathrm{ABC}$. If the bisectors intersect at $F$, then the ratio $\mathrm{AF}: \mathrm{FD}=$

$\mathrm{AD}$ bisect $\angle \mathrm{BAC} \quad \Rightarrow \mathrm{BD}: \mathrm{DC}=\mathrm{AB}: \mathrm{AC}=6: 8$
$=3: 4$

Length of $B C=7$
so $B D=3, C D=4$
and also $\mathrm{BF}$ bisect $\angle \mathrm{ABD}$

$\Rightarrow \mathrm{AF}: \mathrm{FD}=\mathrm{AB}: \mathrm{BD}$
$=6: 3$
$=2: 1$ Ans.

### NMTC 2019 Inter Stage 1 Question 24

In quadrilateral $\mathrm{ABCD}, \mathrm{AB}=10, \mathrm{BC}=33, \mathrm{CD}=10$ and $\mathrm{DA}=15$. If $\mathrm{BD}$ is an integer then $\mathrm{BD}=\ldots$

Difference of two side is less than third side

$\Rightarrow$ in $\triangle \mathrm{ABD}$
$15-10x \Rightarrow x<25$

also in $\triangle B C D, 33-1023$
so $B D=24$

### NMTC 2019 Primary Stage 1 Question 15

A box of dimension $40 \times 35 \times 28$ units is used to keep smaller cuboidal boxes so that no space is left between the boxes. If the box is packed with 100 such smaller boxes of the same size, then dimension of the smaller box is

Dimension of bigger box $=40 \times 35 \times 28$
No. of smaller box $=100$

Volume of smaller box $=\frac{40 \times 35 \times 28}{100}=8 \times 7 \times 7$

### NMTC 2019 Primary Stage 1 Question 21

Small rectangular sheets of length $\frac{2}{3}$ units and breadth $\frac{3}{5}$ units are available. These sheets are assembled and pasted in a big cardboard sheet, edge to edge and made into a square. The minimum number of such sheet required is

$$\text { L.C.M }\left(\frac{2}{3}, \frac{3}{5}\right)=\frac{\text { L.C.M. }(2,3)}{\text { H.C.F }(3,5)}=6$$

Area of square $=6 \times 6$
No. of rectangle $=\frac{\text { Area of Square }}{\text { Area of rectangle }}$

Therefore $\frac{6 \times 6}{\frac{2}{3} \times \frac{3}{5}}=\frac{6 \times 6}{2} \times 5=15 \times 6=90$

### NMTC 2019 Sub junior Stage 1 Question 7

$A B C D$ is a square. $E$ is one fourth of the way from $A$ to $B$ and $F$ is one fourth of the way from $B$ to C. $\mathrm{X}$ is the centre of the square. Side of the square is $8 \mathrm{~cm}$. Then the area of the shaded region in the figure in $\mathrm{cm}^{2}$ is

Draw perpendicular $X P$ from $X$ to $A B$ \& $F Q$ on $P X$.

So required area $=$ Area of $\triangle \mathrm{XPE}+$ Area of $\triangle \mathrm{FQX}+$ Area of quadrilateral $P B F Q$.

Therefore

$$\frac{1}{2} \times 2 \times 4+\frac{1}{2} \times 2 \times 4+4 \times 2=4+4+8=16$$

### NMTC 2019 Sub junior Stage 1 Question 8

$A B C D$ is a rectangle with $E$ and $F$ are midpoints of $C D$ and $A B$ respectively and $G$ is the mid-point of $A F$. The ratio of the area of $A B C D$ to area of $A E C G$ is

Let length of rectangle $A B=C D=\ell$
$\&$ breadth of rectangle $A D=B C=b$
So area of trapezium $\mathrm{AECG}=\frac{1}{2}\left(\frac{\ell}{2}+\frac{\ell}{4}\right) \times \mathrm{b}=\frac{3}{8} \ell \mathrm{b}$

$\frac{\text { Area of } A B C D}{\text { Area of } A E C G}=\frac{\ell b}{\frac{3}{8} \ell b}=\frac{8}{3}(8: 3)$

### NMTC 2019 Sub junior Stage 1 Question 17

In quadrilateral $\mathrm{ABCD}, \mathrm{AB}=10, \mathrm{BC}=33, \mathrm{CD}=10$ and $\mathrm{DA}=15$. If $\mathrm{BD}$ is an integer then $\mathrm{BD}=\ldots$

