Tag: construction of triangles - i

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Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

If $b=3, c=4, \angle B=\dfrac{\pi}{3}$, then the number of triangles that can be constructed is

  1. $0$
  2. $1$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $b=3,c=4,\angle B=\dfrac{\pi}{3}$

We know that as per Sine rule

$\dfrac{\sin \dfrac{\pi}{3}}{3}=\dfrac{\sin C}{4}$

$\dfrac{\sqrt 3}{2\cdot3}=\dfrac{\sin C}{4}$

$\sin C = 2\sqrt{3}$  which is greater than 1

and sin lies between -1 and 1 that means angle C is not possible

Thus Zero triangles can be construct.

Hence A is the correct option

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles
Which of the following pair of sides can form triangle?
  1. $5\ cm, 6\ cm , 4\ cm$
  2. $13\ cm , 12\ cm , 24\ cm$
  3. $2\ cm , 7\ cm , 9\ cm$
  4. $5.6\ cm , 6.5\ cm , 12\ cm$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

As in a triangle, the sum of any two sides should be strictly greater than the third side.

 
In $(A), 5+6 > 4$   $\therefore$ This pair will form triangle.


In $(B), 13+12 > 24$ This pair will form triangle.

In $(C), 2+7\ \ngtr 9$  This pair will not form triangle.

In $(D), 5.6+6.5 > 12$  This pair will form triangle.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Consider $\triangle ABC$ and $\triangle { A } _{ 1 }{ B } _{ 1 }{ C } _{ 1 }$ in such a way that $\overline { AB } =\overline { { A } _{ 1 }{ B } _{ 1 } } $ and M, N, ${ M } _{ 1 }$, ${ N } _{ 1 }$ be the mid points of AB, BC, ${ A } _{ 1 }{ B } _{ 1 }$ and ${ B } _{ 1 }{ C } _{ 1 }$ respectively, then

  1. $\overline { M{ M } _{ 1 } } =\overline { N{ N } _{ 1 } } $
  2. $\overline { C{ C } _{ 1 } } =\overline { M{ M } _{ 1 } } $
  3. $\overline { C{ C } _{ 1 } } =\overline { N{ N } _{ 1 } } $
  4. $\overline { M{ M } _{ 1 } } =\overline { B{ B } _{ 1 } } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given AB = A1B1 and M, N, M1, N1 are midpoints, the triangles are congruent. The vector relationship between the midpoints of sides in congruent triangles preserves the distance between corresponding points. Specifically, M and M1 are midpoints of AB and A1B1, and N and N1 are midpoints of BC and B1C1; the displacement MM1 is equal to BB1.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

The sides  $A B , B C , C A$  of a triangle  $A B C$  have  $3,4$  and  $5$  interior points respectively on them. The number of triangles that can be constructed using these points as vertices is

  1. $205$
  2. $210$
  3. $315$
  4. $216$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total number of points $12$. If no three points are co-linear then total number of the   triangles would be  $^{12}{C _3}$.

But $3$ points on AB , $4$ points on BC and $5$ points on CA are co-linear.

So, total number of triangles formed should be 

$\begin{array}{l} { =^{ 12 } }{ C _{ 3 } }-\left( { ^{ 3 }{ C _{ 3 } }{ +^{ 4 } }{ C _{ 3 } }{ +^{ 5 } }{ C _{ 3 } } } \right)  \\ =220-\left( { 1+4+10 } \right)  \\ =205 \end{array}$



Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Mark the correct alternative of the following.
In which of the following cases can a right triangle ABC be constructed?

  1. $AB=5$cm, $BC=7$cm, $AC=10$cm
  2. $AB=7$cm, $BC=8$cm, $AC=12$cm
  3. $AB=8$cm, $BC=17$cm, $AC=15$cm

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A right triangle must satisfy the Pythagorean theorem: a^2 + b^2 = c^2. For option C, 8^2 + 15^2 = 64 + 225 = 289, which is 17^2. Thus, 8, 15, 17 forms a right triangle.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Mark the correct alternative of the following.
In which of the following cases, a right triangle cannot be constructed?

  1. $12$cm, $5$cm, $13$cm
  2. $8$cm, $6$cm, $10$cm
  3. $5$cm, $9$cm, $11$cm

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Any triangle is right angled if it satisfies $a^2 + b^2 = c^2$, where c is the largest side from option (A)  we take 

$a = 12 cm$
$b = 5 cm$
$c = 13 cm$
we know $(12)^2 + (5)^2 = 169 = (13)^2$
So we can construct a right angled triangle.

Similarly in option (B) we can take 
$a = 8 cm$
$b = 6 cm$
$c = 10 cm$
with these sides also we can construct a right angled triangle.

Consider option (C) then 
$a = 5 cm$
$b = 9 cm$
$c = 11 cm$
$a^2 + b^2 = 25 + 81$   But  $c^2 = 121$
             $= 106$

So we cannot construct right angled triangle from $5 cm, 9 cm, 11 cm$.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Construct a $\triangle PQR$ in which $QR= 4.6\ cm., {\angle Q}={\angle R=50 ^{0}}$. Then the perimeter of the triangle is:

  1. $10.2\ cm$
  2. $13.2\ cm$
  3. $11.8\ cm$
  4. $12.4\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In triangle PQR, if angle Q = angle R = 50 degrees, then angle P = 180 - 100 = 80 degrees. Using the sine rule, PQ/sin(50) = QR/sin(80) = PR/sin(50). PQ = PR = QR * sin(50) / sin(80) = 4.6 * 0.766 / 0.985 approx 3.57. Perimeter = 4.6 + 3.57 + 3.57 = 11.74, which rounds to 11.8.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Construct a right angled $\triangle ABC$ with $\angle B = 90^\circ, BC = 5\ cm$ and $AC = 10\ cm$ and find the the length of side $AB$

  1. $6.2\ cm$
  2. $5\ cm$
  3. $8.7\ cm$
  4. $7.2\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Step 1. Draw a line segment $BC=5\ \ cm$

Step 2. At $B$ draw an angle of $90^{\circ}$ and extend the ray.
Step 3. Now taking $C$ and centre draw an arc of radius $10$ cm intersecting the previous ray at $A$.
Step 4. Join $C$ to $A$.
Now using a ruler measure $AB$
$AB=8.7$ cm

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles
Construct a $\triangle PQR$ such that $\angle P = 30^\circ, \angle Q = 60^\circ$ and $PQ = 10\ cm$.Find the measure of $\angle R$
  1. $30^\circ$
  2. $45^\circ$
  3. $60^\circ$
  4. $90^\circ$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The sum of angles in any triangle is 180 degrees. Given angle P = 30 and angle Q = 60, angle R = 180 - (30 + 60) = 180 - 90 = 90 degrees.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Length of two sides of a $\triangle ABC$ is $AB=6\ cm$ and $BC=7\ cm$. Then, which of the following can represent the third side of the triangle ? Also, construct the triangle formed by these three sides.

  1. $8\ cm$
  2. $13\ cm$
  3. $14\ cm$
  4. $15\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A triangle can be formed if sum of any two sides is greater then the third side.

Here $AB=6$ cm and $BC=7$ cm
Now $AB+BC>AC$
$6+7>AC$
$AC<13$ cm
So only option $A$ is possibel.
Steps of construction

Step 1. Draw a line segment $AB=6\ \ cm$
Step 2. Assuming $A$ a centre draw an arc of radius $8 \ \ cm$
Step 3. Now assuming $B$ as centre draw an arc of $7 \ \ cm$ intersecting the previous arc at $C$.
Step 4. Now join $A$ to $C$ and $B$ to $C$.