A) ASA done clear
B) SSS done clear
C) SAS done clear
D) RHS done clear
View Solution play_arrowquestion_answer2) Which one of the following is true for the given triangle?
A) \[\angle 3=\angle 1+\angle 2\] done clear
B) \[\angle 1=\angle 3+\angle 2\] done clear
C) \[\angle 2=\angle 1+\angle 3\] done clear
D) Both (a) and (b) done clear
View Solution play_arrowA) SAS done clear
B) SSS done clear
C) RHS done clear
D) ASA done clear
View Solution play_arrowquestion_answer4) A triangle can be constructed by taking its sides as
A) \[1.8\text{ }cm,\text{ }2.6\text{ }cm,\text{ }4.4\text{ }cm\] done clear
B) \[2\text{ }cm,\text{ }3\text{ }cm,\text{ }4\text{ }cm\] done clear
C) \[2.4\text{ }cm,\text{ }2.4\text{ }cm,\text{ }6.4\text{ }cm\] done clear
D) \[3.2\text{ }cm,\text{ }2.3\text{ }cm,\text{ }5.5\text{ }cm\] done clear
View Solution play_arrowquestion_answer5) A triangle can be constructed by taking two of its angles as
A) \[{{110}^{o}},\text{ }{{40}^{o}}\] done clear
B) \[{{70}^{o}},{{115}^{o}}\] done clear
C) \[{{135}^{o}},{{45}^{o}}\] done clear
D) \[{{90}^{o}},{{90}^{o}}\] done clear
View Solution play_arrowA) \[3\text{ }cm,\text{ }4\text{ }cm,\text{ }6\text{ }cm\] done clear
B) \[9\text{ }cm,\text{ }16\text{ }cm,\text{ }26\text{ }cm\] done clear
C) \[1.5\text{ }cm,\text{ }3.6\text{ }cm,\text{ }3.9\text{ }cm\] done clear
D) \[7\text{ }cm,\text{ }24\text{ }cm,\text{ }26\text{ }cm\] done clear
View Solution play_arrowquestion_answer7) In which of the following cases, a unique triangle can be drawn?
A) \[AB=4\text{ }cm,\text{ }BC=8\text{ }cm\]and \[CA=2\text{ }cm\] done clear
B) \[BC=5.2\text{ }cm,\text{ }\angle S={{90}^{o}}\]and \[\angle C={{110}^{o}}\] done clear
C) \[XY=5\text{ }cm,\text{ }\angle X={{45}^{o}}\]and \[\angle Y={{60}^{o}}\] done clear
D) An isosceles triangle with the length of each equal side 6.2 cm. done clear
View Solution play_arrowquestion_answer8) Which of the following statements is INCORRECT?
A) If length of any two sides of a triangle are 7 cm and 10 cm, then length of its third side lies between 3 cm and 17 cm. done clear
B) It is possible to construct a unique triangle if all its three angles are given. done clear
C) An angle of \[\left( 7\frac{{{1}^{o}}}{2} \right)\] can't be constructed using compasses and ruler. done clear
D) None of these done clear
View Solution play_arrowStep 1: Draw line XV of length 6 cm. |
Step 2: At X, draw a ray XP making an angle of \[{{30}^{o}}\]with XY. |
Step 3: At V, draw a ray YQ making an angle of \[{{100}^{o}}\] with YX. |
Step 4: The point of intersection of the two rays XY and YQ is Z. |
A) Step 1 done clear
B) Step 2 and Step 4 done clear
C) Step 3 done clear
D) Step 4 done clear
View Solution play_arrowStep 1. Now, extend RQ to S and with P as centre and with a sufficient radius, draw an arc, cutting SO at A and 8. |
Step 2. Along QX, set off \[QP=3.5\text{ }cm.\] |
Step 3. Draw a line segment \[QR=4.2\text{ }cm\]and construct\[\angle RQX={{120}^{o}}\]. |
Step 4. Joint PR. |
Step 5. Joint PC, meeting RQ product at |
M. Then. \[PM\bot QS\] |
Step 6. With A as centre and radius more than half AB, draw an arc. Now with B as centre and with the same radius draw another arc, cutting the previous arc at C. |
A) 1\[\to \]2\[\to \]3\[\to \]4\[\to \]5\[\to \]6 done clear
B) 4\[\to \]1\[\to \]2\[\to \]3\[\to \]5\[\to \]6 done clear
C) 2\[\to \]4\[\to \]3\[\to \]1\[\to \]5\[\to \]6 done clear
D) 3\[\to \]2\[\to \]4\[\to \]1\[\to \]6\[\to \]5 done clear
View Solution play_arrowquestion_answer11) State 'T' for true and 'F' for false.
(1) In a triangle, the measure of exterior angle is equal to the sum of the measure of interior opposite angles. |
(2) The sum of the measures of the three angles of a triangle is\[{{90}^{o}}\]. |
(3) A perpendicular is always at \[{{90}^{o}}\] to a given line or surface. |
A)
(1) | (2) | (3) |
T | T | F |
B)
(1) | (2) | (3) |
T | F | F |
C)
(1) | (2) | (3) |
T | F | T |
D)
(1) | (2) | (3) |
F | T | F |
Step 1. Draw MN of length 3 cm. |
Step 2. At M, draw MX1MN. (L should be somewhere on this perpendicular). |
Step 3. With N as centre, draw an arc of radius 5 cm. (L must be on this arc, since it is at a distance of 5 cm from N). |
Step 4. L has to be on the perpendicular line MX as well as on the arc drawn with centre N. Therefore, L is the meeting point of these two and ALMA/ is obtained. |
A) Only Step 4 done clear
B) Both Step 2 and Step 3 done clear
C) Only Step 2 done clear
D) None of these done clear
View Solution play_arrowStep 1. Take a line 'l' and a point ?A? outside ?l?. |
Step 2. Take any point Son l and join 8 to A |
(i) Now with A as centre and the same radius as in previous step, draw an arc EF cutting AB at G. |
(ii) With the same opening as in previous step and with G as centre, draw an arc cutting the arc EF at H. |
(iii) With B as centre and a convenient radius, draw an arc cutting l at C and BA at D, |
(iv) Now, join AH to draw a line W. |
A) (i)\[\to \](ii)\[\to \](iv)\[\to \](iii) done clear
B) (iii)\[\to \](i)\[\to \](ii)\[\to \](iv) done clear
C) (iii)\[\to \](ii)\[\to \](i)\[\to \](iv) done clear
D) (i)\[\to \](ii)\[\to \](iii)\[\to \](iv) done clear
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