Extend $AD$ to $E$ such that $AD = DE = 5$

$BC = DC$
$DE = AD$
$\angle DBE = \angle ADC$

So by SAS rule

$\triangle \mathrm{BDE} \cong \triangle \mathrm{CDA}$
CPCT
$B E=C A=8$

So $\operatorname{Ar}$ of $(\triangle \mathrm{BDE})=\operatorname{Ar}$ of $(\triangle \mathrm{CDE})$
add $\operatorname{Ar}(\triangle A B D)$ on both sides
$\operatorname{Ar}(\triangle \mathrm{BDE})+\operatorname{Ar}(\triangle \mathrm{ABD})=\operatorname{Ar}(\triangle \mathrm{ABD})+(\triangle \mathrm{ADC})$
So $\quad \operatorname{Ar}(\triangle \mathrm{ABE})=\operatorname{Ar}(\triangle \mathrm{ABC}) \ldots$ (1)
In $\triangle A B E \quad A B=6$
$B E=8$
$A E=10$

### NMTC 2019 Sub junior Stage 1 Question 19

In the given figure, $\triangle \mathrm{ABC}$ is a right angled triangle with $\angle \mathrm{ABC}=90^{\circ} . \mathrm{D}, \mathrm{E}, \mathrm{F}$ are points on $\mathrm{AB}, \mathrm{AC}$, $\mathrm{BC}$ respectively such that $\mathrm{AD}=\mathrm{AE}$ and $\mathrm{CE}=\mathrm{CF}$. Then, $\angle \mathrm{DEF}=$ (in degree).

Let $\angle A=x$

So $\angle \mathrm{ADE}=\angle \mathrm{AED}=\frac{1}{2}(180-\mathrm{x})=90-\frac{\mathrm{x}}{2}$
$\angle \mathrm{C}=90-\mathrm{x}$
$\angle \mathrm{CEF}=\angle \mathrm{EFC}=\frac{180-(90-\mathrm{x})}{2}=\frac{\mathrm{x}+90}{2}=45+\frac{\mathrm{x}}{2}$

So
\begin{aligned} &\angle \mathrm{DEF}=180-\angle \mathrm{AED}-\angle \mathrm{CEF} \ &180-\left(90-\frac{x}{2}\right)-\left(45+\frac{x}{2}\right)=45^{\circ} \end{aligned}

### NMTC 2019 Sub Junior Stage 1 Question 21

The area of a sector and the length of the arc of the sector are equal in numerical value. Then the radius of the circle is

Area of sector = length of are of sector

$\frac{\theta}{360} \times \pi r^{2}=\frac{\theta}{360} \times 2 \pi r$
$r=2$

### NMTC 2019 Junior Stage 1 Question 2

In $\triangle A B C$, the medians through $B$ and $C$ are perpendicular. Then $b^{2}+c^{2}$ is equal to

Let $BG = 2x, GE = x$

$CG = 2y, GF =y$

In $\triangle \mathrm{GCE}$
\begin{aligned} &(2 y)^{2}+x^{2}=\left(\frac{b}{2}\right)^{2} \ &4 y^{2}+x^{2}=\frac{b^{2}}{4} \cdots (i) \end{aligned}
In $\triangle \mathrm{BCG}$
\begin{aligned} &(2 x)^{2}+y^{2}=\left(\frac{c}{2}\right)^{2} \ &4 x^{2}+y^{2}=\frac{c^{2}}{4} \cdots (ii) \end{aligned}
In $\triangle B G E$
\begin{aligned} &(2 x)^{2}+(2 y)^{2}=a^{2} \ &4\left(x^{2}+y^{2}\right)=a^{2} \ &x^{2}+y^{2}=\frac{a^{2}}{4} \cdots (iii) \end{aligned}

Equation (i) + (ii)
\begin{aligned} &5 x^{2}+5 y^{2}=\frac{b^{2}+c^{2}}{4} \ &5\left(x^{2}=y^{2}\right)=\frac{b^{2}+c^{2}}{4} \ &\text { from equation (iii) } \ &5\left(\frac{a^{2}}{4}\right)=\frac{b^{2}+c^{2}}{4} \quad \Rightarrow \quad b^{2}+c^{2}=5 a^{2} \end{aligned}
Option (D).

### NMTC 2019 Junior Stage 1 Question 3

In a quadrilateral $A B C D, A B=A D=10, B D=12, C B=C D=13$. Then
(A) $\mathrm{ABCD}$ is a cyclic quadrilateral
(B) ABCD has an in-circle
(C) ABCD has both circum-circle and in-circle
(D) It has neither a circum-circle nor an in-circle

$AM=\sqrt{10^2-6^2} = 8$

$$\mathrm{CM}=\sqrt{13^{2}-6^{2}}=\sqrt{133}$$

For incircle
\begin{aligned} &A B+D C=A D+B C \ &23=23 \end{aligned}
In circle is possible
for cyclic quadrilateral (circumcircle) theorem should be followed.

\begin{aligned} &A C \times B D=A B \cdot C D+B C \cdot A D \ &(8+\sqrt{133}) \times 12 \neq 10 \times 13+10 \times 13 \end{aligned}
It is not a cyclic quadrilateral

