CBSE 8th Mathematics Practical Geometry Two Marks Questions | QB365

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CBSE 8th Mathematics Practical Geometry Two Marks Questions

Published on: 16/09/2019

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Questions:

Generic 2

15 x 2 = 30

In a parallelogram, the lengths of adjacent sides are known. Do we still need measures of the angles to construct as in the Q-1 above?

In a parallelogram, the lengths of adjacent sides are known. Do we still need measures of the angles to construct?

Can you construct the quadrilateral PLAN, if PL = 6 cm, LA = 9.5 cm, \(\angle \)P = 75°, \(\angle \)L = 150° and \(\angle \)A = 140°?

Can you construct the quadrilateral MIST, if we have 100° at \(\angle \)M instead of 75°. In the quadrilateral MIST, where MI = 3.5 cm, IS = 6.5 cm, \(\angle \)M = 75°, \(\angle \)I = 105° and \(\angle \)S = 120°?

Is it possible to construct a quadruaterat MATH in which MA = 3 cm, AT = 5 cm, TH = 6 cm, HM = 4 cm and diagonal MT = 9 cm? If not, why?

Is it possible to construct a quadrilateral ABCD in which AB = 3 cm, BC = 5 cm, \(\angle \)B = 120°, \(\angle \)C = 105°, \(\angle \)A = 160°? If not, why?

We used some five measurements to draw quadrilaterals so far. Can there be different sets of five measurements (other than seen so far) to draw a quadrilateral? The following problems may help you in answering the question. Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and \(\angle\)B = 80°

Can you construct the quadrilateral PLAN, if PL = 6 cm, \(\angle \)A = 9.5 cm, \(\angle \)P = 75°, \(\angle \)L = 150° and \(\angle \)A = 140°?

Can you construct the above quadrilateral MIST, if we have 100° at M instead of 75°?

10.

Can you construct a quadrilateral PQRS with PQ = 3 cm,RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm?Justify your answer

11.

Can we draw the quadrilateral by drawing \(\Delta \) ABD first and then find the fourth point C?

12.

A student attempted to draw a quadrilateral PLAY, where PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm but could not draw it. What is the reason?

13.

Can you draw a rhombus ZEAL, where ZE = 3.5 cm, diagonal EL = 5 cm? Why?

14.

Can you draw a parallelogram BATS, where BA = 5 cm, AT = 6 cm and AS = 6.5 cm? Why?

15.

Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, \(\angle \)A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique quadrilateral? Give reasons for your answer.

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Answers Key:

Generic 2

15 x 2 = 30

No, the measures of three angles are not necessary in case of a parallelogram as its opposite sides are parallel.

Yes, to construct a unique parallelogram whose two adjacent sides are given, we need the angle included between them, because if angle is not given, then we cannot find a unique parallelogram. So, we cannot construct the parallelogram with the given information.

No, since, using angle sum property of a quadrilateral PLAN, we see that
\(\angle\) P+ \(\angle\)L + \(\angle\)A + \(\angle\)N=75° + 150°+ 140° + \(\angle\)N
= 365°+ \(\angle\)N, which is greater than 360°.
But sum of all the interior angles of a quadrilateral must be equal to 360°. So, construction of quadrilateral PLAN is not possible.

Yes, the quadrilateral MIST can be constructed
with \(\angle\)M = 100°.
By angle sum property of a quadrilateral,
\(\angle\)M+ \(\angle\)I+ \(\angle\)S+ \(\angle\)T=360°
\(\Rightarrow\) 100°+105°+120°+ \(\angle\)T = 360°
\(\Rightarrow\) \(\angle\)T=35°
Thus, we have, MI = 3.5 cm, IS = 6.5 cm, \(\angle\)M =100°, \(\angle\)I = 105° and \(\angle\)S =120° and \(\angle\)T = 35°.

No, here MA = 3 cm, AT = 5 cm and MT = 9 cm, which is not possible.
Since, in any triangle, sum of two sides is always greater than the third side.
\(\therefore\) MA + AT> MT, but 3 + 5 < 9
So, it is not possible to have such a quadrilateral.

No, as we know that the sum of measures of angles of a quadrilateral is 360°.
Here, \(\angle\)A + \(\angle\)B + \(\angle\)C =120° + 105° + 160°
= 385° > 360°
So, it is not possible to have such a quadrilateral.

The given data is sufficient for construction of a quadrilateral ABCD i.e. four sides and one angle are given.

Here, PL = 6 cm, \(\angle \)A = 9.5 cm,\(\angle \)P = 75°, \(\angle \)L = 150° and \(\angle \)A = 140°.
By angle sum property of a quadrilateral,
\(\angle \)N+\(\angle \)P+\(\angle \)L+140° = 360°
\(\Rightarrow\) \(\angle \)N + 75° + 150° + 140° = 365° + \(\angle \)N \(\neq \)360°.
But sum of all the angles of a quadrilateral must be 360°.
So, we cannot construct the quadrilateral PLAN.

Yes,the quadrilateral MIST can be constructed with \(\angle \)M = 100° instead of 75°.
e.g. Construct a quadrilateral MIST, where MI = 35 cm,IS = 6.5 cm, \(\angle \)M = 75°, \(\angle \)I = 105° and \(\angle \)S = 120°.
Obviously, we can construct MIST by taking \(\angle \)M = 100°.

10.

No, we cannot construct a quadrilateral PQRS, because we cannot draw the \(\Delta\)QSP as SQ + PQ \(\ngtr \)SP.

11.

No, we cannot draw the quadrilateral ABCD by drawing \(\Delta\) ABD first, because the sufficient measurements for \(\Delta\)ABD are not given. Consequently, question does not arise to find the fourth point C.

12.

The student could not draw a quadrilateral, because PL+PY\(\because\) 2 cm + 3 cm < 6 cm]
Actually, the sum of the length of any two sides of a triangle is always greater than the third side.

13.

Yes, we can draw a rhombus ZEAL, because all sides of a rhombus are equal and one of the diagonal is given. i.e. ZE = EA = AL = LZ = 3.5 cm and diagonal = 5 cm

14.

Yes, we can draw a parallelogram BATS, where BA = 5 cm, AT = 6 cm and AS = 6.5 cm, because the opposite sides of a parallelogram are equal in length.

15.

No, he cannot construct a quadrilateral ABCD, because sides BC and DC are not given. Although five measurements are given. Yet these are not sufficient to construct a quadrilateral.