Option B is correct

### NMTC 2019 Junior Stage 1 Question 5

In a rhombus of side length 5 , the length of one of the diagonals is at least 6 , and the length of the other diagonal is at most 6 . What is the maximum value of the sum of the diagonals ?
(A) $10 \sqrt{2}$
(B) 14
(C) $5 \sqrt{6}$
(D) 12

Let diagonal are $2x$ and $2y$

$x^{2}+y^{2}=25$

We have to find $2(x+y)_{\max }=$ ?
$$\begin{array}{ll} 2 x \geq 6 & 2 y \leq 6 \ x \geq 3 & y \leq 3 \end{array}$$
By option $(A) 2(x+y)=10 \sqrt{2}$
$$x^{2}+y^{2}=25$$
from here we get $x=y=\frac{5}{\sqrt{2}}$ it is not possible.
$$2 y=7.070 \geq 6$$

By option (B)
\begin{aligned} &2(x+y)=14 \ &x^{2}+y^{2}=25 \ &2 y=6 \text { and } 2 x=8 \end{aligned}
it is possible maximum value which is greater by other two options.

### NMTC 2019 Junior Stage 1 Question 9

The number of acute angled triangles whose vertices are chosen from the vertices of a rectangular box is
(A) 6
(B) 8
(C) 12
(D) 24

From each surface diagonals there are 4 triangles are possible which are equilateral of side $\sqrt{2}a$
each.

But in this manner each triangle is counted thrice therefore $\frac{24}{3} \Rightarrow 8$
Option (B).

### NMTC 2019 Junior Stage 1 Question 21

Circles A, B and C are externally tangent to each other and internally tangent to circle D. Circles A and $B$ are congruent. Circle $C$ has radius 1 unit and passes through the centre of circle D. Then the radius of circle $B$ is units.

In $\triangle MAN$

$(2-r)^{2}=x^{2}+r^{2}$
$4+r^{2}-4 r=x^{2}+r^{2}$
$4(1-r)=x^{2} \Rightarrow 4-4 r=x^{2} \quad \Rightarrow \quad \Rightarrow \quad r \quad \Rightarrow \quad d=\frac{4-x^{2}}{4}$

In $\Delta$ CAN
$(1+x)^{2}+r^{2}=(1+r)^{2}$
$1+x^{2}+2 x+r^{2}=1+r^{2}=2 r$
$x^{2}+2 x=2 r$
$x^{2}=2 r-2 x$
$\Rightarrow \quad x^{2}=2\left(\frac{4-x^{2}}{4}\right)-2 x$
$\Rightarrow \quad x^{2}=\frac{4-x^{2}}{2}-2 x$

$\Rightarrow \quad 2 x^{2}=4-x^{2}-4 x$

$3 x^{2}+4 x-4=0$

$3 x^{2}+6 x-2 x-4=0$

$3 x(x+2)-2(x+2)=0$

$\Rightarrow \quad x=\frac{2}{3}, x=-2 .$

$r=4-\frac{\left(\frac{2}{3}\right)^{2}}{4}=\frac{4-\frac{4}{9}}{4} \Rightarrow \frac{36-4}{36}=\frac{32}{36} \Rightarrow \frac{8}{9} .$

### NMTC 2019 Junior Stage 1 Question 24

In $\triangle \mathrm{ABC}$ shows below, $\mathrm{AB}=\mathrm{AC}, \mathrm{F}$ is a point on $\mathrm{AB}$ and $\mathrm{E}$ a point on $\mathrm{AC}$ such that $\mathrm{AF}=\mathrm{EF}, \mathrm{H}$ is a point in the interior of $\triangle \mathrm{ABC}, \mathrm{D}$ is a point on $\mathrm{BC}$ and $\mathrm{G}$ is a point on $\mathrm{AB}$ such that $\mathrm{EH}=\mathrm{CH}=\mathrm{DH}=\mathrm{GH}=\mathrm{DG}$ = BG. Also, $\angle \mathrm{CHE}=\angle \mathrm{HGF}$. The measure of $\angle \mathrm{BAC}$ in degree is

As $AB= AC$

Therefore

$\quad \angle 1=\frac{180-x}{2}=90-\frac{x}{2}$
$\angle 1=\angle 2=90-\frac{x}{2}$
$\angle 3+60^{\circ}=\angle 1+\angle 2=180-x$
$\angle 3=120-x$
$\angle 4=180-\left(\angle 2+60^{\circ}\right)=30+\frac{x}{2}$

$\angle 5=\angle 4=30+\frac{x}{2}$
$\angle 6=\angle 3=120-x$
$\angle 7=\angle 8=\frac{180-(120-x)}{2}=30+\frac{x}{2}$
$\angle 7+\angle 5=\angle 1$
$30+\frac{x}{2}+30+\frac{x}{2}=90-\frac{x}{2}$
$x=20 .$

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